Irodov 1.155: A Relative Jump, Measured at Landing – Amit & Deepali Sinha Foundation

Problem Irodov 1.155

Two identical buggies, each of mass $M$ , coast one behind the other on a frictionless track, both moving at the same velocity $v_0$ . A man of mass $m$ starts out riding the rear buggy. At some instant he jumps from the rear buggy onto the front one, leaving with velocity $u$ relative to the rear buggy as it moves after he has left it. Find the velocity of each buggy after the man has landed on the front one.

The instant the man leaves the rear buggy: it has recoiled to v_1, the man himself crosses at v_man, and the front buggy is still cruising at v_0, not yet aware of the arrival. — The instant the man leaves the rear buggy: it has recoiled to $v_1$ , the man himself crosses at $v_{\mathrm{man}}$ , and the front buggy is still cruising at $v_0$ , not yet aware of the arrival.

Given

$v_0$
$m$
$M$
$u$

Find

Find the velocity of each buggy once the man has landed on the front one

Solution

Step 1Momentum conservation as the man leaves the rear buggy

The floor is frictionless, so the only horizontal force acting during the jump is the mutual push between the man and the rear buggy: their combined horizontal momentum is conserved across the split. Let $v_1$ be the rear buggy’s velocity once the man is gone.

Writing the man’s ground velocity as $v_{\mathrm{man}} = v_1+u$ and conserving momentum of the man plus rear buggy system,

(M+m)v_0 = M v_1 + m v_{\mathrm{man}} = M v_1 + m(v_1+u) = (M+m)v_1+mu. \tag{1}

Solving (1) for $v_1$ ,

v_1 = v_0-\frac{m}{M+m}u.

Step 2The man’s own velocity in flight

Substituting $v_1$ back into $v_{\mathrm{man}}=v_1+u$ ,

v_{\mathrm{man}} = v_0-\frac{m}{M+m}u+u = v_0+\left(1-\frac{m}{M+m}\right)u.

v_{\mathrm{man}} = v_0+\dfrac{M}{M+m}u

(2)

Step 3Momentum conservation as the man lands on the front buggy

The front buggy has felt nothing yet, so it is still moving at $v_0$ when the man lands on it and the two move off together at a common velocity $v_2$ . Conserving momentum of the front buggy plus man system,

M v_0 + m v_{\mathrm{man}} = (M+m)v_2.

Substituting the boxed $v_{\mathrm{man}}$ from Step 2,

\begin{align*}v_2 &= \frac{M v_0 + m\left(v_0+\dfrac{M}{M+m}u\right)}{M+m} = \frac{(M+m)v_0+\dfrac{mM}{M+m}u}{M+m} \\ &= v_0+\frac{mM}{(M+m)^2}u.\end{align*}

Collecting the two results, with $v_1$ the rear buggy’s final speed and $v_2$ the front buggy’s,

\begin{aligned} v_{\mathrm{rear}} &= v_0-\frac{m}{M+m}u,\\ v_{\mathrm{fore}} &= v_0+\frac{mM}{(M+m)^2}u. \end{aligned}

(3)

Check

Set $u=0$ : the man crosses without any kick relative to either buggy, so no momentum is exchanged and both formulas correctly return $v_{\mathrm{rear}}=v_{\mathrm{fore}}=v_0$ , the untouched initial speed.

Remark

The two jumps are really the same trick applied twice: leaving the rear buggy is a recoil, and landing on the front one is a perfectly inelastic collision, and both conserve only the momentum of the two bodies actually in contact at that instant. Nothing about the front buggy’s own motion mattered until the man actually landed on it, which is why $v_0$ can be substituted for the front buggy’s velocity at the moment of the jump even though it changes right afterward.