Irodov 1.198: Only the Struck Sphere Feels the Impulse

Problem Irodov 1.198

A small ball of mass $m$ , traveling with velocity $v_0$ , strikes one sphere of an initially stationary rigid dumbbell head-on in a perfectly elastic collision. The dumbbell consists of two point spheres, each of mass $m/2$ , joined by a rigid massless rod of length $l$ ; the ball’s velocity is directed straight at one sphere only, perpendicular to the rod, not at the dumbbell’s center. Treating both dumbbell spheres and the incoming ball as points, find the dumbbell’s proper angular momentum $\tilde M$ immediately after the collision, meaning the angular momentum evaluated in the reference frame that translates with the dumbbell’s center of inertia.

The ball of mass m approaches with velocity v_0 aimed at the upper sphere alone, the lower sphere off the line of impact; the two spheres, each of mass m/2, are joined by a rigid rod of length l. — The ball of mass $m$ approaches with velocity $\vect v_0$ aimed at the upper sphere alone, the lower sphere off the line of impact; the two spheres, each of mass $m/2$ , are joined by a rigid rod of length $l$ .

Given

$m$
$v_0$
$l$

Find

\tilde M

, the dumbbell’s angular momentum about its own center of inertia, just after the collision

Solution

Step 1Isolate the struck sphere for the instant of impact

Model the collision as an ordinary one-dimensional elastic collision between the incoming ball, mass $m$ and velocity $v_0$ , and the struck sphere, mass $m_2=m/2$ and initially at rest; call the ball’s own velocity after impact $v_1$ . Momentum and kinetic energy are both conserved,

\begin{align*}mv_0 &= mv_1+m_2u \\ \frac{1}{2}mv_0^2 &= \frac{1}{2}mv_1^2+\frac{1}{2}m_2u^2.\end{align*}

Factoring the energy equation’s difference of squares and dividing it by the momentum equation cancels their common factor.

\begin{align*}m(v_0-v_1)(v_0+v_1) &= m_2u^2 \\ m(v_0-v_1) &= m_2u \\ v_0+v_1 &= u\end{align*}

Substituting $v_1=u-v_0$ into the momentum equation and solving for $u$ gives the struck body’s velocity.

\begin{align*}mv_0 &= m(u-v_0)+m_2u \\ 2mv_0 &= (m+m_2)u\end{align*}

With $m_2=m/2$ ,

\begin{align*}u &= \frac{2m}{m+m/2}v_0 \\ &= \frac{2m}{3m/2}v_0\end{align*}

u = \dfrac{4v_0}{3}

(1)

directed along $\vect v_0$ ; the far sphere keeps its velocity $0$ .

Step 2Locate the center of inertia and its velocity

The two spheres' velocities relative to the center of inertia C: equal in size, opposite in sense, each a distance l/2 from C, the geometry the proper angular momentum sum adds over. — The two spheres’ velocities relative to the center of inertia $C$ : equal in size, opposite in sense, each a distance $l/2$ from $C$ , the geometry the proper angular momentum sum adds over.

The two spheres carry equal mass $m/2$ , so the dumbbell’s center of inertia $C$ sits at the rod’s midpoint, a distance $l/2$ from each sphere. Its velocity is the mass-weighted average of the two sphere velocities right after the collision, using (1),

\begin{align*}V_C &= \frac{(m/2)u+(m/2)(0)}{m} \\ &= \frac{u}{2} \\ &= \frac{1}{2}\cdot\frac{4v_0}{3}\end{align*}

V_C = \dfrac{2v_0}{3}

(2)

Step 3Velocities of the two spheres relative to the center of inertia

The proper angular momentum uses velocities measured in the frame translating with $C$ , so subtract (2) from each sphere’s lab velocity.

\begin{align*}v_1' &= u - V_C = \frac{4v_0}{3}-\frac{2v_0}{3} \\ v_2' &= 0 - V_C = -\frac{2v_0}{3}\end{align*}

v_1' = -v_2' = \dfrac{2v_0}{3}

(3)

The two spheres recede from $C$ with equal and opposite velocities transverse to the rod, exactly the picture of a rigid body spinning about its own center.

Step 4Sum the two spheres’ angular momenta about

C

Each sphere sits a distance $l/2$ from $C$ , and its velocity (3) is perpendicular to that radius, so each contributes $(m/2)\cdot(l/2)\cdot|v_i'|$ to $\tilde M$ , and, because the two spheres are on opposite sides of $C$ with oppositely directed velocities, the two contributions add rather than cancel.

\begin{align*}\tilde M &= \frac{m}{2}\cdot\frac{l}{2}\cdot\frac{2v_0}{3} + \frac{m}{2}\cdot\frac{l}{2}\cdot\frac{2v_0}{3} \\ &= 2\left(\frac{m}{2}\cdot\frac{l}{2}\cdot\frac{2v_0}{3}\right)\end{align*}

\tilde M = \dfrac{lmv_0}{3}

(4)

Check

Dimensionally $[l][m][v_0]=\text{m}\cdot\text{kg}\cdot\text{m/s}=\text{kg}\,\text{m}^2/\text{s}$ , an angular momentum. The stronger check is elasticity itself: the ball’s own post-collision velocity is $v_1=u-v_0=4v_0/3-v_0=v_0/3$ , so the total kinetic energy right after impact is $\frac{1}{2}m(v_0/3)^2+\frac{1}{2}(m/2)(4v_0/3)^2+0=mv_0^2/18+4mv_0^2/9=9mv_0^2/18=\frac{1}{2}mv_0^2$ , exactly the incoming $\frac{1}{2}mv_0^2$ , confirming that (1) was used consistently.

Remark

Treating the struck sphere as an isolated free particle for Step 1’s collision (justified there by the blow’s direction alone) reproduces the right answer here for a second reason too: for a general rigid body struck off center, the mass the incoming ball effectively collides with is $(1/M+d^2/I_{cm})^{-1}$ , not the struck point’s own local mass, but for this equal-mass dumbbell ( $M=m$ , $I_{cm}=m l^2/4$ , $d=l/2$ ) that expression reduces to exactly $m/2$ , the struck sphere’s own mass, which is why the shortcut costs nothing extra. Because the two spheres are equal, the post-collision motion about $C$ is also the clean equal-and-opposite pair (3); unequal spheres would still recoil in opposite senses about $C$ but not at equal speeds. The word “proper” is not just a choice of convenience here: because $\sum_i m_i(\vect r_i-\vect r_C)=\vect 0$ by the very definition of the center of mass, the term carrying $V_C$ drops out of the angular momentum about $C$ identically, whatever frame the individual velocities are measured in, so the lab-frame and center-of-inertia-frame calculations about $C$ must agree; the CM frame is simply the one where that cancellation is easiest to see, since there each sphere’s velocity is already the piece that survives it.