Change in momentum and change in Kinetic energy

AI Thread Summary
A particle thrown vertically upwards with initial speed v will have a change in momentum of 2mv when it returns to its starting point, while the change in kinetic energy remains zero. To analyze this, one must calculate the height reached using the equation v² = u² + 2as, where the final velocity at the peak is zero. The time taken to reach the peak can be determined using t = v/g, allowing for the calculation of the total height. After reaching the peak, the particle falls back down, regaining its initial speed v, confirming that kinetic energy does not change. Understanding these principles clarifies the relationship between momentum and kinetic energy in this scenario.
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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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