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Collision analysis: momentum kept, energy lost

SPH4U Grade 12 Physics · Energy and Momentum

On a level dynamics track, a 500 g cart is released at 1.20 m/s toward a stationary 700 g cart carrying a brass rider. The Velcro pads on their bumpers grab, and the pair rolls away as one. Find the total momentum before impact, the total kinetic energy before impact, the speed the joined carts share afterwards, and the kinetic energy that is left.

Step 1 of 4 · solve for Momentum

Total momentum before impact is the moving cart's alone — the second cart is sitting still and contributes nothing. The balance reads the cart in grams and the photogate reads its speed in m/s, so the 500 g becomes 0.500 kg to match.

Rearranged for p
p=mvp = m v
Your values, in your units
p=(500 g) (1.2 m/s)p = \left( 500\ \text{g} \right) \, \left( 1.2\ \text{m/s} \right)
Converted to base units
p=(0.5 kg) (1.2 m/s)p = \left( 0.5\ \text{kg} \right) \, \left( 1.2\ \text{m/s} \right)
Answer
p=4.34p = 4.34

Carried onward at full precision, not this rounded figure.

Open the Linear Momentum (p = mv) solver →

Step 2 of 4 · solve for Kinetic energy

The kinetic energy before, from the same two numbers. Hold on to it — step 4 is the comparison this whole chain exists for.

Rearranged for Eₖ
Ek=12mv2E_k = \tfrac{1}{2} m v^{2}
Your values, in your units
Ek=12 (500 g) (1.2 m/s)2E_k = \tfrac{1}{2} \, \left( 500\ \text{g} \right) \, \left( 1.2\ \text{m/s} \right)^{2}
Converted to base units
Ek=12 (0.5 kg) (1.2 m/s)2E_k = \tfrac{1}{2} \, \left( 0.5\ \text{kg} \right) \, \left( 1.2\ \text{m/s} \right)^{2}
Answer
Ek=0.36E_k = 0.36

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 3 of 4 · solve for Velocity

Nothing pushes the system along the track, so the momentum from step 1 is still there — now carried by 500 + 700 = 1200 g of cart instead of 500 g.

Rearranged for v
v=pmv = \tfrac{p}{m}
4.34 lb·ft/scarried from step 1
Your values, in your units
v=(0.6 kg⋅m/s)(1,200 g)v = \tfrac{\left( 0.6\ \text{kg·m/s} \right)}{\left( 1{,}200\ \text{g} \right)}
Converted to base units
v=(0.6 kg⋅m/s)(1.2 kg)v = \tfrac{\left( 0.6\ \text{kg·m/s} \right)}{\left( 1.2\ \text{kg} \right)}
Answer
v=50v = 50

Carried onward at full precision, not this rounded figure.

Open the Linear Momentum (p = mv) solver →

Step 4 of 4 · solve for Kinetic energy

Recompute the kinetic energy with the shared speed, then set it beside step 2. The shortfall is the point.

Rearranged for Eₖ
Ek=12mv2E_k = \tfrac{1}{2} m v^{2}
50 cm/scarried from step 3
Your values, in your units
Ek=12 (1,200 g) (0.5 m/s)2E_k = \tfrac{1}{2} \, \left( 1{,}200\ \text{g} \right) \, \left( 0.5\ \text{m/s} \right)^{2}
Converted to base units
Ek=12 (1.2 kg) (0.5 m/s)2E_k = \tfrac{1}{2} \, \left( 1.2\ \text{kg} \right) \, \left( 0.5\ \text{m/s} \right)^{2}
Answer
Ek=0.15E_k = 0.15

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Why this order

This chain exists to break one stubborn belief: that momentum and kinetic energy are two names for the same thing. Steps 1 and 3 are the same equation, p = mv, run before and after the impact — and they agree exactly, because with the track level and friction negligible there is no outside push to change the total. Steps 2 and 4 are also the same equation, Eₖ = ½mv², run before and after — and they do not agree at all. Of the 0.360 J the system started with, 0.150 J is left. The missing 0.210 J went into deforming the Velcro hooks, a faint thud, and a trace of warmth: 58% of the energy, gone, while not one kg·m/s of momentum moved.

Step 3 is deliberately solved from p = mv rather than looked up in a collision formula, because that is where the reasoning lives: the 0.600 kg·m/s you computed in step 1 is literally the number you divide by the combined mass. Nothing new is introduced. And the asymmetry falls out of the exponent — momentum is linear in v, so doubling the mass and halving the speed leaves it untouched, while energy is quadratic, so the same swap costs half the joules. Christiaan Huygens worked this out in the 1660s answering a Royal Society challenge on collisions, and it is still the design brief for a crumple zone: momentum has to go somewhere, but energy can be spent, and an engineer would far rather spend it on sheet metal than on the driver.

Carried values move at full precision, not the rounded figure shown — chaining rounded numbers compounds error.