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

Explanation

Suppose a body of mass m moving with velocity u collides along a straight line with a body of mass M and velocity v. After collision the two masses combine and move with 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)/(m)v`.
Here m = 2000 kg, v = 48 km`cdot h^(-1)` , M = 10000 kg,
V = 15 km`cdot h^(-1)`.
`therefore " "(M)/(m) = (10000)/(2000) = 5 `
u = (1 + 5)`xx 15 - 5 xx 48 = 90 - 240 = - 150"km"cdot h^(-1)`
The negative sign indicates that before collision the car was moving in the direction opposite to that of the truck.