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Chapter 4: Problem 6

A very massive object with velocity \(v\) collides head-on with an object at rest whose mass is very small. No kinetic energy is converted into other forms. Prove that the low-mass object recoils with velocity \(2 v\). [Hint: Use the center-of-mass frame of reference.]

Short Answer

Expert verified
The low-mass object recoils with velocity \(2v\).

Step by step solution

01

Understanding the Problem

We are dealing with a collision between a massive object and a much smaller one where no kinetic energy is lost, meaning it's an elastic collision. The big object's velocity is initially \( v \), and the other object is at rest.
02

Set Up the Center-of-Mass Frame

The center-of-mass (CM) frame is a reference frame where the total momentum is zero. In this frame, the massive object has nearly all the momentum. We will analyze the collision in the CM frame because it simplifies calculations.
03

Calculate Initial Velocities in the CM Frame

The velocity of the center of mass \( V_{cm} \) is given by \( V_{cm} = \frac{Mv + 0}{M+m}\), where \( M \) is the mass of the massive object and \( m \) is the smaller mass. In the CM frame, the massive object's velocity is \( v' = v - V_{cm} \) and the smaller object's initial velocity is \( v_{small}' = -V_{cm} \).
04

Apply Conservation of Momentum and Energy in CM Frame

In an elastic collision, both momentum and kinetic energy are conserved. After the collision, the velocities swap due to the symmetry of the situation in the CM frame. Thus, the small object's velocity becomes equal and opposite to its initial velocity, \( v_{small}'' = V_{cm} - (v - V_{cm}) = 2V_{cm} - v \).
05

Transform Back to Laboratory Frame

To find the final velocity of the small object in the initial lab frame, we convert from the CM frame by adding \( V_{cm} \). Thus, we get: \( v_{small, lab} = V_{cm} + (2V_{cm} - v - V_{cm}) = 2V_{cm} - v = 2v - v = v \). The velocity of the small object in the lab frame becomes twice the bigger object's velocity: \( 2v \).
06

Conclusion

In the lab frame, after the collision, the smaller object moves with velocity \( v' = 2v \). This reflects the idea that the impact effectively doubles the speed of the lighter object after bouncing off the massive object, which maintains nearly all initial momentum.

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Key Concepts

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

Understanding the Center-of-Mass Frame
Imagine two objects in a collision, with one being very massive and the other much smaller. \(
\) To simplify understanding this scenario, we introduce the center-of-mass (CM) frame. This is a special type of reference frame where the total momentum of the system is zero. \(
\)When analyzing collisions, the CM frame can be very helpful. It allows us to study the movement of each object with respect to the collective center of both masses. \(
\)
  • Think of the CM frame as a balancing point where the effects of each object's mass and velocity cancel out, making equations easier to solve.
    • \(
    \)
  • In this particular problem, the massive object carries nearly all the momentum due to its larger mass.
    • \(
    \)
The CM frame provides a symmetrical point of view, making the complex interactions of collisions simpler to understand.
The Principle of Conservation of Momentum
In physics, momentum is a product of an object's mass and velocity. \(
\) For elastic collisions, the principle of the conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. \(
\)
  • Apply the conservation law: \( Mv + m imes 0 = Mv' + mv'' \).
    • \(
    \)
  • Before the collision, the massive object contributes the full momentum \((Mv)\) because the smaller one is at rest.
    • \(
    \)
  • We solve for the velocities after the collision using this principle, staying in the CM frame simplifies the math.
    • \(
    \)
This law is fundamental because it tells us that in the absence of external forces, the momentum of a system is conserved, even if objects within it move or change velocity.
Kinetic Energy Conservation in Elastic Collisions
Kinetic energy, the energy of motion, is also conserved in elastic collisions. This means no energy is lost to sound, heat, or deformation. \(
\)
  • An elastic collision maintains the sum of kinetic energies of involved bodies.
    • \(
    \)
  • The initial kinetic energy \(\left(\frac{1}{2} Mv^2\right)\) remains equal to the final sum \(\left(\frac{1}{2} Mv'^2 + \frac{1}{2} mv''^2\right)\).
    • \(
    \)
  • This equality ensures both momentum and energy are balanced in the CM frame, resulting in realistic post-collision velocities.
    • \(
    \)
The conservation of kinetic energy, alongside momentum, makes elastic collisions unique and predictable. \(
\)Hence, why the small object can leave with a velocity of \(2v\) after adjusting from CM back to the lab frame.
Exploring Reference Frames in Physics
Reference frames in physics are perspectives from which we observe physical phenomena, crucial for analyzing motion. \(
\)
  • By selecting the most fitting reference frame, we ease the complexity of understanding motion, collisions, or other events.
    • \(
    \)
  • In the case of a moving observer (lab frame) versus one situated at the center of mass (CM frame), different insights about the same event are possible.
    • \(
    \)
  • This exercise uses the CM frame for simplicity but converts findings back to the lab frame, aligning results with real-world observation.
    • \(
    \)
Reference frames help us break problems into easier steps by transforming complex interactions, thus ensuring accurate predictions of outcomes.

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Most popular questions from this chapter

At low speeds, every car's acceleration is limited by traction, not by the engine's power. Suppose that at low speeds, a certain car is normally capable of an acceleration of \(3 \mathrm{~m} / \mathrm{s}^{2}\). If it is towing a trailer with half as much mass as the car itself, what acceleration can it achieve? [Based on a problem from PSSC Physics.]

(solution in the pdf version of the book) A firework shoots up into the air, and just before it explodes it has a certain momentum and kinetic energy. What can you say about the momenta and kinetic energies of the pieces immediately after the explosion? [Based on a problem from PSSC Physics.]

A dinosaur fossil is slowly moving down the slope of a glacier under the influence of wind, rain and gravity. At the same time, the glacier is moving relative to the continent underneath. The dashed lines represent the directions but not the magnitudes of the velocities. Pick a scale, and use graphical addition of vectors to find the magnitude and the direction of the fossil's velocity relative to the continent. You will need a ruler and protractor. (answer check available at lightandmatter.com)

A car accelerates from rest. At low speeds, its acceleration is limited by static friction, so that if we press too hard on the gas, we will "burn rubber" (or, for many newer cars, a computerized traction-control system will override the gas pedal). At higher speeds, the limit on acceleration comes from the power of the engine, which puts a limit on how fast kinetic energy can be developed. (a) Show that if a force \(F\) is applied to an object moving at speed \(v\), the power required is given by \(P=v F\). (b) Find the speed \(v\) at which we cross over from the first regime described above to the second. At speeds higher than this, the engine does not have enough power to burn rubber. Express your result in terms of the car's power \(P\), its mass \(m\), the coefficient of static friction \(\mu_{s}\), and \(g\).(answer check available at lightandmatter.com) (c) Show that your answer to part b has units that make sense. (d) Show that the dependence of your answer on each of the four variables makes sense physically. (e) The 2010 Maserati Gran Turismo Convertible has a maximum power of \(3.23 \times 10^{5} \mathrm{~W}\) (433 horsepower) and a mass (including a 50 -kg driver) of \(2.03 \times 10^{3} \mathrm{~kg}\). (This power is the maximum the engine can supply at its optimum frequency of 7600 r.p.m. Presumably the automatic transmission is designed so a gear is available in which the engine will be running at very nearly this frequency when the car is at moving at \(v .\).) Rubber on asphalt has \(\mu_{s} \approx 0.9 .\) Find \(v\) for this car. Answer: \(18 \mathrm{~m} / \mathrm{s}\), or about 40 miles per hour. (f) Our analysis has neglected air friction, which can probably be approximated as a force proportional to \(v^{2}\). The existence of this force is the reason that the car has a maximum speed, which is 176 miles per hour. To get a feeling for how good an approximation it is to ignore air friction, find what fraction of the engine's maximum power is being used to overcome air resistance when the car is moving at the speed \(v\) found in part e.

Two wheels of radius \(r\) rotate in the same vertical plane with angular velocities \(+\Omega\) and \(-\Omega\) about axes that are parallel and at the same height. The wheels touch one another at a point on their circumferences, so that their rotations mesh like gears in a gear train. A board is laid on top of the wheels, so that two friction forces act upon it, one from each wheel. Characterize the three qualitatively different types of motion that the board can exhibit, depending on the initial conditions.
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