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Chapter 7: Problem 67

Two identical pucks are on an air table. Puck A has an initial velocity of \(2.0 \mathrm{m} / \mathrm{s}\) in the \(+x\) -direction. Puck \(\mathrm{B}\) is at rest. Puck A collides elastically with puck B and A moves off at $1.0 \mathrm{m} / \mathrm{s}\( at an angle of \)60^{\circ}\( above the \)x$ -axis. What is the speed and direction of puck \(\mathrm{B}\) after the collision?

Short Answer

Expert verified
In this elastic collision problem, we used the conservation of momentum and the conservation of kinetic energy principles to find the final speed and direction of puck B. After solving the system of equations, we found that the final speed of puck B is \(\sqrt{3}\ \mathrm{m}/\mathrm{s}\), and its direction is approximately 54.7° below the x-axis.

Step by step solution

01

Understand the Initial Conditions

We have two pucks, A and B, that collide elastically. The initial conditions are: - Puck A's initial velocity (\(v_{A, initial}\)) is \(2.0 \mathrm{m}/\mathrm{s}\) in the \(+x\)-direction - Puck B is initially at rest, so its initial velocity (\(v_{B, initial}\)) is \(0\ \mathrm{m}/\mathrm{s}\) After the collision: - Puck A's final velocity (\(v_{A, final}\)) is \(1.0 \mathrm{m}/\mathrm{s}\) at an angle of \(60^\circ\) above the \(x\)-axis We need to find the final speed and direction of puck B (\(v_{B, final}\)) after the collision.
02

Conservation of Momentum

The principle of conservation of momentum states that the total momentum before collision and after the collision must be the same. Since the pucks are identical, we can assume they have the same mass, and let the mass of each puck be \(m\). The momentum in the x-direction before the collision is: $$ m \cdot v_{A, initial} + m \cdot v_{B, initial} = 2m $$ The momentum in the x-direction after the collision is: $$ m \cdot v_{A, final} \cdot \cos{60^\circ} + m \cdot v_{B, final} \cdot \cos{\theta_B} $$ By equating both expressions, we get: $$ 2m = m \cdot (1.0 \ \mathrm{m}/\mathrm{s}) \cdot \cos{60^\circ} + m \cdot v_{B, final} \cdot \cos{\theta_B} $$
03

Conservation of Kinetic Energy

The principle of conservation of kinetic energy states that the total kinetic energy before and after an elastic collision must be the same. The kinetic energy before the collision is: $$ \frac{1}{2}m \cdot v_{A, initial}^2 + \frac{1}{2}mv_{B, initial}^2 = 2m $$ The kinetic energy after the collision is: $$ \frac{1}{2}m \cdot v_{A, final}^2 + \frac{1}{2}mv_{B, final}^2 $$ By equating both expressions, we get: $$ 2m = \frac{1}{2}m \cdot (1.0 \ \mathrm{m}/\mathrm{s})^2 + \frac{1}{2}mv_{B, final}^2 $$
04

Solve for \(v_{B, final}\) and \(\theta_B\)

After simplifying the equations from Step 2 and Step 3, we have the following system of equations: $$ v_{B, final} \cdot \cos{\theta_B} = 1 $$ $$ v_{B, final}^2 = 3 $$ We can solve this system of equations by taking the square root of the second equation to get \(v_{B, final} = \sqrt{3} \ \mathrm{m}/\mathrm{s}\). Then, we can use the first equation to find \(\theta_B\): $$ \cos{\theta_B} = \frac{1}{\sqrt{3}} \Rightarrow \theta_B = \arccos{\frac{1}{\sqrt{3}}} \approx 54.7^\circ $$ So, the final speed of puck B is \(\sqrt{3}\ \mathrm{m}/\mathrm{s}\), and its direction is approximately 54.7° below the x-axis.

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Most popular questions from this chapter

A hockey puck moving at \(0.45 \mathrm{m} / \mathrm{s}\) collides elastically with another puck that was at rest. The pucks have equal mass. The first puck is deflected \(37^{\circ}\) to the right and moves off at $0.36 \mathrm{m} / \mathrm{s} .$ Find the speed and direction of the second puck after the collision.
A cue stick hits a cue ball with an average force of \(24 \mathrm{N}\) for a duration of \(0.028 \mathrm{s}\). If the mass of the ball is \(0.16 \mathrm{kg}\) how fast is it moving after being struck?
In the railroad freight yard, an empty freight car of mass \(m\) rolls along a straight level track at \(1.0 \mathrm{m} / \mathrm{s}\) and collides with an initially stationary, fully loaded boxcar of mass 4.0m. The two cars couple together on collision. (a) What is the speed of the two cars after the collision? (b) Suppose instead that the two cars are at rest after the collision. With what speed was the loaded boxcar moving before the collision if the empty one was moving at \(1.0 \mathrm{m} / \mathrm{s} ?\)
A \(58-\) kg astronaut is in space, far from any objects that would exert a significant gravitational force on him. He would like to move toward his spaceship, but his jet pack is not functioning. He throws a 720 -g socket wrench with a velocity of \(5.0 \mathrm{m} / \mathrm{s}\) in a direction away from the ship. After \(0.50 \mathrm{s}\), he throws a 800 -g spanner in the same direction with a speed of \(8.0 \mathrm{m} / \mathrm{s} .\) After another $9.90 \mathrm{s}\(, he throws a mallet with a speed of \)6.0 \mathrm{m} / \mathrm{s}$ in the same direction. The mallet has a mass of \(1200 \mathrm{g}\) How fast is the astronaut moving after he throws the mallet?
Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of \(0.50 \mathrm{m} / \mathrm{s},\) someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at $1.30 \mathrm{m} / \mathrm{s} .$ (a) What velocity does the other glider have? (b) Is the total kinetic energy of the two gliders after the collision greater than, less than, or equal to the total kinetic energy before the collision? If greater, where did the extra energy come from? If less, where did the "lost" energy go?
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