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Chapter 8: Problem 8

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.

Short Answer

Expert verified
The bullet's speed was approximately 0.28 km/s.

Step by step solution

01

Set Up the Momentum Conservation Equation

Since momentum is conserved during the collision, we can write the equation: \[ (0.015 \, \text{kg}) v + 0 = (3.015 \, \text{kg}) V \] where \(v\) is the speed of the bullet just before impact and \(V\) is the speed of the block and bullet just after impact.
02

Use Energy Conservation for Swing

After the collision, the kinetic energy of the block and bullet system is converted to potential energy as it swings up. Set the gravitational potential energy (PE_G) to zero at the initial level: \[ \frac{1}{2}(3.015 \, \text{kg}) V^2 = (3.015 \, \text{kg})(9.81 \, \text{m/s}^2)(0.10 \, \text{m}) \]
03

Solve for V

Solve the energy equation from Step 2 to find \(V\): \[ \begin{align*} \frac{1}{2}(3.015) V^2 &= (3.015)(9.81)(0.10) \ \frac{1}{2} V^2 &= 0.981 \ V^2 &= 1.962 \ V &= \sqrt{1.962} = 1.40 \, \text{m/s} \end{align*} \]
04

Substitute to Find Bullet Speed

Use the value of \(V = 1.40\, \text{m/s}\) in the momentum equation from Step 1 to solve for \(v\): \[ 0.015 v = 3.015 \times 1.40 \] \[ 0.015 v = 4.221 \] Divide both sides by 0.015: \[ v = \frac{4.221}{0.015} = 281.4 \, \text{m/s} \] Convert this to km/s by dividing by 1000: \( v = 0.2814 \, \text{km/s} \approx 0.28 \, \text{km/s} \).

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Key Concepts

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

Elastic Collision
In physics, an elastic collision is one in which there is no loss of kinetic energy in the system. This means both momentum and kinetic energy are conserved. However, in the original exercise with the bullet and block, not all the kinetic energy is conserved during the collision. This is because the bullet embeds itself into the block rather than bouncing off, resulting in an inelastic collision. Thus, only momentum is conserved immediately after the collision.
In elastic collisions, the two objects would rebound off each other, maintaining their total kinetic energy. Since the bullet and block stuck together, some energy was lost to factors like heat or sound. The key takeaway is that while there is a transfer of momentum in all collisions, perfectly elastic ones are rare in real-life scenarios. Understanding this helps when setting up equations for conservation of momentum, allowing us to correctly model different types of collision events.
Kinetic Energy
Kinetic energy (KE) is the energy an object possesses due to its motion. For the bullet-block scenario, calculating the kinetic energy just after the collision gives insight into how much energy was retained from the original motion of the bullet. The formula for kinetic energy is given by:
\[ KE = \frac{1}{2}mv^2 \]
Where \( m \) is the mass of the object and \( v \) is its velocity.
  • After the collision, the kinetic energy is specifically needed to determine how high the block and bullet can rise.
  • When the block and bullet swing upwards, their kinetic energy is converted into potential energy.
This conversion supports the calculation of \( V = 1.40 \, \text{m/s} \), which represents the speed of the block and bullet right after the collision. This concept shows the constant transformation between kinetic and potential energy in systems. Although no energy is "destroyed," it is transformed to different types including internal energy due to external effects like friction in inelastic collisions.
Potential Energy
Potential energy (PE), particularly gravitational potential energy, is the energy an object possesses due to its position in a gravitational field. For the wood block and bullet system, when it swings upward after the collision, the kinetic energy gets fully converted to potential energy at the highest swing point. This concept is represented by:
\[ PE = mgh \]
Where:
  • \( m \) is the mass.
  • \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
  • \( h \) is the height the object is raised to.
In our problem, when the entire system swings up to a height of 10 cm (or 0.10 meters), this height is crucial for determining how much potential energy is acquired. It's the main reason we can solve for the speed, \( V \), using energy conservation. This approach highlights the symbiotic relationship between potential and kinetic energy, such that they can seamlessly interchange forms depending on movement and position within a gravitational field, offering deep insight into solving real-world physics problems.

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Most popular questions from this chapter

A \(2.0\) -kg brick is moving at a speed of \(6.0 \mathrm{~m} / \mathrm{s}\). How large a force \(F\) is needed to stop the brick in a time of \(7.0 \times 10^{-4} \mathrm{~s}\) ? Since we have a force and the time over which it acts, that suggests using the impulse equation (i.e., Newton's Second Law): Impulse on brick \(=\) Change in momentum of brick $$ \begin{aligned} F \Delta t &=m v_{f}-m v_{i} \\ F\left(7.0 \times 10^{-4} \mathrm{~s}\right) &=0-(2.0 \mathrm{~kg})(6.0 \mathrm{~m} / \mathrm{s}) \end{aligned} $$ from which \(F=-1.7 \times 10^{4} \mathrm{~N}\). The minus sign indicates that the force opposes the motion.

A \(2.0\) -kg block of wood rests on a tabletop. A \(7.0\) -g bullet is shot straight up through a hole in the table beneath the block. The bullet lodges in the block, and the block flies \(25 \mathrm{~cm}\) above the tabletop. How fast was the bullet going initially?

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}

Typically, a tennis ball hit during a serve travels away at about \(51 \mathrm{~m} / \mathrm{s}\). If the ball is at rest mid-air when struck, and it has a mass of \(0.058 \mathrm{~kg}\), what is the change in its momentum on leaving the racket?

Two identical balls collide head-on. The initial velocity of one is \(0.75 \mathrm{~m} / \mathrm{s}-\) EAST, while that of the other is \(0.43 \mathrm{~m} / \mathrm{s}\) -WEST. If the collision is perfectly elastic, what is the final velocity of each ball? Because the collision is perfectly elastic, both momentum and \(\mathrm{KE}\) are conserved. Since the collision is head-on, all motion takes place along a straight line. Take east as positive and call the mass of each ball \(m\). Momentum is conserved in a collision, so we can write Momentum before \(=\) Momentum after $$ m(0.75 \mathrm{~m} / \mathrm{s})+m(-0.43 \mathrm{~m} / \mathrm{s})=m v_{1}+m v_{2} $$ where \(v_{1}\) and \(v_{2}\) are the final values. This equation simplifies to $$ 0.32 \mathrm{~m} / \mathrm{s}=v_{1}+v_{2} $$ Because the collision is assumed to be perfectly elastic, \(\mathrm{KE}\) is also conserved. Thus, $$ \begin{array}{c} \text { KE before }=\text { KE after } \\ \frac{1}{2} m(0.75 \mathrm{~m} / \mathrm{s})^{2}+\frac{1}{2} m(-0.43 \mathrm{~m} / \mathrm{s})^{2}=\frac{1}{2} m v_{1}^{2}+\frac{1}{2} m v_{2}^{2} \end{array} $$ This equation can be simplified to $$ 0.747=v_{1}^{2}+v_{2}^{2} $$ We can solve for \(v_{2}\) in Eq. (1) to get \(v_{2}=0.32-v_{1}\) and substitute that into Eq. (2). This yields $$ 0.747=\left(0.32-v_{1}\right)^{2}+v_{1}^{2} $$ from which $$ 2 v_{1}^{2}-0.64 v_{1}-0.645=0 $$ Using the quadratic formula, $$ v_{1}=\frac{0.64 \pm \sqrt{(0.64)^{2}+5.16}}{4}=0.16 \pm 0.59 \mathrm{~m} / \mathrm{s} $$ from which \(v_{1}=0.75 \mathrm{~m} / \mathrm{s}\) or \(-0.43 \mathrm{~m} / \mathrm{s}\). Substitution back into Eq. \((1)\) gives \(v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\) or \(0.75 \mathrm{~m} / \mathrm{s}\). Two choices for answers are available: $$ \left(v_{1}=0.75 \mathrm{~m} / \mathrm{s}, v_{2}=-0.43 \mathrm{~m} / \mathrm{s}\right) \quad \text { and } \quad\left(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}, v_{2}=0.75 \mathrm{~m} / \mathrm{s}\right) $$ We must discard the first choice because it implies that the balls continue on unchanged; that is to say, no collision occurred. The correct answer is therefore \(v_{1}=-0.43 \mathrm{~m} / \mathrm{s}\) and \(v_{2}=0.75 \mathrm{~m} / \mathrm{s}\), which tells us that in a perfectly elastic, head-on collision between equal masses, the two bodies simply exchange velocities. Hence, \(\overrightarrow{\mathbf{v}}_{1}=0.43 \mathrm{~m} / \mathrm{s}-\) WEST and \(\overrightarrow{\mathbf{v}}_{2}=0.75 \mathrm{~m} / \mathrm{s}-\) EAST. If we recall that \(e=1\) for a perfectly elastic head-on collision, then $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(0.75 \mathrm{~m} / \mathrm{s})-(-0.43 \mathrm{~m} / \mathrm{s})} $$ which gives $$ v_{2}-v_{1}=1.18 \mathrm{~m} / \mathrm{s} $$ Equations ( 1 ) and (3) determine \(v_{1}\) and \(v_{2}\) uniquely.
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