Question

A ball with a mass of `4 kg` moving at `3 m/s` hits a still ball with a mass of `26 kg`. If the first ball stops moving, how fast is the second ball moving? How much kinetic energy was lost as heat in the collision?

Answer 1

`6/13=0.461\ \text{m/s}` & loss of K.E. `=15.231\ J`

Explanation:

Ball of mass `m_1=4kg` moving with `u_1=3\ m/s` strikes another ball of mass `m_2=26\ kg` at rest `u_2=0`. After collision, first ball stops `v_1=0` & second ball moves with the velocity `v_2` then by law of conservation of momentum we have

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

`4(3)+26(0)=4(0)+26v_2`

`v_2=12/26`

`=6/13\ \text{m/s}`

Loss of kinetic energy

`=\text{initial K.E.}-\text{final K.E.}`

`=1/2m_1u_1^2-1/2m_2v_2^2`

`=1/2(4)(3)^2-1/2(26)(6/13)^2`

`=15.231\ J`

Answer 2

The speed is `=0.46ms^-1` and the loss in kinetic energy is `=15.23J`

Explanation:

There is conservation of momentum

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

The mass `m_1=4kg`

The velocity `u_1=3ms^-1`

The mass `m_2=26kg`

The velocity `u_2=0ms^-1`

The velocity `v_1=0ms^-1`

Therefore,

`4*3+26*0=4*0+26*v_2`

`v_2=12/26=0.46ms^-1`

The loss in kinetic energy is

`DeltaKE=KE_1-KE_2`

`=1/2m_1u_1^2-1/2m_2v_2^2`

`=1/2*4*3^2-1/2*26*0.46^2`

`=15.23J`