A 0.145-kg baseball pitched horizontally at 39 m/s strikes a bat and is popped straight up to a height of 31 m. If the contact time between the bat and the ball is 2.35 ms, calculate the average force [exerted by the bat on the ball] during contact. [Let the positive axis lie along the line from the batter to the pitcher, with the batter at the origin.] I'm really having trouble with this problem
To solve this problem, you can use the principle of impulse-momentum to calculate the average force exerted by the bat on the ball during contact.
The principle of impulse-momentum states that the change in momentum of an object is equal to the impulse applied to it. Mathematically, this can be expressed as:
Impulse = Change in Momentum
In this case, the impulse is equal to the average force multiplied by the contact time. The change in momentum can be calculated using the mass of the baseball and its initial and final velocities.
Here are the steps to solve the problem:
1. Find the initial momentum of the baseball:
Momentum = mass × velocity
Given: mass (m) = 0.145 kg
initial velocity (vi) = 39 m/s
Momentum = 0.145 kg × 39 m/s = 5.655 kg·m/s
2. Find the final momentum of the baseball:
Since the baseball is popped straight up, its final velocity (vf) is 0 m/s.
Final momentum = mass × velocity = 0.145 kg × 0 m/s = 0 kg·m/s
3. Calculate the change in momentum:
Change in momentum = Final momentum - Initial momentum
Change in momentum = 0 kg·m/s - 5.655 kg·m/s = -5.655 kg·m/s
4. Calculate the impulse:
Impulse = Change in momentum
Impulse = -5.655 kg·m/s
5. Calculate the average force:
Average force = Impulse / Contact time
Given: contact time = 2.35 ms = 2.35 × 10^-3 s
Average force = (-5.655 kg·m/s) / (2.35 × 10^-3 s)
Average force = -2410.64 N (Note: This value will be negative because the force is exerted in the opposite direction of the positive axis.)
Therefore, the average force exerted by the bat on the ball during contact is approximately -2410.64 N.