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Chapter 7: Problem 61

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with its original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball.

Short Answer

Expert verified
(a) The angle is approximately \(40.58^\text{o}\).(b) The original speed of the cue ball is \(4.76 \, \text{m/s}\).

Step by step solution

01

Set Up the Momentum Equations

Since the collision is between two balls of the same mass, use the conservation of momentum in the x and y-directions. Let the original speed of the cue ball be denoted as \(v_{i}\), and its initial momentum will be \(mv_{i}\). After the collision, the x and y components of momentum for both balls need to be considered.
02

Write the Momentum Equations in the x-Direction

In the x-direction: \[ mv_i = m(3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o}) + m(2.00 \, \text{m/s} \, \text{cos} \, \theta) \]Simplify to: \[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, \theta \]
03

Write the Momentum Equations in the y-Direction

In the y-direction: \[ 0 = m(3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}) - m(2.00 \, \text{m/s} \, \text{sin} \, \theta) \]Simplify to:\[ 3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o} = 2.00 \, \text{m/s} \, \text{sin} \, \theta \]
04

Solve for the Angle \(\theta\)

Isolate \(\theta\) in the y-direction equation: \[ \text{sin} \, \theta = \frac{3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}}{2.00 \, \text{m/s}} \]Calculate: \[ \text{sin} \, \theta = 0.649 \to \theta = \text{arcsin}(0.649) \approx 40.58^\text{o} \]
05

Substitute \(\theta\) into the x-Direction Equation

Use \(\theta = 40.58^\text{o}\) in the x-direction equation: \[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, 40.58^\text{o} \]Calculate: \[ v_i = 3.50 \, \text{m/s} \, (0.927) + 2.00 \, \text{m/s} \, (0.758) \approx 3.24 \, \text{m/s} + 1.52 \, \text{m/s} = 4.76 \, \text{m/s} \]

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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 both momentum and kinetic energy are conserved. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. For example, imagine playing pool, where the cue ball strikes another ball of the same mass. If this is an elastic collision, the energy and momentum are preserved.

During this type of collision, two main principles are adhered to:
  • Conservation of momentum
  • Conservation of kinetic energy
Understanding these principles helps explain how and why the balls move the way they do after colliding.

Elastic collisions are common in atomic and subatomic particles interactions. They exhibit a perfect exchange of energy and momentum, like the stationary ball gaining speed while the cue ball slows down.
momentum in x-direction
Momentum in the x-direction refers to the horizontal components of the momentum. After the pool balls collide, their movements split into x and y components.

The initial momentum of the cue ball is given by: \ p_{initial} = m \, \ v_{i} . In the x-direction, conservation of momentum is applied, resulting in: \ m \, v_i = m \, (3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o}) + m \, (2.00 \, \text{m/s} \, \text{cos} \, \theta). This equation can be simplified to:

\[ v_i = 3.50 \, \text{m/s} \, \text{cos} \, 22.0^\text{o} + 2.00 \, \text{m/s} \, \text{cos} \, \theta \]

By using trigonometry, you can substitute the values to keep momentum conserved in the x-direction, enabling you to find the original speed of the cue ball and the angle \ \theta .
momentum in y-direction
Momentum in the y-direction refers to the vertical components of the momentum after the collision. Like the x-direction, momentum must also be conserved in the y-direction.

Initially, the cue ball only moves horizontally. Hence, the total initial momentum in the y-direction is zero. After the collision, the y-components are given by:

0 = m \ (3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}) - m \ (2.00 \, \text{m/s} \, \text{sin} \, \theta)

This leads to: \[ 3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o} = 2.00 \, \text{m/s} \, \text{sin} \, \theta \]

With this equation, you can find the angle \ \theta by isolating it and solving: \ \text{sin} \ \theta = \frac{3.50 \, \text{m/s} \, \text{sin} \, 22.0^\text{o}}{2.00 \, \text{m/s}} . The calculated angle helps confirm that the momentum in both directions has been correctly conserved.

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Most popular questions from this chapter

A \(0.70 \mathrm{~kg}\) ball is moving horizontally with a speed of \(5.0 \mathrm{~m} / \mathrm{s}\) when it strikes a vertical wall. The ball rebounds with a speed of \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in translational momentum of the ball?

The force of air resistance on a sphere of radius \(R\) can plausibly be argued to have the form $$ \vec{F} \text { drag }=-\frac{1}{2} C \rho R^{2}|v| \vec{v}=-b|v| \vec{v} $$ where \(\vec{v}\) is the vector velocity and \(|\vec{v}|\) is its magnitude (the speed). The density of the air, \(\rho\), is about \(1 \mathrm{~kg} / \mathrm{m}^{3}-1 / 1000\) that of water. The parameter \(C\) is a dimensionless constant. If we drop a steel ball and a styrofoam ball from a height of \(s\). the steel ball reaches the ground when the styrofoam ball is still a bit above the ground. Call this distance \(h .\) Estimate the air resistance coefficient \(C\) as follows: (a) Assume the effect of air resistance on the steel sphere is negligible. Calculate approximately how long the steel sphere takes to fall to the ground \(\left(\Delta t_{\text {ste }}\right)\) and how fast it is traveling just before it hits \(\left(v_{\text {ste }}\right) .\) Express your answers in terms of \(s, g\), and \(m\). (b) Since the steel and styrofoam were not very different, use \(\left\langle\vec{v}_{\text {ste }}\right.\), the average velocity of the steel ball during its fall to calculate an average air resistance force, \((\vec{F}\) drag \()=-b\left\langle\left.\vec{v}\right|^{2}\right.\) acting on the styrofoam sphere during its fall. Express this force in terms of \(b, m\) (the mass of the styrofoam sphere), \(g, s\), and \(h\). (c) The average velocity of the steel ball is \(\left\langle\vec{v}_{\text {ste }}\right\rangle=s / \Delta t_{\text {ste }} .\) The average velocity of the styrofoam sphere was \(\left\langle\vec{v}_{\text {sty }}\right\rangle=(s-h) / \Delta t_{\text {ste }}\) Assume this difference, \(\Delta(\vec{v})\), is caused by the average air resistance force acting over the time \(\Delta t_{\text {ste }}\) with our basic Newton's law formula: $$ \left\langle\vec{F}^{\text {drag }}\right\rangle \Delta t_{\text {ste }}=m \Delta(\vec{v}) . $$ Use this to show that $$ b \cong \frac{m h}{s^{2}} $$ (d) A styrofoam ball of radius \(R=5 \mathrm{~cm}\) and mass \(m=50 \mathrm{~g}\) is dropped with a steel ball from a height of \(s=2 \mathrm{~m}\). When the steel ball hits, the styrofoam is about \(h=10 \mathrm{~cm}\) above the ground. Calculate \(b\) (for the styrofoam sphere) and \(C\) (for any sphere).

Two cars \(A\) and \(B\) slide on an icy road as they attempt to stop at a traffic light. The mass of \(A\) is \(1100 \mathrm{~kg}\), and the mass of \(B\) is \(1400 \mathrm{~kg}\). The coefficient of kinetic friction between the locked wheels of either car and the road is 0.13. Car \(A\) succeeds in stopping at the light, but car \(B\) cannot stop and rear-ends car \(A\). After the collision, \(A\) stops \(8.2 \mathrm{~m}\) ahead of its position at impact, and \(B 6.1 \mathrm{~m}\) ahead; see Fig. 7-27. Both drivers had their brakes locked throughout the incident. Using the material in Chapters 2 and 6 , find the speed of (a) car \(A\) and (b) car \(B\) immediately after impact. (c) Use conservation of translational momentum to find the speed at which car \(B\) struck car \(A\). On what grounds can the use of momentum conservation be criticized here?

When jumping straight down, you can be seriously injured if you land stiff- legged. One way to avoid injury is to bend your knees upon landing to reduce the force of the impact. Suppose you have a mass \(m\) and you jump off a wall of height \(h\). (a) Use what you learned about constant acceleration motion to find the speed with which you hit the ground. Assume you simply step off the wall, so your initial \(y\) velocity is zero. Ignore air resistance. (Express your answer in terms of the symbols given.) (b) Suppose that the time interval starting when your feet first touch the ground until you stop is \(\Delta t .\) Calculate the (average) net force acting on you during that interval. (Again, express your answer in terms of the symbols given.) (c) Suppose \(h=1 \mathrm{~m}\). If you land stiff-legged, the time it takes you to stop may be as short as \(2 \mathrm{~ms}\), whereas if you bend your knees, it might be as long as \(0.1\) s. Calculate the average net force that would act on you in the two cases. (d) The net force on you while you are stopping includes both the force of gravity and the force of the ground pushing up. Which of these forces do you think does you the injury? Explain your reasoning. (e) For the two cases in part (c), calculate the upward force the ground exerts on you.

The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line.
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