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

`m_(1) = 400 kg, u_(1) = 72 kmh^(-1) = 72 x 5/18= 20 ms^(-1)`
`m_(2) = 4000 kg, u_(2) = 9 kmh^(-1) = 9 x 5/18= 2.5 ms^(-1)`
`V_(1) = 18 kmh^(-1) = 18xx5/18 = 5 ms^(-1)`
From Law of Conservation of momentum
`m_(1)u_(1)=m_(2)u_(2)= m_(1)V_(1)=m_(2)V_(2)`
`implies 400xx20+4000xx2.5=400xx(-5)+4000xxV_(2)`
`v_(2)=20,000/4000`
`v_(2) = 5 ms^(-1)= 5xx18/5-= 18 kmh^(-1)`