change in momentum and change in Kinetic energy

eureka_beyond

1. The problem statement, all variables and given/known data
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.


2. Relevant equations
P=mv KE=1/2mv2

3. 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?

radou

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?

eureka_beyond

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?

radou

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).

eureka_beyond

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...

radou

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.