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

(10 M/S) $m(A) = m(B) = 1000 \,kg$. Initial velocity of car $(A)$,$v = 36 \,km/h = 36 \times (5/18) \,m/s = 10 \,m/s$.
Frictional force $F = 100 \,N$.
Since the car $(A)$ moves with a uniform speed,the engine must apply a force equal to the opposing frictional force.
Power $P = F \times v = 100 \,N \times 10 \,m/s = 1000 \,W$.
For the collision,we use the law of conservation of linear momentum:
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
Here,$m_A = 1000 \,kg$,$u_A = 10 \,m/s$,$m_B = 1000 \,kg$,$u_B = 0 \,m/s$,and $v_A = 0 \,m/s$ (as car $(A)$ comes to rest).
$1000 \times 10 + 1000 \times 0 = 1000 \times 0 + 1000 \times v_B$
$10000 = 1000 \times v_B$
$v_B = 10 \,m/s$.