Question

Question 22
An automobile engine propels a 1000 kg car (A) along a leveled road at a speed of 36 kmph. 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 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 getting its engine started. Find the speed of the car (B) just after the collision.

Answer 1

Given, mass of car A, $m_A=1000kg$
mass of car B, $m_B=1000kg$,
Initial velocity of car A, $u_A=36kmph=36\times \frac{5}{18}=10m{s}^{-1}$.
Final velocity of car A, $v_A=0$ and frictional force = 100 N
Initial velocity of car B, $u_B=0$
Let final velocity of car B be $v_B$
Since, the car A moves with a uniform speed, it means that the engine of car applies a force equal to the frictional force
We know, $power=\frac{force\times distance}{time}$ and $\frac{distance}{time}=speed$
$\therefore power=force\times velocity=100\times 10=1000W$

From law of conservation of momentum,
$m_A{u}_A+m_B{u}_B=m_A{v}_A+m_B{v}_B$
$1000\times 10+1000\times 0=1000\times 0+1000\times v_B$
$\Rightarrow v_B=10m{s}^{-1}$

$\therefore$The speed of the car B just after the collision, $v_B=10m{s}^{-1}$