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  4. A ball reaches a racket at 60 m s- 1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 g and the contact time is 0.02 second, the force exerted by the racket on the ball is

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Q. A ball reaches a racket at $60 \, m \, s^{- 1}$ along $+x$ direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as $50 \, g$ and the contact time is $0.02$ second, the force exerted by the racket on the ball is

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Solution:

The change in momentum of the ball, $\Delta \text{p}=\text{mv}-\left(- \text{mv}\right)=2\text{mv}=2\times 0\cdot 05\times 60=6\left(\text{kg m s}\right)^{- 1}$
Force exerted on the bell by the rocket, $\text{F} = \frac{\Delta \text{p}}{\Delta \text{t}} = \frac{6}{0 \cdot 0 2} = 3 0 0 \text{N}$