/*! 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 32 A \(2.0-\mathrm{kg}\) block of w... [FREE SOLUTION] | 91影视

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Chapter 8: Problem 32

A \(2.0-\mathrm{kg}\) block of wood rests on a long tabletop. A \(5.0\) -g bullet moving horizontally with a speed of \(150 \mathrm{~m} / \mathrm{s}\) is shot into the block and lodges in it. The block then slides \(270 \mathrm{~cm}\) along the table and stops. (a) Find the speed of the block just after impact. (b) Find the friction force between block and table assuming it to be constant.

Short Answer

Expert verified
(a) Speed after impact: 0.374 m/s. (b) Friction force: 0.0518 N.

Step by step solution

01

Identify given information

First, we need to summarize the given information: - Mass of the block, \( m_b = 2.0 \) kg- Mass of the bullet, \( m_{b'} = 5.0 \) g = \( 0.005 \) kg- Speed of the bullet before impact, \( v_{b'} = 150 \) m/s- Distance block slides, \( d = 270 \) cm = \( 2.7 \) m- Final velocity of block+bullet, \( v_f = 0 \) m/s (after sliding) We need to find the speed of the block just after the impact and the friction force.
02

Use conservation of momentum for impact speed

We apply conservation of momentum to find the speed of the block just after impact. The total momentum before the collision is the momentum of the bullet. After the collision, the momentum is shared by the bullet and the block:\[ m_{b'} v_{b'} = (m_b + m_{b'}) v \]Plugging in the values:\[ 0.005 \times 150 = (2.0 + 0.005) v \]Calculating gives the velocity \( v \):\[ v = \frac{0.75}{2.005} = 0.374 \text{ m/s} \]
03

Calculate deceleration due to friction

Using the kinematic equation, relate the initial velocity of the block (just after impact) to the final velocity (which is zero) to find the acceleration due to friction. The equation is:\[ v_f^2 = v^2 + 2ad \]Given, \( v_f = 0 \) and \( v = 0.374 \), \( d = 2.7 \) m:\[ 0 = 0.374^2 + 2a \times 2.7 \]Solving for \( a \):\[ a = -\frac{0.374^2}{2 \times 2.7} = -0.0259 \text{ m/s}^2 \]
04

Determine the friction force

Friction force can be found using Newton's second law \( F = ma \):Given the negative acceleration found (which is due to friction):\[ F = (2.0 + 0.005) \times (-0.0259) \]This gives:\[ F = -0.0518 \text{ N} \]The negative sign indicates that the force is opposing the motion.

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Key Concepts

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

Kinematic Equations
Kinematic equations are powerful tools that help us understand motion in physics, particularly when it comes to determining how objects move under certain conditions. These equations relate various motion parameters such as displacement, velocity, acceleration, and time.

Let's take a closer look at some of the basic kinematic equations and how they apply to the motion of our bullet-block system. One of the most common equations used is:
  • \[v_f = v_i + at\]
  • \[s = v_it + \frac{1}{2}at^2\]
  • \[v_f^2 = v_i^2 + 2as\]
Here,
  • \(v_f\) is the final velocity
  • \(v_i\) is the initial velocity
  • \(a\) is the acceleration
  • \(s\) is the displacement
  • \(t\) is the time
In our exercise, the third equation was used to determine the acceleration of the block due to friction after the bullet lodged itself inside. The initial velocity \(v\) was found using the conservation of momentum, which we will discuss later. By setting the final velocity \(v_f\) to zero, since the block comes to rest, we effectively isolated the acceleration \(a\) given the block slides for a known distance \(d\). These equations act as bridges connecting the blocks of information about motion in a straightforward manner.
Friction Force
Friction is a force that opposes the motion of objects moving against each other. It plays a crucial role in this exercise as it is responsible for stopping the block and bullet after impact.

Frictional force arises because of the contact between the moving surfaces and is usually calculated as:
  • \[F = \mu N\]
where:
  • \(\mu\) is the coefficient of friction (a constant that depends on the two surfaces in contact)
  • \(N\) is the normal force (often equal to the gravitational force if no other vertical forces are present)
Since the block was sliding on a flat table, the normal force \(N\) is simply the weight of the combined block and bullet system. In the solution, we used Newton鈥檚 second law to find the friction force directly from the known deceleration and mass. This avoids needing to find \(\mu\), making calculations more straightforward in practical scenarios like ours.

Ultimately, understanding friction helps us predict and calculate how long an object will move until it stops. It highlights how forces in our world work together to bring motion to stability.
Newton's Second Law
Newton's second law of motion states that the force acting on an object is equal to the mass of that object multiplied by its acceleration, represented by the formula:

\[F = ma\]
This fundamental principle allows us to infer the resulting deceleration and force the system experiences. In the context of our block and bullet scenario, it gives us clear insights into how forces apply on an object in motion.

To solve for the friction force in our exercise, we particularly use Newton鈥檚 second law. Once we determined the deceleration of the block from the kinematic equation, we could easily substitute into \(F = ma\) using the combined mass of the block and bullet. The simplicity yet depth of Newton's second law makes it an essential tool in mechanics.

Using this learning, you can approach various scenarios involving motion, always remembering how interconnected force, mass, and acceleration are. This approach not only aids in solving physics problems but also fosters a deeper intuitive understanding of how objects in our universe behave.

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Most popular questions from this chapter

As shown in Fig. \(8-1\), a 15 -g bullet is fired horizontally into a \(3.000-\mathrm{kg}\) block of wood suspended by a long cord. The bullet lodges in the block. Compute the speed of the bullet if the impact causes the block (and bullet) to swing \(10 \mathrm{~cm}\) above its initial level. Consider first the collision of block and bullet. During the collision, momentum is conserved, so Momentum just before \(=\) Momentum just after $$ (0.015 \mathrm{~kg}) v+0=(3.015 \mathrm{~kg}) V $$ where \(v\) is the speed of the bullet just prior to impact, and \(V\) is the speed of block and bullet just after impact. We have two unknowns in this equation. To find another equation, we can use the fact that the block swings \(10 \mathrm{~cm}\) high. If we let \(\mathrm{PE}_{\mathrm{G}}=0\) at the initial level of the block, energy conservation tells us that $$ \begin{array}{l} \text { KE just after collision }=\text { Final } \mathrm{PE}_{\mathrm{G}} \\ \qquad \frac{1}{2}(3.015 \mathrm{~kg}) V^{2}=(3.015 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)(0.10 \mathrm{~m}) \end{array} $$ From this \(V=1.40 \mathrm{~m} / \mathrm{s}\). Substituting this combined speed into the previous equation leads to \(v=0.28 \mathrm{~km} / \mathrm{s}\) for the speed of the bullet. Notice that we cannot write the conservation of energy equation \(\frac{1}{2} m v^{2}=(m+M) g h\), where \(m=0.015 \mathrm{~kg}\) and \(M=3.000 \mathrm{~kg}\) because energy is lost (through friction) in the collision process.

Two identical railroad cars sit on a horizontal track, with a distance \(D\) between their two centers of mass. By means of a cable between them, a winch on one is used to pull the two together. (a) Describe their relative motion. ( \(b\) ) Repeat the analysis if the mass of one car is now three times that of the other. Keep in mind that the velocity of the center of mass of a system can only be changed by an external force. Here the forces due to the cable acting on the two cars are internal to the two-car system. The net external force on the system is zero, and so its center of mass does not move, even though each car travels toward the other. Taking the origin of coordinates at the center of mass, $$ x_{\mathrm{cm}}=0=\frac{\sum m_{i} x_{i}}{\sum m_{i}}=\frac{m_{1} x_{1}+m_{2} x_{2}}{m_{1}+m_{2}} $$ where \(x_{1}\) and \(x_{2}\) are the positions of the centers of mass of the two cars. (a) If \(m_{1}=m_{2}\), this equation becomes $$ 0=\frac{x_{1}+x_{2}}{2} \quad \text { or } \quad x_{1}=-x_{2} $$ The two cars approach the center of mass, which is originally midway between the two cars (that is, \(D / 2\) from each), in such a way that their centers of mass are always equidistant from it. (b) If \(m_{1}=3 m_{2}\), then we have $$ 0=\frac{3 m_{2} x_{1}+m_{2} x_{2}}{3 m_{2}+m_{2}}=\frac{3 x_{1}+x_{2}}{4} $$ from which \(x_{1}=-x_{2} / 3\). Since \(m_{1}>m_{2}\), it must be that \(x_{1}

An \(8.0\) -g bullet is fired horizontally into a \(9.00\) -kg cube of wood, which is at rest on a frictionless air table. The bullet lodges in the wood. The cube is free to move and has a speed of \(40 \mathrm{~cm} / \mathrm{s}\) after impact. Find the initial velocity of the bullet. This is an example of a completely inelastic collision for which momentum is conserved, although \(\mathrm{KE}\) is not. Consider the system (cube \(+\) bullet). The velocity, and hence the momentum, of the cube before impact is zero. Take the bullet's initial motion to be positive in the positive \(x\) -direction. The momentum conservation law tells us that Momentum of system before impact = Momentum of system after impact $$ \begin{aligned} \text { (Momentum of bullet) }+\text { (Momentum of cube) } &=(\text { Momentum of bullet }+\text { Cube }) \\ m_{B} v_{B x}+m_{C} v_{C x} &=\left(m_{B}+m_{C}\right) v_{x} \\ (0.0080 \mathrm{~kg}) v_{B x}+0 &=(9.008 \mathrm{~kg})(0.40 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ Solving yields \(v_{B x}=0.45 \mathrm{~km} / \mathrm{s}\) and so \(\overrightarrow{\mathbf{v}}_{B}=0.45 \mathrm{~km} / \mathrm{s}\) - POSITIVE \(X\) -DIRECTION.

A \(15-\mathrm{g}\) bullet moving at \(300 \mathrm{~m} / \mathrm{s}\) passes through a \(2.0-\mathrm{cm}\) -thick sheet of foam plastic and emerges with a speed of \(90 \mathrm{~m} / \mathrm{s}\). Assuming that the speed change takes place uniformly, what average force impeded the bullet's motion through the plastic? We can determine the change in momentum, and that suggests using the impulse equation to find the force \(F\) on the bullet as it takes a time \(\Delta t\) to pass through the plastic. Taking the initial direction of motion to be positive, $$ F \Delta t=m v_{f}-m v_{i} $$ We can find \(\Delta t\) by assuming uniform deceleration and using \(x=v_{a v} t\), where \(x=0.020 \mathrm{~m}\) and \(v_{a v}=\frac{1}{2}\left(v_{i}+v_{f}\right)=\) \(195 \mathrm{~m} / \mathrm{s}\). This gives \(\Delta t=1.026 \times 10^{-4} \mathrm{~s}\). Then $$ (F)\left(1.026 \times 10^{-4} \mathrm{~s}\right)=(0.015 \mathrm{~kg})(90 \mathrm{~m} / \mathrm{s})-(0.015 \mathrm{~kg})(300 \mathrm{~m} / \mathrm{s}) $$ which yields \(F=-3.1 \times 10^{4} \mathrm{~N}\) as the average retarding force. How could this problem have been solved using \(F=m a\) instead of the impulse equation? By using energy methods?

A \(40000-\mathrm{kg}\) freight car is coasting at a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) along a straight level track when it strikes a \(30000-\mathrm{kg}\) stationary freight car and couples to it. What will be their combined speed after impact?
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