CONSERVATION OF LINEAR MOMENTUM-02-OBJECTIVE SOLVED

OBJECTIVE SOLVED 1. Hail storms are observed to strike the surface of the frozen lake at 300 with the vertical and rebound at 600 with the vertical. Assume contact to be smooth, the coefficient of restitution is : (a) e  1 (b) e  1 3 (c) Ans. (b) e  (d) e = 3. Solution: Components of velocity before and after collision parallel to the plane are equal, So v sin 60 = u sin 30 ...(1) Components of velocity normal to the plane are related to each other v cos 60 =e u (cos 30) ...(2)  cot 60 = e cot 30  e  cot 60 cot 30 1  e  3  e  1 . 3 2. Two particles of masses m1 and m2 in projectile motion have velocities → and → respectively at time t = 0, They collide at time t . Their velocities become → and → at time 2t while still moving 0 v1 v2 0 in air. The value of  →  →    →  →  is : (a) 0 (c) 2 m1  m 2 gt 0 Ans. (c) m1v'1 m 2 v'2 m1v1 m 2 v 2 (b) (b) (d) m1  m 2 gt 0 2 m1  m 2 gt 0 . Solution: Before collision, the velocities of m1 and m2 at t0 are given as →  → → and →  → → v1  v1  gt0 v2  v2  gt0 At t = 2 t0, the velocities of m1 and m2 are expresed as →  →   → and →  →   → so, v1 v1 2gt0 v2 v2 2gt0 →  →  →  →  2(m  → . (m1v1 m2 v2 ) (m1v1 m2 v2 ) 1 m2 )gt0 3. An electron of mass m moving with a velocity v collides head on with an atom of mass M. As a result of the collision a certain fixed amount of energy ΔE is stored internally in the atom. The minimum initial velocity possessed by the electron is : (a) (b) (c) Ans. (c) Solution: 1  mM  v 2  ΔE (d) none of the above is correct. 2  M  m   v  . 4. A bullet of mass 0.01 kg, traveling at a speed of 500 ms-1, strikes a block of mass 2 kg, which is suspended by a string of length 5 m, and emerges out. The block rises by a vertical distance of 0.1 m. The speed of the bullet after it emerges from the block is : (a) 55 ms-1 (b) 110 ms-1 (c) 220 ms-1 (d) 440 ms-1. Ans. (c) Solution: By law of Conservation of Linear momentum mu = mv +MV ...(1) where m = mass of bullet M = mass of block u = velocity of bullet before collision v = velocity of bullet after collision V = velocity of block after collision By law of Conservation of Energy Mgh =  V = 1 MV2 2  V = 1.4 ms-1 Put in (1), we get 5 = 0.01v + 2(1.4)  v = 220 ms-1. v  2.2 ms1 0.01 5. A 6000 kg rocket is set for vertical firing. If the exhaust speed is 1000 ms-1, the amount of gas that must be ejected per second to supply the thrust needed to overcome the weight of the rocket is (g = 10 ms-1). (a) 30 kg (b) 60 kg (c) 75 kg (d) 90 kg. Ans. (b) Solution: M g  Thrust  v dM 0 dt  dM  M 0g = 6000x10 kgs1  dM  60000  60 kgs 1 . dt v 1000 dt 1000 6. A ball of mass m moving with a certain velocity collides against a stationary ball of mass m. The two balls stick together during collision. If E be the initial kinetic energy, then the loss of kinetic energy in the collision is E (a) E (b) 2 E (c) 3 Ans. (b) Solution: From COLM mu = (m+m) v v  u 2 1 2 1 u2 mu2 E (d) 4 . Final E  2mv  2m  2 2 4 4 Initial K.E.  1 mu 2  E 2 Loss  Initial K.E.  Final K.E.  mu 2 2  mu 2 4  mu 2  E 4 2 7. The position of centre of mass of a system consisting of two particles of masses m1 and m2 separated by a distance L apart from m1 (a) (c) Ans. (a) m 2 L m1  m 2 m 2 L m1  m 2 (b) (d) m1L m1  m 2 m1L m1  m 2 . Solution: Let centre of first body be origin and line joining them is taken as x-axis r  m1r1  m2r2  m.0  m 2 L cm 1 m 2 L  m2 m1  m 2 m1 X = m1  m 2 . 8. Two particles of equal mass have velocities →  4 i m/s and →  4 j m/s. First particle has an accel- v1 v 2 eration →  (i  j)m / s2 while the acceleration of the other particle is zero. The center of mass of the two particles moves on a : (a) circle (b) parabola (c) straight line (d) ellipse. Ans. (c) → → → m1v1  m 2 v 2 Solution: v COM  m1  m 2 → →  v1  v 2 2  2( i  j)m / s → Similarly, a COM  a1  a2  1 (i  j)m / s2 2 2  → is parallel to → the path will be a straight line. v COM a COM 9. A 1 kg ball, moving at 12 ms-1, collides head-on with a 2 kg ball moving in the opposite direction at 24 2 ms-1.If the coefficient of restitution is 3 , then the energy lost in the collision is : (a) 60 J (b) 120 J (c) 240 J (d) 480 J. Ans. (c) Solution: Initial momentum = Final momentum m1u1  m2u 2  m1v1  m2 v2 (1) (12) + (2) (-24) = 1v1 + 2v2  v1  2v2  36  v 2  v1  ...(i) Further e    u 2   u1  2   v 2  v1   3   24  12     2  v 2  v1 3 36  v 2  v1  24 ...(ii) Add (i) and (ii) 3v2  12  v 2  4ms 1  v1  28ms 1 K.E.  1 mv2   Loss = Total Initial K.E. – Total Final K.E.    2  Since Total Initial K. E. = 1 1144  1 2576 2 2  E1  72  576  648 J Similarly, total final K.E. E  1 (1)(784)  1 (2)(16) 2 2 2  E2 = 392 + 16  E2 = 408 J  Loss = 648 – 408  Loss = 240 J 10. A man of mass 100 kg. is standing on a platform of mass 200 kg. which is kept on a smooth ice surface. If the man starts moving on the platform with a speed 30 m/sec relative to the platform then calculate with what velocity relative to the ice the platform will recoil? (a) 5 m/sec (b) 10 m/sec (c) 15 m/sec (d) 20 m/sec. Ans. (b) Solution: Let us suppose that the platform recoils with speed V towards right, when the man moves with speed W towards left, both velocity is taken relative to ice. If V1 is the speed of the man relative to the platform, then V  W  V1 .  W  V1  V Since, initially both man and platform are at rest, therefore MV = mW where, M = mass of platform. and m = mass of the man. or, MV  m(V1  V) V  mV1  100  30 = 10 m/sec. M  m (200 100)