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Chapter 7: Problem 74

the ball of mass \(145 \mathrm{g},\) and hits it so that the ball leaves the bat with a speed of \(37 \mathrm{m} / \mathrm{s}\). Assume that the ball is moving horizontally just before and just after the collision with the bat. (a) What is the magnitude of the change in momentum of the ball? (b) What is the impulse delivered to the ball by the bat? (c) If the bat and ball are in contact for \(3.0 \mathrm{ms}\), what is the magnitude of the average force exerted on the ball by the bat?

Short Answer

Expert verified
\(Average\ force = \frac{5.365}{0.003}\) #Solution#: Step 1: The change in momentum of the ball is: \(\Delta momentum = 5.365 kg*m/s\) Step 2: The impulse delivered to the ball is: Impulse = 5.365 kg*m/s Step 3: The average force exerted on the ball is: \(Average\ force \approx 1785 N (rounded\ to\ nearest\ whole\ number)\) Therefore, the average force exerted on the ball by the bat is approximately 1785 N.

Step by step solution

01

Find the change in momentum of the ball

To find the change in momentum, we'll use the formula for momentum: momentum = mass x velocity. We are given the mass of the ball (145 g), which we should convert to kg (1 g = 0.001 kg): \(mass = 145 * 0.001 = 0.145 kg\) The initial velocity of the ball is 0 m/s, and we are given the final velocity (37 m/s). So the change in momentum can be calculated by subtracting the initial momentum from the final momentum: \(\Delta momentum = (mass * final\ velocity) - (mass * initial\ velocity)\) \(\Delta momentum = (0.145 * 37) - (0.145 * 0)\)
02

Calculate the impulse delivered to the ball

The impulse on an object is the product of the force and the time interval during which the force is applied. In this case, the impulse is equal to the change in momentum of the ball. So, we can simply use our result from Step 1 for the impulse: Impulse = \(\Delta momentum\)
03

Find the average force exerted on the ball

To find the average force, we will divide the impulse by the contact time: \(Average\ force = \frac{Impulse}{Contact\ time}\) We are given the contact time of 3 ms (milliseconds), which we should convert to seconds: \(Contact\ time = 3.0 * 10^{-3}s\) Now we can plug in the values and find the average force: \htdocscalculation{}{ mass = 0.145, final_velocity = 37, initial_velocity = 0, delta_momentum = mass * (final_velocity - initial_velocity), contact_time = 3e-3, average_force = delta_momentum / contact_time

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