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Starting with the definitions of momentum and kinetic energy, derive an equation for the kinetic energy of a particle expressed as a function of its momentum

Short Answer

Expert verified
The kinetic energy (KE) of a particle expressed as a function of its momentum (p) is \( KE = \frac{p^2}{2m} \).

Step by step solution

01

Define Momentum

First, define momentum (p) as the product of mass (m) and velocity (v): \( p = m \cdot v \).
02

Define Kinetic Energy

Next, recall the definition of kinetic energy (KE) for a particle: \( KE = \frac{1}{2} m v^2 \).
03

Isolate Velocity

Isolate the velocity (v) in the momentum equation: \( v = \frac{p}{m} \).
04

Substitute Velocity into Kinetic Energy

Substitute the expression for velocity in terms of momentum into the kinetic energy formula: \( KE = \frac{1}{2} m \left(\frac{p}{m}\right)^2 \).
05

Simplify Kinetic Energy Expression

Simplify the kinetic energy expression by canceling out mass where possible and squaring the momentum term: \( KE = \frac{1}{2} m \frac{p^2}{m^2} = \frac{p^2}{2m} \).
06

Final Equation

The final equation expressing kinetic energy as a function of momentum is \( KE = \frac{p^2}{2m} \). This equation shows that the kinetic energy of a particle can be calculated using its momentum and mass.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that represents the quantity of motion an object has. More formally, momentum is defined as the product of an object's mass and its velocity. It's described by the equation:


\( p = m \cdot v \).


Intuitively, you can think of it as a measure of how difficult it would be to stop a moving object; the higher the momentum, the harder it is to stop. Momentum is a vector quantity, which means it has both magnitude and direction, aligning with the direction of the object's velocity. Here are some key points to help you understand momentum better:

  • Conservation of Momentum: In a closed system with no external forces, the total momentum before and after a collision between objects remains constant.
  • Momentum and Force: A change in momentum over a specific time is caused by a force, based on Newton's second law of motion.
  • Impulse: A force applied over a time interval, known as impulse, changes the momentum of an object.

Understanding momentum is critical for solving a variety of problems in physics, from predicting the outcome of collisions to understanding the motion of planets.

Physics
Physics is a branch of science that is concerned with the nature and properties of matter and energy. It encompasses a wide range of phenomena, from the subatomic particles that make up all known matter to the behavior of astronomical objects such as galaxies. Physics principles like momentum, kinetic energy, and mechanical energy are used to explain events occurring in the universe.


Physics is broadly divided into classical and modern physics, with classical physics dealing with forces, mechanics, thermodynamics, and other concepts up through the 19th century, and modern physics building upon these concepts with the theories of relativity and quantum mechanics.

Mechanical Energy
Mechanical energy is the energy associated with the motion and position of an object. It is a combination of kinetic energy, which is energy due to an object’s motion, and potential energy, which is stored energy due to an object's position or state. The total mechanical energy in a closed system is often conserved, meaning it doesn’t change unless non-conservative forces, like friction, are at work.


In the context of our kinetic energy exercise, understanding mechanical energy helps deepen our comprehension of energy conversions within a system. For a moving object, its mechanical energy arises from both its movement (kinetic) and its potential to move (potential energy), and both of these energy forms are intrinsically linked to the concept of work, another key physics idea.

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Most popular questions from this chapter

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large enough to be detected by radar, but there are far greater numbers of very small objects, such as flakes of paint. Calculate the force exerted by a \(0.100\) -mg chip of paint that strikes a spacecraft window at a relative speed of \(4.00 \times 10^{3} \mathrm{~m} / \mathrm{s}\), given the collision lasts \(6.00 \times 10^{-8} \mathrm{~s}\)

Under what circumstances is momentum conserved?

A \(0.0250-\mathrm{kg}\) bullet is accelerated from rest to a speed of \(550 \mathrm{~m} / \mathrm{s}\) in a \(3.00-\mathrm{kg}\) rifle. The pain of the rifle's kick is much worse if you hold the gun loosely a few centimeters from your shoulder rather than holding it tightly against your shoulder. (a) Calculate the recoil velocity of the rifle if it is held loosely away from the shoulder. (b) How much kinetic energy does the rifle gain? (c) What is the recoil velocity if the rifle is held tightly against the shoulder, making the effective mass \(28.0 \mathrm{~kg} ?\) (d) How much kinetic energy is transferred to the rifle-shoulder combination? The pain is related to the amount of kinetic energy, which is significantly less in this latter situation. (e) Calculate the momentum of a \(110-\mathrm{kg}\) football player running at \(8.00 \mathrm{~m} / \mathrm{s}\). Compare the player's momentum with the momentum of a hardthrown \(0.410\) -kg football that has a speed of \(25.0 \mathrm{~m} / \mathrm{s}\). Discuss its relationship to this problem.

Can momentum be conserved for a system if there are external forces acting on the system? If so, under what conditions? If not, why not?

Military rifles have a mechanism for reducing the recoil forces of the gun on the person firing it. An internal part recoils over a relatively large distance and is stopped by damping mechanisms in the gun. The larger distance reduces the average force needed to stop the internal part. (a) Calculate the recoil velocity of a 1.00-kg plunger that directly interacts with a 0.0200-kg bullet fired at \(600 \mathrm{~m} / \mathrm{s}\) from the gun. (b) If this part is stopped over a distance of \(20.0 \mathrm{~cm}\), what average force is exerted upon it by the gun? (c) Compare this to the force exerted on the gun if the bullet is accelerated to its velocity in \(10.0 \mathrm{~ms}\) (milliseconds).

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