Answer: Find Relativistic Collision Momentum & Energy

[bobred]

Homework Statement

A mass m travels at 1.5 x 10^8 m s^-1 and collides with another mass m at rest. The two masses fuse to become M and travel away at v_c. Find an expression for v_c using conservation of relativistic momentum and energy.

Homework Equations

E_a+E_b=E_c and p_a+p_b=p_c. With b at rest p_b=0 so p_c=p_a.

E_a=\frac{mc^2}{1-\sqrt{\frac{v^2}{c^2}}} (1)
E_b=mc^2 (2)

p_a=\frac{mv}{1-\sqrt{\frac{v^2}{c^2}}} (3)E^{2}_{tot}=M^2c^4+p^2_cc^2 (4)

The Attempt at a Solution

Energy conservation
\frac{mc^2}{1-\sqrt{\frac{v^2}{c^2}}}+mc^2=\frac{Mc^2}{1-\sqrt{\frac{v^2_c}{c^2}}}

Momentum conservation
\frac{mv}{1-\sqrt{\frac{v^2}{c^2}}}=\frac{Mv}{1-\sqrt{\frac{v^2_c}{c^2}}}

Inserting the above into eqn 4

\frac{M^2c^4}{1-\frac{v^2_c}{c^2}}=M^2c^4+\frac{M^2v^2_cc^2}{1-\frac{v^2_c}{c^2}}

Am I on the right path? I can't seem to get sensible answer for v_c

 

[vela]

Try using the relation v/c = pc/E.

 

[bobred]

Got it thanks

p_c=\frac{Mv_c}{1-\sqrt{\frac{v^2_c}{c^2}}} (1)

E_c=\frac{Mc^2}{1-\sqrt{\frac{v^2_c}c^2}}} (2)

Dividing (1) by (2)

\frac{p_c}{E_c}=\frac{v_c}{c^2} so v_c=\frac{E_cc^2}{p_c}

James