Question

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

• A. 300 N along + X direction
• B. 300 N along - X direction
• C. 3,00,000 N along + X direction
• D. 3,00,000 N along - X direction

Answer 1

The correct option is B 300 N along - X direction

Here clear that if the ball goes a +x direction and then back to -ve x-direction force act on the ball is in -x-direction. We know that change in momentum = Impulse

= force$\times$time

$(50\times {10}^{-3})60-(-50\times {10}^{-3})60=F\times 0.022\times 50\times 60\times {10}^{-3}=F\times 0.02F=\frac{6}{0.02}=300N-vex-direction$