/*! 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 10 On a frictionless air track, a \... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

On a frictionless air track, a \(0.150 \mathrm{~kg}\) glider moving at \(1.20 \mathrm{~m} / \mathrm{s}\) to the right collides with and sticks to a stationary \(0.250 \mathrm{~kg}\) glider. (a) What is the net momentum of this two-glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the gliders after the collision. (d) Is kinetic energy conserved during the collision?

Short Answer

Expert verified
(a) 0.180 kg·m/s (b) 0.180 kg·m/s (c) 0.450 m/s (d) No, kinetic energy is not conserved.

Step by step solution

01

Calculate Initial Momentum

The initial momentum of the system can be determined using the formula for linear momentum, which is \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. The initial momentum only comes from the moving glider, i.e., \( p_{ ext{initial}} = 0.150 \, \text{kg} \times 1.20 \, \text{m/s} = 0.180 \, \text{kg} \cdot \text{m/s} \).
02

Determining Final Momentum

According to the law of conservation of momentum, the total momentum of a closed system remains constant if no external forces act on it. Hence, the net momentum after the collision is equal to the initial momentum. Thus, \( p_{ ext{final}} = 0.180 \, \text{kg} \cdot \text{m/s} \).
03

Solve for Final Speed

To find the final speed of the combined gliders, use the conservation of momentum formula: \( m_1v_1 + m_2v_2 = (m_1 + m_2)v_f \) where \( v_f \) is the final velocity. Substitute the known values: \( 0.180 = (0.150 + 0.250)v_f \), solve to find \( v_f = \frac{0.180}{0.400} = 0.450 \, \text{m/s} \).
04

Check Kinetic Energy Conservation

Kinetic energy before collision is \( KE_{ ext{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.150 \times (1.20)^2 = 0.108 \, \text{J} \). After collision, \( KE_{ ext{final}} = \frac{1}{2} \times 0.400 \times (0.450)^2 = 0.0405 \, \text{J} \). The kinetic energy is not conserved because \( 0.0405 \, \text{J} eq 0.108 \, \text{J} \). This is due to the inelastic nature of the collision.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Inelastic Collision
In an inelastic collision, the objects involved collide and stick together, as seen with the gliders in our exercise. Unlike elastic collisions, inelastic collisions do not conserve kinetic energy. This is because some of the kinetic energy is transformed into other forms of energy, such as heat or sound, during the collision process. In our example, when the moving glider hits the stationary one, they stick together, forming a combined mass. This behavior confirms the collision is inelastic. While the kinetic energy isn't conserved, the momentum is still maintained before and after the collision.
Momentum
Momentum is a fundamental concept in physics, representing the quantity of motion an object possesses. It is calculated using the formula \( p = mv \), where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity. In the exercise with the gliders, the momentum before and after the collision remains the same due to the conservation principle. Before the collision, only the moving glider contributes to the system's overall momentum. After the collision, the combined mass of the gliders means the system now moves at a reduced velocity but with the same total momentum as initially. This illustrates how momentum takes into account both mass and speed, maintaining balance in a frictionless scenario.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion and is given by the formula \( KE = \frac{1}{2} mv^2 \). In our context, the kinetic energy of the gliders is examined before and after the collision. Before the collision, only the moving glider has kinetic energy, calculated using its mass and speed. However, after the collision, despite both gliders moving together, their combined kinetic energy is less. This loss in kinetic energy is indicative of an inelastic collision, where energy is not conserved. The difference highlights how not all energy remains as motion energy post-collision, as some are transferred to other forms.
Frictionless Surface
A frictionless surface is an idealized concept where no opposing forces slow down or alter the motion of objects. In the given exercise, the frictionless air track ensures that the only forces at play are due to the collision itself. This absence of friction allows us to focus solely on the momentum and energy changes without external interference. The gliders can thus freely translate all momentum and most energy dynamics from the collision into motion. Without resistance, the principles of momentum conservation and energy transformation become clearer and easier to calculate, simplifying the problem-solving process.

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

Combining conservation laws. A \(5.00 \mathrm{~kg}\) chunk of ice is sliding at \(12.0 \mathrm{~m} / \mathrm{s}\) on the floor of an ice-covered valley when it collides with and sticks to another \(5.00 \mathrm{~kg}\) chunk of ice that is initially at rest. (See Figure \(8.39 .\) ) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? (Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

A \(15.0 \mathrm{~kg}\) block is attached to a very light horizontal spring of force constant \(500.0 \mathrm{~N} / \mathrm{m}\) and is resting on a frictionless horizontal table. (See Figure \(8.40 .\) ) Suddenly it is struck by a \(3.00 \mathrm{~kg}\) stone traveling horizontally at \(8.00 \mathrm{~m} / \mathrm{s}\) to the right, whereupon the stone rebounds at \(2.00 \mathrm{~m} / \mathrm{s}\) horizontally to the left. Find the maximum distance that the block will compress the spring after the collision. (Hint: Break this problem into two parts the collision and the behavior after the collision -and apply the appropriate conservation law to each part.)

On a frictionless, horizontal air table, puck \(A\) (with mass \(0.250 \mathrm{~kg}\) ) is moving to the right toward puck \(B\) (with mass \(0.350 \mathrm{~kg}\) ), which is initially at rest. After the collision, puck \(A\) has a velocity of \(0.120 \mathrm{~m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of \(0.650 \mathrm{~m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Forensic scientists can measure the muzzle velocity of a gun by firing a bullet horizontally into a large hanging block that absorbs the bullet and swings upward. (See Figure \(8.52 .\) ) The measured maximum angle of swing can be used to calculate the speed of the bullet. In one such test, a rifle fired a \(4.20 \mathrm{~g}\) bullet into a \(2.50 \mathrm{~kg}\) block hanging by a thin wire \(75.0 \mathrm{~cm}\) long, causing the block to swing upward to a maximum angle of \(34.7^{\circ}\) from the vertical. What was the original speed of this bullet?

Biomechanics. The mass of a regulation tennis ball is \(57.0 \mathrm{~g}\) (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for \(30 \mathrm{~ms}\). (This number can also vary, depending on the racket and swing.) We assume a \(30.0 \mathrm{~ms}\) contact time in this problem. In the 2011 Davis Cup competition, Ivo Karlovic made one of the fastest recorded serves in history, which was clocked at \(156 \mathrm{mph}(70 \mathrm{~m} / \mathrm{s}) .\) (a) What impulse and what average force did Karlovic exert on the tennis ball in his record serve? (b) If his opponent returned this serve with a speed of \(55.0 \mathrm{~m} / \mathrm{s},\) what impulse and what average force did his opponent exert on the ball, assuming purely horizontal motion?

See all solutions

Recommended explanations on Physics Textbooks

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.