**Question:**

An automobile engine propels a $1000 \mathrm{~kg}$ car $(A)$ along a levelled road at a speed of $36 \mathrm{~km} \mathrm{~h}^{-1}$. Find power if the opposing frictional force is $100 \mathrm{~N}$. Now, suppose after travelling a distance of $200 \mathrm{~m}$, this car collides with another stationary car (B) of same mass and comes to rest. Let its engine also stops at the same time. Now car (B) starts moving on the same level road without getting its engine started. Find the speed of car (B) just after the collision.

**Solution:**

Force = 100 N, v = 36 km h-1 = 36 x (5/18) = 10 m s-1

∴ Power, P = Force x velocity = 100 x 10 = 1000 W

According to law of conservation of linear momentum

Momentum of car A + Mementum of car B before collision = momentum of car A + momentum of car B after collision

i.e. 1000 x 10 + 1000 x 0 = 1000 x 0 + 1000 x v

∴ v = 10 m s-1

Thus, speed of car B just after collision =10 ms-1