Pearson - Physics Class 9

1.28 Chapter 1 2 1. T wo variables x and y vary such that xy = constant. (b) the number of C.S.D. is 100, and the zero of the Which of the following graphs represent the above circular scale is 5 divisions below the index line. relationship? (a) (b)  (c) the number of C.S.D. is 50, and the zero of the circular scale is 5 divisions above the index line. y (d) Both (1) and (3) x x 28. If a graph is plotted between the length of a pendu- lum (l) and its time period (T), then the two quanti- (c) (d)  ties are plotted as _____. (a) l along X-axis y y (b) T along Y-axis 1/x 1/x (a) Only (a) is true (b) Only (b) is true 2 2. T he difference between ZRP and HRP of a physical (c) Both (a) and (b) are true balance when 47.86 g of a substance is placed in its (d) Both (a) and (b) are false pans is 3. When 10 mg is added in its pans, the dif- ference between HRP and LRP is obtained as 5. The 2 9. To construct a seconds pendulum having a length of 100 cm, the value of g should be? most accurate mass of the body is _____ g. (a) 47.875 (b) 47.845 (a) π m s−2 (b) 1 00 π m s−2 (c) 47.866 (d) 4 7.854 (c) 1 m s−2  (d) π2ms−2 π2 23. If 17 divisions of the circular scale of a screw gauge are below the index line of the pitch scale, then the 3 0. W hen the jaws of a vernier calipers are in contact, zero error is ______ circular divisions. the eight division of the vernier scale coincides with (a) +17 (b) –17 the seventh division of the main scale. If N is the (c) 83 (d) 34 number of divisions on the vernier scale and the least 24. W hich of the following measurements is most count is x, then the zero error correction is precise? (a) –8x (b) (N – 8) PRACTICE QUESTIONS (a) 5 cm (b) 5 .00 cm (c) – (N – 8)x (d) + 8x (c) 5.000 cm (d) 5.00000 cm 3 1. The percentage errors in the measurement of the 2 5. T he length and breadth of a rectangle were measured length (L) and breadth (B) of a rectangle are l% and using an instrument and the area was determined as 28.83 cm2. The instrument used could be ____. b%, respectively. Then the percentage error in the (a) a vernier caliper whose least count is 0.1 mm calculation of the area will be______. (b) a metre scale (a) (lb)% (b) (l + b)% (c) a vernier caliper whose least count is 0.3 mm (c) (L + l) (B + b) (d) Lb + BL (d) None of the above 32. The least count of a vernier calipers is 0.01 cm and if the zero mark of the vernier scale is to the right of zero of the main scale and the vernier coincidence is 26. T he time period of two pendulums of same length 7 when the jaws are in contact, then the zero error oscillating on different planets A and B is 2 s and 3 s, is______cm. respectively. The ratio of acceleration due to gravity (a) + 6 × 0.01 (b) + 7 × 0.01 on the two planets is ______. (c) – 7 × 0.01 (d) – 6 × 0.01 (a) 9 : 4 (b) 3 : 2 33. The thimble of a screw gauge has 50 divisions. The (c) 2 : 3 (d) 4 : 9 spindle advances 1 mm when the screw is turned 27. If the zero error correction of a screw gauge with through two revolutions. Then the pitch of the screw least count 0.01 mm is +0.05 mm, is­______. (a) the number of C.S.D. is 100, and the zero of the circular scale is 5 divisions above the index line. (a) 0.5 cm (b) 0.15 cm (c) 0.5 mm (d) 0.01 mm

Measurements 1.29 3 4. The ZRP of a physical balance is 10.5 while finding both the instruments are marked in mm. The zero mass of a substance. For a weight 34.23 g the resting error on the vernier calipers is +0.15 mm where as point was found to be 8.5. when 10 mg was removed that on screw gauge is –0.06 mm. If the diameter of a the resting point was 11.0. The most accurate mass of rod lies between 0.9 cm and 1.0 cm, and x and y are the substance is_______ g. the VCD and CSR on the two instruments, relation (a) 34.15 (b) 34.31 between x and y given that the number of C.S.D. = (c) 34.238 (d) 34.222 100, is 35. The smallest weight that can be measured accurately (a) x – 5y = 9 (b) 5x – y = 9 using a physical balance is______ g (grams). (c) x – 5y = 21 (d) 5x – y = 21 (a) 10 (b) 0.001 4 0. The mass of a certain amount of salt determined by (c) 0.1 (d) 0.01 two different persons was found to differ by 200 mg, narvrns_rsimne,_desriarcnaeg9nalrcil7dleea(ecrglsartitdrephlcaiisoeepvmpwrieiepnsssreici)csaoti.tingchindvhdhasientsoaol 3gypnf6s,otcdt.htt sh rihh(iaev((vtscAcla stet2ecateewhhcciiraie))v)sarenarseelrs icveiesclcnogpcetuwe30rfn51Irviznaaeierlfee.uce00narenurnw0awtrgoerrm sardhngi0n,aaouenseedsd1ugcbigartnbcdeehainaegavhatner9clrurueedeletaagr7ehargcgvoolnh,tdnniaeeeedhfpfrdauidliorveieeeisiiecvatmpvrssitrhrsneiznspeeisrpti1ercstebdrecriowihenoatrseomgusncocmentiiiraanelhhtnhscdiamtirmsonzeeecvairoovesriefaeseondt.sesfrrcldhcvcoonpyiaslttnz,reaceehrolheeelrgaert(tesewenerahwn3hsiroddwttcuresi)evriiioccievgcemnvtever((urrhseo0aagirrdbeicssrbudnrfa.suwo))riiaiz1otcuo eeioginnhntenearwrgneendg.rt21erlgeolseoat00Thyaachtudgfof00nnhdahaehoegaivnlsddaiefriuerevuefvprinegttzonreinstneihhresenihesrsrueiunegettrosie1nhmrmoeig.aacnusvcentrzmibbieesmeredtvsesrcewramcoreanetrurroatmhnoinil.hzlcooeaaertpeeeeetrrstffrhrhwdoreyioivste 4rviih(rcs1e4oria.toro)nl hnnr((((2omTaure.iabdc0tasteTsnb)h)h)).te:diro t0 ienaheenohi3nl1cgfTTTwvvTetauefliaoaaeanenthhhhsielntllhtnnsfiircuushieeeewnpuegdtegtfeeerergnehamhotet..ttuvlhtrrecrrisrhtseumibnuupseiehi.srergege?eeloeneaathfrsrhfcrn.mimmtlazeoeeetattewqsratsffpiaartousssawoossspenohnseia.iosfinrsecmfWrtri2ohe1efp0sosre0hal0trel0ihiitasnctepcmhhmiycetsehgenogaletdwerfhmeurlbeetealoahsiatsgelrismeaoehtanrtnetfuhectsoahpeadnlial.annunoinseowdtth3fhttitheoneht:egherg2haelr.tsooiaitgTtotawvhhwtiheoeeetoye-rrrf 3 7. (A2)sc2re0w0 gauge has as many(c3i)rc5u0lar scale divisions as (4) (a1)0 03 : 2 (b) 4 : 9 me00twc00bh01ing5omaefumdmtghmdeeohuscabsrleeawsscmir 3sea80wn.. y5((((gcpAt24cahmcoai))))teui c ucrmrogcnh00e00meut,....?00a00oltabarh0000fesrie1010ntssnth55ehhcmcedoammwrcmelwmdseahmco dnradu estiwibvinilnsiesitiis5htsosce0nrcle.esmf5wiaagsmstlugetrcamnheuoge,guitr((shetnedbhuta))oea? ss frnese00dht..hwto00ahew00hsr10asensct5hacridioemncswswmti.ithnnsIefil5neftiahgctse-mthe lfe 4ing2gu. trh((lPadecaen.ue))nodnlg8 ufd9tp6mthueh:cl.nueom4dTmf su.tch hlrAueeemwripsa.etBxiInofcdismuoxlfuscthmhmoerwiltroeinttrhigmtheheraing((tpbdhhae))aernsr 1ieo21ftcrhd4oe:esqn3cuisdmseesc3npocen:ynd4dis.supTleuhnme- ure is used as PRACTICE QUESTIONS (c) 128 cm. (d) 72 cm. 4 3. If surface tension is defined as force per unit length, which of the following quantities will have thesame units as surface tension? If the readings on the two scales with the tips of the (a) work × volume × speed (b) work × velocity volume mrrmoead,d?ainngdswoinththaerotwd ohesamsclrcdermaelb1w,e2eswt.5wwht5oeeiuntemhcnahmtitrhnhoaegedntdseiiscap5rchse.he9wol6dfosmttbahhermeeetrw,1saewc2rere.hen5aw25tth.s9imes7totmshcumreceamhdwniidsanamgrn5ed.eae9tdae66icrn.hm0og4sfmot,hwerhaar(tcei)s 2t.h9ra7etedm woiamfom crhkeaatn×endrgveeolof cviotylume work volume 5 mm (t2h)e r9o.6d?mm (d) time × velocity 58 mm ((4a) 9N.5omt mpossible to determine tehaoesnticnsocstrurenuwtmsgeoanfutags eavreiesrn–mi(((e0adbcrr).))0kc99Ne6a..d56lomi8ptimnmepmrmom.smsIsamfinbt.dlheTeathodesicdarzemeetwreeortmegerairnuoreogfreaoarnroedtihnlieetsvheberenrtawietieroecnoalf0ip.59er:cs1m4 i.s4T.a+ hnImq0edfu.a1met1sh5ns.ae0cbmiynecemcmaasorcce,mhawaldeenoho?sdfeuorbtxehfleed,uhnoitws of force, velocity and fre- many times has the unit of are the VCD and3 9C.S RThoenletahset tcwouonitns storfuma evenrtns,iefirndcaaliprerlastiaonnd baetswcreeewn x and(ya)gdiovuenblethda t the (b) four times er of C.S.D. = 100. gauge are in the ratio of 5 : 1. The main scales of (c) halved (d) one-fourth –5y=9 (2) 5x – y = 9 – 5y = 21 (4) 5x – y = 21

1.30 Chapter 1 Level 2 4 5. The dimensional formula of a physical quantity is 52. A simple pendulum designed on the moon as a known to be [M1L2T-3]. Write down the units of seconds pendulum is taken to a planet where the this quantity in C.G.S. and S.I. units and calculate acceleration due to gravity on the surface is twice the multiplication factor for conversion from S.I. to that on the Earth. If gearth : gmoon = 6 : 1, find the C.G.S. units. period of oscillation of the pendulum on the planet mentionedabove. 46. The pitch of a screw gauge is 0.5 mm and the num- ber of divisions on the circular scale is 100. When 53. In the figure given above, the main scale is marked a glass plate is held between the studs of the screw in inches. Determine the length of the object held gauge, the main scale reads 1.5 mm and the 69th between the two blocks. If 1 inch = 2.54 cm, find division of the circular scale coincides with the the length in cm. index line of the main scale. What is the thickness of the glass plate? 01 23 45 47. A specific gravity bottle weight 10 g when it is empty. It weighs 100 g and 110 g when filled with an oil and water, respectively. What is the relative density of oil? 4 8. A vernier scale has 10 divisions. It slides over a main 54. Find the ratio of the relative densities of two sub- scale whose least count is 1.0 mm. If the number stances if their masses are equal when the volume of of divisions on the main scale, to the left-hand side of zero of the vernier scale is 4 and the 8th vernier one is 1½ times that of the other. scale division coincides with the main scale, find the measurement. PRACTICE QUESTIONS 49. The fundamental frequency of a stationary wave 5 5. F ind the area of cross-section of a rod, the diameter 11 of which is measured with a screw gauge with read- ings as shown in the figure. formed in a stretched wire is n = 2l m where ‘  ’ is length of the vibrating wire, ‘T’ is the tension Pitch = 1mm in the wire and ‘m’ is its mass per unit length. If the 0 04 30 percentage error in measurement of  , T and m are a%, b% and c%, respectively, then find the maximum 95 25 error in measuring n. 90 20 50. A vernier scale with 10 divisions slides over a main scale whose pitch is 0.5 mm (pitch = 1 M.S.D.). If a 85 15 bob of diameter 9.75 mm is held between the jaws, determine the MSR and V.C.D. if Reading with the tip of Reading with the wire the screw in contact (a) there is no zero error with the stud held between the screwandthestud. (b) the zero error = 0.35 mm 5 6. The working of a common balance or a physical bal- 51. W hen the tip of the screw of a screw gauge is in ance is based on the ‘Principle of moments’. A metre contact with the stud, the zero of the circular scale scale of uniform density is balanced at the centre. If is 3 divisions below the index line. To hold a wire a mass of 2 kg is suspended at an edge of the scale, the thimble is given a little over 3 rotations in the then what mass should be suspended on other side at anti-clockwise direction and with wire held tightly 1/4th length of the scale to maintain the horizontal the 32nd division of the circular scale is now in position of the scale? line with the base line. The pitch of the screw is 0.5 mm and the number of divisions on the circu- 57. D etermine the least count and the zero error of the lar scale are 100. Find the correct diameter of the vernier calipers shown in the figure below. What is wire. the corresponding zero correction?

Measurements 1.31 0 1 cm objects of masses m and m separated by a distance d 0 10 G m1 m2 is given by, F = d2 Where G is the universal gravitational constant. 62. The distance travelled by a body in different time intervals is tabulated as follows. Draw the distance time graph. Time (s) 0 10 20 30 40 Distance (m) 0 40 80 120 160 5 8. A pendulum clock in a museum was found to lose 1 6 3. 1 yotta meter is how many kilometer? minute in every 24 hours. What corrective measure should the curator of the museum undertake? 64. When the studs of a screw gauge are in contact the position of the head of the screw is as shown below, 59. T he corrected length of a rod when measured with Determine the zero error. the help of a vernier calipers is 25.4 mm. N = 100  (Refer the figure below) 1 PSD = 1mm 0 0 1 2 3 4 cm 99 98 97 96 0 10 6 5. Name the part of the physical balance which is used to increases its sensitivity. 66. The ZRP of a physical balance is 10.5 while find- ing mass of a substance. For a weight 34.23 g the resting point was found to be 8.5. When 10 mg was Determine the zero error. removed the resting point was 11.0, find the most accurate mass of the substance. PRACTICE QUESTIONS 60. The following are the values known in a particu- 6 7. If the units of mass, length and time are doubled, lar measurement, 30.56, 12.6, 21.09. What is the then what happens to the unit of ‘relative density’? sum of the values when rounded off to one decimal Discuss. point? 68. An irregular solid, when immersed in water, dis- 61. What is the dimensional formula of (a) volume (b) places 374th litre of water. When immersed in a density and (c) universal gravitational constant ‘G’. given liquid, it displaces 600 g of the liquid. What is The gravitational force of attraction between two the density of the liquid? Level 3 6 9. T he dimensional formula of a physical quantity x is happens to the mass of a body measured in the [M1L2T–2] and that of another quantity y is [M1L1T- balance? 1]. If a third quantity z is directly proportional to the square of y and inversely proportional to x, then find 7 2. A liquid flows into a vessel initially empty at a steady the dimensional formula of z. rate of 70 cm3  s  –1. The pictorial representation (graph) of the increase in the mass of the vessel with 70. A simple pendulum is completely submerged under time is given below. water. Discuss the variation in its time period. (i) W hat does the point A represent? 71. T he purpose of plumb line in a physical balance and (ii) W hat does the horizontal line beyond in the hands of a mason is same. If so, what is that purpose? If it is absent in a physical balance, what B indicate?

1.32 Chapter 1 (iii) F ind the capacity of the vessel. 75. A vernier scale with 10 divisions slides over a main (iv) F ind the density of the liquid. scale whose pitch is 0.5 mm (pitch = 1 MSD). If a bob of diameter 9.75 mm is held between the jaws, 17.0 B determine the M.S.R and VC.D if (a) there is no zero error. (b) the zero error = 0.35 mm. Mass (kg) 76. Using a screw gauge without any zero error, the 0.2 A diameter of a wire was determined as 2.74 mm. If PRACTICE QUESTIONS 0 the screw advances 5 mm when the thimble is given 00 1 2 3 4 5 6 7 10 complete rotations, what are the possible val- ues of the pitch scale reading, circular scale reading 7 3. T he distance between two consecutive threads on and the least count if N is the number of head scale the screw of a screw gauge is 0.5 mm. The number divisions? of divisions on the circular scale is 100. A wire is placed between the studs of the screw gauge. Find 7 7. In a physical balance when the beam is released with the diameter of the wire if the pitch scale shows 14th empty pans, the successive left and right turning division and 40th circular division coincides with the points are obtained as 5, 6, 6 and 18, 18, respec- base line. The given apparatus was detected to have tively. After placing the substance whose mass is to negative zero error. The 90th division on the circular be determined and when the beam is released, the scale coincides with reference line, when the studs resting point obtained is 13. When 10 mg is added are in contact. to the pan, the resting point is obtained as 11. If the mass at LRP is 60 g, find the correct mass of the 74. The divisions on the main scale of a screw gauge are body corrected to a milligram. 1 mm apart and the screw of the spindle advances by 5 main scale divisions when the spindle is given 78. To determine the weight of a solid body lighter than 5 complete rotations. How many divisions are to water, a sinker is used. The solid body of density 0.5 be provided on the circular scale for the least count g cm-3 is tied to the sinker of volume 50 cm3 and the of the instrument is to be 1 mm? What changes are combination is immersed into water. If the volume required of the number if divisions can be only 500 of water displaced is 80 cm3, find the weight of the for the same least count? body.

Measurements 1.33 CONCEPT APPLICATION Level 1 True or false 1.  True 2.  True 3.  False 4.  True 5.  True 6.  True 7.  False Fill in the blanks 8.  proportional 9.  1M.S.D. 10.  –4 11.  + n × L.C. 12.  10 µm 13.  0.005 cm 14.  1⋅9 Match the following 15. A  :  d B  :  h C  :  f D  :  i E  :  g F  :  j G  :  a H  :  e I  :  b J  :  c Multiple Choices 25. (b) 35. (d) 16. (c) 17. (b) 18. (a) 19. (a) 20. (d) 21. (c) 22. (c) 23. (a) 24. (d) 26. (a) 27. (d) 28. (c) 29. (d) 30. (b) 31. (b) 32. (b) 33. (b) 34. (c) 36. (b) 38. (d) 39. (a) 40. (d) 41. (d) 42. (a) 43. (d) 44. (c) Explanation for questions 31 to 45: 3 7. Let the n rotations advance the screw by 5 cm 3 1. Errors are additive, hence, percentage error in area = ∴ Pitch = 5 cm (a% + b%) n 32. The error is positive. But pitch = 0.5 mm = 0.05 cm ∴ Positive zero error = + n × L.C. = + 7 × 0.01. ∴ 5 = 0.05 HINTS AND EXPLANATION n Distance moved by spindle n = 10 Number of rotations 33. Pitch of the screw = Thus, there are 100 threads in 5 cm of the screw, i.e., 100 circular scale divisions. 1 mm = 2 = 0.5mm ∴ L.C.= 0.05 = 0.0005 cm 100 34. Correction = −  ZRP − LRP  × 10mg 3 8. Scale 1 Scale 2 Total  HRP − LRP  5.96 18.51 With rod 12.55 6.04 9.01 − 0.08 = 9.5 mm  10.5 − 8.5 10mg −2 (10mg ) Without rod 2.97  11.0 − 8.5 2.5 = − × = × Movement 9.58 = – 8  mg = – 0  .008 g of screw ∴ Diameter of the rod = 9.5 mm Correct mass = 34.23g – 0.008 g = 34.222 g 1 3 5. zero error of vernier calipers = zero error of screw 3 9. Least count of screw gauge = 100 = 0.01 mm Least count of vernier calipers gauge x L.C. = (N1 – y) L.C. = 5 × 0.01 = 0.05 mm x = N1 – y Let d be the diameter of the rod 9 mm < d < 10 mm x + y = N1 d = 9+ x × 0.05 – 0.15 by vernier caliper d = 9 + y × 0.01 + 0.06 by screw gauge N1 = 3 + 97= 100 5x – 15 = y + 6 5x – y = 21 N = 1 N 1 = 50 2 36. The smallest weight that can be measured accurately using a physical balance is 1 mg ie., 0.001 g.

1.34 Chapter 1 40. l1 : l2 = 2 : 3 4 2. Units of Surface Tension = N m−1 g1 : g2 = 3 : 2 work × volume × speed = (N m) × (m3) × m s−1 T1 1 / g1 4 = N m5 s−1 T2 2 / g2 9 = = (N m) × (m s−1 ) m3 T1 4 2 f1 work × velocity = = N m−1s−1 T2 9 3 f2 volume = = ∴ = 3:2 work × velocity = (N m) × (m s−1 ) rateof changeof volume (m3 s−1 ) 4 1.  A =  −x m3 B = + x work × volume = N m × m s−1 = N m3s2 time velocity s−1 Where l = 100 cm  A < B 4 3. 2 N = 1 N1 = 1 kg1 m1 s1−2 2 (m s−1) = m1 s1−1 TA < TB 2 s−1 = s1−1 TB = 4 ∴ 1 m = 1 m1 TA 3 2 kg m s−2 = 1 kg1 m1 s1−2 ∴ B = TB A TA + x 16  25 1m 1s −2  −x 9 x 7 1m1 1s1−2 = , = 1kg1 = 2kg x= 7 × 100 = 28 cm  1m   1s−1  +2 25  1m   1s1−1  HINTS AND EXPLANATION 1kg1 = 2kg Higher frequency ⇒ lower time period Regular shorter pendulum = pendulum A  1 2 2 1  2 4 2 ∴ length of pendulum A = 100 – 28 = 72 cm = 2 kg 1 × = kg= kg Level 2 4 4. (i) M, L and T in the dimensional formula are (v)  T he mass of water = m3 – m1 replaced by the units of mass, length and time, (vii) The relative density of oil = respectively, in the corresponding system (m2 - m1 )/ v = m2 - m1 (ii) 1 kg m2 s-3 = 107 g cm2 s-3 (m3 - m1 )/ v m3 - m1 45. (i) Find the thickness of the glass plate using the for- (viii) Relative density of the oil = 0.9 mula TR = PSR + (C.S.D. × L.C.) (ii) L.C. = Pitch 4 7. (i) Least count = 1M.S.D. N N (iii) Thickness = 1.845 mm (ii) Measurement = M.S.R. + (V.C.D. × L.C.) 46. (i) The mass of the empty specific gravity bottle = m1 (iii) Measurement = 4.8 mm (ii) T he mass of empty specific gravity (SG) bottle 48. (i) n = 1 Tm + oil = m2 2l (iii) T he mass of empty specific gravity bottle (ii) a% = change in length ×100 + water = m3 length (iv) T he mass of oil = m2 – m1

Measurements 1.35 b% = change in time period block, say y 1 time period 8 Here, 1 M.S.D. = inch change in mass c% = mass Length of the object =(x × 1 + y × 1 ) inch = 16 8  x + 2y  (iii) Error in n =  b + c+ 2a  %  16  inch  2  Convert this value to cm 49. (i) Diameter = x + ly (iii) 5.87 cm Here x is the M.S.R. in mm, l is the least count 5 3. (i) (relative density)A = MA /VA in mm and y is the vernier coinciding division. (relative density)B MB /VB Substitute the given values and estimate x and y (ii) 2 : 3 In case of zero error, 54. (i) Least count = pitch D iameter = x + l (y – z), where z is the V.S.D. for positive zero error. (correction is negative) C orrect reading = observed reading + correc- tion for zero error (ii) (a) 1 9, 5 (b) 20, 2 50. (i) Positive zero error = zero below the index line Correction is positive of the error is negative O bserved reading = P.S.R + (C.C.D. × L.C.) (ii) 31.19 mm2 P.S.R. = 3 55. (i) d1 = 12 of a metre = distance of 2 kg from the support Least count = pitch (ii) d2 = 14 of a metre No.of C.C.D.'s (iii) Find W2 using the theorem of moments, i.e., C orrect diameter = observed reading w1d1 = w2d2 HINTS AND EXPLANATION + correction (iv) Mass = 4 kg (ii) 3.645 mm 5 1. (i) T α 1 5 6. (i) Least count = 1M.S.D. g N Zero correction = – zero error Tp Tm = gm = – [(Main scale coinciding division) gp – (V.S. coinciding division)] G iven, Tm = 2 s, gm = 6ge and gp= 2ge (ii) + 1.2 mm Here the suffixes p, m and e represent planet, 5 7. (i) Time period of a pendulum is proportional to the moon and earth, respectively. square root of its length. (ii) 3.45 s (ii) If the clock loses 1 minute in every 24 hours does it mean that the time period is greater than 52. (i) E xpress 1 M.S.D. in the different portions of the the required one, then should the length of the scale as a fraction of an inch and convert into pendulum be increased or decreased? cm. Count the total number of divisions and convert into cm to arrive at the length of object 58. (i) Correct length = (observed reading) – (zero in cm. Alternately, find the length in inches and error) convert into cm. Zero error is positive if observed reading is (ii) Count the number of M.S.D.’s from left block to greater than the correct length, and negative if 2, say x the observed reading is less than the correct reading 1 M.S.D. = 1 inch 16 (ii) –1.6 mm Count the number of M.S.D.’s from 2 to right

1.36 Chapter 1 59. 30.56 65. Correction = −  ZRP − LRP  × 10 mg  HRP − LRP  12.6 = −  10.5 − 8.5 × 10 mg = −2 × (10 mg ) 21.09  11.0 − 8.5 2.5 --------- = -8 mg = -0.008 g 64.25 --------- Correct mass = 34.23 g - 0.008 g = 34.222 g When rounded off to one decimal point, the value is 6 6. The relative density of a body is ratio of the den- 64.2. sity of the body to the density of water. R.D. has no units. It is a ratio of same physical quantity. 6 0. (a) Volume = length3 = [M0L3T0] If the units of mass, length and time are doubled, (b) Density = mass = mass = [M1L−3 T0 ] there will be the same change in the densities of both volume length3 the body and the water. So, the ratio of the density of body and density of water do not change. Hence, (c) F = Gm1m2 ⇒G = Fd 2 the relative density is a ratio which is a constant for a d2 m1m2 particular substance. ⇒ M 1L1T −2 × L2 = M −1L3T −2  67. An irregular solid when immersed in water displaces M2 3 6 1. distance (m) 4 th litre of water. The volume of the solid = 750 cm3. (l 1 litre = 1000 cm3) 160 The mass of the displaced water = 750 g. 120 HINTS AND EXPLANATION When immersed in a given liquid, it displaces 80 = 600 g. 40 The relative density of the liquid 10 20 30 40 = density of the liquid Time (s) density of water 6 2. 1 yotta metre = 1022 metre = 1019 km = mass of liquid of volume (V) = 600 g mass of water of same volume(V) 750 g 1 63. Error = -2 div = -2 × 100 mm = -0.02 mm Relative density = 0.8 64. The knife-edges in a physical balance increase its ∴ Density of the liquid is 0.8 g cm−3 sensitivity. = 0.8 × 103 kg m−3. = 800 kg m­ Level 3 pitch T = 2π m 6 8. (i) z No.of C.C.D.'s mg D imensional formula of z mg = weight of the body = (dimensional formula of y)2 dimensional formula of x The downward force acting onthe bob. (ii) [M1 L0 T0] W hat is the net downward force when the bob is immersed in water? 6 9. (i) T = 2 π  70. (i) Mass measured is accurate if the beam is g horizontal. Multiplying and dividing by m (m = mass of (ii) If the beam is not horizontal, does it affect the turning points observed? the bob), on the right-hand side,

Measurements 1.37 71. (i) Slope of the graph = mass flow rate S ince y < 10 and 0.05 y < 0.5, 9.75 can be T = 0 represents initial condition written as 9.5 + 0.25 (9.5 mm) + (0.25 mm) = (x mm) + (0.05 y mm) Density = mass x = 9.5 = (MSR) × (L.C.) volume Since 1 MSD = 0.5 mm = mass flow rate The MSR = 9.5 = 19 and 0.05 y 0.25 volumetric flow rate 0.5 (ii) At t = 0 the vessel is empty, and for t > 5 min, y = 5 there is no increase in mass of the vessel. What could be the reasons for the mass of the vessel MSR = 19, VCD = 5 to remain constant? (b) 9.75 = (x + ly) – 0.35 What does the inclined line AB indicate? (x + ly) = 9.75 + 0.35 (iii) 16.8 kg = 10.10 (iv) 21 litres = 10 + 0.1 10 (v) 800 kg m–3 0.5 x = 10 = MSR = = 20 72. (i) Calculate L.C. 0.05y = 0.1 pitch y = 0.1 = 2(= VCD) division on 0.05 = number of circular scale MSR = 20,VCD = 2 (ii) P.S.R. = nth division × pitch 7 5. Pitch = 5 mm = 0.5 mm , least count (iii) Observed reading = P.S.R + (H.S.R. × L.C.) 10 (iv) Zero error = – [H.S.R. + (N – n)× L.C.] = 0.5 = 1 HINTS AND EXPLANATION (v) Correction is positive if error is negative. N 2N (vi) Correct measurement = o bserved reading Diameter = P.S.R. + C.SC.D ∞ L.C. + correction. = xp + y P = p  x + y (vii) Correct measurement = 7.25 mm N  N  7 3. 1 MSD = 1 mm x + y = 5.48 N Pitch = 5 MSD = 1 MSD = 1 mm y y 5 Since N < pitch, x+ N = 5+ 0.48 Least count = P x = 5 N y N 1 mm = 100 N = 0.48 ⇒ y = 0.48N 1µm = Thus, the CSCD depends on the number of divi- N = 500; P = (N) × (L.C.) = 500 × 1 μ m = 0.5 mm sions on the circular scale for example, if the number Pitch = 0.5 mm of circular scale divisions is 100, the CSD is 48. P.S.R = 5 1 74. (a) Least count = 1 MSD L.C.= 2N N = 0.5 mm = 0.05 mm C.SC.D = 0.48 N 10 76. Average of left turning points = 5 + 6 + 6 = 17 3 3 Diameter = x + ly = 5.6 ≈ 5.7 Average of right   9  .75 mm = (x + 0.05y) mm turning points = 18 + 18 = 18 2   = x + 0.05 y

1.38 Chapter 1 Mean of left and right turning points = 5.7 + 18 Where p= (H.R.P −Z.R.P) × 0.01 = 2 H.R.P −L.R.P 23.7 2 = 11.85 ≈ 11.8 = 59.990 + 1.2 × 0.01 2 (Zero-resting point (ZRP) = 11.8 = 59.990 + 0.6 × 0.01 = 59.996 g After placing the substance, the resting point (RP) = 13. 77. Volume of the liquid displaced = 80 cm3 = volume of sinker + volume of solid body. This RP is greater than ZRP, hence, it is higher- resting point (HRP) = 13. Volume of sinker = 50 cm3. ∴ (80 cm3) = (50 cm3) + volume of solid body. When 10 mg is added, RP obtained = 11. ∴ Volume of the solid body = 80 – 50 = 30 cm3. Mass of the body = d × v = 0.5 × 30 = 15 g. This RP is less than ZRP, hence, it is lower-resting point (LRP) = 11. Weight of the body = 15 gwt. Given, mass at LRP = 60 g. Mass at HRP = 59.990 Mass of the body, M = Mass at HRP + P g HINTS AND EXPLANATION

2Chapter Kinematics REMEMBER Before beginning this chapter you should be able to: • Define terms such as rest and motion, scalar and vector, distance and displacement, speed and velocity, acceleration and retardation • Recall the concept of motion and its applications KEY IDEAS After completing this chapter you should be able to: • Understand the term kinematics • Know the types of motion: projectile and uniform circular motion • Define the scalar and vector quantities • Understand the basic terms and to determine the equations of motions relating to these terms • Derive the equations of motion for the objects projected upwards and falling freely under gravity • Discuss the motion of an object on a plane • Explain the motion of objects by graphical representation

2.2 Chapter 2 INTRODUCTION In the present chapter, the focus is on the study of equations which describe the motion of a body and discuss the properties of freely falling bodies, vertically projected bodies and bodies undergoing projectile motion. At first, let us review some important terms and concepts. Kinematics is the branch of mechanics dealing with the study of the motion of particles, without taking into account the forces and energies involved. A particle (point object) is an object without extent. A particle is said to be at rest if its position, relative to the surroundings, does not change with respect to time. A particle is said to be in motion if its position, relative to the surroundings, changes with respect to time. Motion and rest refer to the state of bodies described in relation to its surroundings. Example: Consider any building on the Earth. It is at rest with respect to the Earth. Since the Earth revolves a round the sun as well as it rotates on its own axis, all the objects on the Earth are in motion. Hence, there is no absolute rest. It means the building is at rest with respect to its surroundings but in motion with respect to sun. Types of Motion In day to day life, we come across various types of motion, like the motion of the planet, the motion of wind, etc. Motion can be classified into: 1. Random motion 2. Translational motion 3. Rotational motion 4. Oscillatory or vibratory motion Random Motion In this type of motion, the particles move randomly, i.e., they do not move along a definite path. Example: The motion of dust particles in wind or in air. Translational Motion In this type of motion, every particle of the body has the same displacement. Translational motion can be along a straight line or along a curved path. The motion along a straight line is called rectilinear motion and the motion along a curved path is called curvilinear motion. Example: Rectilinear motion: Train speeding along a straight track. It is also referred to as one dimensional motion. Example: Curvilinear motion: The path of a ball hit by a batsman for a sixer. This type of motion is also referred to as a two or three dimensional motion.

Kinematics 2.3 Rotational Motion If the particles of the body revolve in a circle about the same axis, then the motion is said to be rotational. Examples: 1. Rotation of Earth on its axis 2. Pulley used in drawing water 3. Merry Go round 4. A great wheel 5. Motion of a fan. Oscillatory or Vibratory Motion A to and fro motion about a fixed point is called oscillatory or vibratory motion. Examples: 1. When the string of a guitar is plucked, it performs an oscillatory motion. 2. The motion of the pendulum of a clock. SCALARS AND VECTORS Physical quantities that can be defined using magnitude only are known as scalar quantities. Examples: Distance, speed, mass, density, temperature. Physical quantities that can be defined only if both its magnitude and direction are specified are called vector quantities. Examples: Velocity, acceleration, force, torque. Distance is the length of the path from the initial position to the final position, traced by the particle while in motion. It is a scalar quantity, and is path-dependent. Example: Consider two places A and B. One can reach B from A by three different ways: 1. Along ACDB C1 D 2. Along AOB 3. Along the straight line AB 3 B The length of the path varies. Hence, the distance travelled is not the same in A 2 the three cases though the initial and final positions are the same. S.I. unit of distance is metre (m). O C.G.S. unit of distance is centimetre (cm). FIGURE 2.1 DISPLACEMENT It is the length of the directed straight line connecting the initial and the final positions of a body in motion in a given time interval. It refers to the change of position with reference to direction. Displacement is a vector quantity, and is independent of the path.

2.4 Chapter 2 The magnitude of a displacement is the length of the shortest path from the initial to final position, i.e., it is the length of the line segment joining the initial position and final position. The displacement vector is directed away from the initial position and towards the final position. The unit of distance as well as that of the displacement is centimetre in C.G.S. system and metre in S.I. system. Example: Let us consider the example given earlier for distance. The displacement is always the length of the straight line AB, (the shortest distance) and is directed from A to B (irrespective of the path traversed) Displacement = AB Note AB ≠ BA AB = BA But, Vectors are represented by directed line segments. The arrow of the directed line segment indicates the direction of the vector. The length of the directed line segment drawn to scale represents the magnitude of the vector. Example: If a person moves along a straight path in the east direction for a distance of 3 km (P to Q) and then turns towards right and moves straight and covers a distance Q East of 4 km (towards south and reaches a point R), the total displacement of the P 3 km person is obtained by drawing a straight line from P to R, as shown in the Fig. 2.2. 4 km 5 km If the length of the line segment PQ is 3 cm, the scale taken is 1 cm = 1 km, then the length of the line segment QR would be 4 cm and accordingly R South we get the length of the line segment PR as 5 cm. As the scale taken is 1 cm for 1 km, the magnitude of the total displacement is 5 km and the direction FIGURE 2.2 is nearly along south-east. ∴ PR = PQ + QR ( )Now, the total displacement PR = 5 km along approximate south-east is obtained as ( ) ( )the sum of the displacements 3 km along east PQ and 4 km along south QR . Thus, the two individual displacements PQ and QR are considered as the “component” of the displacement PR . Thus, the vector PR is said to be resolved into PQ and QR , the individual components that are perpendicular to each other. This representation of one vector as a sum of two component vectors is called resolution of a vector.

Kinematics 2.5 SPEED Speed is the rate of distance travelled. It is the ratio of the distance travelled to the time taken to cover that distance. Speed = distance Time It is a scalar quantity. The unit of speed is cm s–1 in C.G.S. system and m s–1 in S.I. system. Instantaneous speed is the speed of a particle at a given instant. It is defined as the ratio of the distance travelled in an extremely small interval of time tending to zero. Average Speed It is the ratio of the total distance to the total time taken. x4 x3 Average speed = Total distance = s distance travelled • • Total time t x2 • t4 x1 • If x1, x2, x3, x4 are the distances travelled in the time intervals t1, t2, t3 and t4, respectively, then Average speed = x1+x 2 +x3 +x4 t1 t2 t3 t1+t2+t3+t4 time taken Average speed is the average of initial and final speed if the particle is in motion such that its speed changes (either increases FIGURE 2.3 or decreases) at a constant rate. Uniform Speed A body is said to be moving with uniform speed if equal distances are covered in equal intervals of time. Example: A bike is moving on the road at 40 kmph speed that means for every one hour it is covering 40 km. Variable Speed A body has a variable speed if it does not cover equal distances in equal intervals of time. Example: A train while departing from railway station increases its speed gradually that means it is moving with variable speed. VELOCITY Velocity is the ratio of displacement to the time interval during which the displacement has occurred. Velocity (v) = Displacement = s Time t It is a vector quantity. The magnitude of velocity is expressed in cm s–1 in C.G.S. system and in m s–1 in S.I. system. The direction of velocity is along the direction of displacement.

2.6 Chapter 2 Uniform Velocity A body is said to have uniform velocity if it undergoes equal displacements in equal intervals of time. Since displacement is a vector, equal displacements implies the body is moving along a straight line path. Thus, a body moving with uniform velocity is in motion along a straight line path with a constant speed. Uniform motion and non-uniform motion In uniform motion, the distance moved by the particle in equal intervals of time is the same whether the duration of time is small or large. In non-uniform motion, the distance moved by the particle in equal intervals of time is not the same. Variable Velocity If a body undergoes unequal displacements in equal intervals of time, then the body is said to possess variable velocity. Velocity is said to be variable if there is a change either in its magnitude or in its direction or both. Instantaneous Velocity It is the velocity of the particle at a given instant. The direction of instantaneous velocity is along the tangent drawn to the curve describing the path at that instant if the body undergoes curvilinear motion. Instantaneous velocity, v = ∆s ∆t ∆s is the change in the displacement in a small interval of time ∆t. In Fig. 2.4, instantaneous velocity, v = ∆s = CB ∆t AC In the case of a body moving with variable velocity (non-uniform motion), the instantaneous velocity at any instant is given by the slope of the tangent drawn at a point in to the displacement time curve in Fig. 2.5. V = CB AC displacement∆s B displacementAC ∆ t time B A Uniform motion C    F I G U R E 2 . 4   Uniform motion time N on-uniform motion F I GU R E 2 . 5   Non-uniform motion

Kinematics 2.7 Average Velocity It is the ratio of total displacement to the total time taken. Average velocity = Total displacement Total time Average velocity is a vector quantity. Example: A 16 m B 9m C 20 m D Consider a body moving along a straight line AD, travelling 16 m in the 1st second, 9 m in the next second and 20 m in the 3rd second. Total displacement = 45 m along AD Total time = 3 s Average velocity = 45 m =15 m s-1 along AD 3 s Note 1. Average velocity of a moving body may be equal to zero but average speed cannot be equal to zero. For example, the average velocity of an athlete completing one round while running along a circular track is zero, though his average speed is not zero. 2. A verage velocity = v+u in case the body is moving with uniform rate of change of velocity. 2 Here, v is the final velocity and u is the initial velocity ACCELERATION Acceleration is defined as the rate of change of velocity. It is a vector quantity. The unit of acceleration is cm s–2 in C.G.S. system and m s–2 in S.I. system. By definition, acceleration, a = Change in velocity = v - u Time t The direction of acceleration is along the direction of change in velocity of the body. Example: A overtaking bus increases its velocity that means it is accelerating. Uniform Acceleration If the change in velocity of the body is equal in an equal intervals of time, then the body is said to move with uniform acceleration. Example: 10 A s –1 20 B s –1 30 C s –1 40 D s –1 m m m m

2.8 Chapter 2 Consider a body moving along a straight line passing through the points A, B, C, D, such that it covers the distances AB, BC and CD in equal intervals of time of one second. Let the velocity of the body at A, B, C and D be 10 m s−1, 20 m s−1, 30 m s−1 and 40 m s−1, respectively. Position Time ‘t’ Δt Velocity v Change in Rate of change (s) (s) (m s−1) velocity of velocity } }A 0 10 Δv (m s−1) Δv/Δt (m s−2) 1 } }B 1 20 10 10 1 } }C 2 10 10 30 10 10 1 D3 40 We find that during the motion of the body, the velocity is not constant, but the rate of change (increase) of the velocity is constant, i.e., the body is moving with uniform acceleration. y Plotting a graph of velocity (along y-axis) and time (along x-axis), we get a straight line as shown in Fig. 2.6. The slope of the graph is positive, Velocity, (m s –1) → 40 • and hence, the acceleration here is considered positive. 30 • 3 However, if the velocity of the body referred to in the above example at 20 • 10 • A, B, C and D is 40 m s−1, 30 m s−1, 20 m s−1 and 10 m s−1, respectively, the graph would once again be a straight line but sloping downwards. The 01 2 x slope of the line here is said to be negative, and hence, the acceleration in such cases where the velocity decreases as time progresses is referred to as Time, (s) negative acceleration or retardation or deceleration. FIGURE 2.6 Whenever the magnitude of velocity decreases, the rate of change of velocity is referred to as retardation or deceleration. It is not necessary that the rate of change of velocity, i.e., the acceleration is always constant. In situations where the rate of change of velocity of a body in motion is not constant, the body is said to be moving with non-uniform acceleration or variable acceleration. Equations of uniformly accelerated rectilinear motion These equations give the relation between velocity, distance, time and uniform acceleration of a body moving along a straight line. Garaptical methods 1. To deduce, v = u + at Consider a body having initial velocity u and velocity after time t Velocity, (m s –1) seconds is v. v -u t slop = v–u Rate of change of velocity = t = v–u v -u t t u ⇒a i.e., a = t v = u + at Time, (s) ⇒ v = u + at

Kinematics 2.9 2. To derive, s = ut + 1 at2: 2 Average velocity = v + u  (2.1)  2 (since acceleration is constant) But average velocity is the ratio of the total displacement to the total time. ∴ Average velocity = s  (2.2) t Comparing (2.1) and (2.2) s = v + u t 2 s = (u + at ) + u (∴ v = u + at) t 2 s = 2u + at t 2 ∴ s= (2u + at ) t 2 s = ut + 1 at2 2 3. To derive, v2 = u2 + 2as Average velocity = s  (2.3) t (2.4) Average velocity = v + u  2 From (2.3) and (2.4) s = v + u t 2 v + u v u s =  2  t but v = u + at; − = t  a    v u  v - u  s =  +   a  2 s = v2 - u2 2s ∴ v2 – u2 = 2as Or v2 = u2 + 2as

2.10 Chapter 2 4. To deduce the equation for the displacement of a body in the nth second: A CB • •• L et AB be the distance travelled by a body in n seconds, and AC be the distance travelled by a body in (n − 1) seconds. Thus, the distance travelled in the nth second is CB = Sn. Sn = AB − AC  (2.5) From s = ut + 1 at2 (2.6) 2 (2.7) AB = un + 1 an2  2 1 AC = u (n − 1) + 2 a (n − 1)2 Substituting (2.6) and (2.7) in (2.5), we get Sn = un + 1 an2 − [u (n − 1) + a (n − 1)2] 2 Sn = un + 1 an2 − [un − u + 1 an2 − an + 1 a] 2 2 2 Sn = un + 1 an2 − un + u − 1 an2 + an − 1 a 2 2 2 Sn = u + an − 1 a 2 Sn = u + a  n − 1  2 Or Sn = u + 1 a [2n − 1] 2 Problem Solving Tactics In deriving the equations, we have assumed that all the vector quantities are in the same direction. However, all vectors need not be in the same direction. Giving due consideration to this, ‘+’ and ‘–’ signs should be assigned to the quantities. Convention I:  In this convention, the direction of displacement is taken as positive. Convention II:  In this convention, a convenient cartesian co-ordinate system is taken with the origin at the initial position of the body, and ‘+’ and ‘−’ signs are accordingly assigned. EXAMPLE A car initially at rest moves with a constant acceleration along a straight road. After its speed increases to 40 km h–1, it moves with a constant speed and finally retards uniformly. The time intervals for the three parts of the journey are in the ratio 1 : 3 : 1. Find the average velocity

Kinematics 2.11 SOLUTION Maximum velocity = 40 km h–1 Let the time taken for the first part of the journey, i.e., during acceleration be t. From definition of average velocity we have, average velocity = s = v + u t 2 Here v is the final velocity = 40 km h–1, u is the initial velocity = 0 and s is the displacement = s1 (say) ∴ s1 =  40 + 0 t ∴  2  s1 = 20 × 1t = 20t km  (1) Since the time intervals are in the ratio of 1 : 3 : 1, if t is the time for the first part, the time taken for second and third parts are 3t and t, respectively. Thus, if s2 is the distance travelled with constant speed, s2 = v × t2 = 40 × 3t = 120t km  (2) Similarly, if s3 is the distance travelled with uniform retardation in the last part of the journey, s3 = v + u t3 .  2  Here v = 0, u = 40 km h–1, t3 = 1 t ∴s3 = (0 + 40) t = 20 t . 2 Total distance travelled = s1 + s2 + s3 = s     s = 20 t + 120 t + 20 t    s = 160 t   Average speed = Total distance Total time    = 160 t = 32 km h−1 5t Since the car is moving along a straight line, therefore, average velocity = 32 km h–1 EXAMPLE A car moves with a constant velocity of 10 m s–1 for 10 s along a straight track, then it moves with uniform acceleration of 2 m s–2 for 5 seconds. Find the total displacement and velocity at the end of the 5th second of acceleration.

2.12 Chapter 2 SOLUTION Let v1 and t1 be the initial velocity and the time for the first part where the car is moving with a constant velocity. t1 = 10 s; v1 = 10 m s–1 ∴ Distance travelled s1 = v1 t1 i.e., s1 = 10 × 10 = 100 m For the second part, i.e., when the car accelerates, a = 2 m s–2, v = ? t = 5 seconds s2 = ? u = 10 m s–1 v = u + at v = 10 + 2 × 5 v = 10 + 10 v = 20 m s–1 s2 = ut + 1 at2 2 s2 = 10 × 5 + × 2 × 52 = 50 + 1 × 25 = 50 + 25 = 75 m Total displacement s = s1 + s2 ∴ s = 100 + 75 = 175 m. EXAMPLE A bike moving along a straight road covers 35 m in the 4th second and 40 m in the 5th second. What is its initial velocity and acceleration (if the acceleration is assumed to be uniform)? SOLUTION Let s4 and s5 be the distances travelled in the 4th second and the 5th second, respectively. s4 = 35 m, s5 = 40 m From the equations of motion, the distance travelled by a body in the nth second is given by, sn = u + a n − 1 2 ∴ 35 = u + a  4 − 1  2 40 = u + a  5 − 1  2

Kinematics 2.13 i.e., 35 = u + 7 a 2 40 = u + 9 a 2 70 = 2u + 7a  (1) (2) 80 = 2u + 9a  Solving equations (1) and (2), we get a = 10 = 5m s−2 ; u = 35 = 17.5 m s–1 2 2 Acceleration Due to Gravity Objects thrown vertically upwards move up to a certain distance and then fall back to the ground. This is due to earth’s gravitational force. Due to gravitational force, all objects are accelerated towards the Earth. This uniform acceleration towards the Earth, irrespective of the mass is known as acceleration due to gravity and is denoted by ‘g’. Equations of Motion of Objects under the Influence of Gravity (Neglecting air Resistance) Here, since acceleration of a body moving vertically, either upward or downward, is due to gravity, ‘a’ is substituted by ‘g’ and the displacement ‘s’ is substituted by ‘h’. Thus, we have v = u + gt s = ut + 1 gt2 2 v2 = u2 + 2gh Where, v is the final velocity, u is initial velocity, g is acceleration due to gravity and t is time taken. ‘g’ is chosen to be positive if the body is moving towards the Earth, i.e., downward and g is negative if the body is moving in upwards direction. NOTE It is purely a matter of choice and convenience that we choose a particular direction as positive. In the above case, for example, the final result will remain the same even if we choose the upward direction as positive and downward as negative or vice-versa. Equations of Motion of a Body Dropped from a Certain Height A freely falling body is one which moves only under the influence of gravity (i.e., no other force acts on it) like air fraction.

2.14 Chapter 2 Let us consider the special case of a freely falling body which is dropped or released from rest for which the initial velocity, u = 0, and acceleration, a = +g. Taking u = 0, s = h and a = + g, we can write the equations of motion for a freely falling body, as 1.  v = u + at u = 0   v = gt h g↓ 2.  s = ut + 1 at2 2 1 h = 2 gt2 3.  v2 – u2 = 2as v2 = 2gh FIGURE 2.7 From h = 1  g  t2; 2 t = 2h , where ‘t’ is the time of descent. g Equations of Motion for a Body Projected Vertically Upwards Consider a body projected vertically upwards with an initial velocity, u. As the body moves up, its velocity decreases, since the Earth pulls the body downwards with an acceleration of ‘g’. The body moves until its velocity becomes zero (at maximum height). Such bodies are called vertically projected bodies. As the direction of motion is against the direction of Maximum height (h max) ; v = 0 acceleration due to gravity (g), the sign of ‘g’ is taken to be negative. Taking s = h and a = –g, we can write the equations of motion for such a body as follows: h 1.  v = u + at g↓  v = u – gt ↑u 2.  s = ut + 1 at2 FIGURE 2.8 2 1 h = ut – 2 gt2 3.  v2 – u2 = 2as v2 – u2 = – 2gh v at maximum height = 0; a = -g; s = h ∴ hmax = u2  (from v2 = u2 − 2gh) 2g For a body projected vertically upward, the time of ascent (ta) is the time taken by it to reach the maximum height. ta = u  (from v = u − gt) g

Kinematics 2.15 Time of Descent As derived earlier, for a body dropped from a height ‘h’, the time of descent is given by, td = 2h g A vertically projected body after reaching its maximum height (hmax) starts falling and behaves as a body dropped from maximum height. The time that it takes to reach the ground from the maximum height, can be given by, td = 2hmax , td = 2u2 / 2 g = u g g g ∴ td = u g This proves that the time of ascent is equal to the time of descent for bodies projected vertically up, and the body returns to the same level from where it is projected. Time of Flight (tf) It is the total time during which a body moving under gravity remains in air above the plane of projection. u u 2u tf = ta + td = g + g = g 2u i.e.,   t r = g EXAMPLE A ball is thrown vertically upwards with a velocity of 20 m s–1. Find the time of flight, neglecting the air resistance (g = 10 m s–2) SOLUTION When the body comes back to the initial position, the displacement s = 0. From s = ut − 1 gt2 2 0 = ut − 1 gt2 2 u2 ut = 2g t = 2u . Here ‘t’ is the time of flight, i.e., ta + td. g ∴ Time of flight, t = 2 × 20 =4s 10 Velocity on Reaching the Ground When a body is dropped from a height ‘h’, its initial velocity is zero and it attains a velocity ‘v’ on reaching the ground.

2.16 Chapter 2 As v2 – u2 = 2gh and u = 0, v = 2 gh  (2.8) We also know that for a body thrown upwards with an initial velocity u; u = 2 ghmax  (2.9) Comparing equations (2.8) and (2.9), we can say that the velocity of the body falling from a certain height h, on reaching the ground is equal to the velocity with which it has to be projected vertically upwards to reach the same height h. This shows that, the upward speed at any point in the flight of a body is same as its downward speed at that point. But note that directions of their velocities are opposite, and hence, the velocities are different though their speeds are the same. EXAMPLE A body falls from a height of 45 m from the ground. Find the time taken by the body to reach the ground. (Take g = 10 m s–2) SOLUTION Given, the initial velocity of the body u = 0 and g = 10 m s–2 Distance travelled by the body, s = 45 m Using the equation, s = ut + 1 , gt2 we get 2 45 = 0 + 1 × 10 t2 2 ∴ t2 = 90 = 9 ⇒t = 3 s 10 PROJECTILE When a body moves under the influence of gravity, it moves along a vertical straight line, only if there is no horizontal component for the initial velocity (neglecting wind velocity and air resistance) Examples: 1. A mango falling from a tree. 2. A ball projected vertically downwards. FIGURE 2.9

Kinematics 2.17 If the initial velocity has both horizontal and vertical components, the path of the body will not be a straight line, but will be a parabola, called trajectory. Examples: 1. A ball kicked horizontally from the top of a building. 2. A missile fired from a cannon. FIGURE 2.10 A billiard ball struck on a billiard table would move horizontally along the surface of the table. Here, changes in the position and the velocity and the corresponding acceleration (deceleration) would be only in the horizontal direction, i.e., either forward (+ve) or in the reverse direction (–ve). The influence of gravity on the motion of the ball is absent. However, when we consider the motion of shot-put or a long jumper, they move under the inf luence of gravity. Such objects which are given horizontal velocity and are allowed to travel under influence of gravity are called projectiles. Any projectile has two dimensional motion, i.e., combination of motion in two different directions. They are as follows: 1. U niform velocity along the horizontal direction, since no force acts on the projectile along the horizontal. Thus, the acceleration is zero along the horizontal direction. 2. Uniform accelerated motion along the vertical since it is acted upon by gravity. The motion of the projectile can be represented graphically by plotting horizontal displacement along X-axis and vertical displacement along the Y-axis. The motion of projectile is considered Similarly the path of a ball hit high by in the absence of air resistance. a batsman would be as shown below. y y Uniform horizontal motion Vertical Displacement ••• •• • • •• • • • • • • • • • • • Combined motion • • Free fall • • • θ under gravity • O• • •••• x O• • • • • • • • • • x    Horizontal Displacement FIGURE 2.11 FIGURE 2.12

2.18 Chapter 2 y Example: Consider a shot-put given an initial velocity making an angle θ with the horizontal. Plot the displacement along the X-axis (horizontal u and the Y-axis (vertical) on graph for every 1 second. θ oR The graph obtained is as shown in Fig. 2.13. This curve is called A parabola. Hence, the trajectory of a projectile is a parabola. x FIGURE 2.13 RANGE (R) It is the maximum distance covered by a projectile along the horizontal. 1. Range is OA as shown in the Fig. 2.13. 2. Range depends on the angle of projection (θ). 3. Time taken by the projectile to reach the highest point from the point of projection = time taken by the projectile to reach the ground from the highest point. The four equations of motion discussed earlier, namely, v = u + at, s = ut + 1 at2, v2 = u2 + 2as and Sn are applicable to bodies that move along a straight 2  n 1 = u + a − 2 path with uniform acceleration. u sin θy However, in the case of projectiles, their motion is not along a straight line path. Thus, the four equations of motion mentioned cannot be directly u applied to projectiles. To know the position of a projectile after certain interval θ of time, we need to know its horizontal and vertical displacements from the u cos θ point of projection in the given time, that is, we need to know the velocity and acceleration of the projectile in both the horizontal and the vertical directions. x Thus, it is necessary to resolve or split the vector (initial velocity of projection of the projectile) into horizontal and vertical directions and take the corresponding component of the velocity. If θ is the angle of projection of a projectile with velocity ‘u’, the horizontal component of velocity is ‘u cosθ’ and the vertical component of velocity is ‘u sinθ’. There exists no change in the horizontal component of velocity of the projectile as gravitational pull does not affect its horizontal motion. Thus, the horizontal component remains ‘u cosθ’ throughout the motion of the projectile. However, the vertical component of velocity, i.e., ‘u sinθ’ changes as the projectile is under gravitational pull in the vertical direction. Thus, the vertical component of velocity decreases in the upward direction, till it becomes zero at the hightest point and then the projectile falls down under gravity. The motion of the projectile is a result of both the vertical and horizontal components and it follows a parabolic path. The equations of motion can be applied independently for acquiring the horizontal and the vertical displacements or velocity of the projectile. For example, the horizontal and the vertical displacement can be obtained by using equation s = ut + 1 at2 in both the directions. 2

Kinematics 2.19 If ‘x’ and ‘y’ are the horizontal and vertical displacements, respectively, at the end of ‘t’ seconds, they are given by, x = (u cosθ)t and y = (u sinθ)t − 1 gt2. 2 UNIFORM CIRCULAR MOTION In the previous topics, we studied about the change in the magnitude of velocity when a → body undergoes rectilinear motion. Now we will consider the situation where a body can be accelerated without changing the magnitude of velocity. V C Consider a person seated in a merry-go-round which is in motion. The → person undergoes circular motion as shown. Examine the direction of motion of the person at every point. V D We observe that any body moving along a circular path changes its direction B → continuously, even though the speed is constant. Thus, the body is continuously → E accelerated. This acceleration acts towards the centre of the circular path and is V V called centripetal acceleration. A If a body moves along a circular path with constant speed such that its → acceleration is uniform, then the body is said to be in uniform circular motion. V FIGURE 2.14 GRAPHICAL REPRESENTATION OF MOTION ALONG A STRAIGHT LINE The motion of a particle (or body) can be analysed by plotting different types of graphs. These graphs are useful to study the rectilinear motion of a body. There are three types of graphs in kinematics, namely, displacement−time graph, velocity−time graph and acceleration− time graph. In plotting these graphs, time is taken along the X-axis as an independent quantity. The dependent quantity, i.e., quantity that changes with time, namely, displacement, velocity or acceleration is taken along the Y-axis. Displacement—Time Graphs Y t→ X 1. The s − t graph is along the X-axis. The displacement is zero for any amount of time. ↑ Thus, this type of graph indicates that the body is at rest, with no displacement. s Example: A bus stopped at bus stop. O FIGURE 2.15 2. T he s − t graph is parallel to X-axis. The displacement is constant and does not Y t→ X change with respect to time. Thus, the body has some initial displacement and is at rest. ↑ Example: A bus stopped at bus stop 2 metres away from the railway station. s O FIGURE 2.16

2.20 Chapter 2 Y 3. T he s − t graph is a straight passing through origin and making an angle with X  the X-axis. The body has equal displacements in equal intervals of time. ↑ Thus, the body is moving with uniform velocity. The initial displacement of the body is zero. s Example: A person riding a bike at the speed of 50 kmph on a straight road O t→ from rest. FIGURE 2.17 Y 4. The s − t graph is a straight line making a positive angle with the horizontal and has a positive intercept on the Y-axis. Thus, the body has uniform ↑ velocity with some initial displacement. s  Example: A bus start 2 m away from TIME’s building and moving at a speed X of 60 kmph. O t→ FIGURE 2.18 Y Q 5. The s − t graph of two bodies P and Q are shown in Fig. 2.19 by lines OP and ↑ P OQ, respectively. Since they are straight lines making positive angle with X-axis and passing through the origin, their initial displacement is zero and sB A they have uniform velocity. But from the Fig. 2.19, it is clear that the OC X displacement of Q is more than that of P in a given time. Thus, the velocity of Q is more than that of P. The magnitude of uniform velocity can be t→ obtained by measuring the slantness (slope) of the lines from the horizontal. FIGURE 2.19 Displacement of P and Q are given by, OA and OB, respectively. Thus, magnitude of the velocity of P is given by, vp = OA/OC. This is the slope of OP. Similarly, magnitude of the velocity of Q is given by, vQ = OB/OC. This is the slope of OQ. As OB > OA, vQ > vP, the time considered (OC) being constant. Y 6. In Fig. 2.20, the s − t graphs of two bodies R and T are shown by thick and T thin lines, respectively. It is clear from the Fig. 2.20 that the bodies are moving ↑ R with uniform velocity. Even though the initial displacement of T is less than s that of R, (As the length of intercept on Y−axis for T is less than that of R), O t→ X the slope of T is greater than that of R. Thus, the magnitude of uniform velocity of T is greater than that of R. FIGURE 2.20 7. The s − t graph of a body is as shown in Fig. 2.21. It is not a straight line, but Y is a curve. It is obvious from the Fig. 2.21 that the displacement of the body is not equal in equal intervals of time. Thus, the body has variable velocity. F E  From the Fig. 2.21, it is clear that OA < AB < BC < CD < DE < EF, which ↑ CD are the successive displacements of the body in successive intervals of time. B s X Thus, the displacement per unit time, i.e., velocity of the body increases with time. Hence, the body is moving with acceleration. A O 12 3 4 56 Example: A train while departing from the station. t→ FIGURE 2.21

Kinematics 2.21  The velocity of the body at any instant of time can be obtained by drawing Y a line that touches the curve at the given point (also called tangent) and measuring the slope of the tangent as shown in the Fig. 2.22. ↑ Example: Speed of a train at 8:00 a.m. is 40 kmph and at 3:30 a.m. is 60 sA kmph. s1 P The displacement of the body at an instant of time t1 is s1. Point P on the C BX graph indicates this position. If the velocity of the body at the instant t1, is to O t1 be calculated, a tangent AC is drawn at the point P, and the slope of the t→ FIGURE 2.22 tangent AC that gives the instantaneous velocity can be calculated by using the expression. AB where BC gives the time interval considered and AB BC gives the displacement of the body in the time interval. 8. The s − t graph of a body is as shown in Fig. 2.23. The graph is a curve. Y v2 This indicates that the velocity of the body is not uniform. The tangents v1 Q at the points P and Q on the curve give the velocity of the body v1 and v2 at the instants t1 and t2, respectively. It is clear from the Fig. 2.23 that ↑P t2 X the slope of the tangent at P is greater than the slope of the tangent at Q. Thus, v2 < v1. This implies that the velocity of the body is decreasing s with time, i.e., the body moves with retardation in the positive direction (direction of displacement) as the displacement of the body increases O t1 with time. t→ Example: A train arrived to the railway station. FIGURE 2.23  9. T he s − t graph of a body is as shown in Fig. 2.24. The graph is a straight Y line making an obtuse angle with the positive X-axis. The body has some initial displacement and this displacement decreases with time. The ↑ decrease in displacement is equal in equal intervals of time. Thus, the body moves in opposite direction to the direction of displacement with s 180° > θ > 90° uniform velocity. X E xample: A bus which is already travelled certain distance on a straight O t→ road returning back with the same constant speed. FIGURE 2.24 10. The s − t graph of a body is as shown in Fig. 2.25. It is a curve, and thus, the Y velocity of the body is non-uniform. The velocity of the body at two different P v1 instants of time t1 and t2 is given by the slope of the tangents at the points P and Q as v1 and v2, respectively, as shown in the Fig. 2.25. It is obvious ↑ v2 Q from the figure that the slope at Q is more than that at P indicating that s v2 > v1. Also, as time increases, displacement of the body decreases as evident from the Fig. 2.25. Thus, the body is moving in opposite direction to the O t1 t2 X direction of displacement with acceleration. t→ FIGURE 2.25

2.22 Chapter 2 Y 11. The s − t graph of a body as shown in the Fig. 2.26 is a curve. The body has initial displacement and as time increases, the displacement decreases. Thus, A v1 the body is moving in opposite direction to the direction of displacement ↑ with non-uniform velocity. The velocity of the body at two instants of time s B v2 ‘t1’ and ‘t2’ is given by the slopes of the tangents drawn to the curve at points A and B as v1 and v2, respectively. It is obvious from the Fig. 2.26 that the O t1 t→ t2 X slope of the tangent at A is greater than the slope of the tangent at B. Thus, v1 > v2. This implies that the body is moving with retardation in the opposite FIGURE 2.26 direction to the direction of displacement. Uses of Displacement–time Graph 1. The position of the body at any intermediate time can be found out. 2. Slope of the tangent to the curve gives instantaneous velocity at that moment. 3. A verage velocity is the slope of the line passing through the initial and final position on the curve. 4. Nature of motion can be determined. 5. T he nature of curve describes the type of motion it was. Y Velocity—Time Graph ↑ Similar to the displacement−time graph, the velocity−time graphs can be any one or the combination of more than one of the following types. v 1. The graph is a straight line along the X-axis. This implies the body is at rest. O t→ X FIGURE 2.27 Y 2. The graph is a straight line parallel to the X-axis. This implies that the velocity does ↑ not change with respect to time. Thus, the graph indicates the uniform velocity. v O t→ X FIGURE 2.28 Y 3. T he graph is a straight line making an acute angle with the positive X-axis and passing through the origin. Thus, the initial velocity of the body is zero and has equal ↑ increase in velocity in equal intervals of time. Thus, the body moves with uniform v acceleration. The magnitude of the acceleration is given by the slope of the line O t→ X (Note: slope of s−t graph gives the velocity). FIGURE 2.29

Kinematics 2.23 4. T he v − t graph of a body is as shown in Fig. 2.30. The graph is a straight line Y X making an acute angle with the positive X-axis and having an intercept on the ↑ positive Y-axis. Thus, the body has some initial velocity u and moves with uniform v acceleration. O t→ FIGURE 2.30 5. The v − t graphs of two bodies P and Q are as shown in Fig. 2.31. The graphs Y P are straight lines making acute angle with the positive X-axis and having intercepts Q on the positive Y-axis. Thus, both the bodies have initial velocities and move ↑ with uniform acceleration. The Y-intercept of P is less than that of Q. This indicates that the initial velocity of P is less than that of Q. But the slope of P is v greater than that of Q. This indicates that the magnitude of uniform acceleration of P is greater than that of Q. O t→ X FIGURE 2.31 6. The v − t graph of a body is as shown in Fig. 2.32. As the curve begins at the Y D origin, the initial velocity of the body is zero and moves with non-uniform or a2 ↑ variable velocity. The acceleration of the body at instants t1 and t2 is given by C the slopes of the tangents drawn to the curve at the points C and D as a1 and a2, v respectively. From the Fig. 2.32, it is obvious that a2 > a1. Thus, the body O a1 X moves with increasing acceleration. t1 t2 t→ FIGURE 2.32 7. The v − t graph of a body is as shown in the Fig. 2.33. The graph is a curve Y F which implies that the body is moving with variable velocity. The acceleration a2 ↑ of the body at instants t1 and t2 is given by the slopes of the tangents drawn to E the curve at points E and F as a1 and a2, respectively. From the Fig. 2.33, it is v a1 obvious that a1 > a2. Thus, the body moves with decreasing acceleration. But O it is not retardation as the velocity of the body is not decreasing with time. As t1 t2 X the curve begins at the origin, the initial velocity of the body is zero. t→ FIGURE 2.33 8. T he v − t graph of a body is as shown in Fig. 2.34. It is a straight line making Y 180° > θ > 90° an obtuse angle with the positive X-axis. There is some initial u velocity of X ↑ the body and has an equal decrease in velocity in equal intervals of time. t→ v Thus, the body moves with uniform retardation, the magnitude of which is O given by the slope of the line. FIGURE 2.34

2.24 Chapter 2 Y 9. The v − t graph of a body is as shown in Fig. 2.35. The graph is not a straight line and is a curve with the velocity of the body decreasing with time. This G H indicates that the body moves with non-uniform retardation. The slopes of ↑ the tangents drawn to the curve at the points G and H gives the retardation v of the body at the instants t1 and t2, respectively. As the slope at G is greater O t1 t → t2 X than the slope at H, the body moves with decreasing retardation. FIGURE 2.35 10. T he v − t graph of a body is as shown in Fig. 2.36. It is a curve indicating Y decrease in velocity with time. Thus, the body has variable retardation. The retardation of the body at t1 and t2 instants of time are given by the slopes of I the tangents drawn to the curve at points I and J, respectively. It is obvious ↑J from the Fig. 2.36 that retardation increases with time. Thus, the body moves with increase in retardation having some initial velocity u and the final v velocity being zero. O t1 t 2 X t → FIGURE 2.36 Uses 1. Nature of motion can be determined. 2. Velocity at any instant can be found out. 3. Area under the curve gives the displacement of the body. 4. Slope of the tangent to the curve gives the instantaneous acceleration. 5. Average acceleration is given by the slope of the line segment joining initial velocity and final velocity. 6. Equations of motions along straight line can be determined. Y C Example: Consider a particle having initial velocity u. It is uniformly at accelerating at the rate a, for time t and covers distance s, gaining Velocity final velocity v. Graphically, it can be represented by the line BC in Dv v – t graph. B u From the Fig. 2.37, it is clear that OA = t, OB = u, AC = v, CD = at, distance travelled = Area of shaded region. OtA time X FIGURE 2.37 s = Area of rectangle OADB + Area of triangle BCD. s = ut + 1 t × at s = ut + 1 at2 2 2 EXAMPLE From the Fig. 2.38 which is a v – t graph of a body, (i) find the deceleration of the body in the region BC. (ii) find the total distance travelled by the body. Given, v is in m s–1 and time in s.

Kinematics 2.25 SOLUTION (10 − 5) m s−1 Y GF = FE (4 − 3) s B Deceleration = slope of BC = = 5 m s–2 10 Total distance travelled = Area of the triangle ABC + Area of rectangle ↑ CD OADE = 1 AC × BC + AD × DE v 2 5 A GF E 1 × 4 × 5 34 5 X  2 0 S= + (5 × 5) = 35 m t→ FIGURE 2.38 Acceleration–Time Graph In this type of graph as shown in Fig. 2.39 acceleration is plotted along Y-axis and time along X-axis. YY a a AB tX tX (i) (ii) FIGURE 2.39 1. a = 0, body moving with constant velocity. 2. ‘a’ is constant, body moving with uniform acceleration. Uses of Acceleration–time Graph 1. Acceleration at any intermediate time can be determined. 2. Area under the curve gives the change in magnitude of velocity that is (v − u). EXAMPLE Y In the Fig. 2.40, find the change in velocity at t = 5 seconds. Given a is in m 10 AB s–2and t is in s. a SOLUTION Here, we find that the acceleration is not constant, and we cannot use the O5 X equation v = u + at. t ∴ Change in velocity = area under the a – t curve. FIGURE 2.40 = 1 × 5 × 10 = 25 m s–1. 2

2.26 Chapter 2 y Graphical Method—Solutions ↑ B Let us consider a body starts from rest and increases its velocity upto 40 ms-1 for some time. Further it travelled some distance with 40 ms-1 velocity and v comes to rest after certain time. 40 A Let the time intervals given for acceleration, uniform velocity and O DE C x retardation be x, 3x and x; represented by OD, DE and EC, respectively. t→ Area under the graph gives the total displacement. FIGURE 2.41 DE Time, s Displacement, m C F B G A BC Velocity, m s-1 H B A C DE FG H Acceleration, m s-2 Time, s G H A Time, s F DE FIGURE 2.42 ∴ S = Area (OAD + ABED + BEC) = 1 (OD) (AD) + (DE) (AD) + 1 (EC) (BE) = 1 x (40) + (3x) (40) + 1 x (40) = 160x 2 2 2 2 km. Total time t = x + 3x + x = 5x h. ∴ Average velocity = s = 160x km h−1 = 32 km h−1. t 5x Given above are graphs that describe the motion of a person on his motorcycle. From the graphs, these are the points that can be concluded about his motion in different intervals of time. A  -  Person starts the motorcycle, u = 0. AB  -  He moves towards right at a constant acceleration, v = at. BC  -  Uniform motion; he is going at constant speed, a = 0. CD  -  He applies brakes and starts to slow down; acceleration is negative (deceleration). D  -  He stops completely. DE  -  He takes rest.

Kinematics 2.27 E  -  Starts his motorcycle again. EF  - He moves toward left at constant acceleration (increasing speed). Velocity and acceleration negative due to the motion in opposite direction FG  -  Moves in opposite direction (to left) at constant speed. GA  -  Applies brakes and starts to slow down (deceleration). H  -  He stops completely. (velocity = 0) NOTE In plotting the graphs, all vectors in the forward direction are taken as positive and in the opposite direction are taken as negative.

2.28 Chapter 2 TEST YOUR CONCEPTS Very Short Answer Type Questions 1. Give the three equations of motion for a particle 16. Why is distance called scalar quantity and displace- moving in one dimension. ment a vector quantity? PRACTICE QUESTIONS 2. A stone is dropped from the top of a tower and 17. If a body does not change its position with respect to allowed to travel freely under gravity. What is the its surroundings, the body is said to be ________. initial velocity of the stone? 1 8. What is uniform circular motion? 3. If the ratio of the final velocities of two bodies fall- ing freely is 4 : 3, then the ratio of the heights from 19. G ive the C.G.S. and M.K.S. units of velocity and which they fall is ________. speed. 4. G ive two examples of projectile motion. 20. Retardation is a _____ quantity (scalar/vector). 5. What is meant by the range of a projectile? 2 1. Which physical quantity is plotted on X-axis for all types of graphs representing motion of a body? 6. E xplain the motion of a body undergoing circular motion with an example. 2 2. W hat is meant by acceleration and retardation? 7. A particle in uniform circular motion has uniform 23. A freely falling body travels with ______________. speed but non-uniform _________. 2 4. W hen is a body said to have zero acceleration? 8. D efine time of ascent and time of descent. 9.  (i) When is a body said to be at rest? 2 5. One body is dropped from the top of a tower and the (ii) When is it said to be in motion? other is projected vertically upwards. Can the accel- eration due to gravity be the same? 10. A body projected with when is it said to be in motion a certain velocity making an angle, other than 90°, to 26. A particle moves from P to Q with a uniform velocity horizontal is known as ________. V1 and Q to P with a velocity V2. Its average velocity is _____________. 11. What are scalar and vector quantities? Give some examples. 2 7. Give examples of bodies moving under the influence of gravity. 1 2. Does the velocity remain the same in case of uniform circular motion? 28. A body moving along a straight line between two places moves with a velocity v1 for first half of time 1 3. W hat are the C.G.S. and M.K.S. units of distance and to travel between the places and with a velocity v2 for displacement? the remaining half time. Then its average velocity is ______. 1 4. A particle in one dimensional motion, moving with constant velocity must have ________ acceleration. 29. W hat is the acceleration of a body when the velocity remains constant? 1 5. What is meant by time of flight? 3 0. W hat is a trajectory? Short Answer Type Questions 31. Derive an expression for maximum height reached 34. D erive an expression for the time taken by a body when a body is projected vertically upwards. which is thrown vertically upwards to reach maxi- mum height. 3 2. A water tank is placed on the top of a building of height 19 m. Water overflowing from the tank was 35. If a runner with a certain initial velocity moves with found to reach the ground in 2 seconds. Find the uniformely acceleration in such away that he cov- height of the tank. (Take g = 10 m s–2) ers 200 cm in the 2nd second and 220 cm in the 4th second, then find his initial velocity and the 33. W hat are the different types of motion? Give an acceleration. example for each.

Kinematics 2.29 36. S how that the time of ascent and the time of descent 43. From the velocity-time graph given below, for aVelocity (m s –1 ) are equal for a body in vertical motion. body projected vertically upwards, OR (i) find the velocity of projection (ii) maximum height attained by the body Show that the time of flight is equal to (2u)/g. +20 37. D erive v = u + at. 02 4 38. The driver of a TGV travelling at a speed of 90 m –20 time (s) s–1 sights a truck on the rail track at a distance of 1 km ahead. Then he applies the brakes to decelerate 44. Determine ‘a’ of the object which the train at the rate of 5 m s–2. What is the distance (a) moves in a straight line with a constant speed of travelled by the train before coming to rest? Will the train collide with the truck? 20 m s–1 for 12 seconds. (b) changes its velocity from 0 m h–1 to 360 m min–1 39. D erive an expression for time of descent. in 4.2 s. 40. A body dropped from the top of a cliff reaches the ground in 5 s. Find the height of the cliff. 45. Give the equations of motion of a body falling under gravity, being dropped from a certain height. 41. Give the equations of motion of a body projected vertically upwards. 42. A train travels from one station to another station at an average a speed of 40 km h–1 and returns back to the first station at an average speed of 60 km h–1. Find the average speed and average velocity of the train? Ignore the stoppage time at the second station. Essay Type Questions 46. Explain the characteristics of the following graphs. 48. Derive s = ut + 1 at2. 2 A.  Displacement–time PRACTICE QUESTIONS B.  Velocity–time 4 9. O btain s = ut + 1 at2 by graphical method. 2 C.  Acceleration–time 47. D erive Sn = u + a (2n – 1). 50. Derive v2 – u2 = 2as. 2 *For Answer Keys, Hints and Explanations, please visit: www.pearsoned.co.in/IITFoundationSeries CONCEPT APPLICATION Level 1 4. If a body moves with constant velocity, its displace- ment depends on the square of time taken. Direction for questions 1 to 7 State whether the following statements are true or 5. When two balls of different masses are thrown verti- false. cally upwards with the same initial speed, the heavier body rises to greater height than the lighter body. 1. A body moves with retardation when it is projected vertically upwards. 6. Equations of motion are applicable only when a body moves with uniform velocity. 2. A body is projected vertically up. On reaching maxi- mum height, its velocity becomes zero. 7. The distance travelled by a freely falling body in every successive second is the same. 3. Velocity-time graph can be used to find the displacement.

2.30 Chapter 2 Direction for questions 8 to 13 Direction for questions 15 to 42 Fill in the blanks. For each of the questions, four choices have been 8. The ratio of velocities acquired by a freely falling provided. Select the correct option. body starting from rest at the end of 1 second and 15. The ratio of magnitude of displacement to distance is 2 seconds is ________. always 9. If a stone is thrown vertically up and it is caught after (a) less than 1 time ‘t’ seconds, then the maximum height reached (b) greater than 1 by it is ________. (c) equal to 1 10. Area under the velocity–time graph gives ________. (d) less than or equal to 1 11. The ratio of magnitude of average velocity to average 16. The ratio of the heights from which two bodies are speed is ________. dropped is 3 : 5, respectively. The ratio of their final velocities is 12. The directions of both displacement and average velocity are ________. (a) 5 : 3 (b)  3 : 5 13. ____________ is produced in a body whenever (c) 9 : 25 (d)  5 : 3 there is a change in its velocity. Direction for question 14 1 7. The variation of the velocity of a particle moving Match the entries in Column A with the appropriate along a straight line is illustrated in the graph given ones in Column B. below. The distance covered by the particle in 4 sec- onds is _________ m. 14. Column A Column B 20 A. Uniform Velocity ( ) a. s t (m v –1)10 s s B. Uniform acceleration ( ) b. t 0 1 23 4 with initial velocity s t PRACTICE QUESTIONS (a) 20 C. Uniform acceleration ( ) c. (b) 40 (s) (b)  35 (d)  55 D. Increasing acceleration ( ) d. t 18. An ant moves from one corner of a hall to the diago- V nally opposite corner. If the dimensions of the floor of hall are 8 m × 6 m, the displacement of the ant is E. Uniform retardation ( ) e. t ________ m. s (a) 14 (b)  10 (c) 28 (d)  2 F. Decreasing acceleration ( ) f. t 19. The figure given below shows the displacement–time at steady rate a graph of the two particles P and Q. Which of the fol- lowing statements is correct? t G. Uniform acceleration ( ) g. a P with initial displacement t s↑ Q H. Body at rest with initial ( ) h. a O displacement t t→ (a) B oth P and Q move with uniform equal speed. (b) P is accelerated and Q is retarded.

Kinematics 2.31 (c) Both P and Q move with uniform speed, but the (c)  (d)  speed of P is more than the speed of Q. a s t (d) B oth P and Q move with uniform speeds but the Ot speed of Q is more than the speed of P. O 20. When brakes are applied, the velocity of a car changes 2 7. The velocity of a body is given by the equation v = from 40 m s−1 to 10 m s−1 in 10 s. The acceleration 6 − 0.02 t, where t is the time taken. The body is produced in it is ________ m s−2. undergoing (a) −3 (b)  3 (a) uniform retardation. (c)  −5 (d)  5 (b)  uniform acceleration. 2 1. If a body starts from rest and moves with uniform (c) non-uniform acceleration. acceleration, then (d)  zero-acceleration. (a) v ∝ t (b)  s ∝ t 2 8. A body starts from rest and moves with uniform (c)  v ∝ s (d)  s ∝ t acceleration for 2 s. It then decelerates uniformly for 22. If a body is projected vertically upwards, then on 3 s and stops. If deceleration is 4 m s–2, the accelera- reaching maximum height, its tion of the body is _______ m s–2. (a) velocity is zero and the acceleration is not zero. (a) 10 (b) 8.7 (b) v elocity is not zero and the acceleration is zero. (c) 4 (d) 6 (c) b oth velocity and acceleration are not zero. 2 9. Density is a __________ quantity. (d) b oth velocity and acceleration are zero. 2 3. The ratio of the times taken by a body moving with (a) scalar (b)  derived uniform acceleration in reaching two points P and Q along a straight line path is 1 : 2. If the body starts (c)  neither (1) nor (2) (d)  Both (1) and (2) from rest, then the ratio of the distances of P and Q from the starting point is 3 0. A particle moves from P to Q with a uniform veloc- ity v1 and Q to P with a velocity v2. If it moves along (a) 4 : 1 (b)  1 : 4 a straight line between P and Q, then its average velocity will be ______. (c)  2 : 3 (d)  3 : 1 2 4. A body with an initial velocity of 3 m s−1 moves with (a) 2v1v2 (b) v1v2 PRACTICE QUESTIONS an acceleration of 2 m s−2. Then the distance trav- v1 + v2 2 elled in the 4th second is _________ m. (c) v1 + v2 (d) zero (a) 10 (b)  6 2 (c)  7 (d)  28 31. If a body is projected vertically up from a point and it returns to the same point, its 2 5. A bus travels the first one-third distance at a speed of (a) average speed is zero, but not average velocity. 10 km h−1, the next one-third distance at a speed of (b) B oth average speed and average velocity are zero. 20 km h−1 and the next one-third distance at a speed (c) average velocity is zero but not average speed. of 30 km h−1. The average speed of the bus is (d) B oth average speed and velocity depend upon the (a) 20 m s−1 (b)  50 m s−1 path. 11 180 (c)  11 m s−1 (d)  30 m s−1 3 2. If a ball thrown vertically up attains a maximum height of 80 m, then its velocity of projection is 26. Which of the following graphs indicates that a body (Take g = 10 m s–2) is undergoing retardation? (a) 40 m s−1 (b) 20 m s−1 (a)  (b)  (c) 50 m s−1 (d) 10 m s−1 v s 3 3. A vertically projected down body travels with Ot Ot (a) uniform velocity. (b) uniform speed.

(1) Linear path (2) Elliptical path (3) Parabolic path (4) Spherical path 40. In the graphs given below which graph indicates that a body is at rest? (1) Y (2) Y (3) Y (4) Y 2.32 Chapter 2 s v v s Ot XO XO t X Ot t X (c) uniform acceleration. 41. Which of the following graphs given below is impossible? (d) uniform retardation. (1) (2) (3) (4) s ss s 34. A particle revolves along a circle with a uniform t tt 0 t speed. The motion of the particle is ______. 42. Is(fu 31)ca9cae2b1.so(ts 1itdv1yeasIaiftntti2vrm2tad)2aeveeilnasbttcwieomrcivtdahelesylatenhiertannarcctatthe(ei2vel)eoreraavavnlt(e1isttaro11aanlgwse2ata,2a21tc)fi2fcttooehhrlreertiatmantiionemntt1oheafactenhtdce2e,aabcovltcede1eylreriarasaanttigiodonenat2a2fcoabrc1eteimiflnoee grrt2a, sttt1uiiomacnndcete2ostb-1efing (a) one dimensional (b) two dimensional (c) translatory (d) oscillatory If u is the initial velocity, of a body projected with the abot d+y ais t( )(3) a1 a 2 a1t1 a 2 t 2 the horizontal, then the maximum43. t1 t 2 t1 t 2 an angle θ with height reached ( ) ( )3 5. (4) 11 ( )2 an d a1t1 + a2t2 2 ttravel, + tthen the initial (a) If a body covers 26 m and 30 m in the 6th 7th seconds (obf i)ts velocity and 2 t + tacceleration of the bo1dy are2 1 2 (1) 4 m s–1, 4 m s–2 (2) 6 m s–1, 4 m s–2 ( a + a ) a t −a t(3) 10 m s–1, 8 m s1–2 u2 u2 sin θ 2(4) 0, 4m s–2 11 2 2 (a) g (b) 2g (c) (d)44. The ratio of maximum heights reached by two bodies projected vertically up is a : b, then the ratio of t + t ( t + t )their initial velociti1es of is2 12 (1) a : b (2) a2 : b ( 43)0.b :Ifa a body cov(4e) rsa :26b m and 30 m in the 6th and 7th (c) u sin θ (d) u2 sin2 θ 45. A particlesemcoovens adlosng oa cfirciutlasr trtarckavofe6l,m trahdieuns sucthhtheat tihne iatrciaolf thveeclirocuclairttyrackacnovdered 2g 2g subtendsaanccaneglleeorfa3t0iºoatntheocfenttrhe.eFinbd othde dyistaanrcee covered by the body. (1) π m m s–1,(24) 1m3π ms–2 (3) 4π(mb) 6 m s–1, 4(4)m6π sm–2 (a) 4 36. If the body is projected up into air with certain angl8e0, then the path followed by it is (c) 10 m s–1, 8 m s–2 (d) 0, 4m s–2 (a) Linear path (b) Elliptical path 4 1. The ratio of maximum heights reached by two bod- CL_9TH_PHY_CH 02.indd 80 ies projected vertically up is a : b, then the ratio of 12/27/13 12:08:02 PM their initial velocities of is 39. If the in(itcia)l vPeloacritayboof laicbodpyahtahs b oth horizontal and (vder)ticSalpchomeproincenatls apnad tiht is projected up with certain angle, then what is the path followed by it. (1 3)7L.in eaWr pahthich one(2o) fElltiphtiecalfpoatlhlowing graph indicates that the body is at rest?(3) Parabolic path (a) a : b (b) a2 : b (4) Spherical path (d) a : b 40. In the graphs given below which graph indicates that a body is at rest? (c) b : a (1) Y (2) Y (3) Y (4) Y 4 2. A particle moves along a circular track of 6 m radius s v t v s X such that the arc of the circular track covered sub- Ot XO XO t X Ot tends an angle of 30º at the centre. The distance cov- PRACTICE QUESTIONS 41. W 3h8ic.h oWf thehfoilclohwingogfraphtshgeiven fboelollwoiws iminpogssibleg? raphs given below is ered by the body is (1) impossible? (2) (3) (4) (a) π m (b) 13π m (d) 6π m s ss (c) 4π m s Level 2 t t tt 0 42. Isfu 4ca3ceb.sos idvyeStthirmaoveewilns twertivthahlsaattnhefanoccthreelearaavtebiroaongedaa1ycfcoeprlertriamotioejnet1ocfattnehdedabcovcdeeylerirsattiiocnaal2lyforutipmeftr2o, t1mandtht2ebeing end of 4 seconds, a body is released from the bal- (1) a21(t1t1goatrf22to)2iutsnudp, wthaerdd(2m)isat(1oatt11nticato22et)n2 tirsavaeclloendstbayntitinindethpeenladsetnstecoof nitds loon. Calculate the time taken by the released body ( a1 itan22 )itial a1t1 a 2 t 2 to reach the ground. t1 (3) velocit(y4). (t1t2 )  (Take g = 10 m s– 2) 43. If 4a4b.o dyAcovberas l2l6 ims adndro30pmpeindthefr6oth mand 7tthh seecotnodspofoitfs traavetlo, twhenetrheoinfitihaleveilgochitty and 47. A pendulum of length 28 cm oscillates such that its accelerat8ion0ofmthe.boAdytarethe same time, another ball is projected (1) horizontally4 m s–1, 4 m s–2 (2) fr06o,m4mms–1s,–2t4hmes–2tower. string makes an angle of 30° from the vertical when (3) (4) Find the time taken by it is at one of the extreme positions. Find the ratio of 10 m s–1, 8 m s–2 44. The ratioboof mthaxitmhume hbeiaglhltss retaocherdebayctwho tbhodeiesgprroojeuctneddve.rtically up is a : b, then the ratio of the distance to displacement of the bob of the pen- their initial velocities of is dulum when it moves from one extreme position to (1) a : b (Take g = 10 m s−2) (2) a2 : b (4 3)5.b :A a person trav(4e)ls at:heb total distance in two parts in the the other. 45. Asubptaerntidcslreaanmtaoinvogelse2aolof n:3g01ºaactwitrhcieutlcahernttraraec.kcFioonfdn6tshmteadrnaisdttiaunsscepsucecohevetdhreadtotbhyfet3hare0cbookfdtymh. e chirc−u1laritnrackthcoevered 4 8. A cannon fires a shell with a speed of 84 m s−1. When (1) π m first part an(2d) 1430π mkm h−1 in(3t)h4eπ msecond part.(4)W6πhmat is the cannon is inclined at 45°, the horizontal distance the average speed of the journey? covered is observed as 630 m. What is the percentage 80 46. A balloon starts rising from the ground, vertically decrease in the horizontal distance observed due to CL_9TH_PHY_CH 02.indd 80 upwards, uniformly at the rate of 1 m s– 1. At the air resistance? 12/27/13 12:08:02 PM

Kinematics 2.33 49. A stone is dropped from a certain height on earth and what is the velocity of the car at the beginning of the last second? it takes 12 seconds to reach the ground. If the same 58. The horizontal component of vector a is equal to stone is dropped from the same height on moon, find the vertical component of vector b . If 30° is the the time that it will take to reach the surface of the angle made by a with its vertical axis and that made moon. Ignore the air resistance. by b with its horizontal axis, then calculate the value  (Given gmoon = 1 gearth) 6 of b in terms of a . 5 0. The distance travelled by a body in the nth second 59. In the given figure, determine the force acting along x-axis if the angle between the force and x-axis is q = is given by the expression (2 + 3n). Find the initial 30° or 60°. velocity and acceleration. Also, find its velocity at the 15 N qX end of 2 seconds. 60. A person travels 6 m towards east, 8 m towards north 51. For a body that is dropped from a height, find the and 16 m towards south. What is the displacement of ratio of the velocities acquired at the end of 1 second, the person? 2 seconds and 3 seconds, respectively. 6 1. The velocity of a retarding body changes from 90 km 52. The ratio of distance described by a body falling h–1 to 36 km h–1. Find the change in its velocity in m s–1. freely from rest in the last second of its motion to that in last but one second of its motion is 5 : 4. Find 62. A person is running along a circular track of area the total time taken by the body to reach the ground. 625 π m2 with a constant speed. Find the distance travelled and displacement in 30 s and 15 s., if he has 53. A ball is thrown vertically upwards with an initial to complete the race in 30 s. velocity such that it can reach a maximum height of 15 m. If, at the same instance, a stone is dropped 6 3. In each of the questions given below, a statement is from a height of 15 m, find the ratio of distances provided. State whether the given statement is true travelled by them when they cross each other. or false. Substantiate your answer by giving the reasons. 54. A body projected vertically up crosses points A and B separated by 28 m with velocities one-third and one- (1) It is not possible for an accelerating body to have fourth of the initial velocity, respectively. What is the zero velocity. maximum height reached by it above the ground? (2) It is possible for a body undergoing linear motion 55. A body is dropped from a certain height. Plot a dis- to have displacement and velocity in opposite PRACTICE QUESTIONS placement–time, velocity–time and acceleration– directions. time graphs of the body. (3) It is not possible for a body undergoing lin- 5 6. Given below is the displacement–time graph of a ear motion to have velocity and acceleration in body moving in a straight line. Find the distance cov- opposite directions. ered in 4 seconds. Also find the displacement of the body at the end of 12 seconds. (4) It is possible for a body to have uniform speed, when it is moving with an acceleration. s (in m) 64. An object travels for 10 s with uniform acceleration {A 8 t ( in s ) along a straight line path. During this period if the 6 10 12 velocity of the object is increased from 5 m s−1 to 25 B 24 m s−1, then find the distance travelled by the body. 0 AO = OB 6 5. A body is projected vertically upward. If its velocity C −8 after 2 s is 25 m s–1, find the velocity of projection. 57. A car moves linearly with uniform retardation. If the  (Take g = 10 m s–2) car covers 40 m in the last 2 seconds of its motion,

25. A body is dropped from a certain height ‘h’ metres. Assuming that the gravitational field is nullified, after the body has travelled h/2 metres such that g = 0, discuss the motion of the body. Find an expression for the time taken by the body to reach the ground. 26. C2a.l3cu4late Cthheatpimteer o2f flight of a body which is thrown upto a height of 5 m from the ground. 27. An object projected vertically up from the top of a tower took 5 s to reach the ground. If the average 6v6e.l oAcitsytoonfe tihs perojbejcetcetdivse5rtimcalsly−1u, pfiwndarditswaitvheraavgeeloscpietyedo.f (given wg i=th1e0qumals−v1e)l.ocity. What is the ratio of centripetal 25 m s–1. Find its time of descent. (Take g = 10 m s–2) force acting on them? 28. 6Irne7m. caaAAisnsesbsuoocmdfoyiannnisgstaodntbrhtol,aiptbqpuutetehdetphfrregoormvjaeevcriatttiialccteeaiorlotncaraoinlhmofhpireeoilizdgnohentinstt‘ahlno’upfmlrvloeieftjilereoecdcst.,iitlye, why the horizontal component of velocity 7 ch4.a nAgebsocdoynitsinpuroojuescltye?d from the ground with a velocity Athebogagedfxrtoy=peurriesn0stddsh,ireoofdinpribssecfotod?udrsyfsrtohhtmehaestcimtemrraetovatetiianlolkenhedneoihgbf/hy2tthtahemnedebbtoraoedndsyoy.tshutFoecihrnrdbetoahcadahnyt 30° with of 10 m s–1 such that it makes an angle 29. is projethcetehdohriozroinzotanl.taWllyh.Wat ihsitchheohnoeriwzoinlltrael avcehlocity at the maximum height? 30. 6ITwf8wht.h oatuCethppaieatnrlocregitutiraitlcoahalhluteeeevnsicetdgohol.hofemtcmtiopitmfyoas5nesoeeomfsnfatmfnfslriogooamhfbntvldioteqhfl4uoeaecmgbiprtooyrdaouryijnenewdcmt.thioilecevhhiinsoisgr2ti0hzaorlmoonwntsagn–l1taahnne 7dd5cv.ii tercAsbrm−utoi1lbacdakfoayrrledoposdymaniairtnshearepcsahatrnooiceoghijfgenlierchn?attg6deod0itihf°her5woagrminirtzoh.douWnt2nhtrdhae?larlehyt soiwpsreititczhhoteinvavetvealelylo,loccitiytyo2f3thme 31. 6 w9i.t hAenquoabljevcetlopcriotyje.Wctehdatviesrttihceallryatiuopoffrocmenttrhipeettoalpfoorfce a7 c6t.i nAg boondtyhtermav?els in a semi-circular path of radius 7 m as a tower took 5 s to reach the ground. If the aver- shown. If the time taken to travel from A to C is 11 s, 32. A boadgye viselporcoijteycotefdthferoombjethcte igs r5oumnds−w1, iftihnda ivtes laovceirtaygoe f 10 m fsi–n1dsuch that it makes an angle 30° with the hspoereidzo. n(Gtailv.eWn,hgat=is1t0hme hs−o1r)i.zontal velocity at the max im um(1)htehieghdits?tance covered. 33. 7 A0.b oIdnyciasseproofjeacntedobhlioqruizeopnrtoaljleyctwileithora hveolroizcoitnytal 2p3ro-m s−1 fr(o2m) thaehdeiisgphlatcoefm5emnt..What is the velocity of thjeecbtioled,ywohnyretahcehihnogrtizhoengtarloucnodm?ponent of veloc- (3) the speed. ity remains constant, but the vertical component of (4) velocity. 34. A bovdelyoctritayveclhsainngeas sceomntii-ncuirocuuslayr? path of radius 7 m as sh ow (n5.) IWf thhaet tisimthee tdaikspelnacteomternatvieflitfrcoommpAletteos the circle? 7 C1.i sA11bosd, yfinisddroped from a certain height and another (1) tbhoedydiisstapnrocejecctoevdehreodri.zontally. Which one will reach B (2) tthhee gdriospulnacdefmirsetn?t. 7 ((342)). vWtaaIhfrneetedlhhotasehicpttiieetnmiysecit.doatikha.mleevspedoalinnospecanilntatycgseoloemffa6ven0en°oltobwiclfiiitqtiyhtucietnohpmetrhohpejeolehcrttoieizlrsoeizntiohstan2el0t,acmwlirachnsl–aed1t? (5) AC 77. A person travels a total distance in three parts in the 35. A pevresrotnicatlradvierelsctaiotno?tal distance in three parts in the ratio 4 : 3ra:t1iow4it:h3a:c1onwstiathntaspceoendstaonft9s0pekemd ohf−19,0 km h−1, 7230. mTw−1oanpda1rt0icmless−o1f remspasescetsiv‘emly’.aWndha4t w‘mil’l abre tmheovavinegrage spee2d0 mof−t1haendjo1u0rnmeys?−, respectively. What is the average PRACTICE QUESTIONS along the circular paths of radii r and 2r, respectively speed of the person while journey? Level 3 83 7 8. Are all physical quantities that have magnitude and 9 seconds. Another ball thrown vertically downwards direction vectors? Give example to support your from the same position with the same speed ‘u’, takes 4 seconds to reach ground. Calculate th1e2/2v7a/1l3ue12o:0f8:‘0u4’P.M TH_PHY_CH 02.indd 83 answer. When is a physical quantity called vector? (Take g = 10 m s–2) 79. A body is dropped from a certain height above the 83. A train leaves station ‘A’ for station ‘B’. The train ground. Its time of descent is 5 s. But at t = 3 s, the travels along a straight line without any halts body is stopped and then released. What is the remain- between the stations. During the first and last 200 ing time the body should travel to reach the ground? m of its journey, the train has uniform acceleration and retardation both equal to 1 m s−2, respectively. 80. A body is dropped from a height of 2 m. It penetrates For the rest of the journey, the train maintains uni- into the sand on the ground through a distance of 10 form speed. Calculate the average speed of the train, cm before coming to rest. What is the retardation of given the distance between the two stations is 4 km. the body in the sand? 8 4. A ball thrown vertically up from the top of a tower 8 1. A car starts from rest and moves with uniform accel- reaches the ground in 12 s. Another ball thrown ver- eration of α m s–2 along a straight line. It then retards tically downwards from the same position with the uniformly at a rate β m s–2 and stops. If ‘t’ is the time same speed takes 4 s to reach the ground. Find the elapsed, they find the average speed of the car. height of the tower. (Take g = 10 m s−2) 82. A ball thrown vertically upwards with a speed ‘u’ from the top of a tower reaches the ground in

3 X–axis 0 12 3 4 5 Time K(si)nematics 2.35 13. From the adjacent figure the velocity-time graph for a body moving in a straight line. Fin 85. Two stones A and B are dropped f(rao)matchceelteorpatoiof ntwoof th e b o(dby) d eucreinlegratthioenfiorsftththerbeoedsyecdounrdinsgotfhietslamstoseticoonn.d of different towers such that they trav(ebl)4d4e.1cemlearantdio6n3.o7f the body idtsumrinogtiothne. last second of its motion. m in the last second of their m(oct)iodni,sprelascpemctievnetlyo. f th e bo(cd) yd iastptlahceemenedntoof f4ths.e body at the end of 4 s. Find the ratio of the heights of the two towers from where the stones were dropped. Y 8 6. A missile is launched from the ground making an 8 angle of 45° with the horizontal. If it is required to hit the target on the ground at 1600 km, with what v6 velocity should it be launched? (Take g = 10 m s−2). (m s–1) 4 Assume the acceleration due to gravity to be uniform throughout the motion of the missile. 2 8 7. If a football kicked from the ground moves with a 0 1 234 X velocity of 20 m s−1 making 30° with the horizontal, t find its vertical displacement and horizontal displace- (in seconds) ment after 1 second. (Take g = 10 m s−2) 88. From the following data, calculate the acceleration in 91. From the below figure, find the displacement at the end of 10 s and also find at what rate the velocity of each time interval of two seconds and plot accelera- the body decreases. tion-time graph. he following data, calcVuelalotecitthye(macsc-e1)lerat0ion i2n.5each5 tim7e.5inte1rv0al o1f2.t5wo seconds and 50 cceleration - time grapTh.ime (s) 40 0 2 4 6 8 10 Velo city (Dmose-1s) the bo0dyCLh_a9v2TH.e5_PuHnY_iCf5oHr0m2.indvd7e.8l5o5 city 1o0r unifo12r.m5 v 30 (m s –1) 20 Time (s) acceleration?0 2 4 6 8 10 10 he body have uni 8fo9.r mThveelobceilotywofriguunriefosrhmowascctheelerdaitsitoannc?e-time graph of 0 1 2 3 4 5 6 7 8 9 10 tfhigeugreivsehnottqwwhueosegtsbhitvoieoedndnieisqs.stuaAensctaieno-dntismB. .eAgnraaplyhseotfhtewgorabpohdiaensdAaannswd eBr. –10 t (in second) he adjacent Analyse the –20 and answer Distance (m) –30 hhaicthisbtohdeydiissttarnacve eltlir navg((abe)w)l lWWoeitdfhhh2biamcysth?oiAsbretoahdnsepyddeieiBssdttar?anatvcteehlletirnaegvnefdlalseotdefrb2?ys?them at the end –40 PRACTICE QUESTIONS –50 Y–axis B 9 2. From the adjacent figure a displacement-time graph A of a body moving in a straight line. Find the distance 15 covered and the displacement of the body at the end 12 of 12 seconds. 9 6 s 10 3 (in m) 5 X–axis 0 2 46 8 10 12 0 12 3 4 5 –5 t (in s) Time (s) –10 he adjacent figur9 e0t.h eFrvoemlocthitey-bteimlowe gfirgauprhe,fofirnda bthoedy moving in a straight line. Find the celeration of the b od y(ad) uarcicnegletrhateiofinrsotftthhreeebosdecyodnudrsinogftihtse mfirosttitohnre.e sec- celeration of the body duornidnsgotfhietslamstosteiocno.nd of its motion. placement of the body at the end of 4 s. Y

2.36 Chapter 2 CONCEPT APPLICATION Level 1 True or false 1.  True 2.  True 3.  True 4.  False 5.  False 6.  False 7.  False Fill in the blanks 8.  1 : 2 9.  g t2 10.  displacement 11. less than or equal to 1 12.  same 13.  Acceleration 8 Match the following 14 A  :  c    B  :  d     C  :  f and d    D  :  g    E  :  b    F  :  h    G  :  a    H  :  e Multiple choice questions 1 5. (d) 16. (b) 17. (d) 18. (b) 19. (c) 20. (a) 21. (a) 24. (a) 25. (b) 26. (b) 27. (a) 28. (d) 2 2. (a) 23. (b) 31. (c) 32. (a) 33. (c) 34. (b) 35. (d) 38. (c) 39. (b) 40. (a) 41. (d) 42. (a) 29. (d) 30. (d) 3 6. (c) 37. (d) HINTS AND EXPLANATION Explanations for questions 30 to 42: 3 6. Parabolic path. 3 0. When a body moves from P to Q and travels back to 3 7. In choice (d), the displacement of the body is con- P the total displacement of the body is zero. Average stant as time passes. velocity is also zero. 38. y 31. When a body is projected up and it returns to the point of projection, the average velocity is zero but s not the average speed. tx 3 2. H max = 80 m= u2 2g In this graph, time remains constant which is impossible. u2 = 80 × 2 × 10 39. Average acceleration = change in velocity = 4 × 2 × 10 time taken × 2 × 10 = 40 m s−1 u Change in Velocity for time t1 = a1t1 Change in Velocity for t2 = a2t2 3 3. A vertically projected down body travels with uni- form acceleration. a1t1 + a2t2 34. The motion of a body along a circle with a uniform ∴ Average acceleration = ( t1 + t2 ) speed can be described as two – dimensional motion. 35. v21 – u21 = 2gh 40. Sn = u + a ( 2n −1) v1 = 0, u1 = usinq 2 u2 sin2 θ u2 sin2 θ n= 6 2g 2g ∴ hmax = = S6 = 26 = u+ a ( 2 × 6 −1) 2

Kinematics 2.37 2 × 26 = 2u + a( 12 −1)  4 1. h1 = a 52 = 2u + 11a (1) h2 b n= 7 u2 2g S7 30 = u+ a ( 2 7 −1) But h= 2 = × 60 = 2u + 13a  (2) Hence, h ∝ u2 Subtract (1) from (2) ∴ h1 u12 h2 u22 60 = 2u + 13a = 52 = 2u + 11a u1 a  u2 b ============ = ⇒ u1 : u2 = a: b 8 = 2a a = 4 m s–2 52 = 2u + 11(u) 42. q = 30º distance covered (l) = rq 2u = 52 − 44 2u = 8 = 30 ×6 = π × 6 m 6 u = 4 m s–1 distance = π m Level 2 43. (i) h or s = uta - 1 gta2 v=0 hm 4 5. (i) Average speed = total distance HINTS AND EXPLANATION 2 }1s t as (ii) 32.73 km h–1 total time taken (ta – 1)s } 1 at2 u 2 46. (i) s = ut + (ii) Snth = h − s(ta −1) (ii) Distance traveled by the balloon in 1 second is 1 m. =  uta − 1 gta2  − u(ta−1 ) − 1 g(ta −)2  (iii) Calculate the distance traveled at the end of 4  2  2  seconds = height at which the body is released. = uta − 1 gta2 − uta + u + g ta2 − 2  g ta + g (iv) Body will have initial velocity in the upward 2 2  2  2 direction = velocity of the balloon. distance covered in u−u + g = g (v) S ubstitute the value of g, u and s in equation S last one second = 2 2 = ut + 12 gt2 and find t. 2h (vi) 1 s g 4 4. (i) t = r q 30˚ (ii) From the value of h, find t using t = 2h  (1) g (iii) T he initial velocity of the dropped ball and the 4 7. (i) Displacement = 2r sinθ initial vertical component of velocity of a hori- D istance = Length of the arc = rθ zontally projected ball is zero. ∴ t1 = t2 (iv) Find time taken by the two balls using 1.

2.38 Chapter 2 (ii) T he angle made by the pendulum with the verti- (iii) Find n, i.e., total time taken cal is = 12 the angle made by the arc described (iv) 5.5 s by the pendulum. (iii) T he length of the arc = radius of the arc × angu- 5 3. (i) Time of motion of A = Time of motion of lar displacement. 1 B= 2 Time of descent of B (iv) Does the length of the arc give the distance trav- eled by the pendulum? (ii) L et x be the distance traveled by the ball. (v) F ind the initial and final positions of the pendu- (iii) L et y be the distance traveled by the stone. lum between its extreme positions. (iv) Time of flight = 2h h=x+y (vi) Is the shortest length between the extreme posi- g tions of the pendulum equal to its length? (v) Time of descent of the stone = 2y . (vii) T he length of the pendulum gives the value of g displacement. (vi) Equate 3 and 4 and obtain the relation between (viii) 22 : 21 x and y. 4 8. 12.5% (vii) Find the ratio of x and y. (viii) 3 : 1 49. (i) Time of descent = 2h u2 g 54. (i) H = 2 g s = ut – ½ gt2 (ii) Calculate the height on earth using equation h = (ii) 576 m 12 gt2 by substituting the values of g and t. 55. Relate equations (iii) g on moon = g on earth. 56. (i) D efinition of distance and displacement. HINTS AND EXPLANATION (iv) F ind the time taken on moon by using the equa- (ii) Distance = total path covered tion h = 12 gt2, by substituting the values of h D isplacement = shortest path covered between ini- and g of moon. tial and final points. (v) 12 6 s (iii) 16 m, 0 5 0. (i) Formula of Sn (ii) Sn = u + a (2n – 1) 2 Sn = (v – a ) 57. (i) Sn = u + a (2n – 1) 2 2 (iii) 9.5 m s–1 v = u + at (ii) Find v after n seconds and relate it to the sum of 51. (i) v = u + gt the distance covered in nth and (n – 1)th second. (ii) W hen the body is dropped, initial velocity = 0. (iii) 40 m s–1 (iii) Substitute values of t in equation v = u + gt and 58.. find the values of v1, v2 and v3, for each value of ‘t’. y →b (iv) Find their ratio. 30° (v) 1 : 2 : 3 x 5 2. (i) Sn = u + g (2n – 1) 2 horizontal, component of a , ⇒ ax = a cosθ and (ii) Using formula of Sn relate Sn to Sn–1 for the ratio given given

Kinematics 2.39 θ = 30 ⇒ ax = a 3 and vertical component of b the body caused by the gravitational pull of the 2 Earth on the body. When the body reaches its maximum height, its velocity is zero, but it is ⇒ by = b sin θ and given q = 60° still under acceleration due to gravity. Hence, it is possible for an accelerating body to have zero ⇒ b sin 30 ⇒ b velocity. 2 (2) T he statement is false. Velocity is defined as Given ax = by ⇒ a3 = b ⇒b= a 3 units. displacement per unit time. Thus, velocity is also 2 2 displacement, but for a unit time. Hence, the direc- tion of displacement and velocity are one and the 5 9. Fx = F cos q = 15 × cos q = 15 cos q same. Thus, it is not possible for a body undergoing 60. AB = 6 m, BD = 8 m linear motion to have displacement and velocity in opposite directions. •C N 8m (3) T he given statement is false. It is possible for a body undergoing linear motion to have veloc- A• 6 m •B W E ity and acceleration in opposite directions. For example, if a body is thrown in upward direction, 8m S its velocity is in the upward direction, whereas D• its acceleration which is due to gravity is in the downward direction. Then the displacement is AD (4) T he given statement is true. It is possible for a body to have uniform speed, when it is moving = ( AB)2 + ( BD)2 = 62 + 82 with an acceleration. For example, if a body is moving with uniform speed along a circular path, = 36 + 64 = 100 = 10 m the direction of velocity of the body changes con- HINTS AND EXPLANATION tinuously, and thus, it has an acceleration directed 6 1. Change in velocity in km h–1 = 36 – 90 = – 54 km h–1 towards the centre of the circular path. –54 km h–1 = –54 × 5/18 m s–1 = –15 m s–1 6 2. Given ground is in circular shape of radius “r” and 6 4. t = 10 s Area = 625π sqm u = 5 m s−1 A = πr 2 = 625π ⇒ r 2 = 625 ⇒ r = 625 = 25 m v = 25 m s−1 s=? In 30 s, the distance travelled by him = circumfer- s = u+ v ence of the circle = 2πr = 2π(25) m = 50 πm = t 2 157.14 m. u+ v 25 + 5 × 10 = 150 m The displacement = 0, because the starting and end- s= 2 ×t = 2 ing points are same. 65. u = ? ⇒ Now, the distance in 15 v = 25 m s–1 g = 10 m s–2 s= distance in 30 s t=2s 2 v = u – gt 25 = u – 10(2) ⇒ u = 25 + 20 = 45 m s–1 = 157.14 = 78.57 m 2 66. In case of vertical projectile, time of ascent = time of descent = u/g. The displacement in 15 ∴ Time of descent = 25/10 = 2.5 s. s = diameter of circular ground 6 7. From height h to h/2 from ground the body travels = 2r = 2 × 25 = 50 m. 63. (1) The statement is false. When a body is pro- jected vertically upwards, its velocity decreases as it moves up. This is due to the retardation of