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Chapter 6: Problem 18

A \(3.00\) -kg steel ball strikes a massive wall at \(10.0 \mathrm{~m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) with the plane of the wall. It bounces off the wall with the same speed and angle (Fig. \(\mathrm{P} 6.18\) ). If the ball is in contact with the wall for \(0.200 \mathrm{~s}\), what is the average force exerted by the wall on the ball?

Short Answer

Expert verified
The average force exerted by the wall on the ball can be calculated as approximately 270 Newtons.

Step by step solution

01

Determine the Initial and Final Momentum of the Ball

The initial momentum of the ball can be found using the equation \( P_{initial} = mv \), where \( m = 3.00 kg \) and \( v = 10 m/s \). We must also break down this momentum into components as the ball is striking at an angle of 60 degrees. Similarly, the final momentum of the ball is also calculated using the same equation and breaking down into components. The key difference is the direction of the ball, which is the opposite for the final momentum components.
02

Calculate the Change in Momentum

To find the change in momentum, we subtract the initial momentum from the final momentum for both components. The resulting difference gives us the change in momentum in both horizontal and vertical direction.
03

Determine the Average Force Exerted by the Wall

We can use the change in momentum and the time of contact to find the average force exerted by the wall on the ball. Using Newton's second law in the form \( F = \Delta p / \Delta t \), where \( \Delta p \) is the change in momentum and \( \Delta t = 0.200 s \) is the time of contact, we find the average force in both directions. Add them up using Pythagorean theorem to get the total force.

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Key Concepts

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

Newton's Second Law
Newton's Second Law is one of the cornerstones of classical mechanics and is crucial in understanding how forces influence objects in motion. It can be stated as, "The force acting on an object is equal to the mass of that object times its acceleration," which is represented by the formula \( F = ma \). However, an alternative and powerful perspective is to consider it in terms of momentum, with the equation \( F = \frac{\Delta p}{\Delta t} \). Here, \( \Delta p \) represents the change in momentum, and \( \Delta t \) denotes the change in time over which this change occurs.

In our given problem, Newton's Second Law is applied to calculate the average force exerted on the ball when it strikes the wall and bounces back. By knowing the change in momentum and the time of contact, we can determine this force. This approach shows how Newton's Law effectively connects momentum with force, providing a comprehensive understanding of interactions between objects.
Components of Momentum
Momentum is a vector quantity, which means it has both magnitude and direction. When solving problems involving angles, it is essential to consider the components of momentum in each direction. In the scenario where our steel ball strikes the wall at an angle, we must resolve the momentum into horizontal and vertical components.

To find these components, we use trigonometric functions:
  • The horizontal component of momentum is calculated as \( P_x = mv \cos \theta \), where \( \theta \) is the angle of impact with respect to the wall.
  • The vertical component is \( P_y = mv \sin \theta \).
Both initial and final momentum of the ball need to be broken down into these components. When the ball bounces back, the direction of the horizontal component changes, which significantly impacts the overall momentum change.

Understanding these components is crucial for calculating the change in momentum accurately, especially in systems where direction plays a vital role.
Average Force
Average force provides an understanding of the overall effect of a force that acts upon an object over a period of time. It is especially useful in situations involving a short duration of interaction, like the collision of the ball with the wall. The formula for average force is given by \( F_{avg} = \frac{\Delta p}{\Delta t} \), where \( \Delta p \) is the change in momentum and \( \Delta t \) is the time interval during which the force acts.

In the exercise, the change in momentum arises from the ball reversing direction after hitting the wall. By understanding the components of momentum as discussed earlier, we can compute the change for both horizontal and vertical values. Then, considering \( \Delta t = 0.200 \) seconds, the average force exerted by the wall can be found.
  • Calculate the individual forces in both directions using the formula.
  • Combine them using the Pythagorean theorem, as they are perpendicular components, to find the total average force.
The concept of average force is fundamental to identifying how much and in what way the wall affects the ball during the brief collision.

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Most popular questions from this chapter

Identical twins, each with mass \(55.0 \mathrm{~kg}\), are on ice skates and at rest on a frozen lake, which may be taken as frictionless. Twin A is carrying a backpack of mass \(12.0 \mathrm{~kg}\). She throws it horizontally at \(3.00 \mathrm{~m} / \mathrm{s}\) to Twin \(\mathrm{B}\). Neglecting any gravity effects, what are the subsequent speeds of Twin A and Twin B?

A cue ball traveling at \(4.00 \mathrm{~m} / \mathrm{s}\) makes a glancing, elastic collision with a target ball of equal mass that is initially at rest. The cue ball is deflected so that it makes an angle of \(30.0^{\circ}\) with its original direction of travel. Find (a) the angle between the velocity vectors of the two balls after the collision and (b) the speed of each ball after the collision.

GP A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\) -direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second piece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\) -plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(m_{\mathrm{g}}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\) - and \(y\) -directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, v_{2}\), and \(v_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\) -components of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\) -components of the momenta of \(m_{1}\) and \(m_{2} .\) (e) Substitute the known momentum components into the general equations of momentum for the \(x\) - and \(y\) -directions, along with the known mass \(m_{3} .\) (f) Solve the two momentum equations for \(v_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(v_{3}\). (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain \(\tan \theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

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 \(1.6 \times 10^{-13} \mathrm{~J}\), find its final kinetic energy and the kinetic energy of the carbon nucleus after the collision.

ecp Measuring the speed of a bullet. A bullet of mass \(m\) is fired horizontally into a wooden block of mass \(M\) lying on a table. The bullet remains in the block after the collision. The coefficient of friction between the block and table is \(\mu\), and the block slides a distance \(d\) before stopping. Find the initial speed \(v_{0}\) of the bullet in terms of \(M\), \(m, \mu, g\), and \(d\).
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