An elastic collision is a collision where total momentum and total kinetic energy is conserved.

This illustration shows two objects A and B traveling towards each other. The mass of A is m_{A} and the moving with velocity V_{Ai}. The second object has a mass of m_{B} and velocity V_{Bi}. The two objects collide elastically. Mass A moves away at a velocity V_{Af} and mass B has a final velocity of V_{Bf}.

Given these conditions, textbooks give the following formulas for V_{Af} and V_{Bf}.

and

where

m_{A} is the mass of the first object

V_{Ai} is the initial velocity of the first object

V_{Af} is the final velocity of the first object

m_{B} is the mass of the second object

V_{Bi} is the initial velocity of the second object and

V_{Bf} is the final velocity of the second object.

These two equations are often just presented in this form in the textbook with little or no explanations. Very early in your science education, you will encounter the phrase “It can be shown …” between two steps of mathematics or “left as an exercise for the student”. This almost always translates into “homework problem”. This “It can be shown” example shows how to find the final velocities of two masses after an elastic collision.

This is a step by step derivation of these two equations.

First, we know total momentum is conserved in the collision.

total momentum before collision = total momentum after collision

m_{A}V_{Ai} + m_{B}V_{Bi} = m_{A}V_{Af} + m_{B}V_{Bf}

Rearrange this equation so the same masses are on the same side as each other

m_{A}V_{Ai} – m_{A}V_{Af} = m_{B}V_{Bf} – m_{B}V_{Bi}

Factor out the masses

m_{A}(V_{Ai} – V_{Af}) = m_{B}(V_{Bf} – V_{Bi})

Let’s call this Equation 1 and come back to it in a minute.

Since we were told the collision was elastic, the total kinetic energy is conserved.

kinetic energy before collision = kinetic energy after collection

½m_{A}V_{Ai}^{2} + ½m_{B}V_{Bi}^{2} = ½m_{A}V_{Af}^{2} + ½m_{B}V_{Bf}^{2}

Multiply the entire equation by 2 to get rid of the ½ factors.

m_{A}V_{Ai}^{2} + m_{B}V_{Bi}^{2} = m_{A}V_{Af}^{2} + m_{B}V_{Bf}^{2}

Rearrange the equation so the like masses are together.

m_{A}V_{Ai}^{2} – m_{A}V_{Af}^{2} = m_{B}V_{Bf}^{2} – m_{B}V_{Bi}^{2}

Factor out the common masses

m_{A}(V_{Ai}^{2} – V_{Af}^{2}) = m_{B}(V_{Bf}^{2} – V_{Bi}^{2})

Use the “difference between two squares” relationship (a^{2} – b^{2}) = (a + b)(a – b) to factor out the squared velocities on each side.

m_{A}(V_{Ai} + V_{Af})(V_{Ai} – V_{Af}) = m_{B}(V_{Bf} + V_{Bi})(V_{Bf} – V_{Bi})

Now we have two equations and two unknowns, V_{Af} and V_{Bf}.

Divide this equation by equation 1 from before (the total momentum equation from above) to get

Now we can cancel out most of this

This leaves

V_{Ai} + V_{Af} = V_{Bf} + V_{Bi}

Solve for V_{Af}

V_{Af} = V_{Bf} + V_{Bi} – V_{Ai}

Now we have one of our unknowns in terms of the other unknown variable. Plug this into the original total momentum equation

m_{A}V_{Ai} + m_{B}V_{Bi} = m_{A}V_{Af} + m_{B}V_{Bf}

m_{A}V_{Ai} + m_{B}V_{Bi} = m_{A}(V_{Bf} + V_{Bi} – V_{Ai}) + m_{B}V_{Bf}

Now, solve this for the final unknown variable, V_{Bf}

m_{A}V_{Ai} + m_{B}V_{Bi} = m_{A}V_{Bf} + m_{A}V_{Bi} – m_{A}V_{Ai} + m_{B}V_{Bf}

subtract m_{A}V_{Bi} from both sides and add m_{A}V_{Ai} to both sides

m_{A}V_{Ai} + m_{B}V_{Bi} – m_{A}V_{Bi} + m_{A}V_{Ai} = m_{A}V_{Bf} + m_{B}V_{Bf}

2m_{A}V_{Ai} + m_{B}V_{Bi} – m_{A}V_{Bi} = m_{A}V_{Bf} + m_{B}V_{Bf}

factor out the masses

2 m_{A}V_{Ai} + (m_{B} – m_{A})V_{Bi} = (m_{A} + m_{B})V_{Bf}

Divide both sides by (m_{A} + m_{B})

Now we know the value of one of the unknowns, V_{Bf}. Use this to find the other unknown variable, V_{Af}. Earlier, we found

V_{Af} = V_{Bf} + V_{Bi} – V_{Ai}

Plug in our V_{Bf} equation and solve for V_{Af}

Group the terms with the same velocities

The common denominator for both sides is (m_{A} + m_{B})

Be careful of your signs in the first half of the expressions in this step

Now we’ve solved for both unknowns V_{Af} and V_{Bf} in terms of known values.

Note these match the equations we were supposed to find.

This was not a difficult problem, but there were a couple spots to trip you up.

First, all the subscripts can get tangled up if you aren’t careful or neat in your handwriting.

Second, sign errors. Subtracting a pair of variables inside parentheses will change the sign on BOTH variables. It is all too easy to carelessly turn – (a + b) into -a + b instead of -a – b.

Last, learn the difference between two squares factor. a^{2} – b^{2} = (a + b)(a – b) is an extremely useful factoring trick when trying to cancel something out of an equation.