Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 103

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right),\) which was at rest momentarily. After the collision, the speed of \(M_{1}\) was reduced to \(.500 \mathrm{~m} / \mathrm{s}\), and it was moving at an angle of \(32^{\circ}\) downward with respect to the horizontal. Assume the collision is elastic. a) What is the speed of \(M_{2}\) after the collision? b) What is the angle between the velocity vectors of \(M_{1}\) and \(M_{2}\) after the collision?

Short Answer

Expert verified
Question: In an elastic collision between two particles M1 and M2, particle M1 initially moves at 2.50 m/s at an angle of 30 degrees downward from the horizontal and particle M2 is at rest. After the collision, particle M1 moves at 1.75 m/s at an angle of 32 degrees upward from the horizontal. Determine the speed and angle of particle M2 after the collision and the angle between the velocity vectors of the particles after the collision. Answer: Follow the step-by-step solution to find the speed (v2) and angle (θ2) of particle M2 after the collision, as well as the angle between the velocity vectors of particles M1 and M2.

Step by step solution

01

Determine the initial momentum of each particle

To have a clear view of the collision, sketch a diagram with the velocities of both particles. For particle M1, we have the initial velocity given to be v1 = 2.50 m/s at an angle 30 degrees downward from the horizontal, while particle M2 is at rest, so its initial velocity is 0 m/s. Now, determine the initial momentum of both particles in x and y coordinates: $$p_{1x} = m_{1}v_{1}\cos(30^{\circ})$$ $$p_{1y} = -m_{1}v_{1}\sin(30^{\circ})$$ $$p_{2x} = 0$$ $$p_{2y} = 0$$
02

Determine the final momentum of the particles

We also need the final momentum of both particles M1 and M2 after the collision. To do this, we must use the final velocities and angles given for the particle M1 and unknowns for the particle M2. We want to find out \(v_{2}, \theta_{2}\), first determine the final momentum of M1 after the collision: $$p'_{1x} = m_{1}v'_{1}\cos(32^{\circ})$$ $$p'_{1y} = -m_{1}v'_{1}\sin(32^{\circ})$$ For M2, we have unknowns as the magnitude of the velocity, \(v_{2}\) and angle \(\theta_{2}\): $$p'_{2x} = m_{2}v_{2}\cos(\theta_{2})$$ $$p'_{2y} = -m_{2}v_{2}\sin(\theta_{2})$$
03

Applying conservation of momentum

Applying momentum conservation along x and y directions respectively, we get: $$p_{1x} + p_{2x} = p'_{1x} + p'_{2x}$$ $$p_{1y} + p_{2y} = p'_{1y} + p'_{2y}$$ Substituting the expressions for momenta, we get $$m_{1}v_{1}\cos(30^{\circ}) = m_{1}v'_{1}\cos(32^{\circ}) + m_{2}v_{2}\cos(\theta_{2}) \qquad (1)$$ $$-m_{1}v_{1}\sin(30^{\circ}) = -m_{1}v'_{1}\sin(32^{\circ}) - m_{2}v_{2}\sin(\theta_{2}) \qquad (2)$$
04

Applying conservation of kinetic energy

For the collision to be elastic, the kinetic energy should also be conserved. So, $$\frac{1}{2}m_{1}v_{1}^2 + \frac{1}{2}m_{2}0^2 = \frac{1}{2}m_{1}v'_{1}^2 + \frac{1}{2}m_{2}v_{2}^2$$ Solving for \(v_{2}\), we get $$v_{2} = \sqrt{\frac{(2m_{1}v_{1}^2 - 2m_{1}v'_{1}^2)}{2m_{2}}} \qquad (3)$$
05

Solving for \(v_{2}\) and \(\theta_{2}\)

Using equation (3), we can calculate the value of \(v_{2}\). Now, knowing the value for \(v_{2}\), we can solve for the angle \(\theta_{2}\) using equations (1) and (2) simultaneously. Substitute the calculated value of \(v_{2}\) into equation (1) and (2), and then solve for \(\theta_{2}\).
06

Finding the angle between velocity vectors of particles after collision

Finally, we can find the angle between the velocity vectors of M1 and M2 after the collision by using trigonometry. $$\cos^{-1}(\cos\theta_{2}\cos(32^{\circ}) + \sin\theta_{2}\sin(32^{\circ}))$$

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

Stuck in the middle of a frozen pond with only your physics book, you decide to put physics in action and throw the 5.00 -kg book. If your mass is \(62.0 \mathrm{~kg}\) and you throw the book at \(13.0 \mathrm{~m} / \mathrm{s}\), how fast do you then slide across the ice? (Assume the absence of friction.)

In waterskiing, a "garage sale" occurs when a skier loses control and falls and waterskis fly in different directions. In one particular incident, a novice skier was skimming across the surface of the water at \(22.0 \mathrm{~m} / \mathrm{s}\) when he lost control. One ski, with a mass of \(1.50 \mathrm{~kg},\) flew off at an angle of \(12.0^{\circ}\) to the left of the initial direction of the skier with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). The other identical ski flew from the crash at an angle of \(5.00^{\circ}\) to the right with a speed of \(21.0 \mathrm{~m} / \mathrm{s} .\) What was the velocity of the \(61.0-\mathrm{kg}\) skier? Give a speed and a direction relative to the initial velocity vector.

Rank the following objects from highest to lowest in terms of momentum and from highest to lowest in terms of energy. a) an asteroid with mass \(10^{6} \mathrm{~kg}\) and speed \(500 \mathrm{~m} / \mathrm{s}\) b) a high-speed train with a mass of \(180,000 \mathrm{~kg}\) and a speed of \(300 \mathrm{~km} / \mathrm{h}\) c) a 120 -kg linebacker with a speed of \(10 \mathrm{~m} / \mathrm{s}\) d) a 10 -kg cannonball with a speed of \(120 \mathrm{~m} / \mathrm{s}\) e) a proton with a mass of \(6 \cdot 10^{-27} \mathrm{~kg}\) and a speed of \(2 \cdot 10^{8} \mathrm{~m} / \mathrm{s}\).

When hit in the face, a boxer will "ride the punch"; that is, if he anticipates the punch, he will allow his neck muscles to go slack. His head then moves back easily from the blow. From a momentum-impulse standpoint, explain why this is much better than stiffening his neck muscles and bracing himself against the punch.

An alpha particle (mass \(=4.00 \mathrm{u}\) ) has a head-on, elastic collision with a nucleus (mass \(=166 \mathrm{u}\) ) that is initially at rest. What percentage of the kinetic energy of the alpha particle is transferred to the nucleus in the collision?
See all solutions

Recommended explanations on Physics Textbooks

Fundamentals of Physics

Read Explanation

Solid State Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Waves Physics

Read Explanation

Conservation of Energy and Momentum

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