A car of mass 2000 kg collides with a truck of mass `10^4 kg` moving at `48 km*h^(-1)`. After collision, the car rides up the truck and the truck-car combination moves at `15 km *h^(-1)`. What was the velocity of the car before collision ?

Explanation

The collision is inelastic as both the bodies move together after collision. Suppose a body of mass m, moving with velocity u, collision inelastically along a straight line with a body of mass M and velocity V. Applying the law of conservation of momentum,
mu+Mv=(m+M)V or, mu=(m+M)V -Mv or, u=`(m+M)/(m) V-(M)/(m)v=(1+M/m) V-M/mv`
Here, m=2000 kg, `v=48 km*h^(-1)` ,M=10000kg,
`V=15 km*h^(-1)`
`therefore M/m=(10000)/(2000)=5`
`u=(1+5)xx15-5xx48=90-240=-150km *h^(-1)`
The negative sign indicates that, before collision, the car was moving in the direction opposite to that of the truck.