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 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.
Mass , (mA) = (mB) = 1000 kg
Initial speed , uA = 36 km / hr.
Opposing frictional force, F = 100 N
Since , the car A moves with a uniform speed , it means that the engine of car applies a force equal to the frictional force
∴ Power = ( Force x displacement ) / time
= F µA = (100 ) x (10 ) = 1000 W
After collision
Final speed of car A, vA = 0
Initial speed of car B, uB = 0
Using conservation of momentum, P, = P,
mAvA + mBuB = mAvA + mBvB
1000 x 10 + 1000 x 0 = 1000 x o + 1000 x vB
∴vB = 10m/s
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Welcome to the NCERT Solutions for Class 9 Science - Chapter . This page offers a step-by-step solution to the specific question from Excercise 0 , Question 22: An automobile engine propels a 1000 kg car (A) along a levelled road at a speed of 36 km h–1. ....
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