Irodov 1.156: Sequential Recoil Beats Simultaneous Jump

Problem Irodov 1.156

A buggy of mass $M$ stands at rest on a horizontal, frictionless surface, carrying two men, each of mass $m$ , on its edge. Each man jumps off horizontally with the same speed $u$ relative to the buggy at the instant he jumps. Find the buggy’s velocity $\vect v_1$ if the two men jump off together, and its velocity $\vect v_2$ if instead the first man jumps off and only then the second man jumps off (each still leaving at speed $u$ relative to the buggy’s instantaneous velocity). Determine which of the two final buggy speeds is larger, and by what factor.

The buggy of mass M carries two men, each of mass m, standing at rest; each will jump off horizontally with velocity u relative to the buggy. — The buggy of mass $M$ carries two men, each of mass $m$ , standing at rest; each will jump off horizontally with velocity $\vect u$ relative to the buggy.

Given

$m$
$M$
$\vect u$ relative to the buggy

Find

(1)

\vect v_1

jumping together, (2)

\vect v_2

jumping one after the other, and which is larger and by what factor

Solution

Step 1Simultaneous jump: one momentum balance

Each man leaves at velocity $\vect v_1+\vect u$ in the ground frame, since $\vect u$ is his velocity relative to the buggy’s own velocity $\vect v_1$ . Conservation of momentum from rest gives

\begin{align*}0 &= M\vect v_1+2m(\vect v_1+\vect u)\\ &= (M+2m)\vect v_1+2m\vect u.\end{align*}

\vect v_1 = -\dfrac{2m}{M+2m}\vect u

(1)

Step 2First man alone: the buggy’s speed after one jump

For the sequential case the first man leaves the buggy (now carrying only the second man, total mass $M+m$ ) while the second man is still aboard. Momentum conservation for this stage, again from overall rest, gives

\begin{align*}0 &= (M+m)\vect v'+m(\vect v'+\vect u)\\ &= (M+2m)\vect v'+m\vect u,\end{align*}

where $\vect v'$ is the buggy’s velocity right after the first man has jumped.

\vect v' = -\dfrac{m}{M+2m}\vect u

(2)

Step 3Second man leaves: the buggy’s final speed

Momentum conservation for the buggy (mass $M$ , final velocity $\vect v_2$ ) and the second man (leaving at $\vect v_2+\vect u$ ) gives

\begin{align*}(M+m)\vect v' &= M\vect v_2+m(\vect v_2+\vect u)\\ &= (M+m)\vect v_2+m\vect u.\end{align*}

Solving for $\vect v_2$ and substituting (2),

\begin{align*}\vect v_2 &= \vect v'-\dfrac{m}{M+m}\vect u\\ &= -\dfrac{m}{M+2m}\vect u-\dfrac{m}{M+m}\vect u\\ &= -m\left(\dfrac{1}{M+2m}+\dfrac{1}{M+m}\right)\vect u\\ &= -\dfrac{m\bigl[(M+m)+(M+2m)\bigr]}{(M+m)(M+2m)}\vect u.\end{align*}

\vect v_2 = -\dfrac{m(2M+3m)}{(M+m)(M+2m)}\vect u

(3)

Step 4Comparing the two speeds

Dividing the magnitude of (3) by that of (1) isolates the ratio $v_2/v_1$ .

\begin{align*}\dfrac{v_2}{v_1} &= \dfrac{m(2M+3m)/[(M+m)(M+2m)]}{2m/(M+2m)}\\ &= \dfrac{2M+3m}{2(M+m)}\\ &= \dfrac{2(M+m)+m}{2(M+m)}.\end{align*}

\dfrac{v_2}{v_1} = 1+\dfrac{m}{2(M+m)} > 1

(4)

Check

As $m\to0$ (negligible men), both $\vect v_1$ and $\vect v_2\to\vect 0$ and the ratio $\to1$ , as it must: two vanishingly light jumpers cannot move the buggy either way, and jumping order stops mattering. In the opposite limit $M\to0$ (a massless buggy), the ratio reaches its largest possible value, $1+m/[2(0+m)]=3/2$ , exactly what the boxed formula gives: with no buggy inertia to share between the two kicks, the order of the jumps matters as much as it possibly can.

Remark

The ratio $1+m/[2(M+m)]$ exceeds 1 for every positive $m$ and $M$ : jumping off one at a time always leaves the buggy moving faster than jumping off together, never merely as fast. The reason is the same one that favors a staged rocket over a single-stage one: the second man’s jump pushes off a buggy that is already lighter, carrying only $M$ rather than $M+m$ , so his fixed relative kick $u$ buys a larger share of momentum than it would if he had to share the recoil with a man still standing beside him. The buggy’s velocity at the moment of that second kick plays no role in the advantage: a fixed kick relative to a moving platform adds the same velocity increment whatever the platform’s own speed (Galilean invariance), so the lighter mass alone accounts for the whole effect.