A man $m = 80 k g$ is standing on a trolley of mass $320 k g$ on a smooth surface. If man starts walking on trolley along rails at a speed of $1 m / s$ , then after $4 s e c$ , his displacement relative to ground is

4 m

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4.8 m

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3.2 m

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6 m

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Solution

The correct option is C $3.2 m$
$v = 1 m / s$ is velocity relative the trolley so if trolley is moving at $V$ opposite to the man
then velocity of man in ground frame will be $v - V$

Note- As initially the trolley and man were at $r e s t .$ so the velocity of $c e n t e r$ of mass should be $z e r o$ all the time until
applied by some external force.
so we should get $0 = \frac{80 (v - V) + 320 \times (- V)}{80 + 320}$ ( $n e g a t i v e$ $V$ because towards $l e f t .$ )

from the formula $V_{c m} = \frac{m_{1} (V_{1}) + m_{2} \times (V_{2})}{m_{1} + m_{2}}$

putting $v = 1$ and on solving we get $V = 0.2 m / s$ $t o w a r d s$ left.$
Displacement of man relative to ground will be $t i m e \times (v - V) = 4 \times 0.8 = 3.2 m e t e r$

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A man m=80 kg is standing on a trolley of mass 320 kg on a smooth surface. If man starts walking on trolley along rails at a speed of 1 m/s, then after 4 sec, his displacement relative to ground is

A man $m = 80 k g$ is standing on a trolley of mass $320 k g$ on a smooth surface. If man starts walking on trolley along rails at a speed of $1 m / s$ , then after $4 s e c$ , his displacement relative to ground is