A ball with a mass of 3 kg is rolling at 18 ms^-1 and elastically collides with a resting ball with a mass of 9 kg. What are the post-collision velocities of the balls?

1 Answer
Feb 24, 2016

In an elastic collision, both momentum and kinetic energy are conserved. If we call the 3 kg ball 1 and the 9 kg ball 2, their final velocities are v_1=-14.4 and v_2=10.8 ms^-1.

Explanation:

Initial momentum:

p=m_1v_1+m_2v_2=3*18+9*0=54 kgms^-1

Initial kinetic energy:

E_k=1/2m_1v_1^2+1/2m_2v_2^2=1/2*3*18^2+1/2*9*0^2=486 J

Because this is an elastic collision, both will be conserved:

Final momentum:

p=54=m_1v_1+m_2v_2=3v_1+9v_2 - call this (1)

Final kinetic energy:

E_k=486=1/2m_1v_1^2+1/2m_2v_2^2=1/2*3*v_1^2+1/2*9*v_2^2 - call this (2)

We now have two equations, (1) and (2), in two unknowns.

Let's double (2), just for neatness:

972=3v_1^2+9v_2^2

We can find a value for v_1 by rearranging (1) and then substitute that into this revised (2) so that we are working with only one variable:

54=3v_1+9v_2

v_1=(54-9v_2)/3

Then:

972=3((54-9v_2)/3)^2+9v_2^2=(54-9v_2)^2/3+9v_2^2

972=(2916-972v_2+81v_2^2)/3+9v_2^2

Let's multiply through by 3 for neatness:

2916=2916-972v^2+81v_2^2+9v_2^2

Rearranging:

90v_2^2-972v_2=0

Solve using the quadratic formula or otherwise, and you get:

v_2=(972+-sqrt(944784-0))/180=(972+-972)/180

There are actually two roots, therefore, 0 and 10.8 ms^-1.

That is, the second ball, with a mass of 9 kg can be either stationary or moving at 10.8 ms^-1.

We should solve (1) with both of these to find the possible values of v_1:

If v_2=0, v_1=54/3=18ms^-1

But wait: this is just the condition before the collision! The 3 kg ball has a velocity of 18 ms^-1 and the 9 kg ball is stationary!

If v_2=10.8, v_1=-14.4 ms^-1.

The minus sign indicates that this velocity is in the opposite direction to the original velocity.

The 3 kg ball approaches at 18 ms^-1 then collides and bounces backward at 14.4 ms^-1, propelling the 9 kg ball forward at 10.8 ms^-1.