A baseball with a mass of 153 g is thrown horizontally with a speed of 40.7 m/s (91 mi/h) at a bat. The ball is in contact with the bat for 1.00 ms and then travels straight back at a speed of 45.8 m/s (102 mi/h). Determine the average force exerted on the ball by the bat. Neglect the weight of the ball (it is much smaller than the force of the bat) and choose the direction of the incoming ball to be positive. (Indicate the direction with the sign of your answer.)
Change in momentum = M(V1 - V2)
where
M = mass of the ball = 153 grams (given) = 0.153 kg
V1 = initial velocity of the ball = 40.7 m/sec. (given)
V2 = final velocity of ball = 45.8 m/sec. (given)
Substituting values,
Impulse = 0.153(40.7 - -45.8)
NOTE the negative sign attached to the velocity of the ball after
being in contact with the bat (-45.8). This simply indicates that
the direction of the ball is opposite that of its initial
direction.
Therefore,
Impulse = 13.2345 kg - m/sec.
Since
Impulse = Force (delta T), then the force can be calculated
by
where
delta T = time ball was in contact with the bat = 1.0 ms
The force is then calculated by
Force = F = Impulse/(delta T)
F = 13.2345/0.001
F = 13234.5 N
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