Question

A 0.120 kg baseball, thrown with a speed of 40.6 m/s, is hit straight back at the pitcher with a speed of 49.9 m/s.a) What is the impulse delivered to the baseball? b) Find the average force exerted by the bat on the ball if the two are in contact for 1x10-3 s.

Answer 1

(A) Impulse will be 1.116 kgm/sec

(B) Force will be equal to 1116 N

Step-by-step explanation:

We have given mass of the basketball m = 0.120 kg

Initial speed of the ball
`v_1=40.6m/sec`

Final speed of the ball
`v_2=49.9m/sec`

(A) Impulse delivered by the ball is equal to the change in momentum

So impulse will be equal to
`=m(v_2-v_1)=0.120* (49.9-40.6)=1.116kgm/sec`

So impulse will be 1.116 kgm/sec

(B) Time is given for which force is exerted
`=10^(-3)sec`

We know that impulse is equal to
`Impluse=force* time`

`1.116=force*10^(-3)`

F = 1116 N