Question is in the description below...

Question

A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizontal rails. If the man starts walking on trolley with a speed of 1m/s, then after 4 seconds his displacement relative to the ground will be

A. 5 m

B. 4.8 m

C. 3.2 m

D. 3.0 m

Answer 1

Correct Answer is (C)

Explanation:

Let V_mg = velocity of man with respect to ground
V_tg = velocity of trolley with respect to ground
V_mt = velocity of man with respect to trolley = 1 m/s
Let M = mass of person = 80 kg
Let M' = the mass of trolley = 320kg
from relative motion 
V_tg = V_mg - V_mt --------------------(1)

Initially trolley and man were at rest initial momentum = 0
According to law of conservation of momentum

V_mgM + V_tgM' = 0

Substituting value of V_tg from equation (1) 

V_mgM + (V_mg - V_mt)M' = 0 

⇒ V_mg(M + M') - V_mtM'

V_mg(320 + 80) = 1(320) 

⇒ V_mg = 320/400 = 0.8 m/s

Thus displacement of man with respect to observer = V_mg × time = 0.8 × 4 = 3.2 m