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A baseball has mass \(0.145 \mathrm{~kg}\). (a) If the velocity of a pitched ball has a magnitude of \(45.0 \mathrm{~m} / \mathrm{s}\) and the batted ball's velocity is \(55.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for \(2.00 \mathrm{~ms}\), find the magnitude of the average force applied by the bat.

Short Answer

Expert verified
The magnitude of the change in momentum of the ball and of the impulse applied to it by the bat is 14.5 kg·m/s, and the magnitude of the average force applied by the bat is 7.25 kN.

Step by step solution

01

Calculate initial and final momentum

First, calculate the ball's initial and final momentum. The initial momentum (before the ball gets hit by the bat) is the product of the ball's mass and its velocity. The final momentum (after the ball gets hit by the bat) is the product of the ball's mass and its final velocity. Be careful with the direction: the velocities before and after the collision have opposite directions, which we can express by giving one of them a negative sign. The initial momentum therefore is \(0.145 \mathrm{~kg} \times 45.0 \mathrm{~m/s} = 6.525 \mathrm{~kg·m/s}\), and the final momentum is \(0.145 \mathrm{~kg} \times -55.0 \mathrm{~m/s} = -7.975 \mathrm{~kg·m/s}\).
02

Calculate change in momentum and impulse

Next, calculate the change in momentum, which is simply the final momentum minus the initial momentum: -7.975 kg·m/s - 6.525 kg·m/s = -14.5 kg·m/s. The magnitude of this change in momentum is 14.5 kg·m/s. This is also the magnitude of the impulse applied to the ball by the bat.
03

Calculate average force

The impulse applied to the ball, which equals the change in momentum calculated in the previous step, also equals the average force applied by the bat times the time interval: \(Impulse = F_{avg} \times \Delta t\). Rearrange this equation to solve for the average force: \(F_{avg} = \frac{Impulse}{\Delta t} = \frac{14.5 \mathrm{~kg·m/s}}{2.00 \mathrm{~ms}} = 7.25 \mathrm{~kN}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Impulse
Impulse in physics represents the change in momentum of an object when it is exposed to a force over a period of time. The interplay between impulse and momentum is crucial in understanding events such as collisions or the act of catching a ball.

In the given exercise, the bat delivers an impulse to the baseball, resulting in a change in the baseball's motion. The formula representing this relationship is \( Impulse = \text{change in momentum} \), and it can be calculated by the difference between the final and initial momentum.

To illuminate this further, consider the baseball's situation before and after being hit by the bat. Before the bat strikes, the ball has a certain velocity and direction. After the impact, not only does the velocity change, but also the direction, indicating a negative sign in the final velocity. This significant shift in velocity—and, by extension, in momentum—epitomizes the impulse provided by the bat.
Average Force
The average force is the force exerted averaged over the time period during which the force is applied. It's essential in scenarios where the force isn't constant but varies with time. Understanding average force helps us describe the effects of force actions over finite times, such as the collision between a bat and a ball.

In our baseball scenario, the batter applies a force to the ball for a very brief contact time, and we're interested in determining the average force throughout this period. To extract the average force from the exercise, the impulse (which equates to the change in momentum) is divided by the time interval during which the impulse is applied, as in \( F_{avg} = \frac{Impulse}{\text{\Delta t}} \).

Real-world Examples

To put it in context, consider a hammer striking a nail: the average force exerted by the hammer during its impact moves the nail. Similarly, a soccer player kicking a ball applies an average force, transferring energy to the ball.
Conservation of Momentum
The conservation of momentum is a fundamental principle in physics stating that if no external forces are acting on a system, the total momentum of that system remains constant. This concept is a cornerstone in analyzing collisions and interactions between objects.

In the context of the exercise, while the bat and ball interact, momentum is conserved. However, it's essential to recognize that within the system of the ball and bat, external forces, such as gravity and air resistance, are minimized for the duration of the contact. Therefore, for this very brief instance, we can consider the system isolated, allowing us to apply the principle of conservation of momentum to determine outcomes of the impact.

A practical application outside textbook exercises is in vehicle collision analysis, where the conservation of momentum aids investigators in reconstructing the event, assuming negligible external forces during the collision.

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Most popular questions from this chapter

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{~kg}\), is moving to the right at \(4.00 \mathrm{~m} / \mathrm{s}\), while the other, of mass \(65.0 \mathrm{~kg}\). is moving to the left at \(2.50 \mathrm{~m} / \mathrm{s}\). What are the magnitude and direction of the velocity of these skaters just after they collide?

The mass of a regulation tennis ball is 57 g (although it can vary slighaly), and tests have shown that the ball is in conthet with the tennis racket for \(30 \mathrm{~ms}\). (This number can also vary, depending on the racket and swing.) We shall assume a \(30.0 \mathrm{~ms}\) contact time. One of the fastest-known served tennis balls was served by "Big Bill" Tilden in \(1931,\) and its speed was measured to be \(73 \mathrm{~m} / \mathrm{s}\). (a) What impulse and what total force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponcnt returned his serve with a speed of \(55 \mathrm{~m} / \mathrm{s},\) what total force and what impulse did he exert on the ball, assuming only horizontal motion?

Two sticky spheres are suspended from light ropes of length \(L\) that are attached to the ceiling at a common point. Sphere \(A\) has mass \(2 m\) and is hanging at rest with its rope vertical. Sphere \(B\) has mass \(m\) and is held so that its rope makes an angle with the vertical that puts \(B\) a vertical height \(H\) above \(A\). Sphere \(B\) is released from rest and swings down, collides with sphere \(A,\) and sticks to it. In terms of \(H,\) what is the maximum height above the original position of \(A\) reached by the combined spheres after their collision?

An ice hockey forward with mass \(70.0 \mathrm{~kg}\) is skating due north with a speed of \(5.5 \mathrm{~m} / \mathrm{s}\). As the forward approaches the net for a slap shot, a defensive player (muss \(110 \mathrm{~kg}\) ) skates toward him in order to apply a body-check. The defensive player is traveling south at \(4.0 \mathrm{~m} / \mathrm{s}\) just before they collide. If the two players become intertwined and move together after they collide, in what direction and at what speed do they move after the collision? Friction between the two players and the ice can be neglected.

A10.0g marble slides to the left at a speed of \(0.400 \mathrm{~m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, clastic collision with a larger \(30.0 \mathrm{~g}\) marble sliding to the right at speed of \(0.200 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{E} 8.48\) ). (a) Find the velocity of each marble (magnitude and dircction) after the collision. (since the collision is head-on, all motion is along a line. (b) Calculate the change in momentum (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the change in kinetic energy (the kinctic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

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