A ballistic pendulum stands along the back wall of the lab: a hardwood block hanging at rest on two long cords, with a spring cannon clamped to a bench and aimed squarely at its face. The block weighed in at 1.240 kg on the balance; the steel slug loaded into the cannon weighed 9.50 g, and a chronograph at the muzzle has clocked this cannon's slugs at 285 m/s. The cannon fires, the slug buries itself in the wood with a crack, and the block — slug and all — swings up as one, pushing a light rider along a curved scale that records the top of the swing. Nothing bounces and nothing falls out; the slug stays embedded in the wood. Find the kinetic energy the slug carried in, the speed the block and slug share the instant the slug stops moving relative to the wood, the kinetic energy that is left at that instant, and the height the block rises to.
Every number in this problem is editable — change any value below and the whole chain recalculates.
Given
m₁ = 9.5 g — Steel slug
v₁ = 285 m/s — Slug speed at the muzzle
m₂ = 1.24 kg — Hardwood block, hanging at rest
Determine
(a)the kinetic energy the slug carries in
(b)the shared speed the instant the slug stops in the wood
(c)the kinetic energy left at that instant
(d)the height the block rises
Step 1 of 4(a) · solve for Kinetic energy
The energy on the way in. Nothing downstream consumes it — it is here so that step 3 has something to be compared against, and the comparison is brutal.
Rearranged for Eₖ
Ek=21mv2
Your values, in your units
Ek=21(9.5g)(285m/s)2
Converted to base units
Ek=21(0.0095kg)(285m/s)2
Answer
Ek=385.82J
Carried onward at full precision, not this rounded figure.
The impact itself. Momentum is conserved through it because the cords are still vertical and gravity has no horizontal component to change the total — so m₁u₁ is shared out over m₁ + m₂.
From here the swing is smooth and lossless, so the surviving kinetic energy converts cleanly into height. This is the number you actually measure, with a rider on an arc scale.
Rearranged for h
h=mgU
2.9334 Jcarried from step 3
Your values, in your units
h=(1.2495kg)(9.80665m/s2)(2.9334J)
Answer
h=239.39mm
Carried onward at full precision, not this rounded figure.
Therefore the slug arrives carrying 385.8 J, the block and slug move off together at 2.17 m/s with only 2.93 J still kinetic — more than 99% of the energy stayed in the wood — and the pair swings up 0.239 m for the rider to mark.
Why this order
This chain exists to stop one specific mistake: using energy conservation through the impact. It is the obvious move — the slug has 386 J, so surely the block leaves with 386 J — and it is wrong by a factor of a hundred and thirty. Step 3 says only 2.93 J survives; more than 99% of the energy went into splintering wood, heating steel and making a bang. Momentum, meanwhile, sails through untouched. The rule that decides which conservation law you may use is a rule about the phase of the problem, not about the apparatus: momentum is conserved during the collision because no outside horizontal force acts, and energy is conserved after it because the swing is gentle and nothing is being destroyed. Mix the two phases and the answer is nonsense.
The second trap is the mass in step 4. Having found 2.93 J in the combined block, students sometimes divide by the slug's mass, or by the block's 1.240 kg, when the thing that rises is the 1.2495 kg pair. It happens to matter very little here and matters enormously with a heavier projectile. Worth noticing too: because Eₖ = ½mv² and U = mgh use the same mass, it cancels and the rise is simply v²/2g — the pendulum never needed to be weighed. That is exactly why the device is useful. Benjamin Robins built the first one in 1740 and, with nothing but a block of wood and a ribbon to mark the swing, measured musket-ball speeds that no clock of his century could have timed directly.
Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.
