Irodov 1.162: Bifilar Pendulum as a Ballistic Scale – Amit & Deepali Sinha Foundation

Problem Irodov 1.162

A bullet of mass $m$ travels horizontally and embeds itself in a block of mass $M$ that hangs level on two identical parallel vertical threads, each of length $l$ , attached to a fixed ceiling (a bifilar suspension, so the block cannot rotate and both threads always stay parallel). After the impact the block, now carrying the bullet, swings so that the threads swerve through an angle $\theta$ from the vertical. Given $m \ll M$ , find (a) the bullet’s speed $v$ just before impact, and (b) the fraction $\eta$ of the bullet’s initial kinetic energy that is converted to heat in the collision.

The bullet m approaching the suspended block M horizontally, the two threads hanging vertical before impact. — The bullet $m$ approaching the suspended block $M$ horizontally, the two threads hanging vertical before impact.

After the bullet embeds, the combined body M+m swings as a rigid pendulum on the two parallel threads, which swerve together through the angle θ. — After the bullet embeds, the combined body $M+m$ swings as a rigid pendulum on the two parallel threads, which swerve together through the angle $\theta$ .

Given

$m$
$M$
$l$
$\theta$
$m\ll M$

Find

(a)

v

, (b)

\eta

Solution

Step 1Momentum conservation across the instantaneous embedding

The bullet embeds in the block in a time far shorter than the pendulum’s swing, so the thread tensions contribute no impulse during the collision itself and the horizontal momentum of the bullet plus block is conserved across it. Writing $V$ for the common speed immediately after impact,

m v = (M+m) V.

Step 2Energy conservation in the pendulum swing

The center of mass rises by $l(1-\cos\theta)$ as the bifilar suspension carries it through angle $\theta$ , so

\frac{1}{2}(M+m)V^2 = (M+m) g l (1-\cos\theta).

Solving for $V$ and using $1-\cos\theta = 2\sin^2(\theta/2)$ ,

\begin{align*}V^2 &= 2gl(1-\cos\theta) \\ &= 4gl\sin^2(\theta/2)\end{align*}

V = 2\sqrt{gl}\,\sin(\theta/2)

(1)

Step 3Bullet speed from the two conservation laws

Solving the momentum relation for $v$ and applying $m\ll M$ , so that $M+m\approx M$ ,

v = \frac{M+m}{m}V \approx \frac{M}{m}V.

Substituting (1) for $V$ gives the bullet’s speed.

v = \dfrac{2M}{m}\sqrt{gl}\,\sin(\theta/2)

(2)

Step 4Heat as the kinetic energy lost in the collision

The pendulum swing that follows conserves mechanical energy, so all the heat is generated during the embedding itself: it is the kinetic energy lost between just before and just after impact. Using $m v = (M+m)V$ to eliminate $V$ ,

\begin{align*}\frac{\dfrac12(M+m)V^2}{\dfrac12 m v^2} &= \frac{(M+m)}{m v^2}\left(\frac{mv}{M+m}\right)^2 \\ &= \frac{m}{M+m}.\end{align*}

The fraction converted to heat is the complement of this surviving fraction,

\eta = 1-\frac{\dfrac12(M+m)V^2}{\dfrac12 m v^2} = 1-\frac{m}{M+m}.

Applying $m\ll M$ once more,

\eta \approx 1-\dfrac{m}{M}

(3)

Check

Dimensionally $\sqrt{gl}$ is a speed, so $v$ in Step 3 is a speed, and $\eta$ in Step 4 is the ratio of two energies and so dimensionless, as required. As $\theta\to0$ the swing is negligible and $v\to0$ , consistent with a vanishingly small bullet speed producing no swerve at all. And the exact (unapproximated) heat fraction $\eta=1-m/(M+m)$ lies strictly between 0 and 1 for any $m,M>0$ , as an inelastic collision must.

Remark

This is the classic ballistic pendulum: a projectile too fast to time directly is measured instead through the height a heavy catcher rises, converting an unmeasurable speed into an easily measured swing angle. The heat fraction $\eta=1-m/(M+m)$ depends only on the mass ratio, never on $l$ , $\theta$ or $g$ , because it is fixed the instant the bullet embeds, before the pendulum geometry ever enters; the $m\ll M$ limit merely says that a light bullet stopped by a heavy block loses almost all its kinetic energy to deformation and heat, keeping almost none as motion of the combined body.