The Ballistic Pendulum

The problem

A 10 g bullet is fired into a 2.00 kg wooden block hanging from a string. The bullet embeds itself, and the block swings up to a height of 15 cm. Find the bullet's speed.

This was a genuine measurement technique before electronic chronographs, and it is the standard exam problem for a reason: it has two stages that obey different conservation laws, and applying the wrong one is the most common error in the whole mechanics course.

The decision that matters

Stage	Duration	Momentum?	Kinetic energy?
1 · Bullet embeds in block	~milliseconds	conserved	not conserved
2 · Block swings up	~half a second	not conserved (gravity, string)	conserved

Stage 1 is a perfectly inelastic collision. It is over before gravity or the string can deliver any appreciable impulse, so momentum is conserved. But the bullet ploughs through wood — enormous friction, deformation, heat — so kinetic energy is emphatically not.

Stage 2 is a smooth swing. The string is always perpendicular to the motion so it does no work, and gravity is conservative, so mechanical energy is conserved. Momentum is not — the string is pulling sideways the whole time.

Working it, backwards

Stage 2 first, because that is where the measurement is. Energy conservation from the bottom of the swing to the top:

\[\tfrac12(m+M)v_A^2 = (m+M)gh \qquad\Longrightarrow\qquad v_A = \sqrt{2gh}\]

The combined mass cancels — a small mercy.

\[v_A = \sqrt{2(9.81)(0.15)} = \sqrt{2.943} = 1.7155\ \text{m/s}\]

Stage 1 second. Momentum conservation through the impact:

\[mv = (m+M)v_A \qquad\Longrightarrow\qquad v = \frac{(m+M)v_A}{m}\]

\[v = \frac{(2.010)(1.7155)}{0.010} = 344.8\ \text{m/s}\]

About Mach 1 — a plausible handgun muzzle velocity, which is the sanity check.

The mistake, and how large it is

Suppose you had used energy conservation for the whole process — bullet's kinetic energy converted to the final height:

\[\tfrac12 mv^2 = (m+M)gh \qquad\Longrightarrow\qquad v = \sqrt{\frac{2gh(m+M)}{m}} = 24.3\ \text{m/s}\]

24.3 m/s instead of 344.8 m/s — wrong by a factor of 14, and in the direction that makes the bullet slower than a thrown ball. The error is not subtle, and its size is the point: energy conservation across the collision is not slightly inaccurate, it is catastrophically wrong.

Where the energy actually went

\[K_{\rm before} = \tfrac12(0.010)(344.8)^2 = 594.5\ \text{J}$$ $$K_{\rm after} = \tfrac12(2.010)(1.7155)^2 = 2.96\ \text{J}\]

99.5% of the kinetic energy is gone — into deforming the bullet, splintering wood, and heat. That is not an anomaly; it is what "perfectly inelastic" means. When a light thing hits a heavy thing and sticks, almost all the kinetic energy is lost, because the surviving motion is constrained to the (large) combined mass.

The general result, for a mass $m$ at speed $v$ sticking to a stationary $M$:

\[\frac{K_{\rm after}}{K_{\rm before}} = \frac{m}{m+M}\]

Only $m/(m+M)$ of the energy survives — here 0.5%. Meanwhile momentum survives entirely, which is exactly why it is the right tool for stage 1.

Common mistakes

Using energy conservation through the collision. See above. This is the whole point of the problem.
Using momentum conservation through the swing. The string exerts a large horizontal force; momentum is not conserved during stage 2.
Forgetting the bullet's mass in $(m+M)$. Small here (0.5%), but it is the same slip that matters when masses are comparable.
Solving forwards. Work backwards from the measurement — that is how the apparatus is actually used.

Collisions & impulse · Linear momentum · Conservation of energy · Center of mass · Collision lab — set $e = 0$ and watch the energy bar drop