Question

A tennis player swings her 1000 g racket with a speed of 10 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 20 m/s. The ball rebounds at 40 m/s.

a. 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.

b. If the tennis ball and racket are in contact for 10 ms, what is the average force that the racket exerts on the ball? How does this compare to the gravitational force on the ball?

Answer 1

m1u1 + m2u2 = m1v1 + m2v2
m1 = mass of racket = 1000g
u1 = initial vel of racket = 10 m/s
m2 = mass of ball = 60 g
u2 = initial velo of ball = -20 m/s (note negative since ball is moving opposite direction of racket.)
v1 = final vel of racket = ?
v2 = final vel of ball = 40 m/s

plug values into above equation and solve for v1
1000 g * 10 m/s + 60 g * (-20 m/s) = 1000 g * v1 + 60 g * 40 m/s
solve for v1
v1 = 6.4 m/s

To calculate average force use impulse-momentum theorem
I = F* (delta t)
I = m (delta v)
equate the above and solve for F
F * (delta t) = m * (delta v)
F = m * (delta v)/(delta t)
F = 0.060 kg * (40 m/s + 20 m/s)/0.010s
F = 360 N

gravitational force on ball is 0.60 x 10N =6N