/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 61 Puck 1 sliding along the \(x\) -... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 \(v_{2f}\): \(v_{2f} = \sqrt{3}v_{1f}\) \(v_{2f} = \sqrt{3}(\frac{2}{3} v_{1i})\) \(v_{2f} = \frac{2\sqrt{3}}{3} v_{1i}\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

If a particle of mass \(5.0 \mathrm{kg}\) is moving east at $10 \mathrm{m} / \mathrm{s}\( and a particle of mass \)15 \mathrm{kg}\( is moving west at \)10 \mathrm{m} / \mathrm{s},$ what is the velocity of the CM of the pair?
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.

The positions of three particles, written as \((x, y)\) coordinates, are: particle 1 (mass \(4.0 \mathrm{kg}\) ) at \((4.0 \mathrm{m}, 0 \mathrm{m}) ;\) particle \(2(\text { mass } 6.0 \mathrm{kg} \text { ) at }(2.0 \mathrm{m}, 4.0 \mathrm{m}) ;\) particle \(3(\mathrm{mass} 3.0 \mathrm{kg})\) at \((-1.0 \mathrm{m},-2.0 \mathrm{m}) .\) What is the location of the \(\mathrm{cm} ?\)
A car with a mass of \(1700 \mathrm{kg}\) is traveling directly northeast $\left(45^{\circ} \text { between north and east) at a speed of } 14 \mathrm{m} / \mathrm{s}\right.\( \)(31 \mathrm{mph}),$ and collides with a smaller car with a mass of \(1300 \mathrm{kg}\) that is traveling directly south at a speed of \(18 \mathrm{m} / \mathrm{s}\) \((40 \mathrm{mph}) .\) The two cars stick together during the collision. With what speed and direction does the tangled mess of metal move right after the collision?
A pole-vaulter of mass \(60.0 \mathrm{kg}\) vaults to a height of $6.0 \mathrm{m}$ before dropping to thick padding placed below to cushion her fall. (a) Find the speed with which she lands. (b) If the padding brings her to a stop in a time of \(0.50 \mathrm{s},\) what is the average force on her body due to the padding during that time interval?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Dynamics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Waves Physics

Read Explanation

Medical Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.