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GR9677 #20
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Alternate Solutions |
daverigie
2009-10-06 08:21:18 | When you look at an elastic collision in the center of mass frame, you can quickly realize that for momentum and energy to be conserved the particles must rebound with the same velocity that they approached one another.
Therefore the magnitude of the velocity change in the particle of mass  must be
where  is the center of mass coordinate.
v}{2} = \dot{\eta})
and
That means we need
From here I would solve by inspection and realize  |  | twistor
2007-11-02 16:21:09 | Here's a quick solution:
Notice that for conservation of momentum we must have:
p = -0.6p + 1.6p and use
 so
so that
Quickly solve to find 
Of course it's never that easy under pressure!... Down with ETS! | |
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Comments |
noether
2009-11-04 16:33:47 | The 'm' should not be squared under the line "from kinetic energy conservation", just the 1.6v.
noether
2009-11-04 16:35:48 |
ok nevermind
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|  | noether
2009-11-04 16:32:55 | The 'm' should not be squared under the line "from kinetic energy conservation", just the 1.6v. |  | daverigie
2009-10-06 08:21:18 | When you look at an elastic collision in the center of mass frame, you can quickly realize that for momentum and energy to be conserved the particles must rebound with the same velocity that they approached one another.
Therefore the magnitude of the velocity change in the particle of mass must be
where is the center of mass coordinate.
and
That means we need
From here I would solve by inspection and realize | | jw111
2008-11-04 23:50:50 | set 4u = 1, v = 1, (change the units)
then conservation laws, (M for mass of surface, k is the velocity)
1 = -0.6 + Mk ...........................(1)
1 = 0.6*0.6 +Mkk ..........(2)
16/10 = Mk
64/100 = Mkk
so k = 64*10 / 16*100 = 4/10
so M = 16*10 / 10*4 = 4
recover the unit M = 4*4u = 16u | | tinytoon
2008-10-05 10:36:00 | I find that the most important thing to remember for elastic collisions is the formula for the relative velocities:
Using this formula and conservation of momentum immediately gives the answer without much work. | | twistor
2007-11-02 16:21:09 | Here's a quick solution:
Notice that for conservation of momentum we must have:
p = -0.6p + 1.6p and use
so
so that
Quickly solve to find
Of course it's never that easy under pressure!... Down with ETS!
blah22
2008-02-14 11:24:27 |
Type in the first formula there, that factor of 2 should be on top, no?
Also...I solved it with the other elastic eqn:
No need to worry about conservation of momentum, just solve for m2 directly. But may be a bit more algebra, depends on preference I suppose.
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WoolfianOperator
2009-11-02 18:48:11 |
You mean solve for  , if you solve for  using =4u one gets 1u. Which doesn't make a whole lot of sense. The equation you are using has as the mass at rest not
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2010-03-26 12:28:25 |
Note that  is -ve compared to  . Plug everything in the equation relating the two velocities and one gets  .
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| | rreyes
2005-11-27 08:39:38 | It may be helpful to memorize the formula valid for elastic collisions. If initial velocity of \"target\" particle is zero, the final velocity of incoming particle is simply given by ![[(m_{inc}-m_{tar})/(m_{inc}+m_{tar})]v_{inc,init}](/cgi-bin/mimetex.cgi?[(m_{inc}-m_{tar})/(m_{inc}+m_{tar})]v_{inc,init}) . One can then solve for  or check each choice- whichever is faster.
Richard
2007-09-25 09:52:31 |
What you really want is the final velocity of the initially stationary particle:
Using this and conservation of momentum, the solution comes out painlessly as  .
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