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57.0- g tennis ball is traveling straight at a player at 21.0 m/ s. The player volleys the ball straight back at 25.0 m/ s. If the ball remains in contact with the racket for 0.060 0 s, what average force acts on the ball?

Respuesta :

Answer:43.7 N

Explanation:

Given

Mass of tennis ball is =57 gm

velocity of ball=21 m/s

The player volleys the ball straight back at 25 m/s(i.e. in opposite of initial direction)

time of contact=0.06 s

We know change in momentum[tex](\Delta P) =Impulse(Fdt)[/tex]

initial momentum[tex]=57\times 10^{-3}\times 21[/tex]

final momentum[tex]=-57\times 10^{-3}\times 25[/tex]

change in momentum

[tex]\Delta P=57\times 10^{-3}\times 21-(-57\times 10^{-3}\times 25)[/tex]

[tex]\Delta P=57\times 10^{-3}\times 46=2.622 kg-m/s[/tex]

Now 2.622[tex]=F_{avg}\times 0.06[/tex]

[tex]F_{avg}=43.7 N[/tex]

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