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A 1.5 kg baseball is pitched at 38 m/s and is hit by the batter. It heads directly back at the pitcher at 27 m/s.

a) Find the impulse applied to the ball.

b) If the bat is in contact with the bat for 0.45 sec, how much force was applied.

Show all your work.

Sagot :

zaniahsims

Answer:

The initial velocity of the ball is +38.0 m/s. The final velocity is -27.0 m/s.  

momentum = mass x velocity

(mass x final velocity) – (mass x initial velocity)

(1.5)(-27) – (1.5)(38) = -97.5

The impulse applied to the ball is -97.5 kg m/s

impulse = force x time interval

force = impulse/time interval

-97.5/0.45 s = - 216.6  

The force applied was 216.6 N.

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