GREPhysics.NET: GR0177 #55

GR0177 #55

Problem

GREPhysics.NET Official Solution

This problem is still being typed.

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

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).

Reply to this comment

Comments

carle257
2010-04-01 20:20:09

Reply to this comment

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.

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

$v_{1} = v_{2}$

ramparts
2009-10-08 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?

Reply to this comment

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 $V/2$

Mindaugas
2007-09-16 01:17:00

It is not said that collision is elastic $\Rightarrow$ energy may not be conserved.

Reply to this comment

Post A Comment!

Bare Basic LaTeX Rosetta Stone
LaTeX syntax supported through dollar sign wrappers $, ex., $\alpha^2_0$ produces $\alpha^2_0$ .
type this...	to get...
$\int_0^\infty$	$\int_0^\infty$
$\partial$	$\partial$
$\Rightarrow$	$\Rightarrow$
$\ddot{x},\dot{x}$	$\ddot{x},\dot{x}$
$\sqrt{z}$	$\sqrt{z}$
$\langle my \rangle$	$\langle my \rangle$
$\left( abacadabra \right)_{me}$	$\left( abacadabra \right)_{me}$
$\vec{E}$	$\vec{E}$
$\frac{a}{b}$	$\frac{a}{b}$

The Sidebar Chatbox...
Scroll to see it, or resize your browser to ignore it...

<a href="http://www6.shoutmix.com/?grephysics">View shoutbox</a>
ShoutMix chat widget

1GHz Smartphone for 1 cent - best deal on a Samsung Galaxy Phone

Free 3D Virtual World. Largest User-Created Fantasy World. Don't Play, Experience. Join Now!

Bored, got time to kill, and want to earn micro-cash by mindlessly clicking on links? Join CrownGPT. Click Offers. Click the Paid to Click button... then click to your heart's delight.

FREE 3D Virtual World Chat - Join Second Life Today!

All-Solutions List | Latest Comments | Unanswered Questions (Cries for Help!) |::| About

This site is not affiliated with ETS, nor is it officially endorsed (yet) by ETS. This site is the work of an individual named Yosun Chang. The solutions, sans user comments and ETS questions, are © 2005 by Yosun Chang except where noted. All rights reserved. / The comments appending particular solutions are due to the respective users. / Educational Testing Service, ETS, the ETS logo, Graduate Record Examinations, and GRE are registered trademarks of Educational Testing Service. The examination questions are © 1997 and © 2001 by ETS.

Bare Basic LaTeX Rosetta Stone