/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 W A S.00-kg steel ball strikes... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

W A S.00-kg steel ball strikes a massive wall at \(10.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta=60.0^{\circ}\) with the plane of the wall. It bounces off the wall with the same speed and angle (Fig. P6.18). If the ball is in con- tact with the wall for \(0.200 \mathrm{~s}\), what is the average force exerted by the wall on the ball? the ball?

Short Answer

Expert verified
The average force exerted by the wall on the ball is -100 N. This indicates the force is directed opposite to the incoming direction of the ball.

Step by step solution

01

Calculate the momentum before and after the impact

To solve this problem, first determine the initial and final momentum of the ball in the direction perpendicular to the wall. The initial and final speeds are the same but in opposite directions, signifying a change in the direction of motion. The speed given is the total speed, so the horizontal component of the speed need to found to calculate the momentum along the plane of the wall. To find this, use the equation \(v_x= v \times \cos(\theta)\) where v is the total speed and \( \theta \) is the angle of incidence. This gives: \(v_x = 10.0 m/s \times \cos(60^{\circ}) = 5.0 m/s\).
02

Compute the change in momentum

Next, calculate the change in momentum. Momentum (p) is given by mass (m) times velocity (v). Due to the sign change in direction, the final momentum will be -mv and the initial momentum will be mv. Therefore change in momentum (\( \Delta p\)) is the difference between the two, giving: \( \Delta p = m \times \Delta v = m \times (v_{final} - v_{initial}) = 2.00kg \times (-5.0 m/s -5.0 m/s) = -20 kg m/s \)
03

Calculate the force exerted by the wall

The force exerted on the ball by the wall is the rate of change of momentum. This can be calculated by dividing the change in momentum (\( \Delta p\)) by the time (t) it changes, using the equation: \( F = \Delta p / t \) . By substituting the given time t = 0.200 s in to the equation you get: \( F = -20 kg m/s / 0.200 s = -100 N \) . The negative sign indicates the direction of the force is opposing the direction of motion.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Momentum Conservation
In physics, when we speak of momentum conservation, we're talking about one of the fundamental principles that states that the total momentum of a closed system of objects (meaning a system not subject to external forces) does not change over time.

Consider the scenario of a steel ball striking a wall. The wall is massive, which we can interpret as having a much larger mass compared to the ball. In this context, we can think of the wall-ball system as being relatively closed since the wall's massive size means its own momentum change can be neglected. The ball's momentum just before impact and just after bouncing back should, therefore, be equal in magnitude but opposite in direction because only the direction of the ball reverses upon collision; its speed remains unchanged as the problem states.

The preservation of momentum in this case helps us understand the dynamics of the collision process. Including an illustration showing the velocity vectors before and after the impact could significantly aid in visualizing how momentum is conserved, especially since the ball maintains the same angle with respect to the wall before and after the bounce.
Impulse-Momentum Theorem
The impulse-momentum theorem is a crucial concept in understanding collisions. In essence, it ties an object's force history—the impulse—to its momentum change. The theorem states that the impulse exerted on an object is equal to the change in momentum of the object.

In the given step-by-step solution, the average force exerted by the wall can be connected to the concept of impulse. The product of the average force (F) applied over the time interval (0.200 s) during which the force acts is the impulse. This impulse is what changes the momentum of the ball. Written mathematically, impulse (J) equals force (F) times time (t), which equals the change in momentum (timestimes):
\[ J = F \times t \]\[ J = \Delta p \]
By calculating the impulse, and knowing the time of contact, we can easily find the average force. This provides a direct application of the impulse-momentum theorem and could be made clear with a diagram or graph depicting the force over time, showcasing how the area under such a graph (impulse) relates to the change in momentum.
Elastic Collision
An elastic collision is an encounter between two bodies where they collide with no net loss in kinetic energy. In the case of our exercise, even though the problem doesn't explicitly state the nature of the collision, we can infer some characteristics of an elastic collision because the ball bounces back with the same speed.

In an ideally elastic collision, not only is momentum conserved, but so is kinetic energy. When the ball hits the wall, it does so at an angle, which means we have to consider both the perpendicular and parallel components of its velocity. However, since the wall is vertical, only the perpendicular component of velocity (and thus momentum) changes.

Presenting an in-depth breakdown of what happens to both the kinetic energy and momentum in the perpendicular and parallel directions during such a collision would enhance understanding. An animation or a sequence of simplified diagrams that contrasts elastic with inelastic collisions, highlighting the conservation of kinetic energy, could greatly assist learners.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A rifle with a weight of \(30 \mathrm{~N}\) fires a \(5.0\)-g bullet with a speed of \(300 \mathrm{~m} / \mathrm{s}\). (a) Find the recoil speed of the rifle. (b) If a \(700-\mathrm{N}\) man holds the rifle firmly against his shoulder, find the recoil speed of the man and rifle.

Most of us know intuitively that in a head-on collision between a large dump truck and a subcompact car, you are better off being in the truck than in the car. Why is this? Many people imagine that the collision force exerted on the car is much greater than that exerted on the truck. To substantiate this view, they point out that the car is crushed, whereas the truck is only dented. This idea of unequal forces, of course, is false; Newton's third law tells us that both objects are acted upon by forces of the same magnitude. The truck suffers less damage because it is made of stronger metal. But what about the two drivers? Do they experience the same forces? To answer this question, suppose that each vehicle is initially moving at \(8.00 \mathrm{~m} / \mathrm{s}\) and that they undergo a perfectly inelastic head-on collision. Each driver has mass \(80.0 \mathrm{~kg}\). Inchuding the masses of the drivers, the total masses of the vehicles are \(800 \mathrm{~kg}\) for the car and \(4000 \mathrm{~kg}\) for the truck. If the collision time is \(0.120 \mathrm{~s}\), what force does the seat belt exert on each driver?

A neutron in a reactor makes an elastic head-on collision with a carbon atom that is initially at rest. (The mass of the carbon nucleus is about 12 times that of the neutron.) (a) What fraction of the neutron’s kinetic energy is transferred to the carbon nucleus? (b) If the neutron’s initial kinetic energy is 1.6 3 10213 J, find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

A man of mass \(m_{1}=70.0 \mathrm{~kg}\) is skating at \(v_{1}=\) \(8.00 \mathrm{~m} / \mathrm{s}\) behind his wife of mass \(\mathrm{m}_{2}=50.0 \mathrm{~kg}\), who is skating at \(\tau_{2}=4.00 \mathrm{~m} / \mathrm{s}\). Instead of passing her, he inadvertently collides with her. He grabs her around the waist, and they maintain their balance. (a) Sketch the problem with before-and-after diagrams, representing the skaters as blocks. (b) Is the collision best described as elastic, inelastic, or perfectly inelastic? Why? (c) Write the general equation for conservation of momentum in terms of \(m_{1}, v_{1}, w_{2}, v_{2}\), and final velocity \(v\) - (d) Solve the momentum equation for \(v_{\gamma}\). (e) Substitute values, obtaining the numerical value for \(v_{f}\), their speed after the collision.

A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to \(5.20 \mathrm{~m} / \mathrm{s}\) in \(0.832 \mathrm{~s}\). What are (a) the magnitudes of the linear impulse and (b) the average total force experienced by a \(70.0-\mathrm{kg}\) passenger in the car during the time the car accelerates?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.