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GR0177 #55
Problem
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Mechanics$\Rightarrow$}Momentum

From conservation of momentum in the horizontal direction, one has $mv = 2 m v' \cos\theta$. Solving for the final velocity, one has $v'=\frac{v}{2\cos\theta}$. Since $|\cos\theta| < 1$ for $\theta > 0$, one finds that $v' > v/2$, as in choice (E).

Alternate Solutions
 carle2572010-04-01 20:20:09 This can be done simply by acknowledging that the x momentum is conserved so $v_{x}=\frac{v}{2}$. Then if they also have a vertical component, the speed must be greater than this value. Thus the answer must be (E).Reply to this comment
shka
2018-07-08 19:07:34
carle257 is using symmetry instead of explicitly writing out the equations for $p_y$ and $p_x$ conservation. His answer is fine and consistent with Yosun\'s, and the fact that Quark is confused by the notation $v_x = v^\\prime cos\\theta$ 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 $p_x = mv_x$ and $p_y = mv_y$, and since $p_x$ and $p_y$ 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
2014-08-12 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
2010-04-01 20:20:09
This can be done simply by acknowledging that the x momentum is conserved so $v_{x}=\frac{v}{2}$. Then if they also have a vertical component, the speed must be greater than this value. Thus the answer must be (E).
 Quark2011-09-26 16:36:49 Linear momentum is conserved but the velocity of each of the particles in the x-direction is not $v_x$=$\frac{v}{2}$. This would only be true if both particles were moving along the x-axis which is certainly not the case in this problem since they've given you the angle $\theta$. As the solution stated, by using conservation of linear momentum in the x-direction, you can solve for the velocity of each particle (not the x-component of the velocity) v'=$\frac{v}{2cos(\theta)}$ which includes a cos($\theta$) term in the denominator. This cos($\theta$) term must be less than 1 since $\theta$ must be between 0 and $\frac{\pi}{2}$. Therefore the velocity (v) must be greater than $\frac{v}{2}$
 Quark2011-09-26 16:39:36 That last line should read "(v') must be greater than $\frac{v}{2}$".
 FutureDrSteve2011-11-06 14:04:33 I'm fairly certain that the x-component 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.
 walczyk2012-10-07 17:32:57 Just thinking about it, and I realize the particles can't deflect at an angle greater than 45 degrees. Since $v' = v/(2cos\theta)$, then we look at the energy: $mv^2 /2 \ge mv^2 /(4cos^2 \theta)$
 berri1042013-10-14 05:26:49 $v'_x=v' cos \theta$
Walter
2008-09-02 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.

$mv_{1 } \, sin \theta - mv_{2} \, sin \theta = 0$

$v_{1} = v_{2}$
 ramparts2009-10-08 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
2006-11-02 19:41:40
another quick way is conservation of energy.

Initial energy is

$(1/2)*m*(V_0)^2$

After you have

$2*(1/2)*m*(V_f)^2$

You solve $V_f=V_0/(sqrt(2))$ which is greater than $V/2$
 Mindaugas2007-09-16 01:17:00 It is not said that collision is elastic $\Rightarrow$ energy may not be conserved.
 eoliv0012010-10-11 16:03:33 I agree. There is no mention of this being an elastic collision. You can not assume that.

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