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Chapter 7: Problem 61

Puck 1 sliding along the \(x\) -axis strikes stationary puck 2 of the same mass. After the elastic collision, puck 1 moves off at speed \(v_{1 f}\) in the direction \(60.0^{\circ}\) above the \(x\)-axis; puck 2 moves off at speed $v_{2 f}\( in the direction \)30.0^{\circ}\( below the \)x\(-axis. Find \)v_{2 \mathrm{f}}\( in terms of \)v_{1 f}$

Short Answer

Expert verified
Answer: The final speed of Puck 2 is \(v_{2f} = \frac{2\sqrt{3}}{3} v_{1i}\), where \(v_{1i}\) is the initial speed of Puck 1.

Step by step solution

01

Write the conservation of momentum equations for x and y axes separately

Since this is an elastic collision, we will need to use the conservation of momentum formula which states that: Total initial momentum = Total final momentum For the x-axis: \(m_1 \cdot v_{1i} = m_1 \cdot v_{1f} \cdot \cos(60^{\circ}) + m_2 \cdot v_{2f} \cdot \cos(30^{\circ})\) For the y-axis: \(0 = m_1 \cdot v_{1f} \cdot \sin(60^{\circ}) - m_2 \cdot v_{2f} \cdot \sin(30^{\circ})\)
02

Simplify the equations

Now let's simplify the above equations using trigonometric values and the fact that the masses are the same: For the x-axis: \(v_{1i} = \frac{1}{2} v_{1f} + \frac{\sqrt{3}}{2} v_{2f}\) For the y-axis: \(0 = \frac{\sqrt{3}}{2} v_{1f} - \frac{1}{2} v_{2f}\)
03

Solve the system of equations for the required value

We need to find \(v_{2f}\) in terms of \(v_{1f}\). Let's use the y-axis equation to solve for \(v_{2f}\): \(v_{2f} = \sqrt{3}v_{1f}\) Now, plug this value of \(v_{2f}\) into the x-axis equation: \(v_{1i} = \frac{1}{2} v_{1f} + \frac{\sqrt{3}}{2} (\sqrt{3}v_{1f})\)
04

Simplify and find the value of \(v_{2f}\) in terms of \(v_{1f}\)

Simplify the above equation: \(v_{1i} = v_{1f}(\frac{1}{2} + \frac{3}{2})\) \(v_{1f} = \frac{2}{3} v_{1i}\) Now replace \(v_{1f}\) with the expression we found in terms of \(v_{1i}\) into the expression we found for \(v_{2f}\): \(v_{2f} = \sqrt{3}v_{1f}\) \(v_{2f} = \sqrt{3}(\frac{2}{3} v_{1i})\) \(v_{2f} = \frac{2\sqrt{3}}{3} v_{1i}\)

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Most popular questions from this chapter

A cannon on a railroad car is facing in a direction parallel to the tracks. It fires a \(98-\mathrm{kg}\) shell at a speed of \(105 \mathrm{m} / \mathrm{s}\) (relative to the ground) at an angle of \(60.0^{\circ}\) above the horizontal. If the cannon plus car have a mass of \(5.0 \times 10^{4} \mathrm{kg},\) what is the recoil speed of the car if it was at rest before the cannon was fired? [Hint: A component of a system's momentum along an axis is conserved if the net external force acting on the system has no component along that axis.]
A system consists of three particles with these masses and velocities: mass \(3.0 \mathrm{kg},\) moving north at \(3.0 \mathrm{m} / \mathrm{s}\) mass $4.0 \mathrm{kg},\( moving south at \)5.0 \mathrm{m} / \mathrm{s} ;\( and mass \)7.0 \mathrm{kg}\( moving north at \)2.0 \mathrm{m} / \mathrm{s} .$ What is the total momentum of the system?
A 1500-kg car moving east at \(17 \mathrm{m} / \mathrm{s}\) collides with a 1800-\(\mathrm{kg}\) car moving south at \(15 \mathrm{m} / \mathrm{s}\) and the two cars stick together. (a) What is the velocity of the cars right after the collision? (b) How much kinetic energy was converted to another form during the collision?
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A 75-kg man is at rest on ice skates. A 0.20-kg ball is thrown to him. The ball is moving horizontally at \(25 \mathrm{m} / \mathrm{s}\) just before the man catches it. How fast is the man moving just after he catches the ball?
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