GR 8677927796770177 | # Login | Register

GR9677 #36
Problem
 GREPhysics.NET Official Solution Alternate Solutions
This problem is still being typed.
Special Relativity$\Rightarrow$}Conservation of Energy

The rest mass for each mass is 4kg. They collide head-on with identical speeds pointing in opposite directions. This implies that the composite mass is at rest. Thus, recalling that the total energy is given by $E=\gamma mc^2$ and that the rest mass is given by $E=mc^2$, one has
$2\gamma mc^2 = Mc^2$, where M is the composite mass.

The particle travels at $v=3c/5$, which yields $\gamma = 5/4$. Plug this in to get $M=10/4 \times 4 = 10kg$.

Alternate Solutions
 There are no Alternate Solutions for this problem. Be the first to post one!
CaspianXI
2009-03-22 16:24:17
The two lumps' energies will be "converted" into mass when they collide and stop because we're given that no energy is radiated. Hence, the final mass must be greater than 8 kg. So, we can immediately eliminate options (A), (B), and (C).

So, if you can't figure it out, you can guess and have a 50% chance of getting it right.
lowder.chris
2007-10-03 21:31:57
Relativity gives me a headache. :)
 bbaker032007-10-16 18:38:25 sorry im looking at the solution and getting confused. are you using the equation $\gamma$ = $\1/sqrt{1-v^2/c^2}$
 grae3132007-10-31 12:44:17 yes, bbaker03, if you plug in $v = 3c/5$ to the equation you wrote for gamma, you get $\gamma = 5/4$
 bbaker032007-11-01 10:24:11 Thanks a lot for your help this site is saving my life on this test. Can't thank you all enough.

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