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​

Answer 1

Answer:

Answer

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×time

(50×10

−3

)60−(−50×10

−3

)60=F×0.02

2×50×60×10

−3

=F×0.02

F=

0.02

6

=300N−vex−direction