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Chapter 10: Problem 44

Bullet Hits Wall A \(30 \mathrm{~g}\) bullet, with a horizontal velocity of \(500 \mathrm{~m} / \mathrm{s}\), comes to a stop \(12 \mathrm{~cm}\) within a solid wall. (a) What is the change in its mechanical energy? (b) What is the magnitude of the average force from the wall stopping it?

Short Answer

Expert verified
(a) The change in mechanical energy is \(3750 \text{ J}\). (b) The magnitude of the average force from the wall stopping the bullet is \(31250 \text{ N}\).

Step by step solution

01

Identify given data

The mass of the bullet is given as \(30 \text{ g} = 0.03 \text{ kg}\). The initial velocity (\(v_i\)) is \(500 \text{ m/s}\). The bullet comes to a stop, so the final velocity (\(v_f\)) is \(0 \text{ m/s}\). The stopping distance (\(d\)) is \(12 \text{ cm} = 0.12 \text{ m}\).
02

Calculate change in kinetic energy

The change in mechanical energy is essentially the initial kinetic energy of the bullet, which is converted to other forms of energy (like heat and deformation energy). The formula for kinetic energy is \(KE = \frac{1}{2}mv^2\). Thus, the change in kinetic energy is:\[\text{Change in KE} = KE_i - KE_f = \frac{1}{2}mv_i^2 - \frac{1}{2}mv_f^2 = \frac{1}{2} \cdot 0.03 \cdot (500)^2 - 0 = 3750 \text{ J}\]
03

Calculate the average stopping force

Using the work-energy principle, the work done by the stopping force is equal to the change in kinetic energy. Work done \(W = F \times d\). Rearranging the formula, we have:\[F = \frac{W}{d} = \frac{3750 \text{ J}}{0.12 \text{ m}} = 31250 \text{ N}\]

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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. This energy depends on two main factors: the mass of the object and its velocity. The formula to calculate kinetic energy is: \(KE = \frac{1}{2}mv^2\), where \(m\) is the mass and \(v\) is the velocity. In the example of the bullet hitting the wall, we used this formula to calculate the kinetic energy of the bullet before it stopped. Since the bullet comes to a stop, its final velocity is zero, meaning its final kinetic energy is also zero. Hence, the change in kinetic energy is equal to its initial kinetic energy, which in our case, was 3750 Joules.
Work-Energy Principle
The work-energy principle is a powerful concept that connects the work done on an object to its change in kinetic energy. According to this principle: \(W = \text{Change in KE}\). Here, the 'work' is the energy transferred to or from an object via the force applied over a distance. For the bullet example, the work done by the wall on the bullet (to stop it) is the same as the negative of the initial kinetic energy of the bullet, which converts to other forms of energy (like heat). The formula we use is \(W = F \times d\), where \(W\) is work, \(F\) is the average force, and \(d\) is the distance over which the force acts. By rearranging this formula, we could solve for the average force.
Average Force
Force is a push or pull exerted on an object, and it can cause the object to accelerate, decelerate, or change direction. Average force is the constant force that would produce the same change in motion as the actual varying force over the given distance. In our bullet example, we determined the average force exerted by the wall to stop the bullet. Using the work-energy principle and knowing the change in kinetic energy and the stopping distance, we found: \(F = \frac{W}{d}\). Here, the work done (\(W\)) was 3750 Joules, and the distance (\(d\)) was 0.12 meters. Plugging these values into our formula, we calculated that the average force from the wall was 31250 Newtons.

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Most popular questions from this chapter

Game of Pool In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

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Closing the Door A student is in her dorm room, sitting on her bed doing her physics homework. The door to her room is open. All of a sudden, she hears the voice of her ex-boyfriend talking to the girl in the room next door. She wants to shut the door quickly, so she throws a superball (which she keeps next to her bed for this purpose) against the door. The ball follows the path shown in Fig. 10-69. It hits the door squarely and bounces straight back. (a) If the ball has a mass \(m\), hits the door with a speed \(v\), and bounces back with a speed equal to \(v\), what is the change in the ball's momentum? (b) If the ball was in contact with the door for a time \(\Delta t\), what was the average force that the door exerted on the ball? (c) Would she have been better off with a clay ball of the same mass that stuck to the door? Explain your reasoning.

Particle \(A\) and Particle \(B\) Particle \(A\) and particle \(B\) are held together with a compressed spring between them. When they are released, the spring pushes them apart and they then fly off in opposite directions, free of the spring. The mass of \(A\) is \(2.00\) times the mass of \(B\), and the energy stored in the spring was \(60 \mathrm{~J}\). Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energies of (a) particle \(A\) and (b) particle \(B\) ?

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