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Chapter 9: Problem 11

Kinetic Energy and Impulse A ball having a mass of \(150 \mathrm{~g}\) strikes a wall with a speed of \(5.2 \mathrm{~m} / \mathrm{s}\) and rebounds straight back with only \(50 \%\) of its initial kinetic energy. (a) What is the speed of the ball immediately after rebounding? (b) What is the magnitude of the impulse on the wall from the ball? (c) If the ball was in contact with the wall for \(7.6 \mathrm{~ms}\), what was the magnitude of the average force on the ball from the wall during this time interval?

Short Answer

Expert verified
The speed after rebounding is 3.68 m/s. The magnitude of impulse is 1.332 Ns. The average force is 175.26 N.

Step by step solution

01

- Convert units

Convert the mass of the ball from grams to kilograms: \(150 \text{ g} = 0.150 \text{ kg}\).
02

- Calculate initial kinetic energy

Use the kinetic energy formula to calculate the initial kinetic energy: \(KE_initial = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.150 \text{ kg} \times (5.2 \text{ m/s})^2 = 2.028 \text{ J}\).
03

- Calculate final kinetic energy

Since the ball retains only \(50\%\) of its initial kinetic energy after rebounding, the final kinetic energy is: \(KE_final = 0.50 \times 2.028 \text{ J} = 1.014 \text{ J}\).
04

- Calculate the final speed

Use the kinetic energy formula inversely to find the final speed: \( KE_final = \frac{1}{2} m v_f^2 \). Rearrange to solve for \(v_f\): \( v_f = \sqrt{\frac{2 KE_final}{m}} = \sqrt{\frac{2 \times 1.014 \text{ J}}{0.150 \text{ kg}}} = 3.68 \text{ m/s}\).
05

- Calculate the change in velocity

The initial velocity is \(5.2 \text{ m/s}\) towards the wall, and the final velocity is \(-3.68 \text{ m/s}\) away from the wall. The change in velocity \( \Delta v\) is: \( \Delta v = v_f - v_i = -3.68 \text{ m/s} - 5.2 \text{ m/s} = -8.88 \text{ m/s}\).
06

- Calculate the impulse

Impulse is the change in momentum: \( J = m \Delta v = 0.150 \text{ kg} \times (-8.88 \text{ m/s}) = -1.332 \text{ Ns}\). The magnitude of the impulse is: \( |J| = 1.332 \text{ Ns}\).
07

- Calculate the average force

The average force is the impulse divided by the time of contact. Convert the time from milliseconds to seconds: \( 7.6 \text{ ms} = 0.0076 \text{ s} \). Then find the force: \( F_{avg} = \frac{|J|}{\Delta t} = \frac{1.332 \text{ Ns}}{0.0076 \text{ s}} = 175.26 \text{ N}\).

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Key Concepts

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

kinetic energy
Kinetic energy is the energy that an object possesses due to its motion. It can be calculated using the formula: \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In our exercise, the ball initially has a kinetic energy of 2.028 J before hitting the wall. After rebounding, its kinetic energy is reduced to 50%, so the final kinetic energy is 1.014 J. To find the final velocity, we rearrange the kinetic energy formula to solve for velocity: \( v_f = \sqrt{\frac{2 KE_{final}}{m}} \). This calculation shows the relationship between an object's mass, velocity, and kinetic energy during motion.
impulse
Impulse measures the change in momentum caused by a force applied over a period of time. It is calculated using the formula: \( J = \Delta p = m \Delta v \), where \( \Delta p \) is the change in momentum, \( m \) is mass, and \( \Delta v \) is the change in velocity. In the given problem, the ball's change in velocity is -8.88 m/s (considering it rebounded back). Hence, the impulse imparted to the wall is calculated to be -1.332 Ns in direction. Since impulse is a vector quantity, its magnitude is \( |J| = 1.332 \text{ Ns} \). Impulse can explain how the abrupt collision changes the ball's momentum.
momentum
Momentum is the product of an object's mass and its velocity, given by the equation: \( p = mv \). For the ball in this exercise, its initial momentum before hitting the wall is \( p = 0.150 \text{ kg} \times 5.2 \text{ m/s} = 0.78 \text{ kg} \cdot \text{ m/s} \). Upon rebounding, with a velocity of -3.68 m/s, the momentum changes. The total change in momentum provides the impulse. Understanding momentum helps in predicting the effects of collisions and how objects will move after such interactions.
contact force
A contact force is exerted when two objects interact physically. In the exercise, the ball exerts a contact force on the wall during the collision. The average force can be found using the formula: \( F_{avg} = \frac{|J|}{\Delta t} \), where \( |J| \) is the magnitude of the impulse and \( \Delta t \) is the time of contact. The ball's impulse was 1.332 Ns, and the contact lasted for 0.0076 seconds. Hence, the average force is \( F_{avg} = 175.26 \text{ N} \). This calculation illustrates how the force was distributed over the short contact period, demonstrating the principles of contact mechanics and force interactions.

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Most popular questions from this chapter

Transporting Boxes Boxes are transported from one location to another in a warehouse by means of a conveyor belt that moves with a constant speed of \(0.50 \mathrm{~m} / \mathrm{s}\). At a certain location the conveyor belt moves for \(2.0 \mathrm{~m}\) up an incline that makes an angle of \(10^{\circ}\) with the horizontal, then for \(2.0 \mathrm{~m}\) horizontally, and finally for \(2.0 \mathrm{~m}\) down an incline that makes an angle of \(10^{\circ}\) with the horizontal. Assume that a \(2.0 \mathrm{~kg}\) box rides on the belt without slipping. At what rate is the force of the conveyor belt doing work on the box (a) as the box moves up the \(10^{\circ}\) incline, (b) as the box moves horizontally, and (c) as the box moves down the \(10^{\circ}\) incline?

Luge Rider A luge and rider, with a total mass of \(85 \mathrm{~kg}\), emerge from a downhill track onto a horizontal straight track with an initial speed of \(37 \mathrm{~m} / \mathrm{s}\). If they slow at a constant rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2}\) (a) what magnitude \(F\) is required for the slowing force, (b) what distance \(d\) do they travel while slowing, and (c) what work \(W\) is done on them by the slowing force? What are (d) \(F,(\mathrm{e}) d\), and (f) \(W\) if the luge and the rider slow at a rate of \(4.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

A Sprinter A sprinter who weighs \(670 \mathrm{~N}\) runs the first \(7.0 \mathrm{~m}\) of a race in \(1.6 \mathrm{~s}\), starting from rest and accelerating uniformly. What are the sprinter's (a) speed and (b) kinetic energy at the end of the \(1.6 \mathrm{~s} ?\) (c) What average power does the sprinter generate during the 1.6 s interval?

Freight Elevator A fully loaded, slow-moving freight elevator has a cab with a total mass of \(1200 \mathrm{~kg}\), which is required to travel upward \(54 \mathrm{~m}\) in \(3.0 \mathrm{~min}\), starting and ending at rest. The elevator's counterweight has a mass of only \(950 \mathrm{~kg}\), so the elevator motor must help pull the cab upward. What average power is required of the force the motor exerts on the cab via the cable?

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