Question

A 0.280-kg volleyball approaches a player horizontally with a speed of 15.0 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.0 m/s.
(a) What impulse is delivered to the ball by the player?
(b) If the player’s fist is in contact with the ball for 0.060 0 s, find the magnitude of the average force exerted on the player’s fist.

Answer 1

Answer:

Impulse = 10.36 kg m/s

average force = 172.667 N

Explanation:

given data

mass = 0.280 kg

speed = 15.0 m/s

speed = 22.0 m/s

to find out

impulse and magnitude of the average force

solution

we know that Impulse is change in momentum that is

initial momentum = mass × speed    ..........1

initial momentum = 0.28 × (15)

initial momentum = 4.2 kg m/s

Final momentum = mass × speed     ..........2

Final momentum = 0.28 × (-22)

Final momentum = -6.16 kg m/s

so now we get Impulse that is

Impulse = 4.2 - (-6.16)

Impulse = 10.36 kg m/s

and

average force will be

average force = impulse ÷ time

average force = $\frac{10.36}{0.060}$

average force = 172.667 N