## law of conservation of linear momentum and energy

1. The problem statement, all variables and given/known data
A body of 2M(KG) mass at velocity V collided with a body of mass M(KG) at rest.
The first body after collision has a velocity ao 1/3V , and the second moves at 4/3V.
Verify both laws of conservation of linear momentum and energy with explanation.

2. Relevant equations
m1v1f + m2v2f= m1v1i+ m2v2i
KE = (1/2)mv^2
PE=mgh

3. The attempt at a solution
KE = (1/2)mv^2
KE before collision = 1/2 *2MV^2+ 1/2* M(0)=MV^2
KE after collision =1/2*2M(1/3V)^2+1/2*m(4/3V)^2=1/9MV62+8/9MV^2=MV^2
KE before collision=KE after collision
there is no height then there is no potential energy
(KE+PE) before collision + (KE+PE) after collision
then mechanical energy is constant at any point

 PhysOrg.com science news on PhysOrg.com >> Heat-related deaths in Manhattan projected to rise>> Dire outlook despite global warming 'pause': study>> Sea level influenced tropical climate during the last ice age

Recognitions:
Homework Help
 Quote by Radwa Kamal 1. The problem statement, all variables and given/known data A body of 2M(KG) mass at velocity V collided with a body of mass M(KG) at rest. The first body after collision has a velocity ao 1/3V , and the second moves at 4/3V. Verify both laws of conservation of linear momentum and energy with explanation. 2. Relevant equations m1v1f + m2v2f= m1v1i+ m2v2i KE = (1/2)mv^2 PE=mgh 3. The attempt at a solution KE = (1/2)mv^2 KE before collision = 1/2 *2MV^2+ 1/2* M(0)=MV^2 KE after collision =1/2*2M(1/3V)^2+1/2*m(4/3V)^2=1/9MV62+8/9MV^2=MV^2 KE before collision=KE after collision there is no height then there is no potential energy (KE+PE) before collision + (KE+PE) after collision then mechanical energy is constant at any point
Welcome to PF.

I will presume that you figured the momentum was conserved. But that said it should be enough to suggest that since the kinetic energy before equals the kinetic after that no energy was lost to other sources, like potential, or friction, or maybe even sound.

 I found it :) m1v1f + m2v2f= m1v1i+ m2v2i momentum before collision=2mv+zero=2mv momentum after collision=2m(1/3v)+m(4/3v)=2/3mv+4/3mv=6/3mv=2mv then Momentum before collision =Momentum after collision

 Tags energy, law of conservation, linear momentum