Q. A man weighing 60 kg is standing on a trolley weighing 240 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on the trolley with a velocity of $1\ ms^{- 1}$, then after 4 s, his displacement relative to the ground is
A) 6 m
B) 4.8 m
C) 3.2 m
D) 2.4 m
Explanation
(c) The trolley shall move backwards to conserve momentum. The backward momentum would be shared by both the trolley and man
Applying conservation of momentum
$60 \times 1(240 + 60)v$ or $60 = 300v$
or $v = \frac{60}{300} = \frac{1}{5}ms^{- 1} = 0.2\ ms^{- 1}$
Speed of man w.r.t. ground
$= (1 - 0.2)ms^{- 1} = 0.8\ ms^{- 1}$
Displacement of man $= 0.8 \times 4$m = 3.2m
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.