Ballistic pendulum: momentum in, energy out

SPH4U Grade 12 Physics · Energy and Momentum

A 9.50 g steel slug leaves a spring cannon at 285 m/s and buries itself in a 1.240 kg hardwood block hanging at rest on two long cords. The block, slug and all, swings up as one. 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.

Step 1 of 4 · 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

E_k = \tfrac{1}{2} m v^{2}

m

v

Your values, in your units

E_k = \tfrac{1}{2} \, \left( 9.5\ \text{g} \right) \, \left( 285\ \text{m/s} \right)^{2}

Converted to base units

E_k = \tfrac{1}{2} \, \left( 0.0095\ \text{kg} \right) \, \left( 285\ \text{m/s} \right)^{2}

Answer

E_k = 386

in

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 2 of 4 · solve for Common final velocity

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.

Rearranged for v

v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}

m1

u1

m2

u2

Your values, in your units

v = \frac{\left( 9.5\ \text{g} \right) \, \left( 285\ \text{m/s} \right) + \left( 1.24\ \text{kg} \right) \, \left( 0\ \text{m/s} \right)}{\left( 9.5\ \text{g} \right) + \left( 1.24\ \text{kg} \right)}

Converted to base units

v = \frac{\left( 0.0095\ \text{kg} \right) \, \left( 285\ \text{m/s} \right) + \left( 1.24\ \text{kg} \right) \, \left( 0\ \text{m/s} \right)}{\left( 0.0095\ \text{kg} \right) + \left( 1.24\ \text{kg} \right)}

Answer

v = 2.17

in

Carried onward at full precision, not this rounded figure.

Open the Perfectly Inelastic Collision solver →

Step 3 of 4 · solve for Kinetic energy

Recompute the kinetic energy with the shared speed and the combined 1.2495 kg. Set it beside step 1 before reading on.

Rearranged for Eₖ

E_k = \tfrac{1}{2} m v^{2}

m

v2.17 m/scarried from step 2

Your values, in your units

E_k = \tfrac{1}{2} \, \left( 1.2495\ \text{kg} \right) \, \left( 2.16687\ \text{m/s} \right)^{2}

Answer

E_k = 2.93

in

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 4 of 4 · solve for Height

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 = \frac{U}{m g}

U2.93 Jcarried from step 3

m

Your values, in your units

h = \frac{\left( 2.9334\ \text{J} \right)}{\left( 1.2495\ \text{kg} \right) \, g}

Answer

h = 239

in

Carried onward at full precision, not this rounded figure.

Open the Gravitational Potential Energy (U = mgh) solver →

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.