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GR0177 #4
Problem
 GREPhysics.NET Official Solution Alternate Solutions
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Mechanics}Conservation of Momentum

From conservation of momentum, one has .

The initial kinetic energy is .

The final kinetic energy is .

The ratio of the final to initial kinetic energy is .

The kinetic energy lost in the collision is 1 minus that ratio. Thus, the answer is , as in choice (C).  Alternate Solutions
 amber2014-10-22 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 1/3Reply to this comment secretempire12012-08-28 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.Reply to this comment Nebula
2015-09-18 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
2014-10-22 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
2012-11-06 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
2012-10-09 02:40:13
how to get the initial kinetic energy if there's no velocity given?
 ezfzx2012-10-13 17:48:25 Since the problem is only looking for a RATIO of energy, the velocity is not needed. secretempire1
2012-08-28 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
2011-04-24 14:21:41
por conservación del momento tenemos:
2mVi+mVi=(m+2m)Vf

2m =3mVf

Vf= 2/3.... archard
2010-04-11 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
2009-10-08 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.
 Dhananjay2013-05-30 00:26:45 is m2 the mass of the moving ball or the ball at rest? herrphysik
2006-09-19 16:14:15
Small typo: You're missing a 2 in the left side of the initial KE equation. por conservación del momento tenemos: 2mVi+mVi=(m+2m)Vf 2m =3mVf Vf= 2/3....     LaTeX syntax supported through dollar sign wrappers $, ex.,$\alpha^2_0$produces . type this... to get...$\int_0^\infty\partial\Rightarrow\ddot{x},\dot{x}\sqrt{z}\langle my \rangle\left( abacadabra \right)_{me}\vec{E}\frac{a}{b}\$