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A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 21.6 m/s. (a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.) kg · m/s (b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist. N

Respuesta :

skyluke89

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

[tex]I= \Delta p = m \Delta v = m (v-u)[/tex]

where

m is the mass of the object

[tex]\Delta v[/tex] is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

[tex]I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s[/tex]

(b) 155 N

The impulse can also be rewritten as

[tex]I=F \Delta t[/tex]

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

[tex]\Delta t[/tex] is the duration of the collision

In this situation, we have

[tex]\Delta t = 0.06 s[/tex]

So we can re-arrange the equation to find the magnitude of the average force:

[tex]F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N[/tex]

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