Question

A ball with a mass of `3 kg` is rolling at `6 m/s` and elastically collides with a resting ball with a mass of `8 kg`. What are the post-collision velocities of the balls?

Answer 1

`-"2.72 m/s"` and `"3.27 m/s"`

Explanation:

From conservation of momentum

`"Initial total momentum = Final total momentum"`

`m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2`

`("3 kg × 6 m/s") + ("8 kg × 0 m/s") = ("3 kg" × v_1) + ("8 kg" × v_2)`

`18 = 3v_1 + 8v_2 color(white)(...)……(1)`

Coefficient of restitution `(e)` is given by

`e = (v_2 - v_1) / (u_1 - u_2)`

For perfectly elastic collision `e = 1`

`1 = (v_2 - v_1)/ (6 - 0)`

`v _2 - v_1 = 6`

Multiply both sides by `3`

`3v_2 - 3v_1 = 18 color(white)(...) ……(2)`

Add equations `(1)` and `(2)`

`18 + 18 = 3v_1 + 8v_2 + 3v_2 - 3v_1`

`36 = 11v_2`

`v_2 = 36/11 = color(blue)"3.27 m/s"`

Substitute `v_2 = 3.27` in equation `(1)`

`18 = 3v_1 + (8 × 3.27)`

`v_1 = (18 - (8 × 3.27))/3 = color(blue)(-"2.72 m/s")`