/*! 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 79 A rifle bullet with mass \(8.00 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A rifle bullet with mass \(8.00 \mathrm{~g}\) strikes and embeds itself in a block with mass \(0.992 \mathrm{~kg}\) that rests on a frictionless, horizontal surface and is attached to an ideal spring (Fig. \(\mathrm{P} 8.79\) ). The impact compresses the spring \(15.0 \mathrm{~cm} .\) Calibration of the spring shows that a force of 0.750 \(\mathrm{N}\) is required to compress the spring \(0.250 \mathrm{~cm}\). (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullct?

Short Answer

Expert verified
The magnitude of the block's velocity just after impact can be determined using the principles of the conservation of momentum and energy combined with the information on the force exerted by the spring given its compression. The initial velocity of the bullet can be determined by setting the total momentum of the system equal to the initial momentum of the bullet and solving for the bullet's initial velocity.

Step by step solution

01

Conservation of Momentum

The law of conservation of momentum states that the total momentum before an event must be equal to the total momentum after the event. In this case, right after the bullet embeds itself in the block, we can consider the bullet-block system as a single object. So, momentum just after collision is \(p_{after} = (m_{bullet} + m_{block})v_{block}\) where \(v_{block}\) is the velocity of the block that we need to find.
02

Determine force exerted by spring

We are given the relationship between the spring compression and the force required to achieve that compression. The information given describes a linear relationship between force and displacement (Hooke's law: \(F = kx\)), and we can determine the spring constant \(k\). Given \(F = 0.750 N\) for \(x = 0.250 cm = 0.00250 m\), we can calculate \(k = F/x\).
03

Calculate Energy Stored in Spring

We calculate the energy stored in the spring at its maximum compression (right before the block moves) using the formula for the elastic potential energy stored in a spring: \(PE_{spring} = 1/2kx^2\), where \(x = 15.0 cm = 0.150 m\). This energy is equal to the kinetic energy that block and bullet have just after impact, as none of the initial kinetic energy was lost (since the collision is fully elastic and there's no friction). Thus, we have \((1/2)(m_{block} + m_{bullet})v_{block}^2 = PE_{spring}\). We can now solve for \(v_{block}\).
04

Find Initial Speed of Bullet

In order to find the initial speed of the bullet, we can use the conversation of momentum again. We know that the momentum of the bullet-block system right after collision is equal to the initial momentum of the bullet. Thus, we have \(m_{bullet}v_{bullet, initial} = (m_{bullet} + m_{block})v_{block}\). We have already calculated \(v_{block}\) previously. So we can now solve for \(v_{bullet, initial}\).

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Key Concepts

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

Hooke's Law
Understanding Hooke's Law allows us to predict how springs behave under force. When you stretch or compress a spring, it exerts a force that is directly proportional to the displacement. This is what we refer to as Hooke's Law, and it can be expressed with the simple equation: \( F = -kx \). In this formula, \( F \) is the force in newtons (N), \( k \) is the spring constant in newtons per meter (N/m), and \( x \) is the spring's displacement from its equilibrium position in meters (m). The negative sign indicates the force exerted by the spring is in the opposite direction of the displacement.

Understanding this relationship is crucial for a wide range of mechanical and engineering applications. For instance, in the textbook problem, the spring exerts a restoring force that slows down and eventually stops the block after the bullet embeds into it. By knowing the force required for a certain displacement, we use Hooke's Law to calculate the spring constant, which is fundamental in determining the system's behavior post-impact.
Elastic Potential Energy
Elastic potential energy is the energy stored in an elastic material, such as a spring, as it is deformed under stress. This form of energy is dependent on the deformation of the material - the more you compress or stretch it, the more potential energy it stores. The energy is termed 'potential' because it has the potential to do work when the material returns to its original shape.

The formula for calculating the elastic potential energy (\(PE_{elastic}\)) in a spring is \( PE_{elastic} = \frac{1}{2}kx^2 \), where \( k \) is the spring constant and \( x \) is the displacement from the equilibrium position. In our textbook exercise, this plays a central role: The bullet compresses the spring, converting its kinetic energy into elastic potential energy. Understanding this concept is crucial because it helps us to conserve mechanical energy in systems where elastic forces are at play. When the spring reaches maximum compression, all the kinetic energy of the bullet-block system is momentarily stored as elastic potential energy before being converted back into kinetic energy.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. Anything that moves has kinetic energy, and it can be transferred during collisions, like in our textbook problem with the bullet and the block. The kinetic energy (\(KE\)) of an object with mass \(m\) moving at velocity \(v\) is given by the equation \( KE = \frac{1}{2}mv^2 \).

In the textbook example, the bullet is moving with a certain velocity and therefore has kinetic energy. When it hits and embeds into the block, this kinetic energy is initially transferred to the block-bullet system and then to the spring in the form of elastic potential energy. Recognizing the relationship between the various forms of energy in a system is instrumental in solving physics problems, particularly those involving conservation of energy and momentum.
Spring Constant Calculation
The spring constant, represented by \( k \), is a value that describes how stiff a spring is. The stiffer the spring, the larger the spring constant, meaning more force is required to compress or stretch the spring a certain distance. You can calculate the spring constant by rearranging Hooke's Law: \( k = \frac{F}{x} \).

In our textbook problem, the force required to compress the spring a certain distance is given, allowing us to calculate the spring constant. The constant \( k \) not only informs us about the spring's stiffness but also determines how much potential energy can be stored in it. As such, accurately calculating the spring constant is integral for predicting the system's behavior, such as how the block will move after the spring has been compressed by the bullet's impact.

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Most popular questions from this chapter

Two figure skaters, one weighing \(625 \mathrm{~N}\) and the other \(725 \mathrm{~N}\) push off against cach other on frictionless ice. (a) If the heavicr skater travels at \(1.50 \mathrm{~m} / \mathrm{s}\), how fast will the lighter one travel? (b) How much kinetic energy is "created" during the skaters' maneuver, and where does this cnergy come from?

In Section 8.5 we calculated the center of mass by considering objects composed of a finite number of point masses or objects that, by symmetry, could be represented by a finite number of point masses. For a solid object whose mass distribution does not allow for a simple determination of the center of mass by symmetry, the sums of Eqs. (8.28) must be generalized to integrals $$ x_{\mathrm{cm}}=\frac{1}{M} \int x d m \quad y_{\mathrm{cm}}=\frac{1}{M} \int y d m $$ where \(x\) and \(y\) are the coordinates of the small picce of the object that has mass \(d m\). The integration is over the whole of the object. Consider a thin rod of length \(I_{4}\), mass \(M\), and cross-sectional area \(A\). I et the origin of the coordinates be at the left end of the rod and the positive \(x\) -axis lie along the rod. (a) If the density \(\rho=M / V\) of the object is uniform, perform the integration described above to show that the \(x\) -coordinate of the center of mass of the rod is at its gcometrical center. (b) If the density of the object varies linearly with \(x\) - that is, \(\rho=\alpha x,\) where \(\alpha\) is a positive constant - calculate the \(x\) -coordinate of the rod's center of mass.

A young ice skater with mass \(40.0 \mathrm{~kg}\) has fallen and is sliding on the frictionless ice of a skating rink with a spced of \(20.0 \mathrm{~m} / \mathrm{s}\). (a) What is the magnitude of her linear momentum when she has this speed? (b) What is her kinetic energy? (c) What constant net horizontal force must be applied to the skater to bring her to rest in \(5.00 \mathrm{~s}\) ?

Two identical \(0.900 \mathrm{~kg}\) masses are pressed against opposite ends of a light spring of force constant \(1.75 \mathrm{~N} / \mathrm{cm}\), compressing the spring by \(20.0 \mathrm{~cm}\) from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless, horizontal table.

In a shipping company distribution center, an open cart of } mass \(50.0 \mathrm{~kg}\) is rolling to the left at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) (lig. \(\mathrm{P} 8.87\) ). Ignore friction between the cart and the floor. \(A 15.0 \mathrm{~kg}\) package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of \(4.00 \mathrm{~m}\) above the bottom of the cart, what are (a) the speed of the packagc just bcfore it lands in the cart and (b) the final speed of the cart?

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