P-3 Allens Made Physics Exercise Solution

JEE-Physics UNIT # 03 (PART – I) CENTRE OF MASS EXERCISE –I (1)(5)  (1)(3) 9 . vCM =  1 m/s 1 1 x   xdm  L kx2  3L Position of centre of mass at t=1s  dm x dx 4 L t = 1s 0 (1)(2)  (1)(8) X CM  1  1  (1)(1)  5  1  6m 1. kx2 L  dx 2 . Equation of line joining the CM of two rods 10. xm1.mx11mm22.x2 xy Let x = distance moved by ring  1 L/2 L/2 x = coordinate  L , L  satisfies this equation. 0= mx  2m(1.2  x)  x = 0.8 m  3 6  m  2m L3,L6  11. Impulse  = pf – pi = 1 × 10 –1× (–25) = 35kg m/s ()  3 /2 3 . x  cos 30  48 60° CM     p = 2p cos 3 = 2mv0 sin  6  /2 1 2 .  /3 4 . Let xp=x shift of plank to the right xp=x  13. For the Ist ball ():h e12 h x  m A x A  m B x B  m c x c  m p x p 4 mA  mB  mC  mP For the IIndball    :h  e 2 h 2 16 40(x  4)  50x  60(x  4)  90x 1 0  40  50  60  90  x= 3 m Impulse on first ball  5 . x  m1.x1  m2.x2 3 m1  m2 = I1 = mv0 (1+e1) = 2 mv0 Impulse on second ball   0  m1a  m 2 x2  x 2   m1a 5 m1  m2 m2 = I2 = mv0(1+e2)= 4 mv0 6. y  m1y1  m 2 y2  I1  3 / 2 mv0  6  5I1 = 6I2 m1  m2 I2 5 / 4 mv0 5 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 m 3m 14. KE = 1 m v 2  1 m v 2  0  4 (1 5 )  4 ( y 2 )  y2 = –5 cm 2 2 2 1 7. CM remains at rest if initially it is at rest.     12m(v2v1 v v1 ) 1 I.(v1 v2 ) ).( 2   2  8.      1  2i  2  (2 cos 30i  2 sin 30j) 15. v v' v v cm m1v1  m2v2 3  m2 m m 2m m1 0= 2mv – mv + mv'  v' = –v   2  2 3  i  2 j Total mechanical energy released  3  3  = 1 (m+m+2m)v2 = 2mv2 2 1

JEE-Physics 16. COLM : 3 × 2 = (3 + 2)v v  6 m/s vC  m1 (1  e) u1   m2  em1  v 2  m (1  1) v 0 = v 5 (m1  m2 )  m1  m2  4m 2 1  3  22  1  5   62  1  480  x2 For collision between A and B : 2 2  5  2 COME : (AB x  1 m vA= 0 + m (1  1)   3v  6v 10 5m  5  25 1 7 . COLM : vB' = 0   m  4m    3v  9v  5m   5   Along horizontal ()  25 u cos u  vB' < vC  B will not collide with C. v Q u cos (B C u cos  Therefore there will be only two collisions. 0 = m(u cos  – v) – 4mv  v = 5   velocity of shell along horizontal w.r.t ground 21.C OME : m1u1 + m2u2 = m1v1 + m2v2 = u cos  – u cos  4 u3 = (u cos ) 1 × u + 0 = 1 × 4  mv2  4 u = mv2 ....(i) 55 () T=2 u sin  e   v2  v1     v2  u / 4  v2 u / 4 g   u2  u1   0u  u Time of flight  x = horizontal displacement   v2 u 5u  u ....(ii) 4 4  4 cos   2u sin  4u2 sin 2   5 u  g   5g 3  5  m3 4  4  5 (i) & (ii) u  m u   0.6kg mm 18. COME : mv cos  = (– v cos ) + v' 2 2 2 2 . At the lowest position ()  3mv cos   m v ' v '  3v cos  COME : M 2 gL  (M  m )v ....(i) 2 2 COME : 1 (M+m)v2 = (M+m)gh ....(ii) v cos mm 2 v 22  M 2gL  M  2 2gh     v   h  m M L (M  m) 1 9 . The ball & the earth froms a system and no external force acts on it. Hence total momentum remains constant.(23. Alongtangent ()  u cos = v sin ...(i) )  20. v v node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 A BC  u A vB vC vA A BC  vB' vC  BC Along normal () e  v cos   cot2  = cot260 1 u sin  = 3 For collision between B and C : (BC u 2 4 . COLM  2mu + 0= 2mv + mu  v = 2  m1  em2  m2 (1  e) vB   m1  m2  u1  (m1  m2 ) u 2 2m m 2m m u  e   v2  v1     u u /2  1 v u(Let)   u2  u1   0 u  2  m  4m    5m  v  0 = – 3/5 v 2

JEE-Physics 2 5 . After 1s , vA= 20–10 × 1 = 10 m/s 1. EXERCISE –II and vB = 0 + 10 × 1 = 10 m/s 2. In the absence of external forces, the linear At the time of collision (  ) momentum of the system remains constant. , VA = VB = 5m/s ( after collision, velocity gets enterchanged. ) ()  Momentum of the coin perpendicular to the 26. e   v2  v1   1   5  v1  v1  5 common normal remains constant.      u2  u1  5  (10) 15   v1 = 20 m/s   Impulse on ball= m(v  u) =1× [20–(10)]=30N-s (4,6) u ()  3 2 7 . v0 v0 v2 v2 (0,0) v1  vy = – 3 (constant) vyt = – 6  t = 2 sec & vxt = –4 2 vX = –2 m/s COLM  2mv0 = 3mv1 v1  3 v0 Which is given by stricker. COME  2  1 m v 2 = 2  1 m ( v 2  v 2 )  1 m v 2 () 2 0 2 1 2 2 1 So initial velocity of stricker = 2ms–1 ()  Final velocity of the striker = 0. ()  2  2  2  2 dv dm dv dm  3   3   2mv02 = 2m v 0  2 m v 2  m v 0 3. Fext = m dt  v rel dt  0 = m dt  2 dt 2 2 v0 m /2 2dm v  2  3     dv  v = 2n2  2 v 2  2 v 2  3  3 v 0   v2 = m 2 0 m 0 4. In the ground frame () 5. Velocity of particle = v 2  v 2  4 v 2  v 2  7 : mAxA + mBxB + mpxp = 0 1 2 0 0 3 v0  40 × 60 + 0 + 40 × xp= 0  xp = –60 (to the left) 93 Hence A & B meet at the right end. ()  2 8 . For second object ()  A B 2v1 = 0 + at; t  2v1 ...(i) u=0 a u=v1 Force exerted by one leg on the ground a 1 2 d () s1 = s2 + d; 1 at2  ut  d N = 1 [Total force]   2 4× node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 1  2v1  2  2v1  2 v 2  2 v12  1 [wt + rate of change of momentum] 2  a   a  1 a 4 [wt  a   v1  d  d  0 +] a 1 = [Mg + n (mv cos 60°)× 2]= 1N 2 9 . Average power ( )  4 W  K  U  1 (x)v2 xg x / 2 6. h1 t =  t   2 t  7. h (h–d) 2g(h  d)  e 2gh  d  1  e2 t <P> = 1 v3   vg COLM : mR(0.8) + mS(0) = mR(0.2) + mS(1.0) 22  0.6 mR = mS  mR > mS 3

JEE-Physics 8 . T.t T.t 1 7 . Initially when the shell is empty the C.M. lies at its geometric centre. Also when the shell is filled with A 3m N.t m sand CM lies at its geometric centre. Bm N.t ( –Nt = m(v – u); Nt – Tt = m(v–0)   Tt = 3m(v–0) v = u/5 )  Impulsive tension ( )Tt = 3mu 5 M1x1  M2x2 M   a   M  a  a 18. a cm  Fnet  (0 .2 )(3 )(1 0 )  2 ms2 2  2  2  3   Total mass 12 9. x   12 M1  M2 M OR Acceleration of 1kg w.r.t. ground 10. (1kg ) (M  0)    M  L    M  4L  L =(0.1)(10)=1ms–2  4 3   4 6   Accceleration of 2 kg w.r.t. ground 4 (2kg ) x M = (0.2)(3)(10)  (0.1)(10)  5 ms2 22 1 1 . x  3M.x  M(x  2)  0  x   1 4M 2 a cm  m1a1  m2a2  (1)(1)  (2)(5 / 2)  2ms 2 m1  m2 12 m vN 1 2 . Nmv = (M + Nm)vf  v f  M  Nm 1 9 . p = change in momentum (  )=2mv y xy t = time between two collision  2(L  d) P (x=y)  1 v 13. 26 x ()  (2,0)  Force exerted on wall () (0,6) p mv2 y' =  t (L  d) Co-ordinate of P (P  )=(3,3) 20. t1 L (time for Ist collision)  v  Speed of 3rd particle (3rd= 3 2 m/s 2L t2  v (time for IInd collision) 1 4 . Let x = displacement of ring to the left.  (x=)  2mx  m(x  L  L cos ) 3L  xcm = 0 t3  v (time for 3rd collision) 3m   x   L (1  cos ) 3 L t(n–1)= (n–1) (time for (nth) collision) v node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 M 1 5 . Mv = 0 + 1 0 v 2  v2 = 10v n 1 6 . COLM  m(v1 cos 45° + v2) + mv2 = 0 h n(n  1) L ti  2 v  v1 = 2 2.v2 COME  Ki + Ui=Kf+ Uf i 1 mgR 1 45° V2 2 1 . In elastic head on collision if the masses of the  0 + 2 = 2 mv22 V1 colliding bodies are equal, the velocities after 1  v 2  v1  v1 2  v2  gR collision are interchanged.(  1  2 22 32  m   ) +   2  2    For Ist bead(),Fd = 1 2Fd mu2  u  2m 4

x  m1x1  m 2x2  m 3x3 JEE-Physics m1  m2  m3 2 8 . COME : – MV + m (v cos 60 –V)=0 v = 10 m/s 22. x  (80  2)  (50  2)  (70  0) 2 9 . x  m x1  M x2  0 80  50  70 m M = 30cm towards right 1( sin 30   sin 30  x)  4x  14 0 P  Displacement of bar (  )=x=0.2 23. + = J J A BA B m 0m R R For A : P – J = mv1 ....(i) 3 0 . xcm= mm  2 for B : J = mv2 ....(ii) J   P  J  xCG = W1 (0)  W2 (R )  mgR  R  m   e   v2  v1     m  = 2J 1 W1  W2 mg   u2  u1    p 0 P   m R R R    xCG – xCM = 2 2 2 4 . mu – I1 = –mu  I1 = 2mu & mu – I2 = 0  I2 = mu (for IInd ball)  I2 = I1/2 3 1 . Velocity before strike (  )  4 u = 2gh 4 Impulse = Ft = m(v–u) 4 4  F = m(v  u)  w(0  2gh ) = 5.21 W t g  0.15 2 5 . Cart frame ():  45° 4 45 3 2 . COLM  m1u1 + m2u2 = (m1 + m2)v 5 × 103 × 1.2 + 0 = (5 + 1) × 103 × v v = 1 m/s Ground frame  : 33. v1= (m1  m2) u1   2m2  u2 (m1  m2 )  m1  m2  44 The velocity of rebound ()=4 5 m/s v2  2m1 u1   m 2  m1  u2 1 m  m 1  m2  vi j  v i v j  (m 2 )  2 2  mv4 26. COLM  m  m v  m   0 2mu 2 For C : vC= u  v 1 1  i 1 1  j 3m 3 v4   2   v  2  mv 3 4 . From COLM Mv xA  2Mv xB  0  v xA  2 v xB so   2v x i  vk A vA m,v node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 135° r1 v 3 5 . 2rsin = 2 ; sin = 4 r/2 m,v Total energy released (  )   r/2  2r 15 B v2 1 1 1 1  1 2  cos = 4 v2 2 2 2 2 v2 = mv2  mv2  mv2  m   1  2 v × 15  2  4 = mv2(3– 2 ) 3 6 . m × 1 = mv1 + (2mv2 × cos60°) 11 10  (15  0) 27. Fext= mv (for first body)Fext = 3  50N  v2 = 2 & v1 = 2 COME  m1u1 + m2u2 = (m1+m2)v 1 10 × 15 + 25 × u2 = (10 + 25)5  u2 = 1 m/s (KE)initial = 2 × 1 × 12 = 0.5 J 5

JEE-Physics 1  1 2 1  12  D 1 (KE)final = (1) ×  2   1    2    2 = mg = gD4  2    2  2 2   = 0.25 + 0.125 = 0.3755 PE of solid cone  KE = 0.5 –0.375 = 0.125 J D   3 7 . Component of velocity of A along common normal = mg 4 =gD4  48  is v cos 60° and this velocity of A after collision PE of solid cylinder   with B is interchanged. Hence A moves along v sin 60° which is normal to common normal. D   = mg 2 = gD4  8  (A vcos60° A v A , vsin60°  : mu +0 = (m + M)v  v   m u  M  m  4 1 . COLM ) KE after collision :  B = 1 (m  M )   m m M 2 u2 m2u2 vcos60° 2     60°  2(m  M) Av 42. pi = – mv , pf = m(v + 2u)  p = 2m(v + u) vsin60°  Force  = p  2m(v  u) 3 8 . At the time of collision both particles have common t t velocity and hence the system has minimum kinetic energy. (KEinitial = 1 mu2 , KEfinal = 1 m(2v  u)2  ) 2 2 COME : mu + 0 = 3mv  v = u/3 1  KE = m[4v2 + u2 + 4uv – u2] 1 KEinitial = mu2 = 3J 2 2  1 m[4v(v  u)]  2mv(u  v) 1 1 u2 2 KEcollision = (3m)v2 = (3m)  1J 2 29 PEcollision = (3–1) = 2J Total energy remains constant and hence KE of system First decreases & then increases. (   ) 3 9 . At the time of maximum compression, node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 () COLM : mu = 2mv (for A & B ) v = u/2 COME : 1 mu2  1 2mv2  1 kx2  x  v m 22 2 2k 4 0 . PE of solid sphere ()  D   = mgR = mg = gD4  12  2 PE of solid cube (  6

JEE-Physics EXERCISE –III  (B) For body 2M : p = pf  p  pf =2p Fill in the blank pp 1 . By applying conservation of momentum  (C) e  v2  v1  MM =0  e = 0 () u1  u2 2p  p M 2M   3 . For 1kg v1 = (2t) i = 4 i ; ai = 2 i = 2 i 0 = mv1  mv2  2mv3 For 2kg v2 = t2 j = 4 j ; a2 = 2t j =4 j  v1  v2 2v v v3  2   2 2 m1a1  m2a2 Total energy released in explosion (A) Acceleration of centre of mass = m1  m2 ()   1 1 1  v 2 3mv2   2 i  8 j  acm= 4 64 68 2 2 2   a cm 3 3  m/s2 mv2 mv2  2m    2 99 3 2  f = macm= 68 N 2 . Area under the F-t curve = impulse=5× 10-3 N-s (B) Velocity of centre of mass () (F-t   m1v1  m2v2 =  4 i  8 j m/s 3 . h = (e2)n h0  h = (0.82)3 h0  h = (0.8)6 h0 v cm m1  m2  3 3 6 . Momentum is conserved in all collision.  16 64 80 | v cm |  ()  99 3 mv  (C) 7 . COLM : mv = (m + Ax)v'  v'= m  Ax  1 2tˆi  2 t2ˆj  2 tˆi  2 t2ˆj v cm 33  12 m1m2 Displacement = 2(m1  m2 ) 8. Loss in KE= (1–e2) (u1 – u2)2 2 2   3  t2  2 2  t3   4 ˆi  16 ˆj vcm dt  2  2  3  3  ˆj  39 5 1   ˆi  0  15  0 Here e = 3 (3)(2) 1 1   42  16  2 20 2(3  2 9   3   9  units KE = )  (15)2 = 120J  Dispalcement    9 Match the Column 4 . As no external force acts on the system velocity of 1 . By applying conservation of momentum centre of mass remain same (     () ) Before collision 2m After collision B vcm=2u/3 A M 2m 4 K=P2 v/3 2v/3 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ m 2m p= 2mK node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 m1u1 + 2m(0) = mv1 + 2mv2 ....(i) 2m(u) 2u Also u = v2  v1 ....(ii) vcm = 3m = 3 In frame of centre of mass velocity of B is 2u/3 2v v 4mv 4p  2u 2u  v2 = and v1 = ; p2 = ,  3 3  3 3 33 and It oscillates from  , mv p 8K K In frame of centre of mass velocity of A is u/3 and p1 =  3   3 ; K2 = 9 ; K1 = 9 It oscillates from   u , u  . In ground frame velocity 2 . Impulse = change in momentum  3 3  () of B 0, 4u  . In ground frame velocity of A  u , u  3   3   (A) For body M : p = pf  p  pf = p 7

JEE-Physics and by conservation of energy K.E. = US x )   B2u/3    m (4–x) = Mx  x = 0.8 m  Co-ordinate where block will leave wedge   2u , 2u  ()  3 3  Ax=4–0.8 = 3.2 u/3   u , u   Time for m will strike the ground is = 22 3 3  10 B0 , 4u  A 3u , u   (m)  3   2  K.E. = US  xf = 3.2 + 4 2 × 10 = 6.8 m 5. (A) Net force on block m acts in downward m1a1  m2a2 a (5m/s2) acm = m1  m2 60 direction 3. (m) (1) a cos 60 = 4a  Acceleration of centre of mass is in downward direction. ( 5 3 aM  8 m/s2 ) (B) Net force acts in downward as well as in N cos 30 = 4maM horizontal direction.  3 Ncos30° 30° 53 30°  N  4(1) 28 N N  N = 5 Newton 4 . Ng = N sin 30 + 40 Ng = 42.5 mg Ng  acm moves both in horizontal & vertical 30° M direction. N Nsin30 ( acm  ) (C) As the mass of monkey & block is same both moves upward. Comprehension # 2 (Byapplying conservation of momentum  )  ()  Centre of mass moves upward 2(6) + 1(4) = 1v2 + 2v1; 16 = v2 + 2v1 ...(i) ( )  By applying newton law of collision ()  (D) Centre of mass of the system does not moves () 1= v2  v1  v2 – v1 = 2 ...(ii) 2 Comprehension # 1 20 14 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 v2 = 3 , v1  3 1 . By applying conservation of momentum ()  20 8 1. Impulse = change in momentum –4  N–s 3 3 mv1 + Mv2 = 0  v1 = –4v2 ...(i) = By applying conservation of energy ( ) 2 . To change the direction of a block impulse should 1 1 v 2  2 v 2  20 ...(ii) be greater than 12 N–s (    2 mv12 + 2 Mv22 = mgh  1 2 12 N-s) 2 v2 = 2m / s ; v1 = 4 2 m/s Comprehension # 3 As horizontal velocity (i.e. velocity along the surface) 2 . When 'm' leaves the wedge 'M' then wedge moves 30 distance 'x' in left side ( Mmremains constant so required time =  6s 5 8

JEE-Physics () Comprehension # 6   = 30 6s  1 . As no external force acts on the two blocks friction acts like an internal forces and the two blocks will 5 Comprehension # 4 move with common velocity.     fM  \\\\\\\\\\\\\\\\\\\\\\ fr  dm By applying conservation of momentum 1 . f = v dt = 2(20) = 40 fr = mg () m = 40 = M0 –2(t)  t = 5 sec (1kg) (15 m/s) = 1v = 2v  v = 5 m/s so (0.1) (50–2t) = 40  t = 5 sec P1 = 5 N–s ; P2 = 10 N–s 2. v = v m – gt= 20 n  4  –gt 20 (0.28) – 5 dp n  m0   3  2 . dt =fext = 5.6 – 5  0.6m/s For block of 1 kg friction fr = mg= 0.4 × 1 × 10 = 4 N Comprehension # 5 3 . v = u + at  5 = 15 – (4) t  t = 2.5 sec   m1 r10  m 2 r20 m1  m2    1 .  1 3ˆi   2 9ˆj  ˆi  6ˆj m rcm  12 Comprehension # 7     f 1  vcm m1v1  m2v2 2m 2   m2 1 3ˆi   2 6ˆj ˆi  4ˆj m / s 1. As macm = f  acm = so scm = at2 =  f t2 m1  4m   12    2. xcm= m1x1  m2x2  x1 + x2 = f Now rcm  v cm t  rcm  rcm0  v cm t 2m t2 2m  x  1ˆi  y  6ˆj  ˆi  4ˆj t x1 – x2 = x0 Therefore f t2 + x0 x1= 4m 2  x  1  t and y= 6+ 4t  y= 4x+ 2  From above equations x2 = ft2 – x0 m1a1  m2a2 4m 2 m1  m2  2.   1 2ˆi  2 2ˆj a cm  Comprehension # 8 12   2 ˆi  2ˆj m / s2 1 . By COLM  3   mA    0 = mA vA + mB vB; vB  mB vA By using vcm  ucm  acm t ; we get t =3s Both velocity are opposite in direction node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 1 () 3 . By using s = ut + at2 for individual partiles  II, IV, V 2  1 2 . If mA = mB  v B  v A  Graph II For Ist particle:s = (3) (1.5) + (2) (1.5)2 = 2.25 m ; mA > mB vB > vA tanB > tanA  Graph IV x2 If mA < mB  vA > vB  Graph IV 3 . vcm is not zero in graph (I), (III) and (VI) 1 For IInd particle = s =(6) (3) + (–2)(3)2 = 9m ((I), (III) (VI)vcm y2 Therefore x = 2.25 +1 = 1.75 m and cm 3 y= 2 9  +6 = 12m cm 3 9

JEE-Physics 1. EXERCISE –IV(A)   (2 )    (a2 ) a a  2 – a – a2 = 0 y 22 8. (i) The centre of mass remains at O as the 2  a2 excluded masses are symmetrically placed. (O  y )   (ii) CM shifts from 0 to 3 diagonally  a  ( 5  1) x UNCUT PART 2 (03) A  (iii) CM shifts along OY a CUT OUT PART (OY ) (–a/2,a/2) x 0a (iv) CM does not shift. a ()  xdm  xydx y=kx2  dm x (v) CM shifts diagonally from 0 to 4. 9. x cm   0  3 a a 4 (04)  ydx (vi) CM doesnot shift. 0 () (m  0)  (2m  a)  (3m  a)  (4m  0) a 1 0 . (i) CM does not shift 2 . xcm =  m  2m  3m  4m 2 () y cm  (m  0)  (2m  0)  (3m  a)  (4m  a)  7a (ii) Plank moves towards right. m  2m  3m  4m 10 ( (iii) x CM  m 1x1  m 2 x2  M x  0 m1  m2  M 3 . Length of rod = (4  2)2  (2  5 )2  13m x cm  (3  2)  (2  4)  14 ; y cm  (3  5)  (2  2)  19  0  50(x  2)  70(x  2)  80x 3 2 5 3 2 5 50  70  80  x = 0.2m (right)  M  a   (M  0)   M  a  a (iv) xm1 = x + 2 = 2.2 m (right)  2   2  3 (v) xm2 = x – 2 = –1.8m (left) 4. x cm   MMM m1v1  m2v2 1 1 . For 0 < t < 1 from vcm = m1  m 2 (M 0)   M  a    M  a   2   2  a y cm   3  2  (1)(1)  (2)v2 M M M 3 6m  v2 = 2.5 ms–1  (2R ) 2   12  4   x2=2.5t 2.5m  3  4R  R  3   R 3  5 R for 1<t<2,2= (1)(1)  2(v2 )  3 5. y cm   = 4R  v2 = 3.5ms–1  (2R )2  1 4  12  3    4R  2  4 R 3  3   node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65    6 . 2 r 2 r   r2   4r  2r  x2 = 2.5 + 3.5 (t –1) 2  2   3 3(4  ) 1 2 . COLM  3mv  30 2m  v  10 2 m / s xcm = = r 2 30m/s 2 r 2    2  m   302 m    562   28     422   35   4    4  7 . x cm  m30m/s    562     422    4     4  3m.v =9cm from left edge 10

JEE-Physics 1 3 . Velocity of mass-1 when string is in normal length. h = 4.9 × 2 1   9.8  2  2  78.4m (1)  2 vA  6g  2g  2 g For the combined mass ()  Now impulsive tension acts on both bodies to come x = ut + 1 × 9.8 t2, t = 4.53 2 to common velocity ( ) vcommon vA  g  Total time of height ( ) 2 Displacement of C.M. when string becomes taut. = 2 + 4.53 = 6.53 sec. ()18. xcm = M x1  m x 2 0 = Mx  m(x  R  r) M m the m m 0  Mm moved by y1cm  m m  cylinder x   m(R  r) 2 Distance Mm Displacment of CM when masses reach the max. height ()  () Formotion along x-axis 0 = m(v1 + v2) + mv2....(i) v2     (x)  2g 2 2 2 y 2m    y cm    1 1 2 2 1 4 . From work energy theorem (  )   mg(R – r) = m (v1+ v2)2 + Mv22 ...(ii) WF + Wg = KE 2g(R  r)  v2  m M(M  m) mg(h3  h1 )  Favg(h2 – h1)–mg(h3–h1)=0  Favg = (h2  h1 ) m m k 2m 1 5 . In the presence of gravity, the CM shifts along vertically downward direction. 19. C v0 A B ( Aftercollision ( ): uA = v0 )  (i) When (vinst) A = (vinst)B mv0 =3mv v =v0/3 For point P : y = r sin (ii) COME x = (/2 – r) cos  1 mv02 = 1 (3m)v2 + 1 kx02 2 2 2 1   x 2   y2 vcm P r 2m  v0  2    r  3  x0    r    k  2   20. u o =v o 1 6 . Let v = velocity of wedge and u = veloicty of par- m1 m2 (m1+m2) ticle relative to wedge COLM  m1u +0 = (m1 + m2)v ....(i) (v= u=)  Energy equation (  ) v 1 m u 2   2  1 (m1  m2 )v2 ..(ii)  2  3 2   u node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 COLM  m1  2 :1  m (v + u cos ) + 4 mv = 0 m2  u cos  = –5v;  u  5v 21. L After Ist collision R R cos 60° ()  17. 1 gt 2    49 t  1 gt2   98  2   2  m 2 2g  2g(1  cos ) (u=) 5  gt(j) 19.6j m  t = 2s u1   / s 1 cos  = 5   (u  gt)j  29.4j m /s 98m u2 For IInd collision ()  78.4 m u1  u2 vf  2  4.9 m / s 49 m/s m=100gm 11

JEE-Physics 2 . 2 2g  2g(1  cos ) 5 m/s 2 m/s 55 27. 3kg For nth collision (n) 2kg 1.6 m/s  2 n 1 cos   4n  1  cos  (A) v cm   (2  5)  (3  2)  i  4 i m/s  5    5   2 3  5   4  n (B) COLM  (2 × 5) + 3(–2) = +2(–1.6) + 3v2  5    v2 = 2.4 m/s cos   1  Put n=0,1,2,3 and get answer (C) e    v2  v1   4  u2  u1  7  4ML  M(L  5R) 4MX  M(X  5R) 28. COLM: implies that vC & vB are opposite to each 2 2 . xcm  5M  5M other. (vCvB )  x = L + 2R 29. mg – T = m  a  g  ...(i)  6  MM (L,0) x g/6 m mm a 23 . In this elastic collision velocity of masses are  T  2 g  2 a ...(ii) m/2 A For solving eq. (i) & (ii) exchanged. (4g  13 mg a & T 18 )  9 So vA = 0  A does not rise 3 0 . Velocity of B when string is in natural length 24. COME  mg(h + x0) = 1 kx 2 0.1kg (B)  2 0 = uB  2gh  2g (h  1)  mg(0.24 + 0.01) T 1 Impulse equation   B = 2 k(0.01 × 0.01) ....(i)  –Tt = m [v – 2g ] ...(i) T T.t  m[v  0] ...(ii) & mg(h + 0.04) = 1 k(0.04)2 ...(ii) A 2 Equation (i) divided by (ii) h= 3.96 m 2g g On solving eq. (i) & (ii) : v  2  2 2 5 . No, KE is not conserved during the short time of  Distance travelled by A before coming to rest, collision. ((A ) )  v2 s =1+  1.25m Av u=0 A vA 1.6v 2g 26. v1 B B B e=1 v0 before collision after collision 31. COLM   (1.6 v )  v A vA = 0.6V ...(i) A BC node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 0v COLM  mv0 = 3mv1 AC v2 v2 COLM  mAv = mA × (0.6V) + mB(1.6 v) v1 = v0 ...(i) 3 mA  4 mB ...(ii) COME KB = 1 mB(1.6 v)2 = 0 = 2mB (0.8 v)2  1 mv02 = 1 3m  v 2  1 2 m  v 2  v2  v0 2 2 2 1 2 0 3 KA = 1 (4mB)v2 = 2mBv2  K B  0.64  64% Therefore velocity of A = v 2  v 2 6 m/s 2 KA 1 2 12

JEE-Physics EXERCISE –IV(B) (b) s  ut  1 at2 , a  0  1  g t2 , t  6a  3 v 2 2  3  gg  u cos2  3R  R t  2v  4   2  g 30° 1. (i)  2g L= 10m (c) 3  cos  = 4/5   = 37° u=50 m/s 3M ucos 4 M 6 . (a) mu – Tt = mv ...(i) B  A R u2 sin2  Tt = mv ...(ii) u (ii) x  2  2g  120m On solving eq. (i) & (ii) v = u/2 2 y  H  u2 sin2   45m 2  u 2g (b)  A 2 . COLM :1 × v1 + 4 × v2 = 5 × 20 cos 60°= 50 ...(i) sin120 sin  20m/s v1  v2 sin   3 ; cos   13 2  44 1kg 4kg 60° mu cos  – Tt = mv ...(i) 60- Tt = mv..(ii) B 60° COME   1 × 1 × v12 + 1 × 4 × v22 On solving eq. (i) & (ii) u cos   v 2 2 2 u (90–) = 1 5  (1 0 ) 2   2 ...(ii) (c) 2 cos  =    = 60°  2  mucos 30° – Tt = mv ..(i) 2  v1 = –10 or 30 m/s & v2 = 15 or 5 m/s mu 3  u  v = 25 m/s Tt  ...(ii)  A Time to fall down to ground 4  2 3 7 . 2mu cos  2mv sec. 2 m,u ()= 2h = v g .98  v  u cos   u  Separation between particles (  )  22 2m = v.t = 44.2m m,u 1  mx0  m1 (x0  h cot )  m2 (x0  h)  cos ;  60 ,  = 120° m  m1  m2 2 2 2 3. xcm =0 8 . COLM  mu = (m + Ax)v  v  mu h(m2  m1 cot ) m  Ax  x0 = (m  m1  m2 ) dx mu 150 150 1    (m  Ax)dx   mudt dt m  Ax 00 4. us  2g(1  cos 60)  2g   3.13 m / s 2  Ax2  4   mx   2  = mut COLM  5uB = 4 × us uB = 5 g = 2.53 m/s  10–2x + 10–3 × 104 x2 = 10–2 × 103 × 150 Energy equation (  ) 2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 1 4  2 x = 105 m 2 m  5  g = mg × 0.8  = 0.4 v v1 e = –  v 2  v1   0.8 9. m M m  u 2  u1  5 . 2mg – T = 2m.a ...(i) 2a 2a v2 v2 T – mg = m.a ...(ii) 2m m (i) COLM  (m + m) × 0 + Mv = (M + 2m)v1 g AB On solving eq. (i) & (ii) a  3 v1   M M  v aa   2m  (a) v B  u2  2as  0  2  g  a  2ag (ii) COME  1 MV2  1 m ( v 2  v 2 )  2  1 m v 2  3  3 2  2 1 2  2 1 13

JEE-Physics Net velocity v0  v 2  v 2 v 2M(M  m) 1 2 . COLM : mv0 = (M + 2m) v1 1 2 (M  2m) mv0  v0 = 1m/s  v1 = 6m 6 2mv0/2 g COME  1 0 . 2mg – T = 2ma; T – mg = ma; a = \\\\\\\\\\\\\\\\\\\\\\\\ 3  1 (2m)  v0 2  1 Mv12 + 2m × g × h  1 (2m)v12 2  2 2 2    h = 0.3 m = L (1 – cos ) AB cos  = 1 – 3  4   = 37° m 2m 15 5 mc 1 3 . For first collision with plate A, final velocity of ball Velocity of m & 2m after falling through a distance A  (x m2m ) v1 = ev0 = e 2gh 0 ...(i) x  2ax  2gx For second collision ( 3 v Impulse equation ( )  mv = 4mv'  v' =  2gx  4 Tt = 2m  v  3   e 2gh0  2 4 2gh2 e = 3  2gx  Tt – T't = m  v  3  Height attained after first collision ()  T't = m(v–0), v  3gx  h1 = e2h0 = 2  2  9 = 4m 8 33 1 4 . Let v = velocity of the ball after collision along the v0 normal 3 11. COLM  mv0 + 0= (m + 2m)v1  v 1  (v=) J = impulse on ball (  ) After collision at highest point v1 = v – (– 2 cos 30°) = v + 3 () Impulse on wedge (  ) vx  1m / s   mv1  v2 J sin 30° = mv1 = 2v1  2m  m v0 v1  v = 4v1 – 3 ...(i) Coefficient of restitution m vy  1m / s   m 2  Normal  2m  ()v1 M A 30° COME  1 m v 2  1 m (v12  v 2 )  1 (2 m ) v 2 v2 2 0 2 2 2 1  v1  J  2  30°  v2  v1  v   v2  24 m/s   u2  u1  1 e=  node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 v 2 2 2 cos 30 2 Max height attained   1.2m 2g  v  3  v1 ...(ii) 22 ()  Solving we get v1  1 For the block ( )vx= 1m/s m/s while for the wedge it has   3 vx = 2m/s For the ball velocity along incline remains constant.  1  ()   2  ( v xwedge  v xblock )t   & ut  at2 block = 1.2  v' = 2 sin 30° = 1 m/s  t = 0.4 sec and  = (2–1)t = 0.4 m = 40 cm  Final velocity of ball  12   1 2  2  m/s  3 3  14

JEE-Physics EXERCISE –V-A 7 . The object will have translation motion without  1 . vcm  m 2v  m v v  rotation, when F is applied at CM of the system. mm 2 (     F   ) 2 . Linear momentum is a vector quantity whereas kinetic energy is a scalar quantity.  (          ) 3. P CM 2m 2 Initial position ( ) x C Just before collision     For this system, position of centre of mass remains If P is the CM then (P) same  x  2m x    x  4 3  m 2  Fsystem  0  8 . On applying law of conservation of linear M 0  5M 12R   M d  5M d  3R   d  7.5 R momentum ( )  M  5M M  5M   16  0     Pf v Pi   12  4ˆi  4  12 ˆi  4v4 4 . In order to shift centre of mass, the system must experience an external force, as there is no external The 4 kg block will move in a direction opposite to force responsible for explosion, hence centre of 12 kg block with a speed of 12 m/s.The mass does not shift. corresponding kinetic energy of 4 kg block (4kg12kg 12 m/s4kg )  = 1  4  122  288 J 5 . Let maximum momentum be p then 2 (p) 9 . m1 d x m2 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 p2  1 kL2  p  L Mk 2M 2 m1 d v Here m1d=m2x  x= m2 3 6. m m 1v2 1 2 1 0 . Since mass  area ()  before collision v2 after collision From COLM () AB mv2  m v 2   m v 2  v2  2v  3  3  Let mass of the bigger disc = 4M 15

JEE-Physics ()  EXERCISE –V(B)  mass of the smaller disc = M 1 . By applying impulse-momentum theorem () () mass of the remaining portion ()     = 4M –M = 3M  m1 v1  m 2 v2  m1 v 2 Now put the cut disc at its place again, centre of  m1  m 2   2 t 0   2 m1  m2 gt0 g mass of the whole disc will be at centre O.  () 2 . Just after collision O 10 14  4  0 vc =  10 m/s Centre of mass of the smaller disc is at its centre 10  4 that is at B. since spring force is internal force, it cannot change (B) the linear momentum of the (two mass + spring) Suppose CM of the remaining portion is at A and system. Therefore vc remains the same. AO is X. Let O as origin ( AOA= X vc  O  )  R 3.   A[ˆi cos(kt)  ˆj sin(kt)]  3M(x)=RM  x = 3 p(t) This suggests that centre of mass of remaining disc     Ak[ˆi sin(kt)  ˆj cos(kt)] will shift from the centre of original disc by a distance F dp of (1/3)R towards left.  dt  F.p F.p  Fp cos  But  0  cos   0   = 90°. (1/3)R  4. A A 1 260° 3 1 m1m2    2 m1  m2 11. Energy loss ( )= u1  u2 2 60° 120° • 2v v • v 2v  0.5 1 2  02  2 J 1st collision IInd collision 2 0.5  1 3 node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 L  x  n Particle with velocity 'v' covers and angle of 120° 0  L  and after collision its velocity become '2v'.  1 2 . k x d x It will cover angle of 240° X cm  xdm  L n   n  1  L dm 0  n  2   x  m1y1  m2y2  m3y3 k  L  d x 5. Y= cm M1 M2 M3 Ycm = 6m(0)  m(a)  m(a)  m(–a)  m(0) a ;Ycm = 10 L 10 m For n=0, xcm = 2 and for n   , xcm = L Multiple choice questions :     1 . As p1  p2  0 so p1'  p2'  0 For (A)  = (a + a ) ˆi + (b + b ) ˆj + c1 kˆ p1'  p2' 1 2 1 2 For (B)  = (a + a ) ˆi + (b + b ) ˆj For (C) p1'  p2' 12 + 1 2 For (D)  p1'  p2' = (c + c ) kˆ 2b ˆj  1 1 p1'  p2' 2 = (a + a ) ˆi 12 16

JEE-Physics But a, b, c, a, b, c  0 1 1 1 2 2 2 Therefore (A) & (D) is not possible to get 15 60   45 45 p1'  p2'  0 15 60 3. Comprehension type questions : Before collision After collision A  vertical component of velocity is zero 1. M ()  h1 v Subjective Questions : h2 60° B 30° C 1 . For body of mass m from A to B 3 33 (ABm)  u = 10 m/s (given) a = –  m g sin   f     m g sin   mg cos    m   m  h1 = tan60°  h = 3m = – [g sin  + g cos ] = – g [sin  +  cos ] 3 1 = – 10 [0.05 + 0.25 × 0.99] = – 2.99 m/s2 h2  h1 = tan30°  h – h = 3  h = 6m v2 – u2 = 2as 33 2 1 2 Velocity of block just before collision at B  v  100  2  2.99  6  8 m/s (B    )   = 2gh = 2  10  3 = 60 ms–1 MC B uA N m 30° 30° mg sin  f  mg cos  60 )Amg After collision (  ):  Let v1 be the velocity of mass m after collision and v2 be the velocity of mass M after collision.Body of mass M moving from B to C and coming to rest. For totally inelastic collision velocity of block along (mv1  normal to BC becomes zero and since there is no Mv2MBC impulse along BC so momentum (velocity) along )  BCrBemCainsunchanged( andsu==0v.25; v = 0, a=– 2.99 m/s2 v2 = 2as node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 – u2 BCBC  (0)2 – v22 = 2(–2.99) × 0.5  )   v22 = 1.73 m/s Speed of block just after collision Body of mass m moving from B to A after collision (  )  (m BA v= 60 cos 30 = 60  3 4 5 m s 1 ) B  2 u = v1; v = +1 m/s 2 . vc2 = vB2 + 2g(h2 – h1) (K.E. + P.E.)initial = (K.E. + P.E.)final + Wfrication  v = 452  2 10  3 = 105 ms–1 1 m v 2  mgh  1 mv2  0  m g s c 2 1 2 17

JEE-Physics 1 v12 + 10 × (6 × 0.05) = 1 Horizontal distance covered by the car 2 (1)2 + 0.25 × 10 × 6 () 2 BS=6 v1 = – 5m/s P = 12 × 53 = 603m h h Since the second ball also struck the trolley  sin  = 6 ()  A h = 6 sin  = 6 × 0.05  In time 12 seconds the trolley covers a distance of 603  Coefficient of restitution( ) (12 s603  e  Relative velocity of seperation For trolley in 12 sec (12s  )   Relative velocity of approach From  5  1.73  0.84 s   u  v  60 3   5.5 3  v  1 2   v  7.8 m/s 8 0  2   2  2 . Consider the vertical motion of the cannon ball To find the final velocity of the carraige after the second impact we again apply conservation of linear ( ) momentum in the horizontal direction  S = ut + 1 – 120 = 50t0 – 5t02 ( at2 ) 2  5t02 – 50t0 – 120 = 0 t02 – 10 t0 – 24 = 0 m vx + (M + m)7.8 = (M + 2m) vf  t0 = – (10)   100  4(1)(24) = 12 or –2  1 × 55 3 + (9 + 1)7.8 = (9 + 2)vf 2  vf = 15.75 m/s The horizontal velocity of the cannon ball remains v2 the same 3. t  vx = 100 cos30° + 5 3  55 3 m/s t +P u = 53m/s – 30°v=53m/s m=9kg   (v2 sin tˆi  v2 cos tˆj) and   v1ˆj C v2 v1 120m A B uT=0    v21 v2  v1  Apply conservation of linear momentum to   v2 sin tˆi  (v2 cos t  v1 )ˆj the cannon ball-trolley system in horizontal direction. If m is the mass of cannon ball and M is the mass of     mv2 sin tˆi  m(v2 cos t  v1 )ˆj p21 mv21 the trolley then ( v2 mwhere  = R M) 4 . The string snaps and the spring force comes into node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 mvx + M × 0 = (m + M) Vx  Vx  mvx play. The spring force being an internal force for m M the two mass-spring system will not be able to change the velocity of centre of mass. This means the location of centre of mass at time t will be v0 t where vx (isthvexvelocityof the(cannonball-trolley)) ( system 1  55 3  Vx = 1  9 = 5.53 m/s tv0t) The second ball was projected after 12 second. x cm  m1x1  m2x2  v0t m1  m2 (12s)  18

JEE-Physics  m1[v0t–A(1–cos t)] + m2x2 = v0tm1 + v0tm2 dp b ... (ii)  m2x2 = v0tm1 + v0tm2–v0tm1+m1A (1–cos t) F = n dt × (Area) = n × (2mv) × a × 2  m2x2 = v0tm2 + m1A(1 – cos t) b b 3b m1 From (i) and (ii) Mg × = n × (2mv) × a × ×  x2 = v0t + m2 A(1 – cos t) 2 24 (b) Given that x1 = v0t – A(1 – cos t) 32  3 × 10 = 100 × 2 × 0.01 × v × 1 × 4  dx1 = v0 – A sin t 6.  v = 10 m/s dt For collision between A & B  d2 x1 = – A2 cos t ... (i) vA  (m  2m) = –3ms–1 dt2 (m  2m) (9) This is the acceleration of mass m1. When the 2(m) spring comes to its natural length instantaneously vB  (m  2m) (9) = +6ms–1 For collision between B and C then (m1  ) d2 x1 = 0 and x2 – x1 =  vC   2m m  (6) = 4ms–1 dt2 0  2m     v 0 t  m1 A (1  cos t) –[v0t – A (1 – cos t)] =  u0cos  m2 0  u0cos Hmax.  m1  1 A (1 – cos t) =0 7. u0  m2  u0 Also when d2 x1 = 0; cos t = 0 from (1) u 2 sin2  dt2 0 Maximum height of first particle Hmax = 2g   =  m1  1 A Speed of 2nd particle at height Hmax given as 0  m2 v 2  u 2  2gHmax  u 2  u 2 sin2   v = u0cos y 0 0 0 y 5. Since the plate is held horzontal therefore net torque By Momentum Conseravtion acting on the plate is   mv0 cos ˆi  zero. (    p f f m v0 cos ˆj  pi  2mv )    vf  v0 cos  ˆi  ˆj 2 m  Angle with horizontal immediately after the node6\\E : \\Data\\2014\\Kota\\JEE-Advanced\\SMP\\Phy\\Solution\\Unit 3 & 4\\01-Centre of mass.p65 3b/4 F  collision = 4 a b/2 Mg b 3b ... (i)  Mg × × F × 24 19