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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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