A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g

Question

A tennis player swings her 1000 g racket with a speed of 12 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 20 m/s. The ball rebounds at 42 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.
m/s

(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?
N

How does this compare to the gravitational force on the ball?
Favg / Wball =

Explanation / Answer

here

Use conservation of linear momentum
m1 *u1 + m2 * u2 = m1 * v1 + m2 * v2
m1 = mass of racket = 1000g
u1 = initial velocity of racket = 12 m/s
m2 = mass of ball = 60 g
u2 = initial velocity of ball = -20 m/s (note negative since ball is moving opposite direction of racket.)
v2 = final velocity of ball = 35 m/s

then put the values in the equation we get

1000 g * 12 m/s + 60 g * (-20 m/s) = 1000 g * v1 + 60 g * 42 m/s

solve for v1

v1 = 8.28 m/s

the rocket move 8.28 m/s

b)

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 * (42 m/s + 20 m/s)/0.01 s

F = 372 N

the average force is 372 N

c)

Favg / Wball = 310 / 0.06 * 9.8

Favg / Wball = 632.56 N

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