GR0177 #4
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Alternate Solutions |
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 | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | secretempire1 2012-08-28 06:47:39 | Formula for kinetic energy ![K=\frac{1}{2} mv^2](/cgi-bin/mimetex.cgi?K=\frac{1}{2} mv^2)
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 ![K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2](/cgi-bin/mimetex.cgi?K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2)
Since , ![K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2](/cgi-bin/mimetex.cgi?K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2)
Total final kinetic energy ![K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2](/cgi-bin/mimetex.cgi?K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2)
Performing the same operation for momentum:
![m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f}](/cgi-bin/mimetex.cgi?m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f})
Solving for will yield ![v_{f} = \frac{2}{3}v_{1}](/cgi-bin/mimetex.cgi?v_{f} = \frac{2}{3}v_{1})
Plug this into the equation for final kinetic energy, and you get ![K_{f} = \frac{2}{3}mv_{1}^2](/cgi-bin/mimetex.cgi?K_{f} = \frac{2}{3}mv_{1}^2)
This implies that 1/3 of the energy is lost. | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) |
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Comments |
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. | ![NEC](/img/nec.gif) | 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 | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | 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... | ![NEC](/img/nec.gif) | CortunaMJM 2012-10-09 02:40:13 | how to get the initial kinetic energy if there's no velocity given?
ezfzx 2012-10-13 17:48:25 |
Since the problem is only looking for a RATIO of energy, the velocity is not needed.
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| ![Answered Question!](/img/answered.gif) | secretempire1 2012-08-28 06:47:39 | Formula for kinetic energy ![K=\frac{1}{2} mv^2](/cgi-bin/mimetex.cgi?K=\frac{1}{2} mv^2)
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 ![K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2](/cgi-bin/mimetex.cgi?K_{i} = \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^2)
Since , ![K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2](/cgi-bin/mimetex.cgi?K_{i} = \frac{1}{2} 2mv_{1}^2 = mv_{1}^2)
Total final kinetic energy ![K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2](/cgi-bin/mimetex.cgi?K_{f} = \frac{1}{2}m_{f}v_{f}^2 = \frac{3}{2}mv_{f}^2)
Performing the same operation for momentum:
![m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f}](/cgi-bin/mimetex.cgi?m_{1}v_{1} + m_{2}v_{2} = m_{f}v_{f})
Solving for will yield ![v_{f} = \frac{2}{3}v_{1}](/cgi-bin/mimetex.cgi?v_{f} = \frac{2}{3}v_{1})
Plug this into the equation for final kinetic energy, and you get ![K_{f} = \frac{2}{3}mv_{1}^2](/cgi-bin/mimetex.cgi?K_{f} = \frac{2}{3}mv_{1}^2)
This implies that 1/3 of the energy is lost. | ![Alternate Solution - Unverified](/img/altsol-unverified.gif) | thesandrus86 2011-04-24 14:21:41 | por conservación del momento tenemos:
2mVi+mVi=(m+2m)Vf
2m =3mVf
Vf= 2/3.... | ![NEC](/img/nec.gif) | 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 ![\frac{1}{3}](/cgi-bin/mimetex.cgi?\frac{1}{3}) | ![NEC](/img/nec.gif) | 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.
Dhananjay 2013-05-30 00:26:45 |
is m2 the mass of the moving ball or the ball at rest?
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| ![NEC](/img/nec.gif) | herrphysik 2006-09-19 16:14:15 | Small typo: You're missing a 2 in the left side of the initial KE equation. | ![Typo Alert!](/img/typo-unfixed.gif) |
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