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

Answer:

(A) Impulse will be 1.116 kgm/sec

(B) Force will be equal to 1116 N

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 $=m(v_2-v_1)=0.120\times (49.9-40.6)=1.116kgm/sec$

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 $Impluse=force\times time$

$1.116=force\times10^{-3}$

F = 1116 N