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Chapter 8: Problem 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.

Short Answer

Expert verified
The bullet's initial velocity was 450 m/s in the positive x-direction.

Step by step solution

01

Understand the System

We have a bullet of mass \(8.0\, \text{g}\) (which is 0.0080 kg) and a wooden cube of mass \(9.00\, \text{kg}\). The bullet is fired into the cube, and post-collision, they move together at \(40 \text{ cm/s}\). This problem involves momentum conservation since the collision is inelastic.
02

Set Up the Momentum Equation

According to the law of conservation of momentum, the momentum of the system before the impact is equal to the momentum of the system after the impact. The equation given is: \[m_B v_{B x} + m_C v_{C x} = (m_B + m_C) v_x\]where \(m_B = 0.0080\, \text{kg}\), and \(m_C = 9.00\, \text{kg}\). \(v_{C x}\) is zero since the cube is initially at rest.
03

Substitute Known Values

Substitute \(m_B = 0.0080\, \text{kg}\), \(m_C = 9.00\, \text{kg}\), and \(v_x = 0.40\, \text{m/s}\) (as speed is given in cm/s, convert to m/s) into the momentum conservation equation:\[0.0080 v_{B x} + 0 = (9.008) \times 0.40\]
04

Solve for Initial Velocity of Bullet

Rearrange the equation to solve for the initial velocity of the bullet, \(v_{B x}\):\[0.0080 v_{B x} = 9.008 \times 0.40\]\[v_{B x} = \frac{9.008 \times 0.40}{0.0080}\]Calculate the result to find the initial velocity \(v_{B x} = 450 \, \text{m/s}\).
05

Interpret the Result

The initial velocity of the bullet, \(v_{B x}\), is \(450\, \text{m/s}\) in the positive x-direction. This means the bullet was traveling at this speed before it got lodged into the cube.

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Key Concepts

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

Inelastic Collision
An inelastic collision is a type of collision where the colliding objects stick together after impact. In this scenario, momentum is conserved, but kinetic energy is typically not. This means that after the collision, the objects move as a single system.
An example of an inelastic collision is when a bullet becomes lodged in a wooden block after impact, as seen in the exercise. Here, the bullet and the block move together post-collision as a result of the inelastic property.
In such collisions, you'll often notice that some kinetic energy is transformed into other forms of energy, such as heat or sound, due to the deformation and other factors during impact.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass, and \( v \) is the velocity of the object.
In the context of inelastic collisions, unlike momentum, kinetic energy is not conserved. This is because some of the kinetic energy is lost to other forms of energy during the collision process.
After an inelastic collision, the combined object (such as the bullet-block pair) usually has less kinetic energy than the sum of the individual kinetic energies before the collision. This loss of energy helps explain why the speed of the system reduces compared to the initial speed of the bullet.
Momentum Equation
The momentum equation is vital for solving problems involving collisions. According to the law of conservation of momentum, the total momentum of a system remains constant if no external forces act on it.
Momentum is calculated by the formula \( p = mv \), where \( m \) is the mass, and \( v \) is the velocity. In an inelastic collision, such as this one, the momentum before and after the collision is the same, even though the system's kinetic energy is not.
In the exercise, the momentum equation is set as follows:
  • Initial momentum of the bullet: \( m_Bv_{B x} \)
  • Initial momentum of the block: \( m_Cv_{C x} \) (zero since the block is at rest)
  • Total momentum after impact: \( (m_B + m_C) v_x \)
The equation is used to link initial conditions of the bullet and the outcome post-collision to find the bullet's velocity.
Initial Velocity Calculation
Calculating the initial velocity of an object in a collision scenario requires using the known masses and final velocity of the system.
In the given scenario, where an 8.0-g bullet is fired into a 9.00-kg wooden block, we are tasked with finding the bullet's initial velocity using the principle of momentum conservation.
To find this, you start by setting up the equation for conservation of momentum:
  • \( 0.0080 \, \text{kg} \times v_{B x} + 9.00 \, \text{kg} \times 0 \) (since the block was at rest)
  • Equals \( (9.008 \, \text{kg}) \times 0.40 \, \text{m/s} \)
By rearranging and solving for the initial velocity \( v_{B x} \), we find \( v_{B x} = \frac{9.008 \times 0.40}{0.0080} \), resulting in the bullet's initial velocity of 450 m/s. This calculation underscores how momentum principles allow us to evaluate pre-collision speeds effectively.

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Most popular questions from this chapter

Two bodies of masses \(8 \mathrm{~kg}\) and \(4 \mathrm{~kg}\) move along the \(x\) -axis in opposite directions with velocities of \(11 \mathrm{~m} / \mathrm{s}-\) POSITIVE \(X\) -DIRECTION and \(7 \mathrm{~m} / \mathrm{s}\) -NEGATIVE \(X\) -DIRECTION, respectively. They collide and stick together. Find their combined velocity just after collision.

Four masses are positioned in the \(x y\) -plane as follows: \(300 \mathrm{~g}\) at \((x=0, y=2.0 \mathrm{~m}), 500 \mathrm{~g}\) at \((-20 \mathrm{~m},-3.0 \mathrm{~m})\), \(700 \mathrm{~g}\) at \((50 \mathrm{~cm}, 30 \mathrm{~cm})\), and \(900 \mathrm{~g}\) at \((-80 \mathrm{~cm}, 150 \mathrm{~cm})\). Find their center of mass.

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 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.

A ball is dropped from a height \(h\) above a tile floor and rebounds to a height of \(0.65 h\). Find the coefficient of restitution between ball and floor. Assign floor quantities the subscript 1 , and ball quantities the subscript 2 . The initial and final velocities of the floor, \(u_{1}\) and \(v_{1}\), are zero. Therefore, $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}}=-\frac{v_{2}}{u_{2}} $$ Since we know both the drop and rebound heights \((h\) and \(0.65 h)\), we can write equations for the interchange of \(\mathrm{PE}_{\mathrm{G}}\) and \(\mathrm{KE}\) before and after the impact $$ m g h=\frac{1}{2} m u_{2}^{2} \quad \text { and } \quad m g(0.65 h)=\frac{1}{2} m v_{2}^{2} $$ Therefore, taking down as positive, \(u_{2}=\sqrt{2 g h}\) and \(v_{2}=-\sqrt{1.30 g h}\). Substitution leads to $$ e=\frac{\sqrt{1.30 g h}}{\sqrt{2 g h}}=\sqrt{0.65}=0.81 $$ Notice that the coefficient of restitution equals the square root of the final rebound height over the initial drop height.
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