Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A \(3.00-\mathrm{kg}\) ball of clay with a speed of \(21.0 \mathrm{~m} / \mathrm{s}\) is thrown against a wall and sticks to the wall. What is the magnitude of the impulse exerted on the ball?

Short Answer

Expert verified
Answer: The magnitude of the impulse exerted on the ball is 63.0 kg m/s.

Step by step solution

01

Identify the given information and write down the formula for impulse

The given information is: - Mass of ball, m = 3.00 kg - Initial velocity, v_initial = 21.0 m/s - Final velocity, v_final = 0 m/s (sticks to the wall) The impulse is calculated using the following formula: Impulse = m * (v_final - v_initial)
02

Calculate the change in velocity

In this step, we'll find the change in velocity as follows: Change in velocity = v_final - v_initial Change in velocity = 0 m/s - 21.0 m/s Change in velocity = -21.0 m/s (since it's just the magnitude, we'll consider the absolute value)
03

Calculate the impulse

Now, using the impulse formula, we'll find the magnitude of the impulse exerted on the ball: Impulse = m * (v_final - v_initial) Impulse = 3.00 kg * |-21.0 m/s| Impulse = 3.00 kg * 21.0 m/s Impulse = 63.0 kg m/s So, the magnitude of the impulse exerted on the ball is 63.0 kg m/s.

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 Vaia!

Key Concepts

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

Conservation of Momentum
Understanding the conservation of momentum is fundamental when dealing with collisions and impulses in physics. The principle articulates that in a closed system with no external forces, the total momentum before an event is equal to the total momentum after the event.

When our aforementioned clay ball strikes the wall, the system consists of the ball and the wall. Since the ball sticks to the wall, we can imagine that, momentarily, they become a single system. If the wall were free to move, the combined system of the wall and clay would move with a momentum equal to that of the clay ball before the collision because momentum is conserved. However, since walls are typically anchored firmly, they don’t move, and thus absorb all the momentum delivered by the ball.

In a theoretical context where the wall could move, if we knew its mass and could measure its post-collision velocity, we could use conservation of momentum to calculate those values, because the momentum before the collision (entirely from the ball) would be equal to the momentum after (shared between the ball and the wall).
Collision
Collisions involve objects coming into contact with each other, which provides a practical depiction of momentum and impulse. In physics, there are two main types of collisions: elastic and inelastic. Elastic collisions are those in which kinetic energy is conserved—objects bounce off each other with no loss of total kinetic energy. Inelastic collisions, in contrast, result in some loss of kinetic energy, and sometimes objects stick together, like the clay ball sticking to the wall.

In our exercise, we are examining a perfectly inelastic collision, where the clay ball doesn't bounce off but rather adheres to the wall. Here, the ball’s final velocity becomes zero—not because it doesn’t move, but because it can't move independently of the wall. The impulse exerted on the clay ball can indicate the magnitude of force during the impact and is a direct result of the change in the ball’s momentum.
Change in Velocity
Velocity can change in two ways: a change in speed, or a change in direction, or both. In the context of our exercise involving the clay ball, the velocity change is exclusively concerning the ball’s speed, which reduces from the initial 21.0 m/s to 0 m/s as it sticks to the wall.

The magnitude of this velocity change is substantial because it signifies a total stop; the ball’s motion ceases in the direction it was initially moving. This abrupt halt is also reflected in the calculation of impulse, which equals the product of the mass and the velocity change. Notice that only the speed's magnitude is considered in the calculation because impulse is a vector quantity and has both magnitude and direction; the negative sign in the change of velocity represents a direction opposite to the initial motion but does not affect the magnitude of the impulse.

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 golf ball of mass 45.0 g moving at a speed of \(120 . \mathrm{km} / \mathrm{h}\) collides head on with a French TGV high-speed train of mass \(3.80 \cdot 10^{5} \mathrm{~kg}\) that is traveling at \(300 . \mathrm{km} / \mathrm{h}\). Assuming that the collision is elastic, what is the speed of the golf ball after the collision? (Do not try to conduct this experiment!)

Stuck in the middle of a frozen pond with only your physics book, you decide to put physics in action and throw the \(5.00-\mathrm{kg}\) book. If your mass is \(62.0 \mathrm{~kg}\) and you throw the book at \(13.0 \mathrm{~m} / \mathrm{s}\), how fast do you then slide across the ice? (Assume the absence of friction.)

Two Sumo wrestlers are involved in an inelastic collision. The first wrestler, Hakurazan, has a mass of \(135 \mathrm{~kg}\) and moves forward along the positive \(x\) -direction at a speed of \(3.50 \mathrm{~m} / \mathrm{s}\). The second wrestler, Toyohibiki, has a mass of \(173 \mathrm{~kg}\) and moves straight toward Hakurazan at a speed of \(3.00 \mathrm{~m} / \mathrm{s} .\) Immediately after the collision, Hakurazan is deflected to his right by \(35.0^{\circ}\) (see the figure). In the collision, \(10.0 \%\) of the wrestlers' initial total kinetic energy is lost. What is the angle at which Toyohibiki is moving immediately after the collision?

The nucleus of radioactive thorium -228 , with a mass of about \(3.78 \cdot 10^{-25} \mathrm{~kg},\) is known to decay by emitting an alpha particle with a mass of about \(6.64 \cdot 10^{-27} \mathrm{~kg} .\) If the alpha particle is emitted with a speed of \(1.80 \cdot 10^{7} \mathrm{~m} / \mathrm{s}\), what is the recoil speed of the remaining nucleus (which is the nucleus of a radon atom)?

Cosmic rays from space that strike Earth contain some charged particles with energies billions of times higher than any that can be produced in the biggest accelerator. One model that was proposed to account for these particles is shown schematically in the figure. Two very strong sources of magnetic fields move toward each other and repeatedly reflect the charged particles trapped between them. These magnetic field sources can be approximated as infinitely heavy walls from which charged particles get reflected elastically. The high- energy particles that strike the Earth would have been reflected a large number of times to attain the observed energies. An analogous case with only a few reflections demonstrates this effect. Suppose a particle has an initial velocity of \(-2.21 \mathrm{~km} / \mathrm{s}\) (moving in the negative \(x\) -direction, to the left), the left wall moves with a velocity of \(1.01 \mathrm{~km} / \mathrm{s}\) to the right, and the right wall moves with a velocity of \(2.51 \mathrm{~km} / \mathrm{s}\) to the left. What is the velocity of the particle after six collisions with the left wall and five collisions with the right wall?

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.

Sign-up for free