Question

An automobile engine propels a $1000kg$ car ($A$) along a levelled road at a speed of $36km/h$. Find power if the opposing frictional force is $100N$. Now, suppose after travelling a distance of $200m$, 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 car $B$ just after collision.

Answer 1

Step 1: Given data

Mass of cars $A$ and $B$$=1000kg$

Frictional force$=100N$

Velocity of the car$=36km/h=\left(36\times \frac{5}{18}\right)m/s=10m/s$

Step 2: Finding the power

The power of the frictional force is calculated below:

$\begin{array}{rcl}\mathrm{Power}& =& \mathrm{Force}\times \mathrm{Velocity}\\ & =& 100N\times 10m/s\\ & =& 1000W\end{array}$

Hence, the power required by the frictional force is $1000W$.

Step 3: Calculate the speed of car $B$ after collision.

According to the law of conservation of momentum, if two objects collide then the total momentum before and after collision remains the same if no external forces are acting on the bodies.

Let the speed of car $B$ after collision be $vm/s$.

The momentum can be obtained by the relation, $\mathrm{momentum}=\mathrm{mass}\times \mathrm{velocity}$

Momentum of car $A$ before collision$=1000kg\times 10m/s=10000kgm/s$

Momentum of car $B$ before collision$=1000kg\times 0m/s=0kgm/s$

Momentum of car $A$ after collision$=1000kg\times 0m/s=0kgm/s$

Momentum of car $B$ after collision$=1000kg\times vm/s=1000vkgm/s$

Considering the law of conservation of momentum:

$$\begin{gathered} 10000kgm/s+0kgm/s=0kgm/s+1000kgm/s \\ 10000=1000v \\ v=10m/s \end{gathered}$$

Hence, the speed of car $B$ after collision is $10m/s$.