Question

(a) A rail road flat car of mass $M$ can roll without friction along a straight horizontal track (figure). Initially, a man of mass $m$ is standing on the car which is moving to the right with speed $v_{0} .$ What is the change in velocity of the car if the man runs to the left so that his speed relative to the car is $v_{rel}$ just before he jumps off at the left end?
(b) If there are $n$ men each of mass $m$ on the car, should they all run and jump off together or should they run and jump one by one in order to give a greater velocity to the car?

Accepted Answer

Law of momentum conservation states that the momentum is conserved for any interaction between objects in an isolated system.

$(a)$ Let $v$ be the velocity of the car relative to the earth, when the man runs to the left, then conservation of linear momentum gives
$(M + m) v_{0} = M v + m (v - v_{r e l})$
$∴ v = v_{0} + \frac{m}{M + m} v_{r e l} \dots (i)$
Change in velocity, $Δ v = v - v_{0} = \frac{m v_{r e l}}{M + m}$

$(b)$ When there are $n$ men on the car and they jump off together, then velocity of the car after jumping off, from Eq. $(i)$ , will be
$v_{1} = v_{0} + \frac{n \times m}{M + n \times m} v_{r e l} \dots (i i)$ , as the total mass of the $v_{rel}$ men is, $n \times m$ .
If the men jump one by one, the mass of the system will go on changing till the last man jumps. Let $v^{'}$ be the velocity when one man jumps off the car, then from Eq. $(i i)$
$v^{'} = v_{0} + \frac{m}{M + n \times m} v_{rel}$

When the second man also jumps off, then
$v^{''} = v^{'} + \frac{m}{M + (n - 1) m} v_{r e l}$

$= v_{0} + \frac{m v_{r e l}}{M + n m} + \frac{m v_{r e l}}{M + (n - 1) m}$

When the last man jumps, then final velocity
$v_{2} = v_{0} + \frac{m v_{r e l}}{M + n m} + \frac{m v_{r e l}}{M + (n - 1) m} + \dots + \frac{m v_{r e l}}{M + m}$

Comparing equations. $(i i)$ and $(i i i)$ , we note that
$v_{2} > v_{1}$
i.e., the men would impart greater velocity to the car by running and jumping off one by one.

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