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Chapter 8: Problem 16

A \(1.0\) -kg ball moving at \(12 \mathrm{~m} / \mathrm{s}\) collides head-on with a \(2.0-\mathrm{kg}\) ball moving in the opposite direction at \(24 \mathrm{~m} / \mathrm{s}\). Determine the motion of each after impact if \((a) e=2 / 3,(b)\) the balls stick together, and (c) the collision is perfectly elastic. In all three cases the collision occurs along a straight line, and momentum is conserved. Hence, $$ \begin{aligned} \text { Momentum before } &=\text { Momentum after } \\ (1.0 \mathrm{~kg})(12 \mathrm{~m} / \mathrm{s})+(2.0 \mathrm{~kg})(-24 \mathrm{~m} / \mathrm{s}) &=(1.0 \mathrm{~kg}) v_{1}+(2.0 \mathrm{~kg}) v_{2} \end{aligned} $$ which becomes $$ -36 \mathrm{~m} / \mathrm{s}=v_{1}+2 v_{2} $$ (a) When \(e=2 / 3\), $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad \frac{2}{3}=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(24 \mathrm{~m} / \mathrm{s}=v_{2}-v_{1}\). Combining this with the momentum equation found above gives \(v_{2}=-4.0 \mathrm{~m} / \mathrm{s}\) and \(v_{1}=-28 \mathrm{~m} / \mathrm{s}\) (b) In this case \(v_{1}=v_{2}=v\), and so the momentum equation becomes $$ 3 v=-36 \mathrm{~m} / \mathrm{s} \quad \text { or } \quad v=-12 \mathrm{~m} / \mathrm{s} $$ (c) Here \(e=1\), and $$ e=\frac{v_{2}-v_{1}}{u_{1}-u_{2}} \quad \text { becomes } \quad 1=\frac{v_{2}-v_{1}}{(12 \mathrm{~m} / \mathrm{s})-(-24 \mathrm{~m} / \mathrm{s})} $$ from which \(v_{2}-v_{1}=36 \mathrm{~m} / \mathrm{s}\). Adding this to the momentum equation yields \(v_{2}=0\). Using this value for \(v_{2}\) then leads to \(v_{1}=-36 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
(a) \(v_1 = -28 \) m/s, \(v_2 = -4 \) m/s; (b) Both \(v = -12 \) m/s; (c) \(v_1 = -36 \) m/s, \(v_2 = 0 \) m/s.

Step by step solution

01

Understand the Problem

We are asked to analyze three different types of collisions for a 1.0 kg ball moving at 12 m/s and a 2.0 kg ball moving in the opposite direction at 24 m/s. We need to determine the post-collision velocities using conservation of momentum and the coefficient of restitution for each scenario.
02

Write Equations for Momentum Conservation

The conservation of momentum equation for this system before the collision is given as: \( 1.0 \times 12 + 2.0 \times (-24) = 1.0 \cdot v_1 + 2.0 \cdot v_2 \). Simplifying this gives \( v_1 + 2v_2 = -36 \).
03

Part (a): Solve with e = 2/3

The coefficient of restitution is given by \( e = \frac{v_2 - v_1}{u_1 - u_2} \). Plugging in values: \( \frac{2}{3} = \frac{v_2 - v_1}{12 - (-24)} \). Simplifying the equation gives \( v_2 - v_1 = 24 \). Solve the system of equations: \( v_1 + 2v_2 = -36 \) and \( v_2 - v_1 = 24 \) to get \( v_2 = -4 \) and \( v_1 = -28 \).
04

Part (b): Solve for Sticking Together

If the balls stick together, their final velocities are equal \( v_1 = v_2 = v \). Substituting into the momentum equation gives: \( 3v = -36 \). Solving for \( v \) gives \( v = -12 \).
05

Part (c): Solve with Perfectly Elastic Collision

In a perfectly elastic collision, \( e = 1 \), therefore \( v_2 - v_1 = 12 - (-24) = 36 \). Together with the momentum equation, solve \( v_1 + 2v_2 = -36 \) and \( v_2 - v_1 = 36 \) to find \( v_2 = 0 \) and \( v_1 = -36 \).

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Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics that states that the total momentum of a system remains constant if it is not influenced by external forces. In the given exercise, we see two balls colliding, and we use this principle to find their velocities after the collision.

To apply the conservation of momentum, we set the total momentum before the collision equal to the total momentum after the collision. Let's denote the first ball's mass as 1.0 kg with an initial velocity of 12 m/s, and the second ball as 2.0 kg traveling at -24 m/s. Here, negative indicates the opposite direction. The momentum equation looks like this:
  • Before the collision: \[ (1.0 ext{ kg})(12 ext{ m/s}) + (2.0 ext{ kg})(-24 ext{ m/s}) \]
  • After the collision (unknown velocities \(v_1\) and \(v_2\)): \[ (1.0 ext{ kg})v_1 + (2.0 ext{ kg})v_2 \]
This results in the equation: \[ v_1 + 2v_2 = -36 \\]By combining this equation with the condition from different coefficients of restitution, we can solve for the unknown velocities in each scenario.
Coefficient of Restitution
The coefficient of restitution \( (e) \) measures the bounciness of a collision between two objects. It is the ratio of relative velocity post-collision to that pre-collision along the direction of impact.

Different values of \(e\) signify different types of collisions:
  • \(e = 1\): A perfectly elastic collision where kinetic energy is conserved.
  • \(0 < e < 1\): A partially elastic collision where some kinetic energy is lost.
  • \(e = 0\): A perfectly inelastic collision where objects stick together after impact.
For example, when \( e = \frac{2}{3} \), the collision is partially elastic. This specific value for \(e\) is used as follows:
  • Relative velocity post-collision (\( v_2 - v_1 \)) divided by relative velocity pre-collision \.i.e\ \(u_1 - u_2\): \[ \frac{v_2 - v_1}{12 - (-24)} = \frac{2}{3} \]
  • Simplifying, we find: \[ v_2 - v_1 = 24 \\]
Thus, combining it with our previous momentum equation helps us determine the specific post-collision velocities.
Elastic and Inelastic Collisions
Collisions in physics are broadly categorized into elastic and inelastic types, based on how they behave after impact. The differences lie in how they handle energy:
  • **Elastic Collision**: Both momentum and kinetic energy are conserved. These collisions are characterized by \(e = 1\). For the given exercise with perfectly elastic conditions, the derivation yields:\[ v_2 - v_1 = 36 \\]When combined with momentum conservation, this allows determination of velocities after collision without energy loss.
  • **Inelastic Collision**: Momentum is conserved, but kinetic energy is not. A completely inelastic collision (\(e = 0\)) results in the objects sticking together.In our exercise, when the balls stick together, we treat their final velocity as equal \(v_1 = v_2 = v\). Thus, we solve: \[ 3 v = -36 \\]leading to: \[ v = -12 ext{ m/s} \\]
Different scenarios reflect the nature of real-world impacts, where some energy loss is inevitable except in theoretical perfectly elastic collisions.

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Most popular questions from this chapter

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?

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.

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.

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}

A rocket standing on its launch platform points straight upward. Its engines are activated and eject gas at a rate of \(1500 \mathrm{~kg} / \mathrm{s}\). The molecules are expelled with an average speed of \(50 \mathrm{~km} / \mathrm{s}\). How much mass can the rocket initially have if it is slowly to rise because of the thrust of the engines? The problem provides mass flow and speed, the product of which is equivalent to the time rate-of-change of momentum. That should bring to mind the impulse- momentum relationship, which, of course, is Newton's Second Law. Since the initial motion of the rocket itself is negligible in comparison to the speed of the expelled gas, we can assume the gas is accelerated from rest to a speed of \(50 \mathrm{~km} / \mathrm{s}\). The impulse required to provide this acceleration to a mass \(m\) of gas is $$ F \Delta t=m v_{f}-m v_{i}=m(50000 \mathrm{~m} / \mathrm{s})-0 $$ from which $$ F=(50000 \mathrm{~m} / \mathrm{s}) \frac{m}{\Delta t} $$ But we are told that the mass ejected per second \((m / \Delta t)\) is \(1500 \mathrm{~kg} / \mathrm{s}\), and so the force exerted on the expelled gas is $$ F=(50000 \mathrm{~m} / \mathrm{s})(1500 \mathrm{~kg} / \mathrm{s})=75 \mathrm{MN} $$ An equal but opposite reaction force acts on the rocket, and this is the upward thrust on the rocket. The engines can therefore support a weight of \(75 \mathrm{MN}\), so the maximum mass the rocket could have is $$ M_{\text {rocket }}=\frac{\text { weight }}{g}=\frac{75 \times 10^{6} \mathrm{~N}}{9.81 \mathrm{~m} / \mathrm{s}^{2}}=7.7 \times 10^{6} \mathrm{~kg} $$
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