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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.

Respuesta :

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 [tex]v_1=40.6m/sec[/tex]

Final speed of the ball [tex]v_2=49.9m/sec[/tex]

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

So impulse will be equal to [tex]=m(v_2-v_1)=0.120\times (49.9-40.6)=1.116kgm/sec[/tex]

So impulse will be 1.116 kgm/sec

(B) Time is given for which force is exerted [tex]=10^{-3}sec[/tex]

We know that impulse is equal to [tex]Impluse=force\times time[/tex]

[tex]1.116=force\times10^{-3}[/tex]

F = 1116 N

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