Irodov 1.170: Only the Normal Component Compresses – Amit & Deepali Sinha Foundation

Problem Irodov 1.170

A ball of mass $m$ , moving translationally with speed $v$ , strikes a second, stationary ball of the same mass $m$ . At the instant of contact the line joining the two centers makes an angle $\alpha=45^\circ$ with the striking ball’s direction of motion. The balls are smooth (frictionless contact) and the collision is elastic overall. Find the fraction $\eta$ of the striking ball’s initial kinetic energy that is stored as potential (deformation) energy at the instant of maximum deformation.

The instant of contact. Ball 1's velocity v makes angle α with the line of centers joining it to ball 2. — The instant of contact. Ball 1’s velocity $\vect v$ makes angle $\alpha$ with the line of centers joining it to ball 2.

The same velocity v resolved at its own vertex, with α again measured from the line of centers: the dashed component vα lies along that line and the dashed vα across it. — The same velocity $\vect v$ resolved at its own vertex, with $\alpha$ again measured from the line of centers: the dashed component $v\cos\alpha$ lies along that line and the dashed $v\sin\alpha$ across it.

Given

$m$ (equal masses)
$v$
$\alpha=45^\circ$
smooth (frictionless) contact

Find

\eta

Solution

Step 1Resolve the striking ball’s velocity along the line of centers

As the figure shows, ball 1’s velocity resolves into a component along the line of centers and a component perpendicular to it,

v_\parallel = v\cos\alpha, \qquad v_\perp = v\sin\alpha.

Ball 2 starts at rest, so it has neither component initially, and being smooth contact can never give it one along $v_\perp$ ’s direction: $v_\perp$ belongs to ball 1 alone, unchanged, for the entire encounter.

Step 2Maximum deformation is a shared normal velocity

Momentum along the line of centers is conserved throughout, with only $v_\parallel$ contributing initially (ball 2 has none, and $v_\perp$ has no component along the normal),

mv_\parallel + 0 = mu+mu,

where $u$ is the common normal-direction velocity of both balls at maximum deformation. Solving,

2mu = mv\cos\alpha \;\Longrightarrow\; u = \frac{v\cos\alpha}{2}. \tag{1}

Step 3Kinetic energy remaining at maximum deformation

At that instant ball 1 still carries its untouched transverse component $v_\perp=v\sin\alpha$ together with the shared normal speed $u$ , while ball 2 carries only $u$ . The total kinetic energy present is

\begin{align*}K_{\text{def}} &= \frac{1}{2}m\left(u^2+v_\perp^2\right)+\frac{1}{2}mu^2 \\ &= mu^2+\frac{1}{2}mv^2\sin^2\alpha.\end{align*}

Substituting (1),

K_{\text{def}} = m\left(\frac{v\cos\alpha}{2}\right)^2+\frac{1}{2}mv^2\sin^2\alpha = \frac{1}{4}mv^2\cos^2\alpha+\frac{1}{2}mv^2\sin^2\alpha.

Step 4The potential energy fraction

The initial kinetic energy is $K_0=\dfrac12 mv^2$ (only ball 1 moves), so the energy converted to potential (deformation) energy is $K_0-K_{\text{def}}$ . Forming the fraction and using $\sin^2\alpha=1-\cos^2\alpha$ ,

\begin{align*}\eta &= \frac{K_0-K_{\text{def}}}{K_0} \\ &= 1-\frac{\dfrac14 mv^2\cos^2\alpha+\dfrac12 mv^2(1-\cos^2\alpha)}{\dfrac12 mv^2} \\ &= 1-\left(\frac{1}{2}\cos^2\alpha+1-\cos^2\alpha\right)\end{align*}

\eta = \dfrac{1}{2}\cos^2\alpha = 0.25

(2)

Check

At $\alpha=0$ (a head-on collision, line of centers along the motion), the boxed formula gives $\eta=1/2$ : with equal masses a head-on elastic collision transfers all the kinetic energy to the target eventually, and at maximum deformation, by symmetry of the equal-mass case, exactly half of it is momentarily stored as potential energy and half remains kinetic. At $\alpha=90^\circ$ (grazing, line of centers perpendicular to the motion) the formula gives $\eta=0$ : the contact force has no component along the incoming velocity at all, so there is nothing to compress.

Remark

The elastic collision as a whole conserves the total kinetic energy, so every joule stored as potential energy at maximum deformation is returned as kinetic energy by the time the balls separate; $\eta$ here answers a question about the middle of the collision, not its outcome, and only the smoothness of the contact (no tangential force) is what lets the transverse component $v\sin\alpha$ be set aside as a spectator throughout.