Relation between kinetic energy and linear momentum

I know that Coefficient of Restitution(e) is defined as $$e = \frac\Biggr)100$$So am i doing something wrong. I think that you have the wrong idea about what the term perfectly inelastic means when it comes to the loss of kinetic energy. Whereas the term perfectly elastic does mean that kinetic energy is conserved, perfectly inelastic does not necessarily mean that all the kinetic energy is lost. For example if two objects, originally moving in the same direction, collide and stick together, the kinetic energy cannot become zero as linear momentum has to be conserved. However, if the objects are moving in opposite direction and the magnitude of their linear momentums is the same then after sticking together the kinetic energy is zero. If you write down the equation which defines the coefficient of restitution and that for the conservation of linear momentum you can derive an equation for the loss of kinetic energy which depends on the initial and final velocities of the colliding objects and the masses of those two objects. To obtain your graph you must have assumed some initial conditions which meant that after the collision the linear momentum of the two objects which were stuck together was not zero, hence the loss kinetic energy was not zero. A lump of chewing gum hitting a wall and sticking to it is an example of the coefficient of restitution being equal to zero but the chewing gum, wall (and Earth) do have some kinetic energy as a result of such a collision. I use t...

It took me quite a long time to click my gears in place and even then I'm not sure it's completely correct. The problem is that I need to understand these concepts (physics concepts; not just these two) with intuition, not only mathematical representations. So $(E_k = \frac mv^2)$ and $(P_l = mv)$ don't tell me much. Hence: Here's how I've been viewing them: • Linear momentum is the moving version of inertia; how much it could resist change in its non-zero velocity. • Kinetic energy is how much a moving object could influence other objects upon contact. So $P_l$ is how much force an object need/can take while $E_k$ is how much it can give. All for moving objects. Am I correct in this view? PS. I'm aware of the similar questions already posted. No, they don't address what I need. So $P_l$ is how much force an object need/can take while $E_k$ is how much it can give. All for moving objects. Momentum & Energy are not forces, I believe I understand what you intended but its important not to mince words. Momentum and Energy are what they are defined to be mathematically and nothing more. Along those lines most of the intuition you will need about these equations comes directly from these equations and conservation laws. As the velocity of an object increases how does its momentum change? Well you know $P_l=mv$ so momentum must increase linearly. How does its $E_k$ change? Again go back to the equation...$\fracmv^2$ ...it increases as a square of the velocity. Then you know that...

Momentum (P) is equal to mass (M) times velocity (v). But there are other ways to think about momentum! Force (F) is equal to the change in momentum (ΔP) over the change in time (Δt). And the change in momentum (ΔP) is also equal to the impulse (J). Impulse has the same units as momentum (kg*m/s or N*s). Created by Sal Khan. When doing momentum problems to obtain the combined velocity, why is it that we do not use the concept of kinetic energy? We can use mv^2/2 to obtain this value, so why would it not work if when we assume v is a constant value after collision and that m is the sum of the two masses, we solve for a different value of v compared to solved with the momentum method? The main reason one cant apply "Work Energy Theorem (Kinetic Energy Conservation, as in this case)" on this problem, because some Kinetic Energy is stored as Potential Energy (Which is due to deformation caused by the collision at the surface of impact) thus the final Kinetic Energy is not equal to initial KE. NOTE: If this question proclaimed an elastic collision between the objects. Then KE(initial)=KE(final), because elastic collision means that the surfaces of the bodies AFTER (NOT DURING) the impact will reform fully to their original shapes, thus no extra PE is stored AFTER the collision. Acceleration = (v-u)/t where v: Final Velocity u: Initial Velocity t: Time elapsed The numerator is change in velocity (v-u), but the whole expression tells you the amount the velocity changes per unit t...

12 Thermodynamics • Introduction • 12.1 Zeroth Law of Thermodynamics: Thermal Equilibrium • 12.2 First law of Thermodynamics: Thermal Energy and Work • 12.3 Second Law of Thermodynamics: Entropy • 12.4 Applications of Thermodynamics: Heat Engines, Heat Pumps, and Refrigerators • Key Terms • Section Summary • Key Equations • 22 The Atom • Introduction • 22.1 The Structure of the Atom • 22.2 Nuclear Forces and Radioactivity • 22.3 Half Life and Radiometric Dating • 22.4 Nuclear Fission and Fusion • 22.5 Medical Applications of Radioactivity: Diagnostic Imaging and Radiation • Key Terms • Section Summary • Key Equations • Teacher Support The learning objectives in this section will help your students master the following standards: • (6) Science concepts. The student knows that changes occur within a physical system and applies the laws of conservation of energy and momentum. The student is expected to: • (C) calculate the mechanical energy of, power generated within, impulse applied to, and momentum of a physical system. Section Key Terms Teacher Support [BL] [OL] Review inertia and Newton’s laws of motion. [AL] Start a discussion about movement and collision. Using the example of football players, point out that both the mass and the velocity of an object are important considerations in determining the impact of collisions. The direction as well as the magnitude of velocity is very important. Momentum, Impulse, and the Impulse-Momentum Theorem Linear momentum is the product...

Momentum Momentum is the product of a moving object's mass and velocity . \[\text\] The symbol for momentum is \(p\) so this can also be written as: \[p=mv\] Momentum is measured in kg ms -1 . Momentum is a vector quantity that depends on the direction of the object. Momentum is of interest during collisions between objects. When two objects collide the total momentum before the collision is equal to the total momentum after the collision (in the absence of external forces). This is the law of conservation of momentum. It is true for all collisions. Watch this video for a practical demonstration of conservation of momentum in elastic collisions.

From a mathematical point of view it seems to be clear what's the difference between momentum and $mv$ and kinetic energy $\frac m v^2$. Now my problem is the following: Suppose you want to explain someone without mentioning the formulas what's momentum and what's kinetic energy. How to do that such that it becomes clear what's the difference between those two quantities? From physics I think one can list essentially the following differences: • momentum has a direction, kinetic energy not • momentum is conserved, kinetic energy not (but energy is) • momentum depends linear on velocity, kinetic energy depends quadratically on velocity I think is is relatively easy to explain the first two points using everyday language, without referring to formulas. However is there a good way to illustrate the 3. point? $\begingroup$ Does the rifle really not have a comparable amount of kinetic energy? It could be that it's just not as harmful as the bullet because it's slower (due to higher mass) and has a much larger surface area that's flat in contrast to the sharp bullet. It really is not clear to me how fast the rifle would go if it weren't held by someone while being fired. $\endgroup$ $\begingroup$ @DrDoolittle: momentum $= mv$, right? and kinetic energy $=mv^2/2$, right? OK, so increase the mass 100 times and divide the velocity by 100. Same momentum, right? $100mv/100 = mv$ But what does it do to the kinetic energy? $100m(v/100)^2/2 = mv^2/100/2$. So by increasing the mass 100 t...

Chapter 4 Conservation laws for systems of particles In this chapter, we shall introduce (but not in this order) the following general concepts: • The linear impulse of a force • The angular impulse of a force • The power transmitted by a force • The work done by a force • The potential energy of a force. • The linear momentum of a particle (or system of particles) • The angular momentum of a particle, or system of particles. • The kinetic energy of a particle, or system of particles • The linear impulse momentum relations for a particle, and conservation of linear momentum • The principle of conservation of angular momentum for a particle • The principle of conservation of energy for a particle or system of particles. We will also illustrate how these concepts can be used in engineering calculations. As you will see, to applying these principles to engineering calculations you will need two things: (i) a thorough understanding of the principles themselves; and (ii) Physical insight into how engineering systems behave, so you can see how to apply the theory to practice. The first is easy. The second is hard, and people who can do this best make the best engineers. 4.1 Work, Power, Potential Energy and Kinetic Energy relations The concepts of work, power and energy are among the most powerful ideas in the physical sciences. Their most important application is in the field of thermodynamics, which describes the exchange of energy between interacting systems. In addition, con...