Question

A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball traveling horizontally at 37.0 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction.) (a) What is the impulse delivered to the ball by the racket? magnitude 5.244 Correct: Your answer is correct. N · s direction Correct: Your answer is correct. (b) What work does the racket do on the ball? (Indicate the direction with the sign of your answer.)

Answer 1

Explanation:

It is given that,

Mass of the ball, m = 0.06 kg

Initial speed of the ball, u = 50.4 m/s

Final speed of the ball, v = -37 m/s (As it returns)  

(a) Let J is the magnitude of the impulse delivered to the ball by the racket. It can be calculated as the change in momentum as :

$J=m(v-u)$

$J=0.06\times (-37-(50.4))$  

J = -5.24 kg-m/s

(b) Let W is the work done by the racket on the ball. It can be calculated as the change in kinetic energy of the object.

$W=\dfrac{1}{2}m(v^2-u^2)$

$W=\dfrac{1}{2}\times 0.06\times ((-37)^2-(50.4)^2)$

W = -35.1348 Joules

Hence, this is the required solution.