Question

A ball with a mass of `3 kg` is rolling at `3 m/s` and elastically collides with a resting ball with a mass of `1 kg`. What are the post-collision velocities of the balls?

Answer 1

Equations of conservation of energy and momentum.

`u_1'=1.5m/s`

`u_2'=4.5m/s`

Explanation:

As wikipedia suggests:

`u_1'=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2=`

`=(3-1)/(3+1)*3+(2*1)/(3+1)*0=`

`=2/4*3=1.5m/s`

`u_2'=(m_2-m_1)/(m_1+m_2)*u_2+(2m_1)/(m_1+m_2)*u_1=`

`=(1-3)/(3+1)*0+(2*3)/(3+1)*3=`

`=-2/4*0+6/4*3=4.5m/s`

[Equations' source]

Derivation

Conservation of momentum and energy state:

Momentum

`P_1+P_2=P_1'+P_2'`

Since momentum is equal to `P=m*u`

`m_1*u_1+m_2*u_2=m_1*u_1'+m_2*u_2'` - - - `(1)`

Energy

`E_1+E_2=E_1'+E_2'`

Since kinetic energy is equal to `E=1/2*m*u^2`

`1/2*m_1*u_1^2+1/2*m_2*u_2^2=1/2*m_1*u_1^2'+1/2*m_2*u_2^2'` - - - `(2)`

You can use `(1)` and `(2)` to prove the equations mentioned above. (I tried but kept getting two solutions, which is not right)