A ball with a mass of `5` `kg` is rolling at `4` `ms^-1` and elastically collides with a resting ball with a mass of `3` `kg`. What are the post-collision velocities of the balls?

Momentum before the collision: `p=mv=5*4=20` `kgms^-1`

Kinetic energy before the collision: `E_k=1/2mv^2=1/2xx5xx4^2=40` `J`

(the stationary ball has a momentum of `0` `kgms^-1` and a kinetic energy of `0` `J` before the collision)

The fact that momentum and kinetic energy are conserved mean that the total momentum after the collision will be `20` `kgms^-1` and the total kinetic energy after the collision will be `40` `J`.

Call the `5` `kg` ball '1' and the `3` `kg` ball '2'.

For momentum:

`m_1v_1+m_2v_2=20` - call this equation (A)

For kinetic energy:

`1/2m_1v_1^2+1/2m_2v_2^2=40` - call this equation (B)

We can substitute in the known masses to make these simpler:

`5v_1+3v_2=20` - (A')

`5/2v_1^2+3/2v_2^2=40` - (B')

To make it even a little neater, let's multiply equation B' by 2:

`5v_1^2+3v_2^2=80` - (B'')

Rearranging equation A', `v_1=4-3/5v_2`

Substituting this value of `v_1` into B'':

`5(4-3/5v_2)^2+3v_2^2=80`

Expanding the square:

`5(16-24/5v_2+9/25v_2^2)+3v_2^2=80`

`80-24v_2+9/5v_2^2+3v_2^2=80`

Subtract 80 from both sides and rearrange:

`24/5v_2^2=24v_2`

Multiplying through by 5 (I hate fractions):

`24v_2^2=120v_2`

Dividing through by `v_2`:

`24v_2=120`

`v_2=0.2` `ms^-1`

Substituting back into A':

`v_1=4-3/5(0.2)=3.88` `ms^-1`

We can substitute these values back into A and B to check them.

`"solution-1:"`
`m_r:"the red ball's mass"`
`m_b:"the blue ball's mass"`
`v_r:"the velocity of red ball before collision"`
`v_b:"the velocity of blue ball before collision"`
`v_r^':"the velocity of red ball after collision"`
`v_b^':"the velocity of blue ball after collision"`

`"momentums before collision"`
`P_r=m_r*v_r=5*4=20 kg*m/s" the red ball's momentum"`
`P_b=m_b*v_b=3*0=0" the blue ball's momentum"`
`Sigma P=P_r+P_b=20+0=20 kg*m/s" thee total momentum"`

`v_r^'=(2*Sigma P)/(m_r+m_b)-v_r`

`v_r^'=(2*20)/(5+3)-4`

`v_r^'=40/8-4`

`v_r^'=5-4=1" " m/s`

`v_b^'=(2*Sigma P)/(m_r+m_b)-v_b`

`v_b^'=(2*20)/(5+3)-0`

`v_b^'=40/8-0`

`v_b^'=5" "m/s`

`"Solution -2:"`

`m_r*v_r+m_b*v_b=m_r*v_r^'+m_b*v_b^'`

`5*4+3*0=5*v_r^'+3*v_b^'`

`20=5*v_r^'+3*color(red)(v_b^')" (1)"`

`v_r+v_r^'=v_b+v_b^'`

`4+v_r^'=0+v_b^'`

`color(red)(v_b^')=4+v_r^'" (2)"`

`"(1) " 20=5*v_r^'+3*(4+v_r^')`

`20=5v_r^'+12+3v_r^'`

`20-12=8v_r^'`

`8=8v_r^'" ; "v_r^'=1" "m/s`

`"use (2)"`

`v_b^'=4+1=5" "m/s`