C. Energy and Momentum · Momentum and Collisions

Using Momentum and Energy Together: Ballistic Pendulums, Springs and Crash Reconstruction

C2.5 C2.6 C1.1About 70 minutes

Learning goals

I can split a problem into stages and choose momentum or energy conservation for each stage.
I can analyse a ballistic pendulum to find the speed of a projectile.
I can analyse a collision through a spring bumper, including its maximum compression.
I can use skid distances and momentum to reconstruct the speeds in a collision.

Words to know

ballistic pendulum
A block hanging from strings that a projectile hits and sticks in; the height of the swing is used to find the projectile's speed.

stage (of a problem)
A part of an event during which one set of conditions holds, such as the collision or the swing.

collision reconstruction
Using physical evidence and conservation laws to work out how a real collision happened.

maximum compression
The greatest shortening of a spring during a collision; it occurs when the colliding objects share the same velocity.

Two laws, two stages

Some problems need both conservation laws, used one after the other. A classic example is the ballistic pendulum. A fast object (a ball of clay, a dart, a pellet) hits a block hanging from strings and sticks in it. The block and object swing up together. From the height of the swing, you can work out how fast the object was going, even if it was far too fast to time by eye.

The secret is to split the event into stages and use the right law in each stage:

During the collision, the objects exert huge forces on each other for a very short time. Momentum is conserved. Kinetic energy is not (the collision is inelastic).
During the swing, there is no collision. Gravity does work, and (ignoring air resistance) mechanical energy is conserved: Ek → Eg. Momentum is not conserved, because the strings and gravity are external forces acting for a long time.

The most common mistake in this topic is to use energy conservation straight across an inelastic collision. Never do that.

The ballistic pendulum in three stages. Momentum is conserved during the very short collision (stage 2). Mechanical energy is conserved during the swing (stage 3), when the block and ball rise a height h.

Worked example

Example 1: The ballistic pendulum

A 0.050 kg ball of clay is thrown horizontally into a 0.95 kg block hanging at rest from long strings, and sticks. The block and clay swing up so their centre rises 0.20 m. Find the speed of the clay before impact, and the percentage of its kinetic energy lost in the collision.

Work backward. Stage 3 (swing): ½(m + M)v′² = (m + M)gh, so v′ = √(2gh) = √(2 × 9.8 × 0.20) ≈ 1.98 m/s.
Stage 2 (collision): momentum is conserved: mv = (m + M)v′, so v = (1.00 ÷ 0.050)(1.98) ≈ 39.6 m/s.
Kinetic energy before: ½(0.050)(39.6)² ≈ 39.2 J. Just after: ½(1.00)(1.98)² ≈ 1.96 J.
Lost: (39.2 − 1.96) ÷ 39.2 ≈ 0.95, or 95%. (Check with the formula from the lesson on elastic and inelastic collisions: M ÷ (m + M) = 0.95/1.00 = 95% ✓.)

The clay was moving at about 40 m/s, and 95% of its kinetic energy became thermal energy, sound and deformation. If you had wrongly used energy conservation from start to finish, you would have found v = √(2gh × 20) ≈ 8.9 m/s, far too small.

Collisions with springs

When two carts collide through a spring bumper, the spring compresses, stores elastic potential energy (Ee = ½kx², from the sibling unit Work, Energy and Simple Harmonic Motion), then pushes the carts apart.

Momentum is conserved at every instant (the spring force is internal).
At maximum compression, the carts are neither approaching nor separating, so they move with the same velocity for an instant, just like a perfectly inelastic collision.
The kinetic energy that "disappears" at that instant is stored in the spring. With an ideal spring, it all comes back: the collision as a whole is elastic.

Worked example

Example 2: A spring bumper

A 2.0 kg cart moving at 3.0 m/s hits a 2.0 kg cart at rest. A light spring (k = 800 N/m) on the second cart is between them. Find the maximum compression of the spring, and the final velocities after the carts separate.

At maximum compression, the carts share one velocity: (2.0)(3.0) = (4.0)v, so v = 1.5 m/s.
Ek before = ½(2.0)(3.0)² = 9.0 J. Ek at maximum compression = ½(4.0)(1.5)² = 4.5 J.
Energy stored in the spring: 9.0 − 4.5 = 4.5 J = ½kx², so x = √(2 × 4.5 ÷ 800) ≈ 0.106 m.
After separation, the spring has returned all its energy, so the collision is elastic. Equal masses swap velocities: the first cart stops and the second moves at 3.0 m/s.

Maximum compression ≈ 0.11 m (11 cm). Finally the first cart is at rest and the second moves at 3.0 m/s.

Crash reconstruction

Collision reconstructionists (often police officers or engineers with special training) work backward from the evidence left after a crash to find how fast vehicles were moving. A simple case is a rear-end collision in which the vehicles lock together and skid to a stop.

Skid stage: kinetic friction does negative work on the locked vehicles. ½(m₁ + m₂)v′² = μ(m₁ + m₂)gd, so v′ = √(2μgd). The skid length d is measured at the scene, and μ is found by test skids on the same road.
Collision stage: momentum is conserved: m₁v₁ + m₂v₂ = (m₁ + m₂)v′.

Real reconstructions use many more measurements (crush depth, rest positions, vehicle data recorders), but momentum and energy are always at the heart of them.

Planning a two-stage crash reconstruction. Start from the evidence at the end (the skid), find the speed just after the collision, then use momentum to find the speed before.

Worked example

Example 3: Was the driver speeding?

A 1200 kg car rear-ends a 1800 kg car that is stopped at a red light. The cars lock together and skid 6.0 m before stopping. Tests show μk = 0.70 for tires on that road. The speed limit is 60 km/h. How fast was the first car moving?

Skid stage: v′ = √(2μgd) = √(2 × 0.70 × 9.8 × 6.0) = √82.3 ≈ 9.07 m/s.
Collision stage: (1200)v₁ + 0 = (3000)(9.07), so v₁ = 2.5 × 9.07 ≈ 22.7 m/s.
Convert: 22.7 × 3.6 ≈ 82 km/h.

About 23 m/s (82 km/h), well over the 60 km/h limit. (In court, a reconstructionist would also report the uncertainty, mainly from μ and d.)

Worked example

Example 4: Down a ramp, collide, up a ramp

A 1.0 kg block slides from rest down a frictionless ramp from a height of 0.80 m. On the level floor it hits an identical block at rest, and they stick. They slide up a second frictionless ramp. How high do they rise?

Stage 1 (ramp down, energy): v = √(2gh) = √(2 × 9.8 × 0.80) ≈ 3.96 m/s.
Stage 2 (collision, momentum): (1.0)(3.96) = (2.0)v′, so v′ ≈ 1.98 m/s.
Stage 3 (ramp up, energy): h′ = v′² ÷ (2g) = (1.98)² ÷ 19.6 ≈ 0.20 m.

They rise 0.20 m, a quarter of the starting height. Halving the speed quarters the kinetic energy, and half of the kinetic energy (the fraction m₂ ÷ (m₁ + m₂)) was lost in the collision.

Just for fun

Why did the ball of clay love the ballistic pendulum?

Because after one collision, they were completely stuck on each other.

Hands-on activity

Build a ballistic pendulum

You need: a small, light cardboard box (for example an empty tissue box) with a crumpled tea towel inside, string, tape, a sturdy horizontal support such as a broom handle laid across the backs of two chairs, a ball of play dough or modelling clay (about 30 to 50 g), a kitchen scale, a ruler and a phone for slow-motion video. Grown-up note: throw gently and make sure no one stands behind the box.

Weigh the box (with the towel) and the clay ball.
Hang the box from the support with two strings of equal length (about 1 m) attached at both ends, so it swings without twisting. Measure the length L from the support to the middle of the box.
Toss the clay ball horizontally into the open side of the box so it sticks. Film from the side, with the ruler held horizontally under the box.
From the video, measure the largest horizontal distance x the box swings out. Find the rise from h = L − √(L² − x²).
Calculate v′ = √(2gh), then the clay's speed v = [(m + M) ÷ m]v′.
Check: estimate the clay's speed directly from the video (distance travelled between frames ÷ time).

What to look for: the two speeds should agree to within about 20%. A gentle toss of a few metres per second may swing the box out tens of centimetres sideways, but it rises only a few centimetres, which is why you measure the sideways swing x and calculate h from it. Air resistance and a box that tilts or spins make the swing a little smaller than ideal.

Your turn

A 0.020 kg dart sticks into a 0.48 kg block hanging from strings. The block rises 0.050 m. Find the dart's speed.

What percentage of the dart's kinetic energy was lost in the previous collision?

A 1.0 kg cart at 4.0 m/s hits a 3.0 kg cart at rest through a spring bumper with k = 1200 N/m. Find the maximum compression.

A 1000 kg car rear-ends a stopped 1500 kg car. They lock and skid 8.0 m with μk = 0.60. How fast was the first car moving?

A 1.0 kg block slides from rest down a frictionless ramp 0.45 m high and sticks to a 2.0 kg block at rest on the floor. How high do the pair rise up a second frictionless ramp?

Explain why you cannot use conservation of mechanical energy to go directly from the clay's speed to the height of the ballistic pendulum's swing.

More practice: warm-up

Which conservation law applies during the collision stage of a ballistic pendulum?

Which conservation law applies during the swing?

A pendulum block rises 0.20 m. How fast was it moving at the bottom?

Write the formula for the speed of a vehicle at the start of a skid of length d.

What type of collision is it when the ball sticks in the pendulum block?

More practice: on your own

A 0.010 kg steel ball launched at 50 m/s sticks in a 0.99 kg box of putty hanging from strings. How high does the box rise?

A ballistic pendulum has strings 1.5 m long. The block rises 0.20 m. Through what angle do the strings swing?

A 2000 kg SUV rear-ends a stopped 1000 kg car. They lock and skid 12 m with μk = 0.50. The speed limit was 50 km/h. Was the SUV speeding?

A 0.16 kg puck moving at 30 m/s is caught in the glove of a 90 kg goalie (with equipment) standing at rest on the ice. Find the goalie's velocity just after the catch.

In Example 2, explain why the carts have the same velocity at the moment of maximum spring compression.

Challenge

Show that the speed of the projectile in a ballistic pendulum is v = [(m + M) ÷ m]√(2gh).

Cart 1 (mass m₁, speed v₁) hits cart 2 (mass m₂, at rest) through a spring of constant k. Show that the maximum compression is x = v₁√[m₁m₂ ÷ (k(m₁ + m₂))]. Check it with Example 2.

A 0.50 kg block moving at 2.0 m/s on a frictionless floor collides elastically and head-on with a 1.5 kg block at rest. The 1.5 kg block then slides up a frictionless ramp. How high does it rise, and what does the 0.50 kg block do?

Check your understanding

Quiz

In a ballistic pendulum experiment, mechanical energy (kinetic plus gravitational potential) is conserved:

In a ballistic pendulum, momentum is conserved:

A pendulum block with an embedded ball rises 0.45 m. What was its speed just after the collision?

A ball of mass m sticks in a block of mass M, and the pair moves off at v′. The ball's original speed was:

A reconstructionist measures the skid length of two locked cars after a collision. The skid length is used to find:

Two carts collide through a spring bumper. The spring is most compressed when:

For parents and teachers

Big idea (C2.5, C2.6, C1.1): many real situations happen in stages. Use momentum for the collision stage and energy (with friction's work if needed) for the stages before and after. This is how a ballistic pendulum measures speed and how collision reconstructionists estimate vehicle speeds.

How to help: Before your teen starts calculating, ask them to name the stages and say which law applies to each. Building the box pendulum together makes the stages very clear.

Common difficulties:

Using energy conservation across an inelastic collision. This is the classic error; it gives answers far too small.
Working forward instead of backward. Start from the measured evidence (height, skid length) and work back.
Forgetting that the mass after sticking is m + M.
Spring bumpers: students often think the spring is most compressed when the first cart stops. It is when both carts share the same velocity.