Change in momentum and change in Kinetic energy

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Homework Help Overview

The problem involves a particle of mass m thrown vertically upwards with an initial speed v, and it seeks to determine the changes in momentum and kinetic energy when the particle returns to its starting point.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion relevant to vertical throws and question how to calculate the velocity when the particle returns to its starting point. Some suggest using specific kinematic equations to find the height reached and the time taken for the particle to fall back down.

Discussion Status

There is an ongoing exploration of different approaches to calculate the necessary parameters, with participants providing guidance on using kinematic equations. Multiple interpretations of the problem are being discussed, particularly regarding the application of equations and the relationships between variables.

Contextual Notes

Participants are navigating through the constraints of the problem, including the need to calculate height and time, while also addressing the definitions of momentum and kinetic energy in the context of the motion described.

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Homework Statement


A particle of mass m is thrown vertically upwards with initial speed v. When the particle returns to it starting point, what are the changes in momentum and kinetic energy of the particle.


Homework Equations


P=mv KE=1/2mv2

The Attempt at a Solution


The answer to this question is change in momentum = 2mv while there's no change in KE, but I really don't understand how to calculate them?
 
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Any ideas? Start by writing down the equations of motion for a vertical throw. Can you find the velocity at the when the body reaches its starting point again?
 
Should I apply v2-u2=2as ? Then I'll know how high does the particle reach, which is v2/20, but how will that help me with the question? Or am I using the wrong equation?
 
First calculate the height the particle reaches (measured from the starting reference point). Then, calculate the time it takes for the particle to fall down along that "height". Then you can calculate the velocity of the particle at the starting point (i.e. momentum and kinetic energy).
 
using 2as=v2-u2, I know s=(v2)/20. using s=1/2at2, t=v/10...using v=u+at, I got v=v?? I don't get it...
 
The velocity of the particle is v(t) = v - gt, when the particle is moving upwards. At the peak, v(t) = 0, and hence t = v/g. Now you can calculate the height the particle reaches. And then the time it takes to fall down again. And then the velocity at that time, and hence prove your point.
 

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