Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 88

A bored boy shoots a soft pellet from an air gun at a piece of cheese with mass \(0.25 \mathrm{~kg}\) that sits, keeping cool for dinner guests, on a block of ice. On one particular shot, his 1.2-g pellet gets stuck in the cheese, causing it to slide \(25 \mathrm{~cm}\) before coming to a stop. According to the package the gun came in, the muzzle velocity is \(65 \mathrm{~m} / \mathrm{s}\). What is the coefficient of friction between the cheese and the ice?

Short Answer

Expert verified
Question: Find the coefficient of friction between the cheese and the ice, given the mass of the cheese, the mass and muzzle velocity of the pellet, and the sliding distance of the cheese after getting hit by the pellet. Answer: To find the coefficient of friction between the cheese and the ice, you need to follow these steps: 1. Calculate the initial momentum of the system. 2. Calculate the final momentum of the system. 3. Use the conservation of momentum to find the final velocity of the cheese after the pellet hits it. 4. Use the work-energy theorem to find the work done in overcoming friction. 5. Calculate the friction force. 6. Find the normal force between the cheese and the ice. 7. Calculate the coefficient of friction by dividing the friction force by the normal force. Plug in the given values into the equations derived in each step and solve for the coefficient of friction.

Step by step solution

01

Calculate initial momentum of the system

The initial momentum of the system consists only of the momentum of the pellet. We have mass of pellet (\(m_p\)) and its muzzle velocity (\(v_p\)). $$ p_{initial} = m_p \times v_p $$
02

Calculate the final momentum of the system

In the final state, the cheese and the pellet move together. Let their final combined velocity be \(v_f\). The mass of the cheese is \(m_c\). Then, $$ p_{final} = (m_p + m_c) \times v_f $$
03

Use the conservation of momentum

Using the conservation of momentum, we can set the initial momentum equal to the final momentum. $$ m_p \times v_p = (m_p + m_c) \times v_f $$ Solve for \(v_f\): $$ v_f = \frac{m_p \times v_p}{m_p + m_c} $$
04

Use the work-energy theorem

The work done by friction during the sliding motion is equal to the change in kinetic energy of the cheese-pellet system. Let \(F_f\) be the friction force and \(d\) be the sliding distance. Then, the work done by friction force is: $$ W = F_f \times d $$ The work-energy theorem states that: $$ W = \Delta KE = KE_{final} - KE_{initial} $$ Since the system comes to a stop, \(KE_{final} = 0\). The initial kinetic energy is: $$ KE_{initial} = \frac{1}{2}(m_p + m_c)v_f^2 $$
05

Calculate the friction force

From the previous step, we have: $$ F_f \times d = \frac{1}{2}(m_p + m_c)v_f^2 $$ Solve for \(F_f\): $$ F_f = \frac{(m_p + m_c)v_f^2}{2d} $$
06

Find the normal force

The normal force (\(F_N\)) is equal to the weight of the cheese and the pellet. $$ F_N = g(m_p + m_c) $$ where \(g = 9.8 \mathrm{~m/s^2}\) is the gravitational acceleration.
07

Calculate the coefficient of friction

The coefficient of friction (\(\mu\)) can be found using the friction force and normal force: $$ \mu = \frac{F_f}{F_N} $$ Substitute the values from Step 5 and Step 6, and solve for \(\mu\). After calculating the values, you will get the coefficient of friction between the cheese and the ice.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

In a severe storm, \(1.00 \mathrm{~cm}\) of rain falls on a flat horizontal roof in \(30.0 \mathrm{~min}\). If the area of the roof is \(100 \mathrm{~m}^{2}\) and the terminal velocity of the rain is \(5.00 \mathrm{~m} / \mathrm{s}\), what is the average force exerted on the roof by the rain during the storm?

A soccer ball rolls out of a gym through the center of a doorway into the next room. The adjacent room is \(6.00 \mathrm{~m}\) by \(6.00 \mathrm{~m}\) with the \(2.00-\mathrm{m}\) wide doorway located at the center of the wall. The ball hits the center of a side wall at \(45.0^{\circ} .\) If the coefficient of restitution for the soccer ball is \(0.700,\) does the ball bounce back out of the room? (Note that the ball rolls without slipping, so no energy is lost to the floor.)

A bullet with mass \(35.5 \mathrm{~g}\) is shot horizontally from a gun. The bullet embeds in a 5.90 -kg block of wood that is suspended by strings. The combined mass swings upward, gaining a height of \(12.85 \mathrm{~cm}\). What was the speed of the bullet as it left the gun? (Air resistance can be ignored here.)

A bungee jumper with mass \(55.0 \mathrm{~kg}\) reaches a speed of \(13.3 \mathrm{~m} / \mathrm{s}\) moving straight down when the elastic cord tied to her feet starts pulling her back up. After \(0.0250 \mathrm{~s},\) the jumper is heading back up at a speed of \(10.5 \mathrm{~m} / \mathrm{s}\). What is the average force that the bungee cord exerts on the jumper? What is the average number of \(g\) 's that the jumper experiences during this direction change?

A rocket works by expelling gas (fuel) from its nozzles at a high velocity. However, if we take the system to be the rocket and fuel, explain qualitatively why a stationary rocket is able to move.
See all solutions

Recommended explanations on Physics Textbooks

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Measurements

Read Explanation

Modern Physics

Read Explanation

Classical Mechanics

Read Explanation

Work Energy and Power

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.

Sign-up for free