A car of mass one metric ton travelling at `32m//s`c dashes into the rear of a truck of mass `8000kg` moving in the same direction with a velocity of `4m//s` After the collision the car bounces back wards with a velocity of `8m//s` What is the velocity of the truck after the impact ?

Answer

`9m//s` .

Explanation

Here, `m_(1) = 1` metric ton `= 1000kg`
`u_(1) = 32 m//s, m_(2) = 8000kg, u_(2) = 4 m//s`
`upsilon_(1) = - 8m//s, upsilon_(2) = ?`
Applying the priniciple of conservation of linear
momentum `m_(1) upsilon_(1) + m_(2) upsilon_(2) = m_(1) u_(1) + m_(2) u_(2)`
`1000 (-8) + 8000 (upsilon_(2)) = 1000 xx 32 + 8000 xx 4`
`8000 upsilon_(2) = 64000 + 8000 = 72000`
`upsilon_(2) =(72000)/(8000) = 9m//s` .