Question

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?

Answer 1

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