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 19 m/s . The ball rebounds at 35 m/s .

Part 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. Express your answer in meters per second. v = nothing m/s Previous AnswersRequest Answer Incorrect;

Part B If the tennis ball and racket are in contact for 12 ms , what is the average force that the racket exerts on the ball? Express your answer in newtons.

Answer 1

Part A.

Using Momentum conservation on system:

Initial Momentum = Final Momentum

Pi = Pf

mr*Vri + mb*Vbi = mr*Vrf + mb*Vbf

Given values are:

mr = mass of racket = 1000 gm = 1 kg

mb = mass of ball = 60 gm = 0.06 kg

Vri = initial speed of racket = 10 m/sec (take this as positive direction)

Vbi = initial speed of ball = -19 m/sec (since travelling in opposite direction of racket)

Vbf = final speed of ball = 35 m/sec (since travelling in same direction of racket)

Vrf = final speed of racket = ?

So,

Vrf = (mr*Vri + mb*Vbi - mb*Vbf)/mr

Vrf = (1*10 + 0.06*(-19) - 0.06*35)/1

Vrf = 6.76 m/sec

Part B.

Average force racket exerts on ball will be

Favg = Change in momentum/time taken

time taken = 12 ms

Change in momentum of ball = mb*(Vbf - Vbi)

Favg = 0.06*(35 - (-19))/(12*10^-3)

Favg = 270 N

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