NCERT Exemplar XVI — task. Science CBSE, Class 9.

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An automobile engine propels a \(1000\ kg\) car (A) along a levelled road at a speed of \(36\ kmh^{–1}\).

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 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.

The formula to calculate the Power is,

\(Power\ =\ Mass \times Velocity\)
\(Power\ =\ Force \times Velocity\)
\(Power\ =\ \frac{Force}{Velocity}\)
\(Power\ =\ \frac{Mass}{Velocity}\)

Power of the engine of car \(A\) \(=\) \(W\)

Conservation of momentum is,

\(m_{A} \times u_{A}\ +\ m_{A} \times v_{A}\ =\ m_{B} \times u_{B} + m_{B} \times v_{B}\)
\(m_{A} \times v_{A}\ +\ m_{B} \times u_{B}\ =\ m_{A} \times u_{A} + m_{B} \times v_{B}\)
\(m_{A} \times u_{A}\ +\ m_{B} \times v_{B}\ =\ m_{A} \times v_{A} + m_{B} \times u_{B}\)
\(m_{A} \times u_{A}\ +\ m_{B} \times u_{B}\ =\ m_{A} \times v_{A} + m_{B} \times v_{B}\)

Speed of the car \(B\) just after collision \(=\) \(ms^{-1}\)

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66. NCERT Exemplar XVI

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