AP Physics 1 · Lesson 3

Elastic and inelastic collisions

Know Know it

In every collision between objects with no net outside force, total momentum is conserved. Kinetic energy is a different story, and it sorts collisions into kinds:

In an elastic collision the total kinetic energy after equals the total before. In an inelastic collision some kinetic energy becomes internal energy, sound or bent material, so the total kinetic energy goes down. In a perfectly inelastic collision the objects stick together and move at one shared velocity, which loses the most kinetic energy that momentum conservation allows.

What is conserved in each kind of collision (no net external force)
KindTotal momentumTotal kinetic energyClue
Elasticconservedconservedobjects bounce apart with no energy lost
Inelasticconserveddecreasessome kinetic energy is lost, whether the objects bounce or stick
Perfectly inelasticconserveddecreases the mostobjects stick and move together

Worked example Easy

On slick ground, a 70 kg player running at 6.0 m/s grabs a 90 kg player standing still, and the two slide off together. Ignore friction during the grab. How fast do they move just afterward?

  1. 2.6 m/sAnswer
  2. 3.4 m/s
  3. 4.7 m/s
  4. 6.0 m/s

How to solve it

  1. Before: p=(70)(6.0)+(90)(0)=420p = (70)(6.0) + (90)(0) = 420 kg·m/s.
  2. After, they move together: p=(70+90)v=160vp = (70 + 90)v = 160v.
  3. v=420/160=2.625v = 420/160 = 2.625 m/s, about 2.6 m/s.

Why each choice is right or wrong

  • A. Correct. Momentum is conserved: (70)(6.0)=(160)v(70)(6.0) = (160)v, so v=420/160≈2.6v = 420/160 \approx 2.6 m/s.
  • B. Incorrect. 3.4 m/s swaps the masses, using the 90 kg player as the runner.
  • C. Incorrect. 4.7 m/s divides the momentum by 90 kg only, leaving the runner’s own mass out of the moving pair.
  • D. Incorrect. 6.0 m/s keeps the runner’s speed, but the same momentum now moves 160 kg, so the speed must drop.

Trap Momentum conserved does not mean kinetic energy conserved

Every collision here conserves momentum, so checking momentum cannot tell you whether a collision was elastic. Compute the total kinetic energy before and after: only if it is unchanged was the collision elastic. If it went up, something such as a released spring added energy.

Explain Explain it

Task verbs: calculate, justify

To classify a collision from data, first use momentum conservation to fill in any missing velocity, then add 12mv2\frac{1}{2}mv^2 for every object before and after. Report both totals and compare them.

For a head-on elastic collision there is a shortcut: the objects separate at the same relative speed at which they approached. Two equal masses simply swap velocities. A light object that hits a much heavier one at rest bounces back; a heavy object that hits a light one keeps going, and the light one shoots ahead faster than the heavy one was moving.

Worked example Medium

A 0.40 kg glider moving right at 0.90 m/s meets a 0.20 kg glider moving left at 0.60 m/s. Velcro makes them stick together. How much kinetic energy does the system lose in the collision?

  1. 0.006 J
  2. 0.048 J
  3. 0.150 JAnswer
  4. 0.198 J

How to solve it

  1. Right is positive: total momentum =(0.40)(0.90)+(0.20)(−0.60)=0.36−0.12=0.24= (0.40)(0.90) + (0.20)(-0.60) = 0.36 - 0.12 = 0.24 kg·m/s.
  2. Stuck together: v=0.24/0.60=0.40v = 0.24/0.60 = 0.40 m/s to the right.
  3. Before: 12(0.40)(0.90)2+12(0.20)(0.60)2=0.162+0.036=0.198\frac{1}{2}(0.40)(0.90)^2 + \frac{1}{2}(0.20)(0.60)^2 = 0.162 + 0.036 = 0.198 J. After: 12(0.60)(0.40)2=0.048\frac{1}{2}(0.60)(0.40)^2 = 0.048 J.
  4. Loss: 0.198 - 0.048 = 0.150 J.

Why each choice is right or wrong

  • A. Incorrect. 0.006 J comes from adding the momenta as if both gliders moved right, giving 0.80 m/s instead of 0.40 m/s.
  • B. Incorrect. 0.048 J is the kinetic energy left after the collision, not the amount lost.
  • C. Correct. They move off at (0.36−0.12)/0.60=0.40(0.36 - 0.12)/0.60 = 0.40 m/s; the kinetic energy falls from 0.198 J to 0.048 J, a loss of 0.150 J.
  • D. Incorrect. 0.198 J is all of the starting kinetic energy, but the stuck gliders still move, so some energy remains.

Justify Your turn

On a frictionless track, a 0.80 kg cart moving at +2.5 m/s hits a 1.2 kg cart at rest. Just after the collision, the 0.80 kg cart moves at -0.50 m/s. Find the velocity of the 1.2 kg cart after the collision, and use the total kinetic energy to say whether the collision was elastic.

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Compare with a model answer

A full answer does this

  • Finds the 1.2 kg cart’s velocity after the collision as +2.0 m/s
  • Shows the total kinetic energy is 2.5 J both before and after the collision
  • Concludes the collision was elastic

Model answer

Momentum: (0.80)(2.5)=(0.80)(−0.50)+(1.2)v(0.80)(2.5) = (0.80)(-0.50) + (1.2)v, so 2.0=−0.40+1.2v2.0 = -0.40 + 1.2v and v=+2.0v = +2.0 m/s. Kinetic energy before: 12(0.80)(2.5)2=2.5\frac{1}{2}(0.80)(2.5)^2 = 2.5 J. After: 12(0.80)(0.50)2+12(1.2)(2.0)2=0.10+2.4=2.5\frac{1}{2}(0.80)(0.50)^2 + \frac{1}{2}(1.2)(2.0)^2 = 0.10 + 2.4 = 2.5 J. The total kinetic energy is unchanged, so the collision was elastic.

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