GR0177 #4



Alternate Solutions 
amber 20141022 15:11:44  Look at the problem thru Center of Mass
Again find the velocity of CM which equals the initial and final velocities of each mass
V = m1v1/ m1+m2 = 2v1/3 = v_final
set the ratio KEf/KEi
0.5 3m 4v1^2/9 / 0.5 2m v1^2 = 2/3
Then 1  2/3 to get the energy lost 1/3   secretempire1 20120828 06:47:39  Formula for kinetic energy
Let and correspond to the mass and velocity of block 1, and for block 2, and and for the resultant combined mass block, where .
Total initial kinetic energy
Since ,
Total final kinetic energy
Performing the same operation for momentum:
Solving for will yield
Plug this into the equation for final kinetic energy, and you get
This implies that 1/3 of the energy is lost.  

Comments 
Nebula 20150918 20:50:37  Don\'t misread this question and think they are asking for the ratio of final kinetic energy to initial kinetic energy, in which you would get answer (E) 2/3. Easy points lost.   amber 20141022 15:11:44  Look at the problem thru Center of Mass
Again find the velocity of CM which equals the initial and final velocities of each mass
V = m1v1/ m1+m2 = 2v1/3 = v_final
set the ratio KEf/KEi
0.5 3m 4v1^2/9 / 0.5 2m v1^2 = 2/3
Then 1  2/3 to get the energy lost 1/3   Dax Feliz 20121106 22:03:16  Does any one know why we subtract the ratio from 1 to get the fractional loss of energy? I think I missed something here...   CortunaMJM 20121009 02:40:13  how to get the initial kinetic energy if there's no velocity given?
ezfzx 20121013 17:48:25 
Since the problem is only looking for a RATIO of energy, the velocity is not needed.

  secretempire1 20120828 06:47:39  Formula for kinetic energy
Let and correspond to the mass and velocity of block 1, and for block 2, and and for the resultant combined mass block, where .
Total initial kinetic energy
Since ,
Total final kinetic energy
Performing the same operation for momentum:
Solving for will yield
Plug this into the equation for final kinetic energy, and you get
This implies that 1/3 of the energy is lost.   thesandrus86 20110424 14:21:41  por conservación del momento tenemos:
2mVi+mVi=(m+2m)Vf
2m =3mVf
Vf= 2/3....   archard 20100411 14:54:59  Another quick way is to remember that kinetic energy can be written as , and since momentum is conserved, the final:initial kinetic energy ratio is = . Which means the fractional kinetic energy loss is   wittensdog 20091008 16:29:07  Here's something that's surprisingly useful. When a moving body hits a second one at rest, and they stick, the fractional loss of kinetic energy is,
m2 / ( m1 + m2)
If you just memorize that, all sorts of problems involving sticky collisions become way faster, even ones where you have falling masses on pendulums and so forth.
Dhananjay 20130530 00:26:45 
is m2 the mass of the moving ball or the ball at rest?

  herrphysik 20060919 16:14:15  Small typo: You're missing a 2 in the left side of the initial KE equation.  

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