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Collision analysis: momentum kept, energy lost

SPH4U Grade 12 Physics · Energy and Momentum

A dynamics track is levelled on the lab bench until a resting cart shows no drift in either direction. At one end a 500 g cart with a Velcro pad on its bumper waits behind a photogate; partway down the track a 700 g cart carrying a brass rider sits stationary, its own Velcro pad facing the oncoming lane. The first cart is given a push and released, and the photogate reads its speed as 1.20 m/s just before impact. The pads grab on contact with an audible rip of hooks, and the pair rolls away as one — no bounce, no separation. 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.

500 g at 1.20 m/s700 g + brass riderVelcro pads grab0.500 m/s together

Every number in this problem is editable — change any value below and the whole chain recalculates.

Given
  • m₁ = 500 g — Moving cart
  • v₁ = 1.2 m/s — Its speed at release
  • m₂ = 700 g — Stationary cart with its brass rider
Determine
  1. (a)the total momentum before impact
  2. (b)the total kinetic energy before impact
  3. (c)the speed the joined carts share afterwards
  4. (d)the kinetic energy that is left
Step 1 of 4(a) · 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.

mvp
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=600 mN⋅sp = 600\ \text{mN}{\cdot}\text{s}

Carried onward at full precision, not this rounded figure.

Open the Linear Momentum (p = mv) solver →

Step 2 of 4(b) · 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.

mvKE
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=360 mJE_k = 360\ \text{mJ}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Step 3 of 4(c) · solve for Velocity

Nothing pushes the system along the track, so the momentum from step 1 is still there — now carried by 500 + 700 = 1,200 g of cart instead of 500 g.

mvp
Rearranged for v
v=pmv = \tfrac{p}{m}
600 mN·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}{\cdot}\text{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}{\cdot}\text{m/s} \right)}{\left( 1.2\ \text{kg} \right)}
Answer
v=500 mm/sv = 500\ \text{mm/s}

Carried onward at full precision, not this rounded figure.

Open the Linear Momentum (p = mv) solver →

Step 4 of 4(d) · solve for Kinetic energy

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

mvKE
Rearranged for Eₖ
Ek=12mv2E_k = \tfrac{1}{2} m v^{2}
500 mm/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=150 mJE_k = 150\ \text{mJ}

Carried onward at full precision, not this rounded figure.

Open the Kinetic Energy solver →

Answer

Therefore 0.600 kg·m/s goes into the impact and every bit of it comes out — the joined carts roll away at 0.500 m/s — while of the 0.360 J of kinetic energy only 0.150 J survives: the missing 0.210 J, 58% of the start, went into the Velcro hooks, the thud and a trace of warmth.

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 dissipated, and an engineer would far rather it crumple sheet metal than the driver.

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