A man weighing 80 kg is standing in a trolley weighing 320 kg. The trolley is resting on frictionless horizont

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 the trolley with a speed of 1 m/s, then after 4 s his displacement relative to the ground will be:

Accepted Answer

Since the system (man + trolley) is initially at rest and there is no external force in the horizontal direction, the total linear momentum of the system is conserved .

1.  **Define Variables:**
    *   Mass of man ($m$) = $80	ext{ kg}$
    *   Mass of trolley ($M$) = $320	ext{ kg}$
    *   Velocity of man relative to trolley ($v_{rel}$) = $1	ext{ m/s}$
    *   Let $v_T$ be the velocity of the trolley relative to the ground.
    *   Let $v_m$ be the velocity of the man relative to the ground.

2.  **Relative Velocity Relation:**
    $v_{rel} = v_m - v_T \implies v_m = v_T + v_{rel} = v_T + 1$.

3.  **Conservation of Momentum:**
    $P_{initial} = 0$
    $P_{final} = m v_m + M v_T = 0$
    Substitute $v_m$: $80(v_T + 1) + 320 v_T = 0$
    $80 v_T + 80 + 320 v_T = 0$
    $400 v_T = -80 \implies v_T = -0.2	ext{ m/s}$.

4.  **Calculate Man's Velocity w.r.t. Ground:**
    $v_m = -0.2 + 1 = 0.8	ext{ m/s}$.

5.  **Calculate Displacement:**
    Displacement = $v_m 	imes t = 0.8	ext{ m/s} 	imes 4	ext{ s} = 3.2	ext{ m}$.

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