GR0177 #55



Alternate Solutions 
carle257 20100401 20:20:09  This can be done simply by acknowledging that the x momentum is conserved so . Then if they also have a vertical component, the speed must be greater than this value. Thus the answer must be (E).  

Comments 
shka 20180708 19:07:34  carle257 is using symmetry instead of explicitly writing out the equations for and conservation. His answer is fine and consistent with Yosun\'s, and the fact that Quark is confused by the notation makes me think he doesn\'t understand how right triangles work. You can ALWAYS solve for the components of the velocity, by definition of linear momenta and , and since and are conserved separately. Whether or not you choose to fuss with the angles is completely up to you. Since the angles and masses are identical, if you paid attention in PHYS 100 you can literally just read off the answer. Thanks Quark, for obfuscating a perfectly good solution with your utter misunderstanding of first principles. Maybe you should crack open Giancoli a few more times before applying to CERN, since I\'m pretty sure they like to conserve momentum linear momentum there.   aziza 20140812 12:59:18  If you run the collision backward in time, you see it as two objects mass m combining into one of mass 2m, so this is inelastic. Thus the final kinetic energy of the system should be greater than initial, giving each of the two masses velocity > v/2 .   carle257 20100401 20:20:09  This can be done simply by acknowledging that the x momentum is conserved so . Then if they also have a vertical component, the speed must be greater than this value. Thus the answer must be (E).
Quark 20110926 16:36:49 
Linear momentum is conserved but the velocity of each of the particles in the xdirection is not =. This would only be true if both particles were moving along the xaxis which is certainly not the case in this problem since they've given you the angle .
As the solution stated, by using conservation of linear momentum in the xdirection, you can solve for the velocity of each particle (not the xcomponent of the velocity)
v'=
which includes a cos() term in the denominator. This cos() term must be less than 1 since must be between 0 and . Therefore the velocity (v) must be greater than

Quark 20110926 16:39:36 
That last line should read "(v') must be greater than ".

FutureDrSteve 20111106 14:04:33 
I'm fairly certain that the xcomponent of each particle would in fact be v/2. It's the total velocity that is greater than v/2, thanks to the addition of the vertical component of velocity.

walczyk 20121007 17:32:57 
Just thinking about it, and I realize the particles can't deflect at an angle greater than 45 degrees. Since , then we look at the energy:

berri104 20131014 05:26:49 

  Walter 20080902 14:57:08  Just a trivial matter  you need to apply conservation of momentum vertically to ascertain that because the particles move away at equal angles and have equal masses their velocities must be the same.
ramparts 20091008 10:57:03 
No you don't  there's symmetry. There is absolutely no reason for one direction (up or down) to be preferred. Yes, you can do the math to check, but what's the point?

  Healeyx76 20061102 19:41:40  another quick way is conservation of energy.
Initial energy is
After you have
You solve which is greater than
Mindaugas 20070916 01:17:00 
It is not said that collision is elastic energy may not be conserved.

eoliv001 20101011 16:03:33 
I agree. There is no mention of this being an elastic collision. You can not assume that.

 

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