Irodov 1.154: A Perpendicular Jump Keeps Its Speed – Amit & Deepali Sinha Foundation

Problem Irodov 1.154

Two identical buggies, labeled 1 and 2, each carrying one man, roll without friction on two parallel rails, moving toward each other. Buggy 1’s mass alone is $M$ and its man’s mass is $m$ ; the same for buggy 2. As the buggies pass alongside each other, the two men jump across, each in the direction perpendicular to the rails, so man 1 lands in buggy 2 and man 2 lands in buggy 1. Afterward buggy 1 is at rest, and buggy 2 continues in its original direction with speed $v$ . Find the two buggies’ initial velocities $v_1$ and $v_2$ (signed along the common line of the rails).

The two buggies with their men, moving toward each other with velocities v_1 and v_2 along the rails, just before the men jump sideways and change places. — The two buggies with their men, moving toward each other with velocities $\vect v_1$ and $\vect v_2$ along the rails, just before the men jump sideways and change places.

Given

Find

v_1

v_2

Solution

Step 1Each man keeps his along-track speed through the jump

The jump is directed perpendicular to the rails, so it changes only the sideways component of a man’s velocity; the component along the rails, the only one that matters for what happens next, is untouched by the jump itself. A man therefore arrives at the far buggy still carrying the along-track speed of the buggy he left.

Step 2Buggy 1 absorbs the incoming man

Momentum conservation along the rails for buggy 1 plus the man who lands in it gives one equation relating $v_1$ and $v_2$ .

Mv_1+mv_2 = (M+m)\cdot0

Mv_1 = -mv_2

(1)

Step 3Buggy 2 absorbs the incoming man

The same bookkeeping for buggy 2, mass $M$ at $v_2$ , catching the man who left buggy 1 still moving at $v_1$ , gives the second equation; the combination ends at the given speed $v$ .

Mv_2+mv_1 = (M+m)v \tag{2}

Step 4Solving the pair

Substituting $v_2=-Mv_1/m$ from (1) into (2) eliminates $v_2$ and leaves one equation in $v_1$ .

\begin{align*}M\left(-\frac{Mv_1}{m}\right)+mv_1 &= (M+m)v \\ v_1\cdot\frac{m^2-M^2}{m} &= (M+m)v \\ v_1\cdot\frac{(m-M)(m+M)}{m} &= (M+m)v \\ v_1(m-M) &= mv\end{align*}

v_1 = -\dfrac{mv}{M-m}

(3)

Back-substituting into (1) gives $v_2$ .

v_2 = -\frac{M}{m}v_1 = -\frac{M}{m}\left(-\frac{mv}{M-m}\right)

v_2 = \dfrac{Mv}{M-m}

(4)

Check

In the limit of vanishingly light men, $m\to0$ with $M$ fixed, catching a passenger of negligible mass cannot change a buggy’s own along-track velocity by an appreciable amount: buggy 1 must already have been moving near its final velocity of $0$ , so $v_1\to0$ , and buggy 2 must already have been near its final $v$ , so $v_2\to v$ . The boxed results give exactly that, $v_1=-mv/(M-m)\to0$ and $v_2=Mv/(M-m)\to v$ as $m\to0$ , confirming both the signs and the magnitudes by a physical argument independent of how (1) and (2) were solved.

Remark

Treating the exchange as two separate one-dimensional perfectly inelastic collisions only works because the jump is perpendicular to the rails: that is what lets each man carry his old along-track velocity into the new buggy unchanged, decoupling the sideways motion of the jump from the along-track momentum balance this problem is actually about. A jump at any other angle would leave the two collisions coupled through the jump speed itself.