A car of mass 400 kg and travelling at `72 km h^(-1)` crashes into a truck of mass 4000 kg and travelling at `9 km h^(-1)` in the same direction. The car bounces back at a speed of `18 km h^(-1)`. The speed of the truck after the impact is 

Options

A. `9 km h^(-1)`
B. `18 km h^(-1)`
C. `27 km h^(-1)`
D. `36 km h^(-1)`

Answer

B

Explanation

Here, `m_1 = 400 kg, m_2 = 4000 kg `
`u_1 = 72 km h^(-1) = 72 xx 5/18 ms^(-1) = 20 ms^(-1)`
`u_2 = 9 km h^(-1) = 9 xx 5/18 = 2.5 ms^(-1)`
`v_1 = -18 km h^(-1) = -18 xx 5/18 = -5 ms^(-1)`
`v_2 = ?`
According to law of conservation of linear momentum , we get
`m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2v_2`
`400 xx 20 + 4000 xx 2.5 = 400 xx (-5) + 4000 xx v_2`
`20 + 25 = 5 + 10v_2`
`45 = 5 + 10v_2`
`v_2 = 5ms^(-1) = 5 xx 18/5 km h^(-1) = 18 km h^(-1)`.