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A tennis player swings her 1000 g racket with a speed of 15.0 m/s. She hits a 60 g tennis ball that was approaching her at a speed of 18.0 m/s. The ball rebounds at 40.0 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.
_________m/s
Part B) If the tennis ball and racket are in contact for 7.00 , what is the average force that the racket exerts on the ball?
_________N

Answer :

Manetho

Answer:

a) 16.32 m/s

b) 640 N

Explanation:

A) mass of rocket m_r = 1000 g = 1 kg

initial speed of rocket u_r = 15 m/sec

initial speed of ball is u_b = 18 m/sec

final speed of ball is v_b = 40 m/sec

Let m_b be the mass of the ball= 60 g and v_r be the final velocity of the rocket

from law of conservation of momentum

momentum of the system remains zero

m_r×(u_r-v_r)+m_b(16-42) = 0

1×(15-v_r) = -0.060(18-40)

15-v_r = -1.32

v_r = 15+1.32 = 16.32 m/sec.

B) Average force that the rocket exert's on the ball is  F_avg can be calculated as

contact time t=7.00 ms

 F_avg = m(v-u)/t = 0.06×(40+18)/0.007   = 640 N

Answer:

Explanation:

mass, M = 1000 g = 1 kg

U = 15 m/s

mass of ball, m = 60 g = 0.06 kg

u = - 18 m/s

v = 4 m/s

(a) Let the final speed of the racket is V.

Use conservation of momentum

initial momentum of racket + ball = final momentum of racket + ball

1 x 15 - 0.06 x 18 = 1 x V + 0.06 x 4

15 - 1.08 = V + 0.24

V = 13.68 m/s

(b) Average force = rate of change of momentum of racket

F = 1 (13.68 - 15) / 0.007

F = 188.57 N

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