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GR9277 #37
Problem
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\prob{37}

A $\pi^0$ meson (rest-mass energy 135 MeV) is moving with velocity $0.8c\hat{k}$ in the laboratory rest frame when it decays into two photons $\gamma_1$ and $\gamma_2$. In the $\pi^0$ rest frame, $\gamma_1$ is emitted forward and $\gamma_2$ is emitted backward relative to the $\pi^0$ direction of flight. The velocity of $\gamma_2$ in the laboratory rest frame is

1. $-1.0c\hat{k}$
2. $-1.2c\hat{k}$
3. $+0.8c\hat{k}$
4. $+1.0c\hat{k}$
5. $+1.8c\hat{k}$

Special Relativity$\Rightarrow$}Maximal Velocity

The maximal velocity of any object, even light itself, is the speed of light. Moreover, light always travels at light speed (c). This is true in all frames, and in fact, it is one of the two postulates of Special Relativity (the other being the equivalence of inertia frames).

There's no need to chunk out the addition of velocity formula for this. The only possibilities are choices (A) and (D). Since $\gamma_2$ is emitted backwards, according to the coordinate system in the diagram, its velocity would be $-c \hat{k}$, as in choice (A).

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Abou Ghazala
2018-03-11 20:16:17
$\\gamma_2$ is just a photon\r\nit follows from the introductory concepts that\r\n1- the speed of light is indepentant of the speed of the light source\r\n2-speed of light is the same in any inertial reference frame\r\nthese remain choices (A) and (D)\r\nthe velocity of $\\gamma_1$ photon is in opposite direction to that of $\\gamma_2$ photon\r\n$\\gamma_1$ photon moves in positive x-direction with speed =+1.0c\r\n$\\gamma_2$ photon moves in negative x-direction with speed =-1.0c as in choice (A)
e2ka
2009-10-07 11:47:39
Argh! So easy, yet missed. Don't be like me and see "1.0" and "c" and quickly write down answer D, which happened to be the first 1.0c I saw...
 kroner2009-10-07 20:35:22 I did that too...
 gravity2010-11-08 00:52:04 This is frustrating and is a stupid trick done by ETS. I was all like, "haha, ETS, you're trying to trick me with your stupid question, I'm $not$ falling for that. BAM! D!"
MP
2007-10-29 21:44:57
 ramparts2009-08-09 20:25:36 This is true, but it's a non-issue, since the answer is wrong :P

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}\$ $\frac{a}{b}$