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 m s-1, then after 4 s, his displacement relative to the ground is

Question

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 m s-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

Tardigrade Admin · Accepted Answer

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 = 300 v

or

υ = \frac{60}{300} = \frac{1}{5} m s^{- 1} = 0.2 m s^{- 1}

Speed of man w.r.t. ground

= (1 - 0.2) m s^{- 1}

= 0.8 m s^{- 1}

Displacement of man

= 0.8 \times 4 = 3.2 m

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 $1 m s^{- 1}$ , then after $4 s$ , his displacement relative to the ground is

$6 m$

$4.8 m$

$3.2 m$

$2.4 m$

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 1ms−1, then after 4s, his displacement relative to the ground is

6m

4.8m

3.2m

2.4m

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 $1 m s^{- 1}$ , then after $4 s$ , his displacement relative to the ground is

$6 m$

$4.8 m$

$3.2 m$

$2.4 m$