A 0.060-kg handball is thrown straight toward a wall with a speed of 12 m/s. It

Question

A 0.060-kg handball is thrown straight toward a wall with a speed of 12 m/s. It rebounds straight backward at a speed of 6.7 m/s.
(a) What impulse is exerted on the wall?


(b) If the ball is in contact with the wall for 3.0 ms, what average force is exerted on the wall by the ball?
N

(c) The rebounding ball is caught by a player who brings it to rest. During the process, her hand moves back 0.60 m. What is the impulse received by the player?
magnitude N·s
direction

(d) What average force was exerted on the player by the ball?
N

Explanation / Answer

a) Impulse = 0.060( 6.7 - [-12])= 1.122 kg m/s

b) force= impulse /tim e= 1.122/ 3 x 10-3= 374 N

c) let's calculate the retardation of ball in hand of palyer and time taken

v^2- u^2=2as

0 - 6.7^2 = 2(a)(0.60)

a=37.408 m/s^2 apprx

t= (v-u)/a= (0- 6.7)/ 37.408= 0.179 sec apprx

impulse = fx t= 0.060( 37.408)(0.179)=0.402 N s aprx

d) forc e=  0.060( 37.408 ) = 2.244 N apprx

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