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Chapter 7: Problem 60

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.

Short Answer

Expert verified
Answer: The final speed of the second puck is 0.54 m/s, and its direction is 64.24 degrees to the left.

Step by step solution

01

Identify the given information

We are given: - Initial velocity of the first puck: \(v_1 = 0.45 \mathrm{m/s}\) - The first puck's angle after collision: \(37^{\circ}\) - Final velocity of the first puck: \(v'_1= 0.36 \mathrm{m/s}\) - Initial velocity of the second puck: \(v_2 = 0 \mathrm{m/s}\) - Mass of both pucks are equal: \(m_1=m_2\)
02

Apply conservation of momentum

Since the collision is elastic, both momentum and kinetic energy are conserved. For the conservation of momentum, we have: $$m_1v_1 + m_2v_2 = m_1v'_1 + m_2v'_2$$ Since \(m_1 = m_2\) and \(v_2 = 0\), this simplifies to: $$v_1 = v'_1 + v'_2$$
03

Apply conservation of kinetic energy

For the conservation of kinetic energy, we have: $$\frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 = \frac{1}{2}m_1(v'_1)^2 + \frac{1}{2}m_2(v'_2)^2$$ Since \(m_1 = m_2\) and \(v_2 = 0\), this simplifies to: $$v_1^2 = (v'_1)^2 + (v'_2)^2$$
04

Analyze the angles

The first puck's angle after the collision is given as \(37^{\circ}\), and it's moving to the right. We can break its velocity into horizontal and vertical components: $$v'_{1x} = v'_1\cos{37^{\circ}}$$ $$v'_{1y} = v'_1\sin{37^{\circ}}$$ Since the final velocity of the second puck is unknown, we can write the angle it makes after the collision as \(θ\), and we can break its velocity components as: $$v'_{2x} = v'_2\cos{θ}$$ $$v'_{2y} = v'_2\sin{θ}$$
05

Apply conservation of momentum for horizontal and vertical components

As there was no initial vertical velocity, we have: $$v'_{1y} + v'_{2y} = 0$$ Substituting the expressions for \(v'_{1y}\) and \(v'_{2y}\), we get: $$v'_1\sin{37^{\circ}} + v'_2\sin{θ} = 0$$ For the horizontal component, momentum conservation gives us: $$v_1 = v'_{1x} + v'_{2x}$$ Substituting the expressions for \(v'_{1x}\) and \(v'_{2x}\), we get: $$0.45 = v'_1\cos{37^{\circ}} + v'_2\cos{θ}$$
06

Solve for the final speed and direction of the second puck

We now have a system of equations to solve for \(v'_2\) and \(θ\). Using the given \(v'_1 = 0.36 \mathrm{m/s}\), we can solve the equations: 1. Using the vertical momentum equation: $$v'_2\sin{θ} = -v'_1\sin{37^{\circ}}$$ 2. Using the horizontal momentum equation: $$v'_2\cos{θ} = 0.45 - v'_1\cos{37^{\circ}}$$ 3. Squaring both equations and summing them (since \((v'_2\sin{θ})^2 + (v'_2\cos{θ})^2 = (v'_2)^2\)), we find the value of \(v'_2\): $$v'_2 = 0.54 \mathrm{m/s}$$ 4. Using the vertical momentum equation again, we can find \(θ\): $$θ = \arctan{\frac{-v'_1\sin{37^{\circ}}}{v'_2}}$$ $$θ = 64.24^{\circ}$$
07

Final Answer

The speed of the second puck after the collision is \(0.54 \mathrm{m/s}\), and its direction is \(64.24^{\circ}\) to the left.

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