Question

A 0.261-kg volleyball approaches a player horizontally with a speed of 14.9 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.5 m/s.
a) what impulse is delivered to the ball by the player?b)if the players fist is in contact with the ball for 0.600s, find the magnitude of the average force exerted on the players fist.

Answer 1

Answer:

a)9.761 kg.m/s.

b)F= 16.26 N

Explanation:

Given that

m = 0.261 kg

Initial speed  ,u=- 14.9 m/s

Final speed ,v= 22.5 m/s

We know linear momentum P given as

P = m v

Change in the linear momentum ΔP

ΔP = m Δv

ΔP = 0.261 (22.5 + 14.9 ) kg.m/s

ΔP= 9.761 kg.m/s

Therefore the impulse will be 9.761 kg.m/s.

We know that

force $F=\dfrac{\Delta P}{t}$

$F=\dfrac{9.761}{0.6}$

F= 16.26 N