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Derive a formula expressing the kinetic energy of an object in terms of its momentum and mass.(answer check available at lightandmatter.com)

Short Answer

Expert verified
The kinetic energy in terms of momentum and mass is \( KE = \frac{p^2}{2m} \).

Step by step solution

01

Understand the relationship between momentum and velocity

The momentum \( p \) of an object is defined as the product of its mass \( m \) and velocity \( v \). Therefore, the formula is \( p = mv \). This equation will be important for deriving the kinetic energy formula in terms of momentum.
02

Recall the formula for kinetic energy

The kinetic energy \( KE \) of an object is given by the formula \( KE = \frac{1}{2} mv^2 \). This equation represents the energy due to motion based on mass and velocity.
03

Express velocity in terms of momentum

Using the momentum formula \( p = mv \), solve for velocity: \( v = \frac{p}{m} \). This allows us to replace velocity in terms of momentum in the kinetic energy equation.
04

Substitute velocity in kinetic energy formula

Substituting \( v = \frac{p}{m} \) into the kinetic energy formula, we have: \( KE = \frac{1}{2} m \left( \frac{p}{m} \right)^2 \).
05

Simplify the expression

Simplify the equation \( KE = \frac{1}{2} m \left( \frac{p}{m} \right)^2 \) as follows: \( KE = \frac{1}{2} m \times \frac{p^2}{m^2} \). This simplifies further to \( KE = \frac{1}{2} \times \frac{p^2}{m} \).
06

Write the final expression

The kinetic energy in terms of momentum and mass is \( KE = \frac{p^2}{2m} \). This formula relates the kinetic energy to momentum and mass without directly involving velocity.

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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 relates directly to motion. It is the measure of the quantity of motion an object possesses and is a vector, meaning it has both magnitude and direction. The calculation of momentum is straightforward: you multiply the mass \(m\) of an object by its velocity \(v\). Therefore, you get the formula \(p = mv\).

Understanding momentum helps to grasp how objects interact when they collide or bounce off each other. Since momentum depends on both mass and velocity, these parameters greatly influence how momentum behaves in various situations:
  • If an object has a large mass and is moving fast, it will have a higher momentum.
  • Similarly, a small object moving at high speed can also possess significant momentum.
  • Momentum conservation plays a crucial role in understanding collisions in closed systems.
Recognizing the concept of momentum aids in understanding the changes in motion occurring from external forces.
Mass
Mass is an intrinsic property of an object that tells us how much "stuff" it contains. Unlike weight, which can change based on location, mass is constant no matter where the object is. It is measured in kilograms in the metric system.

Mass plays a key role in various equations in physics, one of which is the equation for momentum that we discussed earlier. The mass also influences an object's kinetic energy and how difficult it is to change that object's state of motion through acceleration or deceleration.
  • In terms of momentum \(p = mv\), if the mass \(m\) of an object increases, keeping velocity \(v\) constant, the momentum will also increase.
  • Mass is also found in the kinetic energy formula \(KE = \frac{1}{2} mv^2\), showing its impact on the energy needed to move an object at a certain velocity.
  • The more massive an object is, the more energy is required to change its motion (speed up or slow down).
Mass essentially is a measure of an object's resistance to acceleration when a net force is applied. While in many everyday situations mass is directly measured, in physics, it often serves as a linchpin in understanding interactions.
Velocity
Velocity tells us how fast an object is moving in a specific direction. Unlike speed, which is only about how fast something is traveling, velocity considers direction as a crucial component. Due to its vector nature, velocity can indicate changes in motion not just by how quickly or slowly an object is moving, but also by shifts in direction.

In the context of momentum and kinetic energy, velocity helps determine their magnitude:
  • In the momentum equation \(p = mv\), a higher velocity leads to a higher momentum if mass \(m\) stays constant.
  • For kinetic energy \(KE = \frac{1}{2} mv^2\), even small changes in velocity have a large impact because the velocity term is squared, thus amplifying the energy.
  • Velocity influences how objects move, collide, and transfer energy.
Velocity lets us explore many phenomena, such as motion through space, effects in transport, and the dynamics of systems in physics. It's a critical concept because understanding velocity as a function of speed and direction aids in comprehensive analysis of motion.

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Most popular questions from this chapter

(solution in the pdf version of the book) In each case, identify the force that causes the acceleration, and give its Newton'sthird-law partner. Describe the effect of the partner force. (a) A swimmer speeds up. (b) A golfer hits the ball off of the tee. (c) An archer fires an arrow. (d) A locomotive slows down.

(solution in the pdf version of the book) A car is accelerating forward along a straight road. If the force of the road on the car's wheels, pushing it forward, is a constant \(3.0 \mathrm{kN}\), and the car's mass is \(1000 \mathrm{~kg}\), then how long will the car take to go from \(20 \mathrm{~m} / \mathrm{s}\) to \(50 \mathrm{~m} / \mathrm{s} ?\)

Find the angle between the following two vectors: $$ \begin{array}{l} \hat{\mathbf{x}}+2 \hat{\mathbf{y}}+3 \hat{\mathbf{z}} \\ 4 \hat{\mathbf{x}}+5 \hat{\mathbf{y}}+6 \hat{\mathbf{z}} \end{array} $$ \mathrm{\\{} ~ h w h i n t ~ \\{ h w h i n t : a n g l e b e t w e e n \\} ( a n s w e r ~ c h e c k ~ a v a i l a b l e ~ a t ~ l i g h t a n d m a t t e r . c o m ) ~

A flexible rope of mass \(m\) and length \(L\) slides without friction over the edge of a table. Let \(x\) be the length of the rope that is hanging over the edge at a given moment in time. (a) Show that \(x\) satisfies the equation of motion \(d^{2} x / d t^{2}=g x / L .\) [Hint: Use \(F=d p / d t\), which allows you to handle the two parts of the rope separately even though mass is moving out of one part and into the other.] (b) Give a physical explanation for the fact that a larger value of \(x\) on the right-hand side of the equation leads to a greater value of the acceleration on the left side. (c) When we take the second derivative of the function \(x(t)\) we are supposed to get essentially the same function back again, except for a constant out in front. The function \(e^{x}\) has the property that it is unchanged by differentiation, so it is reasonable to look for solutions to this problem that are of the form \(x=b e^{c t}\), where \(b\) and \(c\) are constants. Show that this does indeed provide a solution for two specific values of \(c\) (and for any value of \(b\) ). (d) Show that the sum of any two solutions to the equation of motion is also a solution. (e) Find the solution for the case where the rope starts at rest at \(t=0\) with some nonzero value of \(x\).

A plane is flown in a loop-the-loop of radius \(1.00 \mathrm{~km}\). The plane starts out flying upside-down, straight and level, then begins curving up along the circular loop, and is right-side up when it reaches the top. (The plane may slow down somewhat on the way up.) How fast must the plane be going at the top if the pilot is to experience no force from the seat or the seatbelt while at the top of the loop? (answer check available at lightandmatter.com)

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