Question

A ball with a mass of `3 kg` is rolling at `5 m/s` and elastically collides with a resting ball with a mass of `9 kg`. What are the post-collision velocities of the balls?

Answer 1

The velocity of the first ball is `=-2.5ms^-1`
The velocity of the second ball is `=2.5ms^-1`

Explanation:

In an elastic collision, we have conservation of momentum and conservation of kinetic energy

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

and

`1/2m_1u_1^2+1/2m_2u_2^2=1/2m_1v_1^2+1/2m_2v_2^2`

Solving the above 2 equations for `v_1` and `v_2`, we get

`v_1=(m_1-m_2)/(m_1+m_2)*u_1+(2m_2)/(m_1+m_2)*u_2`

and

`v_2=(2m_1)/(m_1+m_2)*u_1+(m_2-m_1)/(m_1+m_2)*u_2`

Taking the direction as positive `rarr^+`

`m_1=3kg`

`m_2=9kg`

`u_1=5ms^-1`

`u_2=0ms^-1`

Therefore,

`v_1=-6/12*5+18/12*(0)=27/7=-2.5ms^-1`

`v_2=6/12*5-6/12*(0)=90/7=2.5ms^-1`

Verificaition

`m_1u_1+m_2u_2=3*5+9*0=15`

`m_1v_1+m_2v_2=-3*2.5+9*2.5=6*2.5=15`