Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 49

A projectile of 1.0-kg mass approaches a stationary body of \(5.0 \mathrm{kg}\) at \(10.0 \mathrm{m} / \mathrm{s}\) and, after colliding, rebounds in the reverse direction along the same line with a speed of $5.0 \mathrm{m} / \mathrm{s} .$ What is the speed of the 5.0 -kg body after the collision?

Short Answer

Expert verified
Answer: The final velocity of the 5.0-kg body after the collision is 3.0 m/s.

Step by step solution

01

Write the linear momentum conservation equation

The linear momentum conservation equation states that the total momentum before a collision is equal to the total momentum after the collision. In this case, we have two bodies (projectile and stationary body), so we can write the equation as: \(m_1v_1 + m_2v_2 = m_1v_1' + m_2v_2'\) where \(m_1\) and \(m_2\) are the masses of the projectile and the stationary body respectively, \(v_1\) and \(v_2\) are their initial velocities, and \(v_1'\) and \(v_2'\) are their final velocities after the collision.
02

Input the known values

Now, let's plug in the values given in the exercise: \(m_1=1.0\,\mathrm{kg}\), \(v_1=10.0\,\mathrm{m/s}\), \(m_2=5.0\,\mathrm{kg}\), \(v_2=0\,\mathrm{m/s}\) (stationary), and \(v_1'=-5.0\,\mathrm{m/s}\) (since it rebounds in the reverse direction). We want to find \(v_2'\). \((1.0\,\mathrm{kg})(10.0\,\mathrm{m/s}) + (5.0\,\mathrm{kg})(0\,\mathrm{m/s}) = (1.0\,\mathrm{kg})(-5.0\,\mathrm{m/s}) + (5.0\,\mathrm{kg})v_2'\)
03

Solve for \(v_2'\)

We will now solve for \(v_2'\) in the equation we derived in Step 2: \(10\,\mathrm{kg\cdot m/s} = -5\,\mathrm{kg\cdot m/s} + 5.0\,\mathrm{kg}\cdot v_2'\) Now, add \(5\,\mathrm{kg\cdot m/s}\) to both sides of the equation: \(15\,\mathrm{kg\cdot m/s} = 5.0\,\mathrm{kg}\cdot v_2'\) Finally, divide both sides by \(5.0\,\mathrm{kg}\) to get the final velocity \(v_2'\): \(v_2' = \frac{15\,\mathrm{kg\cdot m/s}}{5.0\,\mathrm{kg}}\) \(v_2' = 3.0\,\mathrm{m/s}\) The final velocity of the 5.0-kg body after the collision is \(3.0\,\mathrm{m/s}\).

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

A submarine of mass \(2.5 \times 10^{6} \mathrm{kg}\) and initially at rest fires a torpedo of mass \(250 \mathrm{kg} .\) The torpedo has an initial speed of \(100.0 \mathrm{m} / \mathrm{s} .\) What is the initial recoil speed of the submarine? Neglect the drag force of the water.
Two identical gliders on an air track are held together by a piece of string, compressing a spring between the gliders. While they are moving to the right at a common speed of \(0.50 \mathrm{m} / \mathrm{s},\) someone holds a match under the string and burns it, letting the spring force the gliders apart. One glider is then observed to be moving to the right at $1.30 \mathrm{m} / \mathrm{s} .$ (a) What velocity does the other glider have? (b) Is the total kinetic energy of the two gliders after the collision greater than, less than, or equal to the total kinetic energy before the collision? If greater, where did the extra energy come from? If less, where did the "lost" energy go?
Use the result of Problem 54 to show that in any elastic collision between two objects, the relative speed of the two is the same before and after the collision. [Hints: Look at the collision in its \(\mathrm{CM}\) frame - the reference frame in which the \(\mathrm{CM}\) is at rest. The relative speed of two objects is the same in any inertial reference frame.]
Prove Eq. $(7-13) \Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}=M \overrightarrow{\mathbf{a}}_{\mathrm{CM}} .\( [Hint: Start with \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{cxt}}=\lim _{\Delta t \rightarrow 0}(\Delta \overrightarrow{\mathbf{p}} / \Delta t),\( where \)\Sigma \overrightarrow{\mathbf{F}}_{\mathrm{ext}}$ is the net external force acting on a system and \(\overrightarrow{\mathbf{p}}\) is the total momentum of the system.]
A stationary 0.1-g fly encounters the windshield of a \(1000-\mathrm{kg}\) automobile traveling at \(100 \mathrm{km} / \mathrm{h} .\) (a) What is the change in momentum of the car due to the fly? (b) What is the change of momentum of the fly due to the car? (c) Approximately how many flies does it take to reduce the car's speed by \(1 \mathrm{km} / \mathrm{h} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Physics of Motion

Read Explanation

Waves Physics

Read Explanation

Measurements

Read Explanation

Scientific Method Physics

Read Explanation

Classical Mechanics

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