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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 2.0-kg object is at rest on a perfectly frictionless surface when it is hit by a 3.0-kg object moving at \(8.0 \mathrm{m} / \mathrm{s}\) If the two objects are stuck together after the collision, what is the speed of the combination?
A 0.020-kg bullet is shot horizontally and collides with a \(2.00-\mathrm{kg}\) block of wood. The bullet embeds in the block and the block slides along a horizontal surface for \(1.50 \mathrm{m} .\) If the coefficient of kinetic friction between the block and surface is \(0.400,\) what was the original speed of the bullet?
A block of wood of mass \(0.95 \mathrm{kg}\) is initially at rest. A bullet of mass \(0.050 \mathrm{kg}\) traveling at \(100.0 \mathrm{m} / \mathrm{s}\) strikes the block and becomes embedded in it. With what speed do the block of wood and the bullet move just after the collision?
In a circus trapeze act, two acrobats actually fly through the air and grab on to one another, then together grab a swinging bar. One acrobat, with a mass of \(60 \mathrm{kg},\) is moving at \(3.0 \mathrm{m} / \mathrm{s}\) at an angle of \(10^{\circ}\) above the horizontal and the other, with a mass of $80 \mathrm{kg},\( is approaching her with a speed of \)2.0 \mathrm{m} / \mathrm{s}$ at an angle of \(20^{\circ}\) above the horizontal. What is the direction and speed of the acrobats right after they grab on to each other? Let the positive \(x\) -axis be in the horizontal direction and assume the first acrobat has positive velocity components in the positive \(x\)- and \(y\)-directions.
A jet plane is flying at 130 m/s relative to the ground. There is no wind. The engines take in \(81 \mathrm{kg}\) of air per second. Hot gas (burned fuel and air) is expelled from the engines at high speed. The engines provide a forward force on the plane of magnitude \(6.0 \times 10^{4} \mathrm{N} .\) At what speed relative to the ground is the gas being expelled? [Hint: Look at the momentum change of the air taken in by the engines during a time interval \(\Delta t .\) ] This calculation is approximate since we are ignoring the \(3.0 \mathrm{kg}\) of fuel consumed and expelled with the air each second.
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