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Jennifer hits a stationary 0.04-kg ball, and it leaves her racket at 41.3 m/s. Time-lapse photography shows that the ball was in contact with the racket for 26.5 ms. What average force, in N, did the ball
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Jennifer hits a stationary 0.04-kg ball, and it leaves her racket at 41.3 m/s. Time-lapse photography shows that the ball was in contact with the racket for 26.5 ms. What average force, in N, did the ball exert on the racket?

User Simon
by
3.9k points

1 Answer

5 votes

Answer:

62.33N

Step-by-step explanation:

We know that the relationship for the impulse and momentum is given as

Ft=mv

given data

mass m=0.04kg

t= 26.5ms= 26.5/1000= 0.0265seconds

velocity v= 41.3m/s

substituting in the expression we have

Ft=mv

make F subject of formula we have

F=mv/t

F=(0.04*41.3)/0.0265

F=1.652/0.0265

F=62.33N

User Ian McGrath
by
5.1k points
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