A 500.0-g baseball moves to the left striking a bat at 20.0 m/s. The bat is in contact with the ball for 0.002 s, and it leaves in the opposite direction at 40.0 m/s. (a) Solve for the momentum of the ball before and after it was hit by the bat. (b) What was the average force on the bat? (c) What was its impulse?
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Given: m = 500.0 g = 0.5 kg (mass of the baseball) v1 = -20.0 m/s (velocity of the baseball before hitting the bat) v2 = 40.0 m/s (velocity of the baseball after hitting the bat) t = 0.002 s (time of contact between the bat and the baseball)
(a) The momentum of the ball before it was hit by the bat can be calculated using the formula p = mv, where p is the momentum, m is the mass, and v is the velocity. Therefore, p = (0.5 kg)(-20.0 m/s) = -10.0 kg m/s (Note: the negative sign indicates that the ball is moving to the left).
The momentum of the ball after it was hit by the bat can also be calculated using the same formula. Therefore, p = (0.5 kg)(40.0 m/s) = 20.0 kg m/s (Note: the positive sign indicates that the ball is moving to the right).
(b) The average force on the bat can be calculated using the formula F = Δp/Δt, where F is the force, Δp is the change in momentum, and Δt is the time interval. Since the ball is in contact with the bat for 0.002 s, we can use this as the time interval. Therefore, Δp = 20.0 kg m/s - (-10.0 kg m/s) = 30.0 kg m/s, and Δt = 0.002 s. Substituting these values into the formula, we get:
F = 30.0 kg m/s / 0.002 s = 15,000 N
Therefore, the average force on the bat is 15,000 N.
(c) The impulse can be calculated using the formula J = Δp, where J is the impulse and Δp is the change in momentum. Therefore, J = 20.0 kg m/s - (-10.0 kg m/s) = 30.0 kg m/s.
Therefore, the impulse is 30.0 kg m/s.