Question

A tennis player swings her 1000 g racket with a speed of 11 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18 m/s. The ball rebounds at 44 m/s. (a) What is the magnitude of the change in momentum of the tennis ball? (Remember that momentum is a vector quantity.) kg m/s (b) How fast is her racket moving immediately after the impact? You can ignore the interaction of the racket with her hand for the brief duration of the collision. HINT: Think about Newton's Third Law for the ball and the racquet. m/s

Answer 1

- 3.72 Ns.

9.44 m/s

Step-by-step explanation:

mass of racket, M = 1000 g = 1 kg

mass of ball, m = 60 g = 0.06 kg

initial velocity of racket, U = 11 m/s

initial velocity of ball, u = 18 m/s

final velocity of ball, v = - 44 m/s

Let the final velocity of the racket is V.

(a) Momentum is defined as the product of mass and velocity of the ball.

initial momentum of the ball = m x u = 0.06 x 18 = 1.08 Ns

Final momentum of the ball = m x v = 0.06 x (- 44) = - 2.64 Ns

Change in momentum of the ball = final momentum - initial momentum

= - 2.64 - 1.08 = - 3.72 Ns

Thus, the change in momentum of the ball is - 3.72 Ns.

(b) By use of conservation of momentum

initial momentum of racket and ball = final momentum of racket and ball

1 x 11 + 0.06 x 18 = 1 x V - 0.06 x 44

12.08 = V - 2.64

V = 9.44 m/s

Thus, the final velocity of the racket afetr the impact is 9.44 m/s .