contestada

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.

Respuesta :

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.