A collision in an isolated system in which kinetic energy and momentum are conserved throughout the collision.

See also: inelastic collision

Solving an elastic collision of equal masses in 1 dimension

In an elastic collision, both momentum and energy are conserved. The 1 dimensional case is mildly interesting: all collisions between points are 1 dimensional (an off-centre collision between balls is not 1 dimensional; see below).

So say we have 2 equal masses in an elastic collision on the line. All we know is that total momentum (m(v1+v2)) must be conserved, and so too must total kinetic energy (m*(v12/2+v22)). The velocities after the collision are unknowns, so we have 2 equations in 2 unknowns. We could solve them. We'd be likely to make a mistake, though.

So let's solve them without knowing anything! First, note we have a quadratic equation and a linear equation in our system, so we'll have 2 solutions (or a double solution, but we'll see that only happens if v1=v2). When this happens in Physics, it's invariably the case that one of the solutions is "unphysical" -- it cannot happen.

What's the infeasible solution (use a technical term, to sound more professional)? Easy! Suppose both velocities remained unchanged. Obviously total momentum and kinetic energy would be unchanged! The reason this solution is unphysical is that it involves the 2 objects passing right through each other, which is generally frowned upon by experimental physicists.

So now we have a hint how to get the other solution. The balls have equal mass, so if we just exchange v1 and v2 we're guaranteed to conserve kinetic energy and momentum. Hence this is the (only) other solution, therefore it describes what happens.

Here's another way of thinking about it: if the points were indistinguishable (say, both painted the same shade of pink, etc.), we'd expect not to be able to distinguish the result of the collision from no collision occurring. And this is indeed what we find.

Since we don't make mistakes when we do physics, we daringly expand the problem to involve a collision between two entities with different masses (still with only one dimension).


o  --->  <-- o
mA  vA0   vB0 mB
<-- o    o --->
 vA mA   mB vB 

Now we get the following:
I) mA * vA + mB * vB = mA * vA0 + mB * vB0 (conservation of momentum)
II) 1/2 * mA * vA2 + 1/2 * mB * vB2 = 1/2 * mA * vA02 + 1/2 * mB * vB02 (phew, conservation of energy)
Note that we can drop the factor 1/2 everywhere by multiplying both sides by two.

Now for an example: You are rolling a bowling ball towards another, lying still. The two collide elastically. The ball you tossed is somewhat heavier than the other. In fact,
mA = 5 kg
vA0 = 30 kph
mB = 3 kg
vB0 = 0 kph

What happens next?

I) vA = (mA * vA0 + mB * vB0 -  mB * vB)/mA)
      = (150 kg/kph - vB * 3 kg) / 5 kg
      = 30 kph - 3/5 * vB
... which may not come as a surprise.

Now insert this in (II), which we leave as an exercise for the clever student, and you get two solutions. Again, one solution is uninteresting - you missed, and the balls continue with their starting velocities. The other solution is (unless I'm completely wrong)

vB = 31.25 kph
vA = 11.25 kph

In general for this example, if the tossed ball is heavier, they both continue in the same direction. If the tossed ball is lighter, it bounces back, giving a smaller speed to the other ball.

An overview

An elastic collision is a collision where both the momentum and the kinetic energy of the objects are conserved. There are two equations to be remembered when analyzing collisions

Kinetic Energy = 1/2mv^2 = 1/2m(v.v)

Momentum = mv

In these equations, m stands for the mass of the object being measured, and v its velocity. In the first equation the "." refers to the dot product of v with itself. This is used when in two or more dimensions, where v is a vector.

Collision between two objects of equal mass

The most common example of an elastic collision is a collision between two billiard balls, call them ball A and ball B. Ignoring friction, if one was to measure the energy and momentum of the balls before and after they collide one would find that they were equal. Even more specifically, a collision between two billiard balls is an example of an elastic collision between two objects of equal mass. These collisions are interesting because the resulting objects will always move away at right angles from each other. To prove this we simply look at our two equations. Using the principle of relativity, we will center our frame of reference around the start position and velocity of one of the balls, ball B. This allows us to cancel one of the terms in the first half of the equation

1/2m*vA1^2 = 1/2m*vA2^2 + 1/2m*vB2^2

m*vA1 = m*vA2 + m*vB2

Where m is the mass of a ball, vA1 is velocity of A before collision, vA2 is velocity of A after collision, etc. By manipulating these equations we can show that

vA2 . vB2 = 0

From linear algebra, and the definition of the dot product, this means that vA2 and vB2 are perpendicular. This could mean a couple different outcomes. The trivial solution is that the balls missed each other and vB2 remained 0. The next solution is a direct collision where ball A hit ball B directly along its center of mass. This means that vA2 is now zero and vB2 is equal to vA1. These two are the only solutions possible in one dimension. However in two or more dimensions, there is the possibility that the collision was not straight on, and both will move off with a lesser velocity than the original velocity of ball A. However the dot product guarantees us that their velocities will be at right angles to each other. Think of this the next time you're playing pool.

Direct collision between two objects of unequal mass

Moving up in complexity, we can consider the case of a direct collision, meaning two objects headed straight at one another along one dimension, between two objects of non-equal mass. We will continue to call these ball A and ball B, but they now have mass mA and mB, respectively. We continue to use the principle of relativity to remove one of the terms from the first half of the equation, and our new equations are

1/2mA*vA1^2 = 1/2mA*vA2^2 + 1/2mB*vB2^2

mA*vA1 = mA*vA2 + mB*vB2

To simplify matters, we can multiply the equations by 1/mA and substitute u = mB/mA. We can also multiply the energy equation by 2 to cancel out the 1/2's. Doing this grants us

vA1^2 = vA2^2 + u*vB2^2

vA1 = vA2 + u*vB2

We will ignore the trivial solution where no change occurs and the balls velocities remain the same as at the beginning. We can then rearrange these equations to get both vA2 and vB2 in terms of u and vA1, which can be measured at the start of the incident. We end up getting

vA2 = (2/(u+1) - 1)*vA1

vB2 = 2/(u+1)*vA1

If we take a look at Sverre's example, at the start of the incident we are given

vA1 = 30

mA = 5

mB = 3

u = mB/mA = 3/5

And plugging these into our equations gives us

vA2 = 7.5

vB2 = 37.5

As was expected the tossed ball, A, moves forward at a slower speed. The light ball, B, that was hit moves forward at a faster speed than A was originally moving at. If you want to check this simply plug all the variables back into the original two equations, you should get identity.

As the equations are used to describe more and more complex situations, the math can become more obfuscated however the principles remain the same. These two equations based on the conservation of energy and the conservation of momentum can predict everything you would want to know about elastic collisions.

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