Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.

A 0.263-kg volleyball approaches a player horizontally with a speed of 17.4 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.8 m/s.

Required:
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.

Sagot :

Answer:

(A) J = -10.57 kg-m/s   (B) 17.61 N

Explanation:

Given that,

Mass of a volleyball, m = 0.263 kg

Initial speed of volleyball, u = 17.4 m/s

Final speed of volleyball, v = -22.8 (in opposite direction)

(a) We need to find the impulse delivered to the ball by the player.

Impulse = change in momentum

J = m(v-u)

Put all the values,

J = 0.263(-22.8-17.4) kg-m/s

= -10.57 kg-m/s

(b) The time of contact with the ball, t = 0.6 s

We need to find the magnitude of the average force exerted on the players first.

Impulse, J = Ft

[tex]F=\dfrac{J}{t}\\\\F=\dfrac{10.57 }{0.6}\\\\F=17.61\ N[/tex]

So, the magnitude of the average force exerted on the player is 17.61 N.

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for choosing Westonci.ca as your information source. We look forward to your next visit.