DC Pandey Mechanics Volume 1

490 — Mechanics - I (s)U = 1CV 2 7. [T ]= [padbEc ] = [ML−1T−2 ]a [ML−3 ]b[ML2T−2 ]c 2 Equating the powers of both sides, we have ⇒ (farad) (volt)2 = Joule or [(farad) (volt)2 (kg)−1 ] = [(Joule)(kg)−1 ] a+ b+ c=0 ... (i) = [ML2T−2 ][M −1 ] = [L2T−2 ] − a − 3b + 2c = 0 ... (ii) −2a − 2c = 1 ... (iii) Solving these three equations, we have Subjective Questions a= − 5,b= 1 1. [a] = [ y] = [L] 62 ⇒ [ωt ] = [M 0L0T0 ] and c = 1 3 ∴ [ω ] = 1 = [T-1 ] [t] 8. [Y ] = [ML−1T−2 ] θ is angle, which is dimension less.  MgL  =  MLT−2L   πr2l   L2L  2. 1N = 105 dyne   1 m2 = 104 cm2 = [ML−1T−2 ] ∴ 2.0 × 1011 N Dimensions of RHS and LHS are same. Therefore, m2 the given equation is dimensionally correct. 2.0 × 1011 × 105 dyne 9. [E ] = k [m ]x [n ] y [a ]z 104 cm2 = where, k is a dimensionless constant. ∴ [ML2T−2 ] = k [M ]x [T−1 ] y [L ]z = 2.0 × 1012 dyne/cm 2 Solving we get, 3. 1 dyne = 10−5 N x = 1, y = 2 1 cm = 10−2 m and z = 2 ∴ E = k m n2a2 ∴ 72 dyne/cm = 72 × 10−5 N 10. Since dimension of 10−2 m Fv = [Fv ] = [MLT−2 ][LT−1 ] = [ML2T−3 ] = 0.072 N m So, β  should also be [ML2T−3 ]  x2  E J 4. h = γ = s−1 = J-s [β] ML2T −3 [x2 ] = [ ] [h] = [ML2T-2 ][T] [β ] = [ML4T−3 ] = [ML2T-1 ] Ans. 5. St = ut + 1 at2 + u (t − 1) + 1 a (t − 1)2 and Fv + β  will also have dimension [ML2T−3 ], 2 2 x2  =  u + at − 1 a so LHS should also have the same dimension 2 [ML2 T−3 ]. Equation is dimensionally correct. So [α ] = [ ML2T −3 ] [t2 ] 6. (a) Young's modulus = F / A [α ] = [ML2T−1 ] Ans. ∆l / l 11. [ b] = [V] = [L3 ] Hence, the MKS units are N/m2. [a] (c) Power of a lens (in dioptre) [V ]2 = [P] = [ML−1T−2 ] =1 f (in metre) ⇒ [a] = [ML5T−2 ]

Chapter 4 Units and Dimensions — 491 12.  a = [M 0L0T 0 ]  dx   RTV  is dimensionless. 14.   a2 − x2  ⇒ [a] = [RTV] = [ML2T−2 ][L3 ] 1 sin −1  ax  has the dimension of [ L−1 ].  a  = [ML5T−2 ]   [b] = [V] = [L3 ] dx  is dimensionless. 15.   2ax − x2  13. (a) Magnetic flux, φ = =  F  Bs   (s)  qv an sin −1  x 1  Therefore, a −  should also be [as F = Bqv ] dimensionless. Hence, n = 0 ∴ [ φ ] =  Fs  =  MLT−2L2  16. (a) [density] = [F ]x [L ] y [T ]z  qv   QLT−1      ∴ [ML−3 ] = [MLT−2 ]x [L ] y [T ]z = [ML2T−1Q−1 ] Equating the powers we get, (b) [Rigidity Modulus] = F =  MLT−2  x = 1, y = − 4 and z = 2  A    ∴ [density ] = [FL−4 T2 ]  L2  = [ML−1T−2 ] In the similar manner, other parts can be solved.

5. Vectors INTRODUCTORY EXERCISE 5.1 3. A − 2B + 3C = (2$i + 3$j) − 2(i$ + $j) + 3k$ 3. Apply R = A2 + B2 + 2AB cos θ = ($j + 3k$ ) 4. Apply S = A2 + B2 − 2AB cos θ ∴ |A − 2B + 3C| = (1)2 + (3)2 5. R = S ⇒ A2 + B2 + 2AB cos θ = 10 units = A2 + B2 − 2AB cos θ Solving we get, cos θ = 0 or θ = 90° 4. (a) Antiparallel vectors INTRODUCTORY EXERCISE 5.2 (b) Perpendicular vectors (c) A lies in x - y plane and B is along positive 1. A or |A| = (3)2 + (−4)2 + (5)2 = 5 2 units z- direction. So, they are mutually perpendicular vectors Directions of cosines are, (d) cos α = Ax = 3 A 52 B 3 ^j cos β = Ay = −4 and θ A 52 45° −3 ^i O 4 ^i A cos γ = Az = 1 A2 θ = 135° 2. y Fx O x INTRODUCTORY EXERCISE 5.3 60° 3. A ⋅ B = AB cos θ Fy 4. 2A = 4$i − 2$j F = 10 N − 3B = − 3$j − 3k$ $i $j k$ Fx = 10 cos 60° = 5 N (along negative x-direction) Fy = 10sin 60° = 5 3 N (along negative y-direction) ⇒ (2A) × (−3B) = 4 −2 0 0 −3 −3 = i$(6 − 0) + $j(0 + 12) + k$ (−12 − 0) = 6i$ + 12$j − 12k$ Exercises Single Correct Option 11. τ = r × F 6. A ⋅ B = AB cos θ $i $j k$ =3 2 3 = (3) (5) cos 60° = 7.5 2 −3 4 9. A × B = AB sinθ = $i (8 + 9) + $j (6 − 12) + k$ (−9 − 4) 0 ≤ sinθ ≤ 1 ∴ 0 ≤ A × B ≤ AB Ans. = (17i$ − 6$j − 13k$ ) N - m Ans. 10. W = F ⋅ s = 9 + 16 = 25 J

Chapter 5 Vectors — 493 12. (0.5)2 + (0.8)2 + c2 = 1 Ans. 26. |A × B| = 3 A ⋅ B 13. R = A2 + A2 + 2 AA cos θ ⇒ AB sin θ = 3 AB cos θ or tan θ = 3 R = A atθ = 120° Ans. ⇒ θ = 60° Ans. 14. (2i$ + 3$j + 8k$ ) ⋅ (4i$ − 4$j + α k$ ) = 0 Ans. |A + B | = A2 + B2 + 2AB cos 60° Ans. ∴ 8 − 12 + 8α = 0 = A2 + B2 + AB ∴ α=1 27. (B × A) is ⊥ to both A and B 2 ∴ (B × A) ⋅ A = 0 15. A ⋅ B = 9 + 16 − 25 = 0 28. R = P2 + Q 2 + 2PQ cos θ ∴ Angle between A and B is 90°. Substituting the values of P, Q and R we get, 16. A + B = 7 cos θ = 0 and A − B = 3 ⇒ θ = 90° ∴ A = 5 and B = 2 29. Q 17. P ⊥ Q = P ⋅ Q = 0 18. s = rB − rA Ans. 135° x√2 P x√2 90° = (−2i$ + 6$j + 4k$ ) − (3$j − k$ ) R 135° = − 2$i + 3$j + 5k$ 20. A + B + C = 0 |P | = |Q | = x |R| = 2 x ⇒ C = − (A + B) or |C| = −(A + B) 30. (P + Q) ⋅ (P − Q) or A ~ B ≤ C ≤ A + B = P2 + PQ cos θ − PQ cos θ − Q 2 21. Resultant is always inclined towards a vector of = P2 − Q2 larger magnitude. Since dot product may be positive (if P > Q ), negative (if Q > P) or zero (if P = Q ). Therefore 22. A = |A | = (3)2 + (6)2 + (2)2 angle between (P + Q) and (P − Q) may be acute, obtuse or 90°. =7 31. Scalar triple product of three vectors α = cos−1  Ax  = cos−1  37 A = volume of parallel applied = 0 2 3 −2 = angle of A with positive x-axis. Similarly, β and γ angles. ∴ 5 n 1 =0 −1 2 3 23. R = A + B = 12i$ + 5$j Solving this equation, we get R = |R | = (12)2 + (5)2 n = 18 = 13 Ans. 32. (a + b) × (a − b) R$ = R = a × a − a × b+ b× a − b× b R = 2 (b × a) As a × a and b × b are two null vectors 24. Component of A along B and −a × b = b × a = A cos θ = A⋅B = 2+ 3 = 5 B 1+ 1 2 25. C 2 = A2 + B2 + 2AB cos θ At θ = 90°; C 2 = A2 + B2

494 — Mechanics - I 33. R = (3P)2 + (2P)2 + 2 (3P) (2P) cos θ ...(i) (b) If A and B are parallel or antiparallel to each other then their cross product is a null vector and 2R = (6P)2 + (2P)2 + 2 (6P) (2P) cos θ ...(ii) they may be said to be equal otherwise for any other angle, Solving these two equations we get, A × B = −B × A cos θ = − 1 or θ = 120° 2 (c) A2 + B2 + 2AB cos θ 34. R R = 2x cos θ = A2 + B2 − 2AB cos θ θα 2x ∴ cos θ = 0 x or θ = 90° Ans. x 2x sin θ (d) A + B = C only at θ = 0° 2x sin θ = x ∴ θ = 30° Subjective Questions or α = 120° Ans. 35. A ⋅ B = 0 1. θ = cos−1 aa⋅bb ∴ A⊥B 2. If the angle between A and B is θ, the cross product A⋅C = 0 will have a magnitude ∴ A ⊥C |A × B| = AB sin θ A is perpendicular to both B and C and B × C is also or 15 = 5 × 6 sin θ perpendicular to both B and C. Therefore, A is or sin θ = 1 parallel to B × C. 2 36. R = 8N B cos θ = 8N Thus, θ = 30° B θ or 150° A 5. The angle between the force F and the displacement B sin θ r is 180°. Thus, the work done is A W =F⋅r A + B = 16 ...(i) = Fr cos θ B sin θ = A = (12 N) (2.0 m) (cos 180°) and B cos θ = 8 = − 24 N -m = − 24 J Squaring and adding these two equations, we get 6. A × B is perpendicular to both A and B. So this is B2 = A2 + 64 ...(ii) parallel or antiparallel to C. Now, cross product of two parallel or antiparallel vectors is zero. Hence, Solving Eqs. (i) and (ii), we get Ans. A = 6 N and B = 10 N C × (A × B) = 0. 37. (A + B) ⊥ (A − B) 7. A × B is perpendicular to A. Now, dot product of ∴ (A + B) ⋅ (A − B) = 0 two perpendicular vectors is zero. Hence, ∴ A2 + BA cos θ − AB cos θ − B2 = 0 A ⋅ (A × B) = 0 or A = B Ans. 8. S = S 1 + S 2 + S 3 38. A ~ B ≤ C ≤ A + B = [(5cos 37°)$i + (5sin 37° )$j] + 3$i + 2$j = (4i$ + 3$j) + 3i$ + 2$j 39. A ~ B ≤ C ≤ A + B = (7i$ + 5$j) m Match the Columns 5m S 1. (a) AB sin θ = | AB cos θ | Ans. 7m ∴ tan θ = ± 1 S = (7)2 + (5)2 = 74 m ⇒ v = 45° or 135°

Chapter 5 Vectors — 495 tanθ = 5 =3 5m 7 tan θ = By = 3 = 1 or θ = tan−1 75 Bx 6 2 9. Suppose α is the angle between A and B and β the ∴ θ = tan−1  12 angle between C and B. But α =/ β. 16. (a) a + b + c = P = $i − 2$j Given that, (b) a + b − c = $i + 4$j = Q A⋅B =C⋅B ⇒ AB cosα = CB cosβ (c) Angle between P and Q is, θ = cos−1  P ⋅ Q ⇒ A cosα = C cosβ  PQ  or A =/ C as α =/ β 17. R = [(10cos 30°)i$ + (10cos 60°)$j] 10. A + B = R = 3i$ + $j , A − B = S = i$ + 5$j + [(10cos 60°)i$ + (10cos 30°)$j] Now, angle between R and S is = 10(cos 30° + cos 60° )i$ + 10(cos 60°+ cos 30°)$j θ = cos−1  R⋅S  = 10(2cos 45° cos15°)$i + 10(cos 45° cos15°)$j RS = 20 cos 45° cos15° $i + 20 cos 45° cos15° $j 11. Ratio of their coefficients should be same. = Rx$i + Ry$j ∴ 2 = 3 = −4 3 −a b Here, Rx = Ry = 10 2 cos15° ∴ a = − 4.5 ∴ |R| = Rx2 + Ry2 = 20cos15° and b = − 6 Since, Rx = Ry , therefore, θ = 45° i$ $j k$ 18. Their x components should be equal and opposite. 12. A × B = 2 4 −6 ∴ 6cosθ = − 4 10 2 or θ = cos−1 − 32 = i$ (8 − 0) + $j (−6 − 4) + k$ (0 − 4) 19. Component of A along another vector (say B ) is = 8$i − 10$j − 4k$ given by, A cos θ = A ⋅ B Area of parallelogram = |A × B| B = (8)2 + (10)2 + (4)2 = 13.4 units Ans. 20. Find A × B and then prove that (A × B) ⋅ A = 0 and 13. See the hint of Q-23 of objective type problems. (A × B) ⋅ B = 0. 2 −3 7 It means (A × B) ⊥ to both A and B. 14. A ⋅ (B × C) = 1 0 2 21. B 0 1 −1 R R = x cos θ = 2 (0 − 2) − 3 (0 + 1) + 7 (1− 0) x θ x α =−4 −3+ 7=0 2 A 15. (a) A + B = R R = x cos θ Ans. Rx = Ax + Bx or x = x cos θ ∴ Bx = Rx − Ax = 10 − 4 = 6 m Similarly, 2 ∴ cos θ = 1 By = Ry − Ay = 9 − 6 = 3 m 2 (b) B = Bx2 + By2 = (6)2 + (3)2 or θ = 60° ∴ α = 150°

496 — Mechanics - I AB + BC + CA = 0 ∴ ci$ + (a sin B$j − a cos Bi$) 22. y + (−b cos Ai$ − b sin A$j) = 0 3P ∴ (c − a cos B − b cos A) $i 4P 2P + (a sin B − b sin A) $j = 0 O x y P R = (Pi$ + 2P$j − 3P$i − 4P$j) C = (−2Pi$ − 2P$j) b a C |R | = (2P)2 + (2P)2 AB x =2 2P Ans. A cB 23. R2 = P2 + Q 2 + 2PQ cos θ ....(i) Putting coefficient of $j = 0, we find that ...(ii) a=b S 2 = P2 + Q 2 − 2PQ cos θ sin A sin B Adding these two equations we get, Now taking B as origin and BC as the x-axis, we can R2 + S 2 = 2 (P2 + Q 2) prove other relation. 24. From polygon law of vector addition we can see that,

6. Kinematics 6.1INTRODUCTORY EXERCISE 5. (a) On a curvilinear path (a path which is not straight 2. (a) v⋅ a = 6 − 4 − 4 = −2 m2/s3 line), even if speed is constant, velocity will change (b) Since dot product is negative. So angle between due to change in direction. v and a is obtuse. (c) As angle between vand a at this instant is obtuse, ∴ a≠0 speed is decreasing. (b) (i) On a curved path velocity will definitely change (at least due to change in direction). ∴ a≠0 6.2INTRODUCTORY EXERCISE (ii) In projectile motion, path is curved, but acceleration is constant (= g). 1. v and a both are constant vectors. Further, these two (iii) Variable acceleration on curved path is vectors are antiparallel. definitely possible. 2. a is function of time and v and a are neither parallel 6. (a) T = 2πR = 25.13 s Ans. nor antiparallel. v 6.3INTRODUCTORY EXERCISE (b) (i) Since speed is constant. Therefore, 1. Distance may be greater than or equal to magnitude average speed = constant speed of displacement. = 1.0 cm/s Ans. 2. Constant velocity means constant speed in same (ii) Average velocity = s = 2 R = 4 2R t (T /4) T direction. Further if any physical quantity has a constant value (here, the velocity) then its average d value in any interval of time is equal to that constant value RS 90° 3. Stone comes under gravity R ∴ F = mg or a= F =g Ans. m 4. In 15 s, it will rotate 90° or 1 th circle. = (4)( 2)(4.0) = 0.9 cm/s Ans. 25.13 Ans. 4 (iii) Velocity vector will rotate 90° d ∴ |∆v| = v2 + v2 − 2vv cos 90° RS = 2v 90° |∆v| 2v 4 2v R ∆t (T /4) T ∴ |aav | = = = ∴ Average speed = Total distance = d = (4 2)(1.0) = 0.23 cm/s2 Total time t 25.13 = (2πR/4) = πR = π (2.0) 6.4INTRODUCTORY EXERCISE 15 30 30 1. Average speed = d = d1 + d2 =  1π5 cm/s Ans. t t1 + t2 Average velocity = Total displacement = s = v1t1 + v2t2 Total time t t1 + t2 = 2 R = ( 2)(2.0) = 2 2 cm/s Ans. = (4 × 2) + (6 × 3) Ans. t 15 15 2+3 = 5.2 m/s

498 — Mechanics - I 2. 2 m/s C d/4 D 3d/4 B s ut + 1 at2 1 d1 4 m/s 6 m/s t 2 A ...(i) 6. uav = = = u+ at Ans. ...(ii) t2 Ans. Ans. tt 7. v = u + at 2t = d1 i.e. v-t function is linear. In linear function, (d /4) + (3d /4) = t average value = (final value + initial value) 46 2 ⇒ 3d = t ∴ vav = vf + vi = v1 + v2 16 2 2 Average speed = total distance 8. t = 2h = 2 × 125 = 5 s total time g 10 2t + 16t d1 + d 3 s 125 = = vav = t = 5 = 25 m/s (downwards) 2t 2t = 11 m/s 9. (a) S = ut + 1 at2 3 2 6.5INTRODUCTORY EXERCISE = (2.5 × 2) + 1 (0.5) (2)2 2 1. St = (displacement upto t second) = 6 m = distance also − [displacement upto (t − 1) sec] (b) v = u + at = (ut + 1 at2) − [u (t − 1) + 1 a (t − 1)2 ] 7.5 = 2.5 + 0.5 × t ⇒ t = 10 s 22 = u + at − 1 a (c) v2 = u2 + 2as 2 ∴ (7.5)2 = (2.5)2 + (2) (0.5)s ut 1  u 1  ⇒ s = 50 m = distance also 2  2  2. st = + at2 − (t − 1) + a (t − 1)2 10. (a) h = u2 = (50)2 = 125 m = (u)(1) + 1 (a)(2t) − 1 (a)(1)2 2g 2 × 10 22 (b) Time of ascent = u = 50 = 5 s The first term is (u) (1), which we are writing only u. g 10 Dimensions of u are [LT−1 ] and dimensions of 1 (c) v2 = u2 + 2as = (50)2 + 2(−10)  1225 (which is actually 1 second) are [T] ∴ [(u)(1)] = [LT−1 ][T ] Solving, we get speed v ≈ 35 m/s = [L]= st 6.6INTRODUCTORY EXERCISE Therefore dimension of (u) (1) are same as the 1. (a) a = dv = 5 − 2t dimensions of st. Same logic can be applied with dt other terms too. At t = 2 s 5. In 4 s, it reaches upto the highest point and then a = 1 m/s2 Ans. Ans. changes its direction of motion. 33 s = ut + 1 at2 = (40) (6) + 1 (−10) (6)2 (b) x = ∫ vd t = ∫ (10 + 5t − t2) dt 22 00 = 60 m Ans.  t3  3 10 t =u2+ 1 g − t0)2 =  + 2.5 t2 − 3  2g 2  d = | s | 0− 4 + | s | 4−6 (t 0 = (40)2 + 1 × 10 × (6 − 4)2 = 100m Ans. = (10 × 3) + (2.5) (3)2 − (3)3 2 × 10 2 3 = 43.5 m

2. (a) Acceleration of particle, Chapter 6 Kinematics — 499 a = dv = (6 + 18t) cm/s2 6.7INTRODUCTORY EXERCISE dt 1. u = 2$i At t = 3 s, ^j 2 m/s2 a = (6 + 18 × 3)cm/s 2 = 60cm/s 2 60° (b) Given, v = (3 + 6t + 9t2)cm/s ^i 2 m/s or ds = (3 + 6t + 9t2) a = (2cos 60°) i$ + (2sin 60° ) $j dt = ($i + 3 $j) or ds = (3 + 6t + 9t2) dt t=2s s 8 (3 + 6t + 9t2) dt (i) v = u + at = (2i$) + (i$ + 3 $j) (2) ds = = 4i$ + 2 3 $j ∫ ∫∴ 05 ∴ s = [3t + 3t2 + 3t3 ]85 or s = 1287 cm ∴ | v| = (4)2 + (2 3)2 3. (a) Position, x = (2t − 3)2 = 2 7 m/s Velocity, v = dx = 4 (2t − 3)m/s (ii) s = u t + 1 a t2 = (2i$) (2) + 1 (i$ + Ans. dt 22 3 $j) (2)2 and acceleration, a = dv = 8 m/s2 = (6$i + 2 3 $j) dt At t = 2 s, ∴ |s | = (6)2 + (2 3)2 x = (2 × 2 − 3)2 =4 3m Ans. = 1.0 m 2. v = (2i$ + 2t$j) v = 4 (2 × 2 − 3) (i) a = d v = (2$j) m/s2 = constant = 4 m/s dt and a = 8m/s 2 ∴ v = u + a t can be applied. 11 (ii) s = ∫ vdt = ∫ (2$i + 2t$j) dt = [2t$i + t2$j ] 1 0 (b) At origin, x = 0 00 or (2t − 3) = 0 = (2i$ + $j) m Ans. ∴ v=4×0=0 vt 4. (a) At t = 0, x = 2.0 m Ans. 3. ∫ dv = ∫ adt = ∫ (2ti$ + 3t2$j) dt 2$i 0 (b) v = dx = 2t + 6t2 Solving we get v = 2i$ + t2i$ + t3$j dt At t = 1 s, v = (3i$ + $j ) m/s At t = 0, v = 0 Ans. (c) a = dv = 2 + 12t tt t dt ∫ dr = ∫ vdt = ∫ [(2 + t2)i$ + t3$j] dt 00 0 At t = 2 s Ans. ⇒ r =  + t3  i$ + t4 $j a = 26 m/s2  2t   3 4 5. v ∝ t3/ 4 At t = 1 s, r = 7 i$ + 1 $j a = dv ⇒ a ∝ t−1/ 4 34 dt ⇒ s ∝ t7/4 Therefore, the co-ordinates are  7 , 41 m s = ∫ vdt 3

500 — Mechanics - I 6.8INTRODUCTORY EXERCISE 4. Applying sine law in ∆ ABC , we have 1. (b) v = Slope of x - t graph N 2. Distance travelled = displacement 115500°m-q/s B = area under v - t graph C q Net velocity = v Acceleration = Slope of v - t graph 20 m/s 30° 3. Acceleration = Slope of v - t graph Distance travelled = displacement AE = area under v - t graph 150 = v = 20 4. (a) Average velocity = s = xf − xi sin 30° sin θ sin (150° − θ) tt From first and third we have, sin (150° − θ) = 20sin 30° = 1 = x10 sec − x0 sec = 100 = 10 m/s 150 15 10 10 ∴ 150° − θ = sin−1  115 (b) Instantaneous velocity = Slope of x - t graph 5. From 0 to 20 s, displacement s1 = area under v - t graph Ans. = + 50 m (as v is negative) From 20 to 40 s, N 150°-1q50 m/s B displacement s2 = area under v - t graph = − 50m (as v is negative) Total distance travelled = s1 + |s2| = 100 m AE Average velocity = s = s1 + s2 tt = 50 − 50 = 0 40 6.9INTRODUCTORY EXERCISE Hence, we can see that direction of 150 m/s is 1. vA = Slope of A = 2 = 0.4 m/s 150° − θ or sin −1  115  , east of the line AB 5  vB = Slope of B = 12 = 2.4 m/s (b) 150° − θ = sin−1  115 = 3.8° 5 ∴ vAB = vA − vB = − 2 m/s (downwards) ∴ θ = 146.2° 2. aA = aB = g ∴ aAB = aA − aB = 0 Ans. From first and second relation, v = 150 sinθ 3. (a) t = 400 = 40 s Ans. sin 30° 10 BC 10 m/s Ans. Substituting the values we get, Ans. 400 m v = 167 m/s A 2 m/s ∴ t = AB v (b) BC = (2 m/s) (t) = 2 × 40 = 80 m = 500 × 1000 min = 49.9 min 167 × 60 ≈ 50 min

Chapter 6 Kinematics — 501 5. Let vr = velocity of river vb = vb2r − vr2 12.5 = ω = ω vbr = velocity of river in still water and Now, K (iii) vb vb2r − vr2 ω = width of river ω = 10 Given, tmin = 10 min or vbr K (i) Solving these three equations, we get A vbr = 20 m/min, vr = 12 m/min vbr ω and ω = 200 m 6. A ω 2√2 m/s For minimum time Drift in this case will be, 45° 2 m/s x = vrt 3 m/s ∴ 120 = 10vr K (ii) Shortest path is taken when vb is along AB. In this case, vr B 5 m/s Net velocity vbr vb of boatman A (a) t = ω = 20 = 10 s Shortest path 22 (b) Drift = 5t = 50 m Exercises LEVEL 1 3. Average velocity = s Assertion and Reason t 1. a = − 2 (i$ − $j) = − 2v = area of v-t graph i,e. a and v are constant vectors and antiparallel to t each other. So motion is one dimensional. First retarded then accelerated in opposite direction. =  1 v0t0 = v0 2 2. In the given situation, v-t graph is a straight line. t0 2 But s-t graph is also a straight line. 4. It is not necessary that if v = 0 then acceleration is vs also zero. If a particle is thrown upwards, then at highest point v = 0 but a =/ 0. 5. If acceleration is in opposite direction of velocity v0 Þ then speed will be decreasing even if magnitude of t acceleration is increasing. v = constant = v0 6. a = 2t ≠ constant t Therefore, velocity will not increase at a constant rate. s = v0t 7. If a particle is projected upwards then s = 0, when it returns back to its initial position.

502 — Mechanics - I ∴ Average velocity = s = 0 v F2 t a1 a2 But its acceleration is constant = g 8. At point A, sign of slope of s-t graph is not changing. Therefore, sign of velocity is not changing. F1 mg mg v 9. v1 = ds1 = (2 − 8t) Upward Downward dt journey journey v2 = ds2 = (−2 + 8t) Acceleration in downward journey dt mg − F2 F2 ∴ v12 = v1 − v2 = (4 − 16t) a2 = m = g − m v12 does not keep on increasing. Since, a1 > a2 ⇒ T1 < T2 10. a = d v 3. v = 2πR = (2π ) (1) =  3π0 cm / s T 60 dt So, direction of a and d v is same. 11. t1 = 2 (h/2) = h In 15 s, velocity vector (of same magnitude) will g g rotate 90°. Velocity of second particle at height h is, ∴ |∆v| = | vf − vi | = v2 + v2 − 2vv cos 90o 2 = 2 v = π 2 cm/s Ans. v = u2 − 2g (h/2) 30 = gh − gh = 0 4. h = u2 or u ∝ h or h is its highest point. 2g 2 5. vav = d = d1 + d2 t t1 + t2 u gh = h ∴ t2 = g = g g 21 = (18) (11) + (42) (v) 60 ⇒ t1 = t2 ⇒ v = 25.3 m/s Ans. 12. a = mg − F F 6. v = dx = 32 − 8t2 m dt = g − F = depends on m a v = 0 at t = 2 s m mg a = dv = −16 t Since m1 ≠ m2 a1 ≠ a2 dt Ans. ∴ t1 ≠ t2 at t = 2s, a = −32 m/s2 or 7. a = bt Single Correct Option ∴ dv = bt dt 1. Packet comes under gravity. Therefore only force is mg. vt a = F = mg = g (downwards) or ∫ dv = ∫ (bt) dt mm v0 0 2. Air resistance (let F ) is always opposite to motion ∴ v = v0 + bt2 2 (or velocity) Retardation in upward journey t t bt 2   a1 = F1 + mg = g + F1 ∫ ∫s= vdt =  v0 + 2 dt m m  0 0 = v0t + bt3 6

Chapter 6 Kinematics — 503 8. t = 2h = 2 × 5 = 1 s 11. vx = dx = 5 dt g 10 vy = dy 3 dt t0 At 45°, = (4t + 1) Ans. h, t 2 ⇒ vx = vy Ans. t0 t =1s 1 12. h = 1 gT 2 2 Let t0 is the interval between two drops. Then At T second, distance fallen 3 2 t0 = t d = 1 g  T  = h ∴ t0 = 0.5 s 2 3 9 2nd drop has taken t0 time to fall. Therefore distance fallen, ∴ Height from ground = h − d = 8h 9 d = 1 gt02 =  12 (10) (0.5)2 2 13. It should follow the path PQR. = 1.25 m R ∴ Height from ground = h − d Ans. = 5 − 1.25 = 3.75 m 9. 1 gt2 = 20 (t −1) + 1 g(t − 1)2 Q P 22 Solving this equation we get, ∴ t = 1.5 s PQR = 5 a Now, d = 20 (t − 1) + 1 g(t − 1)2 ∴ t= 5a Ans. 2 u = 11.25 cm Ans. 14. At 4 s 10. vx = dx = (8t − 2) u = at = 8 m/s dt s1 = 1 at2 = 1 × 2 × 42 = 16 m 2 2 xt ∴ ∫ dx = ∫ (8t − 2) dt From 4 s to 8 s 14 2 a = 0, v = constant = 8 m/s x − 14 = (4t2 − 2t) − (12) ∴ s2 = vt = (8)(4) = 32 m From 8 s to 12 s ∴ x = 4t2 − 2t + 2 ...(i) s3 = s1 = 16 m vy = dy = 2 ∴ sTotal = s1 + s2 + s3 = 64 m Ans. dt yt 15. Retardation is double. Therefore retardation time ∴ ∫ dy = ∫ 2dt will be half. 42 Let t0 = acceleration time ∴ y − 4 = 2t − 4 Then t0 = retardation time or y = 2t 2 ∴ t= y t0 + t0 = t 2 2 Substituting this value of t in Eq. (i) we have, ∴ t0 = 2t 3 x = y2 − y + 2 Ans.

504 — Mechanics - I Now, s = s1 + s2 Now applying  21 (a)(t0)2  21 (2a) t0  2 v = u + at , we have 2 = + v = (+ 20) + (−10) (3) 3 at02 = − 10 m/s 4 = ∴ Velocity is 10 m/s, downwards. =  43 (a) 23t 2 20. a = dv = 0.2 v2 dt = at2 Ans. v2 3 ∴ ∫ − 5v−2 = ∫ dt 10 0 16. At the time of overtaking, ∴ 5v = 2  v  10 s1 = s2 ∴ 2ut + 1 at2 = ut + 1 (2a)t2 ∴ 5− 5 =2 v 10 22 ∴ t = 2u or v = 2.0 m/s Ans. a 21. d = d1 + d2 (2u) 2au 1 (a) 2au 2 ∴ s1(or s2 ) = + 2 = v12 + v22 2a1 2a2 = 6u2 Ans. a = (10)2 + (20)2 2 × 2 (2 × 1) 17. t = 80 m = 8 s. Ans. = 225 m 30 m/s 3 Now in vertical direction 22. s1 = s2 t = 2ur ∴ (40) t − 1 × 10 × t2 = (40) (t − 2) ar 2 − 1 × 10 × (t − 2)2 or ur = tar (ar = g = 10 m/s2) 2 2 = (8/3) (10) Solving this equation, we get 2 t=5s = 40 m/s Ans. Then s1 = (40) (5) − 1 × 10 × (5)2 3 2 18. s = net area of v-t graph = 75 m Ans. At 2 s, net area = 0 23. d = 1 g (T − t)2 ∴ s=0 and the particle crosses its initial position. 2 Free fall 19. Total height = 15 + u2 T–t d 2g = 15 + (10)2 H 2 × 10 th or h = 20 m Initial velocity u = 2gh = 2 × 10 × 20 ∴ h = H − d = H − 1 g (T − t)2 2 = 20 m/s

Chapter 6 Kinematics — 505 24. x = t2 29. F = 3t2 − 32 2 a = F = (0.3 t2 − 3.2) m ∴ vx = dx = t y = x2 = t4 dt 28 v5 5 ∴ vy = dy = t3 ∫ dv = ∫ adt = ∫ (0.3 t2 − 3.2) dt dt 2 10 0 0 At t = 2 s v − 10 = − 3.5 ∴ v = 6.5 m/s Ans. vx = 2 m/s 30. v2 = u2 − 2g  2h and vy = 4 m/s ∴ (10)2 = u2 − gh ∴ v = vx i$ + vy $j = (2i$ + 4$j) m/s 25. x = t + 3 ∴ u2 = (100 + gh) ∴ x = (t + 3)2 Now, h = u2 = 100 + gh or v = dx 2g 2g dt h=5+ h (as 2g = 20 m/s2) = 2 (t + 3) 2 ∴ v-t equation is linear Ans. ∴ h = 10 m Ans. Ans. 26. ∆v = vf − vi = area under a-t graph Subjective Questions vi = 0 1. (a) dv is the magnitude of total acceleration. ⇒ ν f = area dt = 40 + 50 = 90 m/s While d | v| represents the time rate of change of dt 27. v2 = 25 + 25 s speed (called the tangential acceleration, a or v2 = (5)2 + 2 (12.5) s component of total acceleration) as | v| = v. Now compare with v2 = u2 + 2as (b) These two are equal in case of one dimensional 28. Retardation during upward motion motion. a1 = 10 + 2 2. (a) x = 2t ⇒ t = x = 12 m/s2 2 2 m/s2 y = t2 or y =  2x 2 v ∴ x2 = 4 y is the trajectory Ans. a1 a2 (b) r = xi$ + y$j = (2t$i ) + (t2$j) 10 m/s2 2 m/s2 v v = d r = (2i$ + 2t$j) units Ans. 10 m/s2 dt Ans. Acceleration during downward motion, (c) a = d v = (2$j) units a2 = 10 − 2 = 8 m/s2 dt t = 2s 3. CN a B or t ∝ 1 a DW E ∴ t1 = a2 = 8 = 2 Ans. A t2 a1 12 3 S

506 — Mechanics - I (a) Distance = AB + BC + CD Let t be the time when particle collides with ground. = (500 + 400 + 200) = 1100 m Then using the equation s = ut + 1 at2 (b) Displacement = AD = (AB − CD )2 + BC 2 2 we have, −60 = (20) t + 1 (−10) t2 = (500 − 200)2 + (400)2 2 = 500 m Solving this equation, we get (c) Average speed = Total distance t=6s Total time (a) Average speed = Total distance = 1100 = 55m/min Total time 20 = 20 + 20 + 60 (d) Average velocity = AD 6 Ans. t = 16.67 m/s = 500 = 25 m/min (along AD) 20 (b) Average velocity = s = 60 t6 4. (a) The distance travelled by the rocket in 1 min = 10 m/s (downwards) Ans. (= 60 s) in which resultant acceleration is vertically upwards and 10 m/s2 will be 6. Average velocity = Total displacement h1 = (1/ 2) × 10 × 602 = 18000 m = 18 km …(i) Total time ∴ 2.5 v = vt0 + 2vt0 + 3vT and velocity acquired by it will be t0 + t0 + T v = 10 × 60 = 600 m/s …(ii) Solving, we get Now, after 1 min the rocket moves vertically up T = 4 t0 Ans. with velocity of 600 m/s and acceleration due to gravity opposes its motion. So, it will go to a 7. Retardation time is 8 s (double). Therefore, height h2 till its velocity becomes zero such that retardation should be half or 2 m/s2. 0 = (600)2 − 2gh2 s1 = acceleration displacement or h2 = 18000 m [as g = 10 m/s2 ] …(iii) = 1 a1 t12= 1 × 4 × (4 )2= 32 m = 18 km 2 2 So, from Eqs. (i) and (iii) the maximum height s2 = retardation displacement reached by the rocket from the ground = 1 a2 t22 2 h = h1 + h2 = 18 + 18 = 36 km Ans. (b) As after burning of fuel the initial velocity from = 1 × 2 × (8)2 = 64m 2 Eq. (ii) is 600 m/s and gravity opposes the motion of rocket, so the time taken by it to reach (a) aav = vf − vi = 0−0 = 0 Ans. the maximum height (for which v = 0), t 12 Ans. 0 = 600 − gt (b) and (c) d = s = s1 + s2 = 96 m ∴ Average speed = Average velocity or t = 60 s i.e. after finishing fuel the rocket further goes up =d t for 60 s, or 1 min. Ans. 5. h = u2 = 2 = 20 m or s = 96 = 8 m/s t 12 (20) 2g 2 × 10 20 m/s 8. T = 2πR = (2)  272  2221 = 6 s h v1 t=2s=T 60 m 3 ∴ Particle will rotate by 120°.

Chapter 6 Kinematics — 507 (a) vav = s = 2R sin θ (θ = 120° ) Solving this equation we get, Ans. t 2 t = v0 + t0 g2 t 12. (a) Velocity = slope of s-t graph S ∴ Sign of velocity = sign of slope of s-t graph R 120° (b) Let us discuss any one of them, let the portion R cd, slope of s-t graph (= velocity ) of this region is negative but increasing in magnitude. = 3 R = 21 3 m/s Ans. Therefore velocity is negative but increasing. t 44 Therefore sign of velocity and acceleration both are negative. 13. Displacement s = net area of v-t graph |∆v | v2 + v2 − 2vv cos 120° = 40 + 40 + 40 − 20 = 100 m t t (b) aav = = Distance d = |Total area | = 3 v = 3 m/s2 Ans. = 40 + 40 + 40 + 20 = 140 m Ans. t2 (a) Average velocity = s = 100 = 50 m/s 9. At minimum distance, their velocities are same, t 14 7 ∴ vA = vB or uA + aA t = uB + aB t (b) Average speed = d = 140 = 10 m /s Ans. t 14 or (3 + t) = (1 + 2t) or t = 2 s 14. v1 = speed of person Minimum distance, dmin = Initial distance − v2 = speed of escalator extra displacement of A upto this instant due to its v1 = l and v2 = l greater speed t1 t2 = 10 − (sA − sB ) ∴ t= l = l = 10 + sB − sA v1 + v2 l + l t1 t2 = 10 +  uB t + 1 aB t2 −  uA t + 1 aA t2 Ans. 2 2 = t1 t2 = 90 × 60 = 36 s t1 + t2 90 + 60 (1) (2) 1 (2) (2)2 (3) 1 (2)2  = 10 + + 2 − (2) + 2 (1)  15. At time t = t1, slope of s-t graph (= velocity ) is =8m Ans. positive and increasing. Therefore velocity is positive and acceleration is also positive. 10. Given h1 − h2 = 10 m At time t = t2, slope is negative but decreasing (in magnitude). Therefore velocity is negative but h2 decreasing in magnitude. Hence, acceleration is h1 2 positive. 16. Comparing with v = u + at, we have, u = 40 m/s and a = − 10 m/s2 1 au ∴ 1 gt2 − 1 g (t − 1)2 = 10 X = – 60m X=0 X = + 60m 22 t=0 t1 Solving equation, we get t = 1.5 s Ans. 11. At the time of collision, t3 t2 s1 = s2 Distance of 60 m from origin may be at x = + 60 m and x = − 60 m. ∴ v0 t − 1 gt2 = v0 (t − t0 ) − 1 g (t − t0 )2 2 2

508 — Mechanics - I From the figure, we can see that at these two points 18. (a) aav = vf − vi = v12 − v6 t 12 − 6 particle is at three times, t1, t2 and t3. For X = + 60 m or t1 and t2 = (−10) − (20) = − 5 m/s2 Ans. 6 s = ut + 1 at2 2 (b) ∆v = vf − vi = net area of a-t graph ⇒ 60 = (+40) t + 1 (−10) t2 ∴ v14 − v0 = 40 + 30 + 40 − 20 = 90 2 But v0 = 0 Solving this equation, we get ∴ v14 = 90 m/s Ans. t1 = 2s and t2 = 6 s Ans. 19. s rf − ri t t For X = − 60 m or t3 (a) vav = = s = ut + 1 at2 = (6i$ + 4$j) − ($i + 2$j) 2 4 ∴ − 60 = (+ 40) t + 1 (−10) t2 = (1.25$i + 0.5$j) m/s Ans. 2 ∆v vf − vi Solving this equation, we get positive value of t as (b) a av = t = t t3 = 2 (2 + 7) s Ans. = (2$i + 10$j) − (4 $i + 6$j) 4 17. Displacement = Area under velocity-time graph Hence, sOA = 1 × 2 × 10 = 10 m = (−0.5$i + $j ) m/s2 Ans. 2 or (c) We cannot calculate the distance travelled from sAB = 2 × 10 = 20 m the given data. or and sOAB = 10 + 20 = 30 m 20. In falling 5 m, first stone will take 1 s (from or sBC = 1 × 2(10 + 20) = 30 m h = 1 gt2) 2 2 sOABC = 30 + 30 = 60 m First stone, t = 2h ...(i) g sCD = 1 × 2 × 20 = 20 m 2 sOABCD = 60 + 20 = 80 m Second stone, (t − 1) = 2 (h − 2s) ...(ii) g s (m) 80 Solving these equations, we get Ans. 60 h = 45 m 21. If we start time calculations from point P and apply the equation, aa 30 10 u=0 P v = at0 +ve 2468 t (s) 1 at02 Ans. 2 Between 0 to 2 s and 4 to 6 s motion is accelerated, s = ut + 1 at2 hence displacement-time graph is a parabola. 2 Between 2 to 4 s motion is uniform, so displacement-time graph will be a straight line.  − 1 at02 = (+ at0) t + 1 (−a) t2 Between 6 to 8 s motion is decelerated hence 2 2 displacement-time graph is again a parabola but inverted in shape. At the end of 8 s velocity is zero, Solving we get, t = ( 2 + 1) t0 therefore, slope of displacement-time graph should = 2.414 t0 be zero. The corresponding graph is shown in above figure. ∴ Total time, T = t + t0 = (3.414) t0

Chapter 6 Kinematics — 509 22. (a) h = 15 + u2 After next 2 s 2g (a) v2 = v1 + a2 t2 = 15 + (5)2 = (4i$) + (2i$ − 4$j) (2) 2 × 10 = 16.25 m = (8$i − 8$j) m/s Ans. (b) r2 = r1 + v1t2 + 1 a 2 t22 2 (b) t = 2h = 2 × 16.25 = 1.8 s g 10 Ans. = (6$i + 4$j) + (4$i ) (2) + 1 (2$i − 4$j) (2)2 2 23. (a) and (b) 60 = u (6) + 1 × a × (6)2 ...(i) = (18$i − 4$j) m 2 ∴ Co-ordinates are, 15 = u + (a) (6) ...(ii) x = 18 m Solving these two equations, we get and y = − 4 m u = 5 m/s 1 and a = 5 m/s2 Ans. 26. x = uxt + 2 ax t2 3 ∴ 29 = (0) (t) + 1 × (4.0) t2 (c) (5)2 = (2)  53 s (v2 = 2 as) 2 ∴ s = 7.5 m Ans. ∴ t2 = 14.5 s2 or t = 3.8 s 24. s = s0 + ut + 1 at2 (a) y= uy t + 1 ay t2 2 2 at t = 0, s = s0 = 2 m ...(i) = (8) (3.8) + 1 × 2 × 14.5 2 at t = 10 s, = 44.9 m ≈ 45 m Ans. s = 0= s0 + ut + 1 at2 (b) v = u + at 2 = (8$j) + (4.0i$ + 2.0$j) (3.8) = (15.2i$ + 15.6$j) or 0 = 2 + 10u + 50a or 10u + 50a = −2 ...(ii) v = u + at ∴ Speed = | v| = (15.2)2 + (15.6)2 ∴ 0 = u + 6a (at t = 6 s) = 22 m/s Ans. ∴ u + 6a = 0 ...(iii) 27. dv = − 3m/s Solving Eqs. (ii) and (iii) we get, Ans. ds m u = − 1.2 m/s and a = 0.2 m/s2 a = v ⋅ dv ds Now, v = u + at Ans. ∴ v = (−1.2) + (0.2) (10) = (10) (−3) = − 30 m/s2 = 0.8 m/s Ans. 28. v = 3t2 − 6t ⇒ v = 0 at t = 2 s 25. After 2 s For t < 2 s, velocity is negative. At t = 2 s, velocity is zero and for t > 2 s velocity is positive. v1 = u + a1 t1 = 0 + (2i$) (2) = (4i$) m/s 3. 5 3. 5 r1 = ri + 1 a 1 t12 ∴ s1 = ∫ vdt = ∫ (3t2 − 6t) dt 2 00 = (2$i + 4$j) + 1 (2$i ) (2)2 = 6.125 m 2 = displacement upto 3.5 s = (6i$ + 4$j) m 22 s2 = ∫ vdt = ∫ (3t2 − 6t) dt 00

510 — Mechanics - I =−4m 33. Kinetic energy will first increase and then decrease. = displacement upto 2 s v = gt (in downward journey) –4m O 6.125 m ∴ KE = 1 mv2 = 1 mg2 t2 22 Hence, KE versus t graph is a parabola. ∴ d = distance travelled in 3.5 s 34. Speed first increases. Then just after collision, it = 4 + 4 + 6.125 becomes half and now it decreases. = 14.125 m Ans. Pattern of velocity is same (with sign). In the answer, downward direction is taken as positive. Average speed = d = 14.125 Further, v = gt t 3.5 Hence, v-t graph is straight line. = 4.03 m/s Ans. 35. Area of a-t graph gives change in velocity Average velocity = s1 = 6.125 ∴ ∆v = vf − vi = area of a-t graph t 3.5 Since, vf = vi = 1.75 m/s Ans. a B 29. a = 4 or dv = 4 A dt v v + C Et ∴ vt O ∴ ∫ vdv = ∫ 4 dt – ∴ 62 D v2 − 18 = 4t − 8 ∴ Net area of a-t graph should be zero Ans. 2 ∴ Area OABC = Area CDE Substituting the values we get, v = 8t + 20 tE = (2 + 3) s a = dv = 4 dt 8t + 20 36. (a) a = slope of v-t graph At t = 3s a = 0.603 m/s2 Ans. 30. v-s equation is v (m/s) (b) s = rf − ri = net area of v -t graph ∴ rf = ri + net area v = − 2 s + 20 20 = 10 + 10 + 20 + 10 − 10 − 10 3 = 30 m ∴ dv = − 2 per second (c) (i) For 0 ≤ t ≤ 2 s ds 3 30 s (m) u = initial velocity = 0 At s = 15 m, v = 10 m/s s0 = initial displacement = 10 m a = v ⋅ dv = − 20 m/s2 Ans. a = slope of v-tgraph = + 5 m/s2 ds 3 1 31. a = slope of v-t graph and s = area under v-t graph. ∴ s = s0 + ut + 2 at2 = 10 + 2.5 t2 Further, for t ≤ 2s, velocity (i.e. slope of s-t graph) (ii) For 4 s ≤ t ≤ 8 s is positive but increasing. Therefore, s-t graph is as u = initial velocity = 10 m/s under. = velocity at 4 s s s0 = (10 m) + area of v-t graph upto 4 s = 10 + 10 + 20 = 40 m a = slope of v-t graph = −5 m/s2 t ∴ s = s0 + u (t − 4) + 1 a (t − 4 )2 2 Same logic can be applied with other portions too. = 40 + 10 (t − 4) − 2.5 (t − 4)2 Ans. 32. Same as Q.No. 31.

Chapter 6 Kinematics — 511 37. (a) a1 = a2 = −g (b) For automobile ∴ a12 = a 1 − a 2 = 0 Ans. s2 = 1 a2 t2 = 1 × 3.5 × (7.39)2 (b) u21 = u2 − u1 = 20 − (−s) Ans. 2 2 = + 25 m/s = 95.5 m (c) u12 = − u21 = − 25 m/s ∴ Initial distance between them, Ans. Since, a12 = 0 = s2 − s1 ∴ u12 = constant = 35.5 m = − 25 m/s Ans. (c) v1 = a1 t = (2.2) (7.39) = 16.2 m/s and v2 = a 2t = (3.5) (7.39) = 25.9 m/s (d) a12 = 0. Therefore, relative motion between them is uniform with constant velocity 25 m/s. 40. Net velocity is along AB or at 45° if, ∴ t = d = 20 = 0.8 s Ans. B v 25 38. (a) When the two meet, Net 18 m/s 4 cos q 1g 45° A (4 sin q + 2) 2 m/s = constant 12 m 2 4 cosθ = 4 sinθ + 2 5m Solving this equation we get, Ans. s2 = s1 + 7 θ ≈ 24.3° or (2t) = (18t) − (4.9t2) + 7 E 41. (a) sinθ = 200 = 0.4 500 N Solving we get, Ans. Q 500 km/h t = 3.65 s q s2 = 2 × 3.65 = 7.3 m Net ∴ Height = 5 + 7.3 velocity = v = 12.30 m q Ans. P 200 km/h (b) vball = u − gt = 18 − 9.8 × 3.65 = − 17.77 m/s ∴ Velocity of ball with respect to elevator ∴ θ = sin−1 (0.4), west of north (b) v = (500)2 − (200)2 = velocity of ball − velocity of elevator = 100 21 km/h = (−17.77) − (2) ∴ t = PQ = 1000 = 10 h = − 19.77 m/s v 100 21 21 ≈ − 19.8 m/s Ans. 42. v Negative sign indicates the downward Ans. direction. 39. (a) For truck s1 = 1 a1 t2 2 v0 60 = 1 × 2.2 × t2 t 2 t1 t2 ∴ t = 7.39 s Ans.

512 — Mechanics - I v0 = x ⇒ t1 = v0 2. dv = a = − 4v + 8 t1 x dt v0 = y ⇒ t2 = v0 ∴ da = − 4 dv t2 y dt dt + =  1 + 1 =4 = − 4 (−4v + 8)  x y t1 t2 v0  ...(i) = 16v − 32 Further, ∴  ddat  i = 16 vi − 32 area = displacement ∴ 1 v0 × t = s = (16) (0) − 32 2 = − 32 m/s2 But numerically, t = s = 4 units Further, ∴ v0 = 2units ∫ ∫v dv t Substituting in Eq. (i) we get, = dt 0 8 − 4v 0 1 + 1 = 2. xy Solving this equation we get, v = 2 (1 − e−4t ) LEVEL 2 v (m/s) Single Correct Option 2 1. Let velocity of rain is 5 m/s ^j t (s) 37° 37° Hence, v-t graph is exponentially increasing graph, 37° terminating at 2 m/s. ^i 5 m/s 3. For collision, (i) (ii) (iii) rA = rB (at same instant) vR = ai$ − b$j ∴ (ri + vt )A = (ri + vt )B In first case, vM = (−4$i − 3$j) ∴ vRM = vR − vM = ( a + 4) i$ + (−b + 3) $j ⇒ (5i$ + 10$j + 5k$ ) t = 30i$ + (10i$ + 5$j + 5k$ ) t It appears vertical Equating the coefficients of x ∴ a+ 4=0 5t = 30 + 10t or a = − 4 ⇒ t = − ve In second case, vM = (4$i + 3$j) So, they will never collide. ∴ vRM = vR − vM = (a − 4) $i + (−b − 3$j) 4. d vy = 2t = − 8i$ + (−b − 3$j) dt ∴ vy = t2 or dy = t2 dt It appears at θ = tan−1  87 or y = t3 ...(i) 3 Ans. ∴ −b − 3 = − 7 ∴ b=4 and x = v0t ⇒ t=x and speed of rain = a2 + b2 v Substituting in Eq. (i) we have, = 32 m/s y = x3 3v03 Ans.

Chapter 6 Kinematics — 513 5. dt = (2αx + β) 10. B 2 m/s dx 60° ∴ dx = v =  1    dt  2αx + β v a = dv = −2α  1  2 ⋅ dx 30° dt  2αx + β dt A = − 2α (v2) (v) = − 2αv3 Ans. v cos 60° = 2 m/s ...(i) 6. f = v ⋅ dv = a − bx ∴ v = 4 m/s Ans. dx 11. 90 km/h = 90 × 5 = 25 m/s vx 18 180 km/h = 108 × 5 = 30 m/s or ∫ vdv = ∫ (a − bx) dx 00 18 ∴ v = 2ax − bx2 At maximum separation their velocities are same. At other station, v = 0 Ans. ∴ Velocity of motorcycle = 25 m/s ⇒ x = 2a or at = 25 or t = 5 s b But thief has travelled up to 7s. Further acceleration will change its direction when, s1 = displacement of thief f = 0 or a − bx = 0 or x = a b = v1 t1 = 25 × 7 = 175 m At this x, velocity is maximum. s2 = displacement of motorcycle 1 Using Eq. (i), = 2 × a2 t22 vmax = 2a  ab − b  ab 2 = a Ans. = 1 × 5 × (5)2 b 2 7. a = F = −kx2 = 62.5 m mm ∴ Maximum separation ∴ v ⋅ dv = − kx2 = s1 − s2 = 112.5 m Ans. dx m 12. Relative velocity of A with respect to B should be v 0 or v dv = − kx2 dx along AB or absolute velocity components ∫ ∫ m 0a perpendicular to AB should be same. v2 = ka3 ∴ 2u sin θ = u sin 30° 2 3m 3 ∴ v = 2ka3 ∴ θ = sin−1  43 Ans. 3m Ans. 8. a = v ⋅ dv = (4) (− tan 60°) 13. Deceleration is four times. Therefore, deceleration ds time should be 1 th. 4 = − 4 3 m/s2 v 9. x1 = 1 gt2 = 0.5 gt2 2 x1 + x2 = 1 g (2t)2 = 2 gt2 vmax = 4t 2 a1 a2 x2 = 2gt2 − x1 = 1.5 gt2 4t t t ∴ x2 − x1 = gt2 or t= x2 − x1 vmax = (a1) (4 t) = (1) (4 t) = 4 t g

514 — Mechanics - I Area of v-t graph = displacement More than One Correct Options ∴ 200 = 1 (5t) (4 t) 1. (a) a = − α v 2 ∴ dv = − α v or t = 20 s dt Total journey time = 5t = 22.4 s Ans. t 1 0 14. Area of v-t graph = displacement ∫ ∫or dt = − α v−1/ 2 ⋅ dv 1 0 v0 2 ∴ 1032 = (56 + t0 ) (24 ) or t0 = 30 s ∴ t = 2 v0 Ans. α v (m/s) 24 t0 (d) a = − α v ∴ v ⋅ dv = − α v ds t2 t1 t (s) s ds = − 1 0 56 s ∫ ∫∴ v1/ 2 dv 0 α v0 24 Deceleration time t1 = 4 = 6 s ∴ s = 2v03/ 2 Ans. 3α ∴ Acceleration time t2 = 56 − t0 − t1 = 20 s ∴ Acceleration = 24 = 1.2 m/s2 Ans. 2. a = − 0.5t = dv 20 dt 15. vQ = − 20i$ vt y ∴ ∫ dv = ∫ −0.5t dt 16 0 x P 20 cm/s ∴ v = 16 − 0.25t2 3m OP = 3 m s = ∫ vdt = 16t − 0.25t3 OQ = 4 m 3 60° Q v = 0 when 16 − 0.25t2 = 0 O 4 m 20 cm/s or t = 8 s Ans. vP = − 20 cos 60i$ − 20 sin 60° $j = − 10i$ − 10 3 $j So direction of velocity changes at 8 s. Up to 8 s Assuming P to be at rest, distance = displacement vQP = vQ − vP = − 10i$ + 10 3 $j ∴ At 4 s Now, ∴ tan θ = 10 3 = 3 or θ = 60° 10 d = 16 × 4 − 0.25 × (4)3 = 58.67 m Ans. 3 where, θ is the angle of vQP from x-axis towards positive y -axis. S8s = 16 × 8 − (0.25) (8)3 3 vQP = 85.33 m N ON = OQ = 4 m OP = 3 m S 10s = 16 × 10 − (0.25) (10)3 3 60° M \\ PN = 1 m P = 100 cm = 76.67 cm 85.33 m 8s 0s 60° 60° 10 s O Q 76.67 m Shortest distance = PM = PN sin 60° Distance travelled in 10 s, = (100) 3 = 50 3 cm 2 d = (85.33) + (85.33 − 76.67) = 94 m Ans.

Chapter 6 Kinematics — 515 At 10 s v = 16 − 0.25 (10)2 = − 9 m/s v1 = t1 s1 t2 = 1 a1 t1 ∴ Speed = 9 m/s + 2 Ans. s2 = 1 (t1 + t3 ) vmax 2 3. If a = constant = 1 (t1 + 2t2) (2a1 t1) Then |a | is also constant or d v = constant 2 dt v2 = (t1 s2 t3 ) = a1 t1 4. rB = 0 at t = 0 + Ù v N ( j) vmax 2a1 t3 = 2t2 A t1 tmax = 2a1t1 a2 = 2a2t2 B 4 km 3 km t E (^i) t3 rA = 3i$ + 4$j From the four relations we can see that at t = 0 v2 = 2v1 Ans. vA = (−20$j) and 2s1 < s2 < 4s1 vB = (40 cos 37° )$i + (40 sin 37° )$j = (32i$ + 24$j) 6. In the complete journey, vAB = (−32i$ − 44$j) km/h Ans. vB s = 0 and a = constant = g (downwards) ...(i) 7. a = F = α t 37° mm or a ∝ t At time t = 0, i,e. a-t graph is a straight line passing through rAB = rA − rB = (3i$ + 4$j) origin. ∴ At time t = t , If u = 0, then integration of Eq. (i) gives, v = αt2 or v ∝ t2 rAB = (rAB at t = 0) + vAB t 2m = (3 − 32t)i$ + (4 − 44 t)$j Ans. Hence in this situation (when u = 0 ) v-t graph is a parabola passing through origin. 5. a1 t1 = a2 t2 8. a-s equation corresponding to given graph is, vmax = a1 t1 = a2 t2 a=6− s s1 = Area of v-t graph 5 = 1 (t1 + t2 ) (a1 t1) ∴ v ⋅ dv =  6 − 5s 2 ds v or v vdv = s  6 − 5s ds vmax ∫ ∫ a1 00 t1 a2 or v = 125 − s2 t2 5 t At s = 10m, v = 10 m/s Ans.

516 — Mechanics - I Maximum values of v is obtained when ∴ t0 = t − t1 − t2 =  t − v − βv dv = 0 which gives s = 30 m α ds 1 αt12 1 2 12 × 30 − (30)2 Now, l = 2 + vt0 + 2 βt 2 5 ∴ vmax = 2  v  v 2  β  β = 180 m/s Ans. = 1 (α )  v  + v t − v − + 1 ( β ) 2 α α 2 9. vav = s and vav = d = vt − v2 − v2 t t 2α 2β Now, d ≥ |s| ∴ vav ≥ | vav | ∴ t = l + v 1 + 1 v 2  α β 10. v = at and x = 1 at2 (u = 0) i.e. v-t graph is a straight For t to be minimum its first derivation with respect 2 to velocity be zero or, line passing through origin and x-t graph a parabola l α +β passing through origin. v2 2αβ 11. For minimum time 0 = − + ∴ tmin = b ∴ v = 2lαβ v α +β bv 14. x = t2 ⇒ vx = dx = 2t u dt ⇒ For reaching a point exactly opposite ⇒ ax = dvx = 2 dt u b v Net velocity y = t3 − 2t ⇒ vy = dy = 3t2 − 2 dt ⇒ ay = dv y = 6t dt Net velocity = v2 − u2 (but v > u ) At t = 0, vx = 0, vy = − 2, ax = 2 and ay = 0 ∴ t= b ∴ v = − 2$j and a = 2i$ net velocity or v ⊥ a 12. For t < T , v = −ve At t = 2, vy = 0, vx ≠ 0`. 3 For t > T , v = +ve At t = T , v = 0 Hence, the particle is moving parallel to x-axis. ∴ Particle changes direction of velocity at t = T 15. (14)2 = (2)2 + 2 as S = Net area of v-t graph = 0 ∴ 2 as = 192 units At mid point, v2 = (2)2 + 2a  2s a = Slope of v-t graph = constant = 4 + 192 = 100 13. v = αt1 ⇒ t1 = v 2 α ∴ v = 10 m/s v = βt2 ⇒ t2 = v β Ans. v XA : AY = 1: 3 v ∴ XA = 1 s and AY = 3 s t 44 t1 t0 t2 v12 = (2)2 + 2a  4s 

Chapter 6 Kinematics — 517 = 4 + 192 = 52 3. S = 25 × 2.13 − 5 × (2.13)2 4 = 30.56 m Ans. ∴ v1 = 52 ≠ 5 m/s Ans. 4. At maximum separation, their velocities are same 10 = 2 + at1 (v = u + at) ∴ 25 − 10t = 10 + 5t or t = 1 s ∴ t1 = 8 14 = 10 + at2 Maximum separation = 2 + S2 − S1 a = 2 + [25 × 1 − 5 × (1)2 ] − [10 × 1 + 2.5 (1)2 ] ∴ t2 = 4 or t1 = 2 t2 Ans. a S1 = (2t) + 1 a (t2) = distance travelled in first half = 9.5 m Ans. 2 5. uA + aAT = uB − aBT 1 S2 = 2 (2t) + 2 a (2t)2 Putting T =4s S3 = S2 − S1 = distance travelled in second half we get 4 (aA + aB ) = uB − uA ....(i) We can see that, Ans. Now, SA = SB ∴ Ans. S1 ≠ S3 uA t + 1 aA t2 = uB t − 1 aB t2 2 2 2 Comprehension Based Questions ∴ t = 2 (uB − uA) = 2 × 4 = 8 s (aA + aB ) 1. Velocity of ball with respect to elevator is 15 m/s 6. SA = SB (up) and elevator has a velocity of 10 m/s (up). 5t + 1 aAt2 = 15t − 1 aB t 2 Therefore, absolute velocity of ball is 25 m/s 2 2 (upwards). Ball strikes the floor of elevator if, 25 m/s + or 10 + aA t = 30 − aB t 2 10 m/s2 ∴ (5 + aA t) − (15 − aB t) = 10 or vA − vB = 10 m/s 2m 1 – 7. 8 = 6 + aAT = 6 + 4aA ∴ aA = 0.5 m/s At 10 s, 50 m 10 m/s, 5 m/s2 vA = uA + aA t = (6) + (0.5) (10) = 11 m/s Ans. S1 = S2 + 2 Ans. Match the Columns ∴ 10t + 2.5t2 = 25t − 5t2 + 2 1. In (a) and (b), if velocity is in the direction of Solving this equation we get, t = 2.13 s acceleration (or of the same sign as that of acceleration) then speed increases. And if velocity 2. If the ball does not collides, then it will reach its is in opposite direction, then speed decreases. maximum height in time, In (c) slope of s-t graph (velocity) is increasing. Therefore speed is increasing. In (d) slope of s-t t0 = u = 25 = 2.5 s graph is decreasing. Therefore, speed is decreasing. g 10 2. If v ⋅ a = 0, speed is constant because angle between Since, t < t0, therefore as per the question ball is at its maximum height at 2.13 s. v and a in this case is 90°. If v ⋅ a = positive then speed is increasing because hmax = 50 + 2 + 25 × 2.13 − 5 × (2.13)2 angle between v and a in this case is acute. If v ⋅ a = negative then speed is increasing because = 82.56 m Ans. angle between v and a in this case is obtuse.

518 — Mechanics - I 3. In portion AB, we can see that velocity is positive Subjective Questions and increasing. Similarly for other parts we can 1. aav = ∆v = vf − vi +ve draw the conclusions. ∆t ∆t –ve 4. (a) Average velocity = S = Area of v-t graph = 2g hf + 2g hi Ans. ∆t t Time = 20 = 5 m/s = 2 × 9.8 × 2 + 2 × 9.8 × 4 12 × 10−3 4 (b) Average acceleration = vf − vi = 1.26 × 103 m/s2 time Note vf is upwards (+ve) and vi is downwards (− ve). = v4 s − v1 s = 0 − 5 = − 5 m/s2 2. v dv = ads 4 −1 3 3 v 12 m (c) Average speed = d = |Total area | t Time ∴ ∫0vdv = ∫0 a ds = 20 + 10 = 5 m/s ∴ v2 = area under a -s graph from s = 0to s = 12 m. 6 2 (d) Rate of change of speed at 4 s = |a | = 2 + 12 + 6 + 4 = |slope of v-t graph | = 5 m/s2 = 24 m2/s2 5. (a) x = 0 at t = 2s or v = 48 m/s = 4 3 m/s Ans. (b) v = dx = 10 t and a = dv = 10 3. Let AB = BC = d dt dt BD = x v = a at 1s and BB′ = s = displacement of point B. (c) v = 10 t AN ∴ Velocity is positive all the time. (d) v = 0 at t = 0 s 6. vx = dx = 2t − 2 vt BN dt s ax = dvx = 2 A BD C dt 1 at 2 2 vy = dy = 2t − 4 dt CN From similar triangles we can write, dv y …(i) ay = dt = 2 1 at2 2 …(ii) vt = s = (a) It crosses y -axis when, x = 0 d+x x d−x ⇒ t =1s From first two equations we have, At this instant vy is −2 m/s. 1 + d = vt (b) It crosses x-axis when, y = 0 xs ⇒ t=2s or d = vt − 1 xs At this instant vx is + 2 m/s (c) At t = 0, vx = − 2 m/s and vy = − 4 m/s From last two equations we have, ∴ Speed = vx2 + v2y = 2 5 m/s 1 at2 2 (d) At t = 0, ax = 2 m/s2 d −1= and ay = 2 m/s2 ∴ a = ax2 + a2y xs or d = 1 at2 + 1 2 = 2 2 m/s xs

Chapter 6 Kinematics — 519 Equating Eqs. (i) and (ii) we have, ∴ 1 (3t)(0.2 t) = 14 Ans. 2 vt −1= 1 at2 + 1 2 Solving this equation, we get 3t = 20.5 s ss vt − 1 at2 Note Maximum speed 0.2 t is less than 2.5 m/s. or 2 = 2 6. Let t0 be the breaking time and a the magnitude of s deceleration.  v2 1  2a t2 ∴ s = t − 2 80.5 km/h = 22.36 m/s, 48.3 km/h = 13.42 m/s. Comparing with s = ut + 1 at2 we have, In the first case, 2 56.7 = (22.36 × t0) + (22.36)2 …(i) Initial velocity of B is + v and acceleration − a. 2a 22 and 24.4 = (13.42 t0) + (13.42)2 …(ii) 2a 4. Let us draw v-t graph of the given situation, area of Solving these two equations, we get which will give the displacement and slope the t0 = 0.74 s Ans. acceleration. and a = 6.2 m/s2 v 7. Absolute velocity of ball = 30 m/s y 10 m/s yd 30 m/s d x 2 2m xx 1 t 28 m t1 t2 t3 s2 − s1 = x d + 1 yd …(i) (a) Maximum height of ball from ground 2 …(ii) = 28 + 2 + (30)2 = 76 m 2 × 9.8 s3 − s2 = xd + yd + 1 yd 2 (b) Ball will return to the elevator floor when, Subtracting Eq. (i) from Eq. (ii), we have s1 = s2 + 2 s3 + s1 − 2s2 = yd or 10t = (30t − 4.9 t2) + 2 or s3 + s1 − 2 s1s3 = yd (s2 = s1s3 ) Solving, we get t = 4.2s Ans. Dividing by d2 both sides we have, ( s1 − s3 )2 = y 8. (a) Average velocity d2 d 5 = Displacement = ∫0v dt = slope of v-t graph = a. Hence proved. Time 5 5. Area of v-t graph = displacement ∫5 − t2) dt (3t v (m/s) =0 5 = − 0.833 m/s 0.2t (b) Velocity of particle = 0 at t = 3 s t 2t t (s) i.e. at 3 s, particle changes its direction of motion. Average speed = Total distance Total time

520 — Mechanics - I = (Distance from 0 to 3 s) + (Distance from 3 s to 5 s) ∴ Total time taken Time t = t1 + t2 = (200 sec θ − 120 tan θ) ∫d0-3 = 3 − t2)dt = 4.5 m For t to be minimum, dt = 0 (3t dθ 0 or 200 sec θ tan θ − 120 sec2 θ = 0 ∫d3-5 = 5 2 − 3t) dt = 8.67 m (t 3 or θ = sin−1 (3/5) ∴ Average speed = 4.5 + 8.67 5 (b) tmin = 200 sec θ − 120 tan θ (where, sin θ = 3) = 200 × 5 − 120 × 3 5 = 2.63 m/s Ans. 9. (a) a = 2t − 2 (from the graph) 44 vt t = 250 − 90 = 160 s = 2min 40 s Ans. Now, ∫0 dv = ∫0a dt = ∫0(2t − 2) dt 11. Given that | vbr | = vy = dy = u …(i) ∴ dt v = t2 − 2t 4 4(t2 − 2t) dt = 6.67 m | vr | = vx = dx =  2cv0 y …(ii) dt v dt = ∫ ∫(b) s = 22 10. (a) From Eqs. (i) and (ii) we have, dy = uc dx 2v0 y 3 m/s 120 m vbr 4 m/s θ c vr y y x y y dy = uc x y2 = ucx 3 cos θ ∫ ∫or or Ans. dx Ans. 0 2v0 0 v0 (4 – 3 sin θ) At y = c , x = cv0 2 4u Time to cross the river or xnet = 2x = cv0 2u 120 40 t1 = 3 cos θ = cos θ = 40 sec θ 12. a = v dv = v (slope of v-s graph) ds  40  At s = 50 m Drift along the river x = (4 − 3 sin θ)   v = 20 m/s and dv = 40 = 0.4 per sec  cos θ ds 100 = (160 sec θ − 120 tan θ) ∴ a = 20 × 0.4 = 8 m/s2 To reach directly opposite, this drift will be covered by walking speed. At s = 150 m Time taken in this, v = (40 + 5) = 45 m/s and dv = 10 = 0.1per sec 53 ds 100 ∴ a = 45 × 0.1 = 4.5 m/s2 a- s graph θ From s = 0 to s = 100 m v = 0.4 s 4 and dv = 0.4 t2 = 160 sec θ − 120 tan θ ds 1 ∴ a = v. dv = 0.16 s = 160 sec θ − 120 tan θ ds

Chapter 6 Kinematics — 521 i.e a - s graph is a straight line passing through origin y xx(a − x) of slope 0.16 per (sec)2. 0 a2 At s = 100 m, a = 0.16 × 100 = 16 m/s2 dy = From s = 100 m to s = 200 m 0 ∫ ∫or dx v = 0.1s + 30 dv = 0.1 or y = x2 − x3 …(iii) ds 2a 3a2 ∴ a = v dv = (0.1s + 30)(0.1) = (0.01s + 3) This is the desired equation of trajectory. ds (b) Time taken to cross the river is i.e. a-s graph is straight line of slope 0.01 (sec)-2 and intercept 3 m/s 2. t= a = a At s = 100 m, a = 4 m/s2 vx v and at s = 200 m, a = 5 m/s2 (c) When the boatman reaches the opposite side, Corresponding a-s graph is as shown in figure. x = a or vy = 0 [from Eq. (i)] Hence, resultant velocity of boatman is v along a (m/s2) positive x-axis or due east. (d) From Eq. (iii) y = a2 − a3 = a 2a 3a2 6 At x = a (at opposite bank) Hence, displacement of boatman will be 16 s = x $i + y $j or s = a$i + a $j 6 5 14. (a) Since, the resultant velocity is always 4 perpendicular to the line joining boat and R, the boat is moving in a circle of radius 2ω and centre at R. s (m) R 100 200 2ω ω 13. (a) Let vbr be the velocity of boatman relative to QS river, vr the velocity of river and vb is the absolute velocity of boatman. Then, Y Nj P vr (b) Drifting = QS = 4ω2 − ω2 = 3ω. O XW E (c) Suppose at any arbitrary time, the boat is at vbr point B. A i R 60° S a vb = vbr + vr θ S Q Given, | vbr | = v and | vr |= u v vnet Now θθ v u = vy = dy = x(a − x) v …(i) B dt a2 …(ii) and v = vx = dx = v P dt Dividing Eq. (i) by Eq. (ii), we get Vnet = 2v cos θ dy = x(a − x) or dy = x(a − x) dx dθ = Vnet = v cos θ dx a2 a2 dt 2ω ω

522 — Mechanics - I or ω sec θ dθ = dt 16. From the graph, v a = 22.5 − 22.5 ⋅ s or t = ω 60° 150 ∫ ∫∴ dt sec θ dθ  22.5 s 0 v0 ∫ ∫v 60 150 v ⋅ dv = 0 22.5 − × ds ω ∴ t = v [ ln (sec θ + tan θ )]600° 0 or t = 1.317ω Ans. ∴ v2 = 22.5 × 60 − 22.5 × (60)2 v 2 150 2 ∴ v = 46.47 m/s Ans. 15. For 0 < s ≤ 60 m 17. (a) ux = 3m/s v = 12 s + 3 = 3 + s ax = − 1.0 m /s2 60 5 dv =  15 ⋅ ds Maximum x-coordinate is attained after time dt dt t = ux = 3 s = 1 (v) = 1  3 + 5s = 3 + s …(i)  ax  5 5 5 25 At this instant vx = 0 and or a = 3 + s Ans. 5 25 vy = uy + ayt = 0 − 0.5 × 3 = − 1.5 m /s ∴ v = (− 1.5 $j) m/s i.e. a-s graph is a straight line. (b) x = uxt + 1 axt2 2 At s = 0, a = 3 m/s2 = 0.6 m/s2 5 = 3 × 3 + 1 (− 1.0) (3)2 = 4.5 m 2 and at s = 60 m, a = 3.0 m/s2 For s > 60 m y = uy t + 1 ay t2 = 0 − 1 (0.5) (3)2 =– 2.25 m v = constant 2 2 ∴ a=0 ∴ r = (4.5 i$ – 2.25 $j) m Ans. Therefore, the corresponding a-s graph is shown in figure. 18. (b) v = vx i$ + vy $j and a = ax i$ + ay $j` a (m/s2) v⋅ a = vxax + vyay Further v= vx2 + v 2 ∴ y 3.0 v⋅ a = vx ax + vy ay v vx2 2 + v y 0.6 dv dt s (m) = = at 60 120 or component of a parallel to v From Eq. (i), dv = v dt 5 = tangential acceleration. v dv = 1 t 19. (a) vbr = 4 m/s, vr = 2 m/s ∫ ∫or dt tan θ = BC = | vr | = 2 = 1 3 v 50 AB | vbr | 4 2 ln  v3 = t D 5 v = 3 et/ 5 or 60 t1 et/ 5 dt θ vb ds = 3 ∫ ∫∴ 00 B C 200 m √5 1 vbr vb θ 60 = 15 (et1/ 5 − 1) or t1 = 8.0 s 100 m θ 2 Time taken to travel next 60 m with speed 15m/s will be 60 = 4 s A vr 15 ∴ Total time = 12.0 s Ans. In this case, vb should be along CD.

Chapter 6 Kinematics — 523 ∴ vr cos θ = vbr sin α 21. In order that the moving launch is always on the D straight line AB, the components of velocity of the current and of the launch in the direction θ perpendicular to AB should be equal, i.e. vb θ B vbr α u C vr β 2  2 = 4 sin α α  5  v or sin α = 1 A 5 B ∴ α = θ = tan−1  21 α v (b) t1 = 200 = 200 = 50 s uβ | vbr | 4 DC = DB sec θ = (100) 5 = 50 5 m A 2 u sin β = v sin α …(i) | vb | = | vbr | cos α − | vr | sin θ S = AB = (u cos β + v cos α )t1 …(ii)  2  1 6 = 4  5  − 2  5  = 5 m /s Further BA = (u cos β − v cos α )t2 …(iii) t1 + t2 = t …(iv) ∴ t2 = t1 + DC = 25 + 50 5 = 200 s Solving these equations after proper substitution, 2 | vb | 6 3 we get 5 u = 8m/s and β = 12° Ans. or t2 = 4 Ans. 22. Here, absolute velocity of hail stones v before t1 3 colliding with wind screens is vertically downwards 20. vb = velocity of boatman = vbr + vr and velocity of hail stones with respect to cars after and vc = velocity of child = vr collision vH′ C is vertically upwards. Collision is elastic, hence, velocity of hail stones with respect to ∴ vbc = vb − vc = vbr cars before collisionvHC and after collision vH′ C will make equal angles with the normal to the wind screen. vbc should be along BC. i.e. vbr should be along BC, where, tan α = 0.6 = 3 (v ′HC)1 0.8 4 A β or α = 37° Ans. C (vHC)1 v 2β β Child C 0.8 km A 90° – 2β α B –v1 α1 1.0 km 0.6 km α1 vbr (vHC)1 = velocity of hail stones – velocity of car 1 α = v − v1 B Boat From the figure, we can see that β + 90° − 2β + α 1 = 90° Further t = BC = 1 h | vbr | 20 or α 1 = β or 2β = 2α 1 = 3 min Ans.

524 — Mechanics - I In ∆ ABC, tan 2β = tan 2α 1 = v1 …(i)  k vy + g −kt v …(ii) m or =e m Similarly, we can show that  k + g m v0 sin θ0 v2 tan 2α 2 = v From Eqs. (i) and (ii), we get or vy = m  k v0 sin θ0 + g − kt −  …(ii) k m m g Ans. e  v1 = tan 2α 1 = tan 60° = 3 = 3 v2 tan 2α 2 tan 30° 1/ 3 (b) Eq. (i) can be written as ∴ v1 = 3 dx = v0 cos θ0 − k t v2 dt m Ans. e dvx kv cos θ k x t −kt dt m m 0 dx = v0 cos θ0 m 23. ax = = − = − vx ∫ ∫or e dt 0 ∫ ∫∴ dvx = − k dt or vx mv0 cos θ0 [1 − k t dvx = − k t or x = − m ] e dt k vx m v0 cos θ0 vx m0 θ0 −kt xm = mv0 cos k or vx = v0 cos θ0 e m …(i) at t = ∞. Y 24. In the first case, BC = vt1 and w = ut1 BC θ uu kv mg α X Av v Similarly, In the second case, ay = dv y = − kv sin θ − g = −  k vy + g u sin α = v and w = (u cos α ) t2 dt m m Solving these four equations with substitution, we get proper vy dv y t Ans. v0 sin θ0 vy + w = 200 m, ∫ ∫or k = − dt u = 20m/min, m v =12m/min g 0 and α = 36°50′ or m ln  k vy + g  vy =−t k m  v0 sin θ0

7. Projectile Motion INTRODUCTORY EXERCISE 7.1 1. v1 ⋅ v2 = 0 y u ⇒ (u 1 + a1t)⋅ (u 2 + a2t) = 0 ⇒ (10i$ − 10t$j)⋅ (−20i$ − 10t$j) = 0 a Tx O ⇒ −200 + 100t2 = 0 S 4. ∆v = vf − vi R ⇒ t = 2 sec 2. It is two dimensional motion. y ^j 3. The uniform acceleration is g. 4. u = (40i$ + 30$j) m/s, a = (−10$j) m/s2, t = 2 sec u ^i Now, v = u + at and s = ut + 1 at2 α x 2 O α u 5. ux = uy = 20 m/s, ay = −10 m/s2 sy = uyt + 1 ayt 2 = (u cos α $i − u sin α $j) − (u cos α $i + u sin α $j) 2 = (−2u sin α ) $j ⇒ −25 = 20t − 1 × 10 × t2 Therefore, change in velocity is 2u sin α in 2 downward direction. Solving this equation, we get the positive value of, 5. (a) R = u2 sin 2θ, H = u2 sin2θ t = 5 sec g 2g and T = 2usinθ Now apply, v = u + at and sx = uxt g INTRODUCTORY EXERCISE 7.2 Here, u = 20 2 m/s and θ = 45° 1. u = 40i$ + 40$j (b) v = u + at where, u = (20$i + 20$j) m/s a = − 10$j and a = (−10$j) m/s2 t=2s 40 m/s (c) Horizontal component remains unchanged (= 20i$) and vertical component is reversed in 40Ö2 m/s a = g = 10 m/s2 direction (= − 20$j) v = (20i$ − 20$j) m/s Ans. 45° 6. (a) Since, acceleration is constant. Therefore, 40 m/s a av = a = g = (−10$j) = ∆v ∆t (a) Apply v = u + at as a = constant ∴ ∆v = (−10$j) (∆t) = (−10$j) (3) (b) Apply s = ut + 1 at2 = (−10$j) (3) = (−30$j) m/s 2 ∴Change in velocity is 30 m/s, vertically 3. Average velocity downwards. Ans. s =R (ux T ) = t T = T = ux ut + 1 at2 2 =u = u cos α (b) vav = s = t + 1 at t 2

526 — Mechanics - I = (20$i + 20$j) + 1 (−10$j) (3) and g (1 + tan 2 θ) = c 2 2u2 = (20i$ + 5$j) m/s ∴ a (1 + b2) = c 2u2 ∴ | vav | = (20)2 + (5)2 ∴ u = a (1 + b2) = 20.62 m/s Ans. 2c 7. T = 2u sin θ = 2 × 20 × 1 2s INTRODUCTORY EXERCISE 7.3 g 10 2 =2 R = u2 sin 2 θ = (20)2 sin 90° 1. T = 2u sin (α − β) g 10 g cos β = 40 m = (2) (20 2) sin (45° − 30°) = 1.69 s (10) (cos 30°) Now the remaining horizontal distance is R = g u2 β [sin (2α − β) − sin β] (50 − 40) m = 10 m. Let v is the speed of player, cos2 then = (20 2)2 [sin (2 × 45° − 30° ) − sin 30° ] vT = 10 or v = 10 = 10 = 5 m/s Ans. (10) cos2 30° T 22 2 = 39m 8. (a) Initial velocity is horizontal. So, in vertical 2. T = 2u sin (α + β) direction it is a case of free fall. g cos β t = 2h = 2 × 100 = 20 sec g 10 = 2 × 20 2 sin (45° + 30° ) ≈ 6.31 s (10) cos 30° (b) x = uxt = 20 20 m R = g u2 β [sin (2α + β) + sin β ] (c) vx = ux = 20 m/s cos2 θ vx vy = uy + ayt (20 2)2 = 0 − 10 20 = (10) cos2 30° [sin (2 × 45° + 30° ) + sin 30° ] = −10 20 m/s vy v = 145.71 m v= vx2 + v 2 = 49 m/s 3. Using the above equations with α = 0° y θ = tan −1  v y  = tan −1 ( 5) T = 2 (20) sin 30° = 2.31 s   (10) cos 30°  vx  (20)2 9. R = u2 sin 60° = 3 km R = (10) cos2 30° [sin 30° + sin 30° ] g = 53.33 m ⇒ u2 = Rmax = 3 4. (a) Horizontal component of velocities of g sin 60° passenger and stone are same. Therefore = 2 3 km relative velocity in horizontal direction is zero. Hence the relative motion is only in vertical Since, Rmax < 5 km direction. So, it can't hit the target at 5 km. (b) With respect to the man, stone has both velocity 10. Comparing with, components, horizontal and vertical. Therefore path of the stone is a projectile. gx2 tan 2 y = x tan θ − 2u2 (1 + θ) 5. (a) a1 = a2 = g (downwards) tan θ = b, g = a ∴ Relative acceleration = 0

Chapter 7 Projectile Motion — 527 (b) u 12 = u 1 − u 2 ∴ d = |u 12| t = (20$j) − (20$i + 20$j) = 20 × 2 = 40m = (−20i$) m/s 6. The range is maximum at, or 20 m/s in horizontal direction (c) u 12 = (−20i$) m/s is constant. α=π +β (given in the theory) 42 Therefore relative motion is uniform. = 45° + 30° = 60° Ans. 2 Exercises LEVEL 1 T = 2u sinθ ⇒ T ∝ u sin θ g Assertion and Reason 1. In the cases shown below path is straight line, even T1 = (usinθ)1 = 2 T2 (u sin θ)2 1 if a is constant R = uxT uu aa ∴ R1 = (ux )1 ⋅ T1 =  21  21 =1 R2 (ux )2 T2 3. T = 2u sinθ ⇒ u sin θ = gT ∴ R1 = R2 g2 H = u2 sin2θ = (gT /2)2 9. d v= 10 m/s2 2g 2g  dt  or H ∝ T 2 But d| v| ≠ 10 m/s2 4. u dt 90° 10. vx = ux v vy = ± u2y − 2gh and v = vx2 + v2y and u2 = ux2 + u2y 5. In projectile motion along an inclined plane, Single Correct Option normally we take x and y-axis along the plane and 1. a = g = constant for small heights. perpendicular to it. In that case, ax and ay both are 2. Hθ = u2 sin2θ non-zero. 2g 7. T = 2u sinθ = 4 H 90− θ = u2 sin2 (90 − θ) = u2 cos2θ 2g 2g g ∴ (u sin θ) = 20 m/s Ans. ∴ Hθ = sin2 θ ∴ H = u2 sin2θ = (20)2 = 20 m H 90 − θ cos2 θ 2g 20 8. H = u2 sin2 θ 3. Rθ = R45° 2 2g u2 sin 2θ  u2  or u sin θ ∝ H ∴ = 1   or sin 2θ = 1 (u sin θ)1 = 4H = 2 g 2 g 2 (u sin θ)2 H ∴ 2θ = 30° or θ = 15°

528 — Mechanics - I 4. At 45°, range is maximum. At highest point, it has Other stone should be projected at 90° − θ or 30° only horizontal component of velocity or from horizontal. 20 cos 45° ≈ 14 m/s. ∴ H2 = u2 sin2 30° 2g 5. F ⋅ v = 0 ⇒ F ⊥ v Hence path is parabola. = (52.2)2 (1/4) 20 6. vx = ux = 34 m Ans. ∴ v cos 30° = u cos 60° or v = u 11. Using s = ut + 1 at2 in vertical direction 3 2 Velocity has become 1 times. Therefore, kinetic ∴ −70 = (50 sin 30° ) t + 1 (−10) t2 3 2 energy will become 1/3 times. On solving this equation, we get 7. T1 = 2u sinθ , T2 = 2u cos θ t=7s Ans. g g 12. S = H 2 + R2/4 R = 2 (u sinθ) (u cosθ) g = 2  g2T1  g2T2 SH g = 1 gT1T2 Ans. R/2 2 8. Rmax = v02 at θ = 45° Average velocity = S = S = 2S g t T /2 T R 13. Velocity of train in the direction of train is also 30 m/s. So there is no relative motion in this direction. In perpendicular direction, d = R = u2 sin 2θ g ∴ Amax = π Rm2ax = (30)2 sin 90° = 90 m Ans. 10 Ans. 9. Maximum range is obtained at 45° 14. For maximum value of y u2 = 1.6 or u = 4 m/s g dy = 10 − 2t = 0 dt T = 2u sin 45° = 2 × 4 × (1/ 2) ⇒ t=5s g 10 ymax = (10) (5) − (5)2 = 0.4 2 s = 25 m Number of jumps in given time, u2 g cos2 n = t = 10 2 = 25 15. R = β [sin (2α + β) + sin β ] T 0.4 2 u = 50 m/s, g = 10 m/s2, α = 0°, ∴ Total distance travelled = 1.6 × 25 = 40 m Ans. (50)2 10. H1 = u2 sin2θ R = 10 cos2 30° [sin (2 × 0 + 30° ) + sin 30° ] 2g (2500)  1 12 ∴ 102 = (u2) sin2 60° = 10 × (3/4) 2 + 20 = 1000 m ∴ u = 52.2 m/s 3 Ans.

Chapter 7 Projectile Motion — 529 16. T = 2u sinθ = (2) (8) sin 53° = 1.28 s and vertical component of B g 10 v2 = 5 2 cos 45° = 5 m/s R = u2 sin 2θ = (80)2 sin (106° ) = 615.2 m Since, v1 = v2, so they may collide g 10 Now the second condition is, Distance travelled by tank, R1 + R2 ≥ d (d = 15 m) d = (5) T = (5) (1.28) = 6.4 m ∴ R1 = (10)2 sin 60° = 8.66 m ∴ Total distance = (615.2 + 6.4) m 10 = 621.6 m Ans. R2 = (5 2)2 sin 90° = 5 m 10 Subjective Questions Since, R1 + R2 < d, so they will not collide. 1. At 45° y 5. v1 ⋅ v2 = 0 when v1 ⊥ v2 vy = ± vx = ± ux = ± 60 cos 60° ∴ (u 1 + a1t) ⋅ (u 2 + a2t) = 0 = ± 30 m/s ∴ (3i$ − 10t$j) ⋅ (4i$ − 10t$j) = 0 Now, vy = uy + ay t x or t = 0.12 s ∴ t = vy − vy ay Now in vertical direction they have no relative motion and in horizontal direction their velocities = (± 30) − 60sin 60° are opposite. −10 ∴ d = 3t + 4t = 7t ∴ t1 = 2.19 s and t2 = 8.20 s Ans. = (7) ( 0.12) m 2. Vertical component of initial velocity ≈ 2.5 m Ans. u = 20 2 sin 45° = 20 m/s 6. y = x 1 − Rx  tanα Now, apply s = ut + 1 at2 (to find t ) in vertical 2 y direction, with P = (6, 3) s = 15m, u = 20 m/s and a = − 10 m/s2 Ans. P 3. vx = ux = 20 cos 60° = 10 m/s Given, v=u 3m x 2 6 m 18 m ∴ 4 v2 = u2 or 4 (vx2 + v2y ) = u2 ∴ tanα = yR x (R − x) ∴ 4 [(10)2 + v2y ] = (20)2 = (3) (24) = 2 or vy = 0 (6) (18) 3 Hence, it is the highest point ∴ t = T = u sin θ ∴ α = tan−1  23 Ans. 2g Ans. = (20) sin 60° 7. y -coordinate of particle is zero 10 Ans. when, = 3s 4t − 5t2 = 0 4. For collision to take place, relative velocity of A ∴ t = 0 and 0.8 s x = 3t with respect to B should be along AB or their at t = 0, x = 0 vertical components should be same. and at t = 0.8 s, x = 2.4 m Vertical components of A v1 = 10 sin 30° = 5 m/s

530 — Mechanics - I 8. T = 2u sin (α − β) (α = 60° , β = 30° ) 11. Horizontal component of velocity always remains g cos β constant = 2 × 10 × sin 30° y ∴ 40 cos 60° = v cos 30° (10) cos 30° or v = 40 m/s Ans. 3 =2s 3 12. (a) If they collide in air then relative velocity of A vx = ux = 10 cos 60° = 5 m/s x with respect to B should be along AB or their vy = uy + ay t vertical components should be same. = (10 sin 60° ) + (−10) (2/ 3) ∴ 20 sin θ = 10 or θ = 30° Ans. = 5 3 − 20 = − 5 m/s (b) x = |SA| + |SB | 33 = (20 cos 30° ) (t) + 0 ∴ =v= vx2 + v 2 = 20 × 3 × 1 y 22 = 25 + 25 =5 3m Ans. 3 13. vx = ux = 7.6 m/s = 10 m/s 3 vy = u2y − 2gh 9. vy = uy + ay t ∴ uy = v 2 + 2gh y = (10 sin 30° ) + (−g cos 30°) T = (6.1)2 + 2 × 10 × 9.1 10 m/s x = 14.8 m/s (a) H = u2y = (14.8)2 ≈ 11 m y 2g 2 × 10 30° (b) R = 2uxuy = 2 × 7.6 × 14.8 ≈ 23 m 30° g 10 Ans. = 5 − 10 × 3 × 2 (c) u = ux2 + u2y 23 = (7.6)2 + (14.8)2 = − 5m/s 10. In vertical direction, = 16.6 m/s Ans. SB (d) θ = tan −1  uy  SA    ux  10 m = tan−1  174..68 ≈ tan−1 (2), SA = SB + 10 below horizontal. Ans. ∴ (10) t − 1 gt2 = − 1 gt2 + 10 ...(i) 14. ux2 + u2y = (2 gh)2 = 4gh 22 ∴ t =1s B AC SA SB hh In horizontal direction, D d = |SA| + |SB | = (10) t + 10 t = 20t O = 20 m (as t = 1 s) 2h

Chapter 7 Projectile Motion — 531 vy = u2y − 2gh ...(ii) Adding Eqs. (i) and (ii) and by putting R = u2 sin 2θ , vx = ux ...(iii) g Now for the projectile ABC, vx and vy are the initial we get the result. components of velocity. 17. (a) t = Sx = 10 ∴ 2h = range = 2vxvy = 2uxvy gg ux 5 or ux = gh ...(iv) =2s vy (b) Vertical components of velocities are zero. Using Eqs. (ii) and (iv) for rewriting Eq. (i) we have, ∴ h = 1 gt2  gvhy  2 2  = 1 × 9.8 × (2)2 + (v 2 + 2gh) = 4 gh y 2 ∴ v4y − (2gh) v2y + g2h2 = 0 = 19.6 m (c) Sx = uxt = 7.5 × 2= 15 m ∴ v 2 = 2gh ± 4g2h2 − 4g2h2 18. ux = 20 km / h = 20 × 5 = 5.6 m/s y 2 18 ∴ Now, = gh uy = 12 km/h = 12 × 5 = 3.3 m/s vy = gh 18 tAC = time of projectile ABC Ans. Using S = ut + 1 at2 in vertical direction, we have = 2vy = 2 h 2 gg −50 = (3.3) t + 1 (−10) t2 15. Let v is the velocity at time t 2 Using vy = uy + ay t Solving this equation we get, ∴ v sin β = usin α − gt ...(i) t ≈ 3.55 s ...(ii) vx = ux At the time of striking with ground, ∴ v cos β = u cos α vx = ux = 5.6 m/s From Eqs. (i) and (ii), we get vy = uy + ay t  ucos α  = (3.3) + (−10) (3.55)  cos β    sin β = usinα − gt = 32.2 m/s On solving we get, ∴ Speed = (32.2)2 + (5.6)2 u = gt cosβ sin (α − β) ≈ 32.7 m/s 16. R − a = u2 sin 2α 19. x = (u sin θ) T g y=  1 gT 2 − (u cos θ) T 2 Multiplying with b we have, y 1 gT 2 − (u cos θ) T bR − ab = bu2 sin 2α 2 g ...(i) = tan θ = x (u sin θ) T R + b = u2 sin 2β g ∴ (2u sin θ) (tan θ) = gT − (2u cos θ) T = 2u [sinθ tanθ + cosθ ] Multiplying with a we have, ...(ii) g aR + ab = au2 sin 2β = 2u sec θ g g

532 — Mechanics - I u cos q (ii) yA = yB 1 qu 2 ∴ 10 + (u1 sin θ1) t − gt2 = 20 u sin q 1 2 yR + (u2 sin θ2) t − gt2 q or (u1 sin θ1 − u2 sin θ2) t = 10 xq LEVEL 2 Now, R = (x) sec θ Single Correct Option = (u sinθ T ) sec θ 1. Relative acceleration between two is zero. = (u sin θ)  2u sec θ (sec θ) Therefore, relative motion is uniform.   g u 12 = u 1 + u 2 = (20$j) − (20 cos 30° i$ + 20 sin 30° $j) = 2u2 sinθ sec2 θ g = (10$j − 10 3 $i ) 20. (a) Acceleration of stone is g or 10 m/s2 in |u 12 | = (10)2 + (10 3)2 = 20 m/s downward direction. Acceleration of elevator is ∴ d = |u 12| t = 24 m Ans. 1 m/s2 upwards. Therefore, relative acceleration of 2. T = 2h stone (with respect to elevator) is 11 m/s2 g downwards. Initial velocity is already given Ans. relative . d = (u) T = u 2h g (b) T = 2ur sin θr = (2) (2) sin 30° ar 11 ∴ d2 = 2hu2 g = 0.18 s . Ans. (c) vS = vSE + vE u2 cos2 1 m/s 3. R = g β [sin (2α + β) + sin β] 2 m/s 2 m/s u = 10 m/s, g = 10 m/s2, β = 30° and α = 60° 30° Ö3 m/s 4. y = x tan θ − gx2 2u2 cos2 Elevator Stone relative to elevator θ 3 m/s 2Ö3 m/s ∴ dy = (tanθ) −  g  dx  cos2 θ x  u2 60° ∴ dy versus x graph is a straight line with negative dx Ö3 m/s slope and positive intercept. Absolute velocity of stone dy = (2 βx) ⋅ dx d2 y  d2x  dx  2  (d) In that case relative acceleration between stone dt dt dx2 dt  and elevator will be zero. So, with respect to 5. and = 2β  dt 2 +  elevator path is a straight line (uniform) with  constant velocity of 20 m/s at 30°. But path, with respect to man on ground will remain d2 y = α = ay unchanged. dt2 21. (i) xA = xB d 2x = ax = 0 and dx = vx dt2 dt ∴ 10 + (u1 cos θ1) t = 30 − (u2 cos θ2) t ∴ (u1 cos θ1 + u2 cos θ2) t = 20 ∴ α = 2β ⋅ vx2 or vx = α Ans. 2β

Chapter 7 Projectile Motion — 533 6. R = u2 sin 2 (60° ) 8. ay = 0 = d2 y dt2 2 2g = 3 u2 dx = (2 y + 2) ⋅ dy 4g dt dt H = u2 sin2 60° = 3u2 d 2x d2 y  ddyt 2 2g 8g dt2 dt2 = (2 y + 2) ⋅ + 2 ∴ AB =  R2 2 + H 2 = 21u2 ∴ ax = a = (2 y + 2) (0) + 2 (5)2 Ans. 8g = 50 m/s2 B 9. tanβ = H = 2H H R/2 R A = (2u2 sin2 α )/2g = tan α R/2 (2u2 sinα cos α )/g 2 7. ∴ β = tan −1  tan α  Ans. 2 10Ö3 m/s Wedge 10. a1 = a2 = g (downwards) 10Ö3 sin 60° = 15 m/s 30 m/s 10 m/s Ans. 12 10Ö3 m/s 10Ö3 m/s 1 20Ö3 m 2 dmin 20Ö3 m/s 30° 10Ö3 cos 60° = 5Ö3 m/s Particle 20 m/s 10Ö3 m/s 15 m/s 60° ∴ a12 = 0 ∴ Relative motion between them is uniform. 5Ö3 m/s Relative velocity v21 dmin = 20 3 sin 30° Particle with respect to wedge T = 2u sin (α − β) = 10 3 m Ans. g cos β = 2 (10 3) ⋅ sin (60° − 30° ) = 2 s More than One Correct Options 10 cos 30° 1. α + β = 90° or β = 90° − α h1 = u2 sin2α and h2 = u2 cos2 α 2g 2g 10Ö3 m/s At t1 = 2u sin α and t2 = 2u cos α Rest 60° g g 30° R1 = R2 = 2u2 sin α cosα = R g

534 — Mechanics - I 2. Since u = 0, motion of particle is a straight line in the 6. Horizontal component of velocity remains unchanged direction of anet . X OA = 20 m = X AB 2 aH ∴ tOA = tAB =1s 2 g anet For AB projectile T = 2 s = 2uy g t = 2h = 2 × 49 ∴ uy = 10 m/s g 9.8 H = u2y = (10)2 = 5 m 2g 2 × 10 = 3.16 s Ans. ∴ Maximum height of total projectile, remains = 15 + 5 = 20 m 3. Horizontal component of velocity Ans. tO B = tOA + tAB = 1 + 2 = 3 s unchanged For complete projectile ∴ v cos θ = v′ cos (90 − θ) T = 2 (tOA) + tAB or v′ = v cot θ = 4 s = 2uy In vertical ( y) direction, g vy = uy + ay t ∴ uy = 20 m/s ∴ t = vy − uy ux = AB = 40 = 20 m/s ay tAB 2 = −v1 sin (90 − θ) − v sinθ Comprehension Based Questions −g 1. At Q, component parallel to OB becomes zero = (v cot θ) ⋅ cos θ + v sin θ g y x = v cosec θ g 4. ux = vx = 10 m/s uQ uy = v 2 + 2gh 90° 30° y 30° 30° = (10)2 + (2) (10) (15) = 20 m/s Ans. vx = ux + ax t 0 = (u cos 30° ) + (−g sin 30°) t Angle of projection, ∴ t = u cot 30° = 3u θ = tan −1  uy  = tan −1 (2) gg Ans.   Ans.  ux  2. PQ = range = 2 (PM ) = 2a cos 30° T = 2uy = (2) (20) = 4 s g 10  3  = (2) (4.9)    2 R = uxT = (10) (4) = 40 m H = u2y = (20)2 = 20m = 4.9 3 m = u2 sin 2 (60° ) 2g 2 × 10 9.8 5. d v = a = constant = g ∴ 4.9 3 = u2 ( 3 /2) dt 9.8 ∴ u = 9.8 m/s d2v = da = 0 = constant dt2 dt

Chapter 7 Projectile Motion — 535 u (c) R75° = R15° = 10 m 60° M Q 60° (d) H 30° = u2 sin2 30° 2g P 30° 30° a 120° 90° v = u2 = 20 = 2.5 m 8g 8 30° 30° O 4. T = 2uy ⇒ T ∝ uy Note Velocity at P is making an angle of 60° with g horizontal and velocity at Q is making an H = u2y angle of 60° with horizontal. That is the reason 2g PQ = range. Because under this condition, points P and Q lie on same horizontal line. ⇒ H ∝ u2y Match the Columns R = uxT ⇒ R ∝T 1. (a) ∆x = (ux )2 (t − 1) ⇒ R ∝ ux ⇒ tan θ = uy = (10) (2 − 1) = 10 m ux = 1 gt2 = 1 × × (2)2 = ⇒ tan θ ∝ uy 2 2 (b) y1 10 20 m By doubling uy, tan θ will become two times, notθ. y2 = 1 g (t − 1)2 = 5 m 5. (a) aav = a = (−g$j) = ∆v 2 ∆t ∴ ∆y = y1 − y2 = 15 m ∴ ∆v = (−g$j) ∆t (c) vx1 = 0 , vx 2 = 10 m/s = (−g$j)  T  2 ∴ vx 2 − vx1 = 10 m/s = (−g$j)  u sin θ (d) vy1 = gt = 10 × 2 = 20 m/s  vy2 = g (t − 1) = 10 × 1 = 10 m/s  g ∴ vy1 − vy2 = 10 m/s = (−u sin θ) $j 2. H = u2y = 20 m (b) vav = s t 2g ∴ uy = 20 m/s A T = 2uy = 4 s g R = uxT = 40m SH ∴ ux = 40 = 10 m/s O R/2 T 3. 10 = u2 sin 2 (15° ) = (R/2)2 + H 2 T /2 g ⇒ u2 = 20 m (c) ∆v = (−g$j) (∆t) g (a) Rmax = u2 = 20 m = (−g$j) T g  θ u2 = (−g$j)  2u sin 2g g (b) H max = when thrown vertically = 10 m = (−2u sin θ) $j

536 — Mechanics - I (d) vav = S =R or R ∝ ux (asT → same) t T R for C is maximum. Therefore ux is greatest for C. u = u2y + ux2 (uy → same) R is least for A. Therefore ux and hence u is least for A. T Subjective Questions OB 1. ux = v0 cos θ, uy = v0 sin θ, ax = − g sin θ, S=R ay = g cos θ = uxT At Q , vx = 0 T ∴ ux + axt = 0 = ux = u cos θ 6. (a) x1 = (ux1) t = (30) (2) = 60 m x2 = (130) + (ux2) (t − 1) x v0 θ P = 130 + (−20) (1) y = 110 m θ Sy ∴ ∆x = 50 m h (b) y1 = uy1t − 1 gt2 Q 2 = (30) (2) − 1 (10) (2)2 θ 2 O = 40 m y2 = 75 + uy2 (t − 1) − 1 g (t − 1)2 or t = v0 cos θ …(i) 2 g sin θ = 75 + 20 × 1 − 1 × 10 × (1)2 sy = h cos θ 2 ∴ uyt + 1 ayt2 = h cos θ = 90 m 2 ∴ ∆y = 50 m ∴ (v0 sin θ)  v0 cos θ   g sin θ  (c) vx1 = 30 m/s vx2 = − 20 m/s 1  v0 cos θ  2 ∴ vx1 − vx2 = 50 m/s   (d) vy1 = uy1 + ayt + (g cos θ) = h cos θ 2  g sin θ  = (30) + (−10) (2) Solving this equation we get, = 10 m/s v0 = 2gh Ans. 2 + cot2 θ vy2 = uy2 + ay (t − 1) = 20 + (−10) (1) 2. Let vx and vy be the components of v0 along x and y = 10 m/s directions. ∴ vy1 − vy2 = 0 (vx )(2) = 2 7. H = u2y H ∝ u2y ∴ vx = 1 m/s 2g or vy (2) = 10 Since H is same. Therefore, uy is same for all three. or vy = 5 m/s T = 2uy v0 = vx2 + v 2 g y ∴ or t ∝ uy = 26 m/s Since uy is same. Therefore T is same for all. tan θ = vy/vx = 5/1 Ans. R = uxT ∴ θ = tan−1 (5)

Chapter 7 Projectile Motion — 537 Note We have seen relative motion between two Now (t2 − t1 )ux = t2vx or vx = t2 − t1 …(i) particles. Relative acceleration between them ux t2 is zero. Further h = uyt − 1 gt2 3. v1 = (u cos α ) i$ + (u sin α − gt) $j 2 v2 = (v cos β) i$ + (v sin β − gt) $j or gt2 − 2uyt + h = 0 These two velocity vectors will be parallel when, or gt2 − 4 gh t + 2h = 0 the ratio of coefficients of $i and $j are equal. t1 = 4 gh − 16gh − 8gh = (2 − 2) h ∴ u cos α = u sin α − gt 2g g v cos β v sin β − gt 2) h Solving we get, and t2 = (2 + g t = uv sin (α − β) g(v cos β − u cos α ) Ans. Substituting in Eq. (i) we have, vx = 2 4. At height 2 m, projectile will be at two times, which ux 2 + 1 Ans. are obtained from the equation, 6. (a) Time of descent t = 2H = 2 × 400 2 = (10 sin 45° )t + 1 (− 10)t2 10 2 g or 2 = 5 2t − 5t2 = 8.94 s or 5t2 − 5 2t + 2 = 0 Now vx = ay = 5 y or or t1 = 5 2 − 50 − 40 dx = 5  1 gt2 = 5 5t2 10 dt 2 and Now = 5 2 − 10 x t t2 dt 10 dx = 5 5 ∫ ∫∴ 00 t2 = 5 2+ 10 or horizontal drift 10 x = 5 5 (8.94)3 = 2663 m ≈ 2.67 km. d = (10 cos 45° )(t2 − t1) 3 10  2 10  (b) When particle strikes the ground   = 2  10  = 4.47 m vx = 5 y = ( 5)(400) = 400 5 m/s Distance of point of projection from first hurdle vy = gt = 89.4 m/s = (10 cos 45° )t1 ∴ Speed = vx2 + v2y = 899 m/s ≈ 0.9 km/s Ans. = 10  5 2− 10  7. At t = 0, vT = (10$j) m/s   2  10  vST = 10 cos 37° k$ − 10 sin 37° i$ = (8k$ − 6i$) m/s =5− 5 ∴ vS = vST + vT = (−6$i + 10$j + 8k$ ) m/s = 2.75 m Ans. (a) At highest point vertical component (k$ ) of vS 5. 2h = u2y will become zero. Hence, velocity of particle at highest point will become (– 6i$ + 10$j) m/s . 2g (b) Time of flight, T = 2vZ = 2 × 8 = 1.6 s y g 10 t1 vx 2h t2 x = xi + vxT uy h = 16 − 6 × 1.6 = − 4.5 m h π ux x y = (10) (1.6) = 16 mand z = 0 or uy = 2 gh Therefore, coordinates of particle where it finally lands on the ground are (− 4.5 m, 16 m , 0).

538 — Mechanics - I At highest point, t = T = 0.8 s C B 2 37° vx cos 37° ∴ x = 16 − (6) (0.8) = 0.3 m A 37° π Ans. Using vy = uy + ayt, we have y = (10) (0.8) = 8.0 m vy = u sin α − g (x/u cos α ) (Q t = x/u cos α ) and z = vZ2 = (8)2 = 3.2 m = 5 5 − 10 × 5 = 0 2g 20 5 (5 5 × 2/ 5) Therefore, coordinates at highest point are, Thus, at C, the particle has only horizontal (0.3 m, 8.0 m, 3.2 m) component of velocity 8. |v21x | = (v1 + v2) cos 60° = 12 m/s vx = u cos α = 5 5 × (2/ 5) = 10m/s |v21y | = (v2 − v1) sin 60° = 4 3 m/s Given, that the particle does not rebound after collision. So, the normal component of velocity ∴ v21 = (12)2 + (4 3)2 = 192 m /s (normal to the plane AB) becomes zero. Now, the particle slides up the plane due to tangential v21 C component vx cos 37° = (10) 45 = 8 m/s Ay 30° B x 45° 105° 30° Let h be the further height raised by the particle. O Then BC = (v21) t = 240 m (Given) mgh = 1 m (8)2 or h = 3.2 m AC = 70 m Ans. 2 Hence, AB = (240)2 + (70)2 = 250 m Height of the particle from the ground = y + h 9. (a) Let, (x, y) be the coordinates of point C ∴ H = 1.25 + 3.2 = 4.45 m Ans. 10. For shell uz = 20 sin 60° = 17.32 m/s √5 50 kg 1 4 m/s α 2 x = OD = OA + AD 40 kg 10 kg (20 cos 60° + v) ∴ x = 10 + y cot 37° = 10 + 4 y …(i) v 33 As point C lies on the trajectory of a parabola, ∴ z = uzt − 1 gt2 = (17.32 × 2) –  1 × 9.8 × 4 we have 2 2 y = x tan α − gx2 (1+ tan2 α) …(ii) or z = 15 m ⇒ uy = 0 2u2 ∴ y=0 For ux conservation of linear momentum gives, Given that, tan α = 0.5 = 1 2 50 × 4 = (40) (v) + 10(20 cos 60° + v) Solving Eqs. (i) and (ii), we get x = 5 m and or v = 2 m /s y = 1.25 m. Hence, the coordinates of point C are (5 m, ∴ ux = (20 cos 60° ) + 2 = 12 m /s 1.25 m). Ans. ∴ x = uxt = (12)(2) = 24 m (b) Let vy be the vertical component of velocity of ∴ r = (24i$ + 15k$ ) m Ans. the particle just before collision at C.

8. Laws of Motion 8.1INTRODUCTORY EXERCISE No friction will act between sphere and ground because horizontal component of normal reaction 1. w1 = weight of cylinder from rod (on sphere) will be balanced by the horizontal normal reaction from the wall. w2 = weight of plank N 1 = normal reaction between cylinder and plank 5. FV N 2 = normal reaction on cylinder from ground N 3 = normal reaction on plank from ground FH O f1 = force of friction on cylinder from ground AT f2 = force of friction on plank from ground N1 w FBD of rod w1 N1 N3 In the figure N2 w2 T = tension in the string, W = weight of the rod, f1 f2 FV = vertical force exerted by hinge on the rod FH = horizontal force exerted by hinge on the rod 2. N = normal reactions 6. In the figure f1 N1 B w = weights N 1 = Normal reaction at B, f1 = force of friction at B, N 2= normal reaction at A, f2 = force C N3 N4 of friction at A w2 N1 N3 w = weight of the rod. N2 w w1 A N2 f2 3. FBD of the rod T 7. Force F F x y F1 4 cos 30° 4 sin 30° = 2 N =2 3N Nw 4. N2 N3 F2 −4 cos 60° 4 sin 60° = 2 3 N A =−2N N1 C N4 F3 0 −6 N f B F4 4 N 0 N3 W2 W1 8. T1 cos 45° = w FBD of rod FBD of sphere and T1 sin 45° = 30 N (i ) (ii ) T1 In the figure 45° N 1 = normal reaction between sphere and wall, T2 = 30 N N 2 = normal reaction between sphere and ground N 3 = normal reaction between sphere and rod and w Ans. N 4 = normal reaction between rod and ground ∴ w = 30 N f = force of friction between rod and ground