Law of conservation of linear momentum and energy

[Radwa Kamal]

Homework Statement

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.

Homework Equations

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

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

 

[LowlyPion]

Homework Statement

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.

Homework Equations

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

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.

 

[Radwa Kamal]

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