[1] A proton (mp=938.3 MeV/c^2) is moving with a velocity of 0.8 c. It collides

Question

[1] A proton (mp=938.3 MeV/c^2) is moving with a velocity of 0.8 c. It collides with a helium atom (3728 MeV/c^2) at rest. After the collision, the helium atom is moves with a velocity of 0.35 c. Assume this is a 1D collision where the two particles move along the x-axis. Using momentum conservation with the classical definition of momentum p= mv, calculate (a) the momentum of the proton after the collision and (b) the speed of the proton after the collision.

[2] Repeat problem 1, but instead of the classical formula, use the correct relativistic definition of momentum. >> p= mv/sqrt(1-v^2/c^2)

Explanation / Answer

mp*0.8c = final momemtum of proton + m of helium *0.35c

final momentum of proton = 938.3 *0.8 - 3728 *0.35 =-554.16 Mev/c negitive sign indicates the particle moving in opposite direction to helium.

b) m*v =-554.16

v =-554.16/3728 = 0.148c

2)

a) mp*0.8c/(sqrt(1-0.8^2)) =  final momemtum of proton + m of helium *0.35c/(sqrt(1-0.35^2))

final momentum of proton = 938.3*0.8/(0.6) - 3728 *0.35/(0.9367) = 1251.06 - 1392.975 = -141.915 Mev/c negitive sign indicates the particle moving in opposite direction to helium.

b) 938.3 *v/sqrt(1-(v^2/c^2))= -141.915

6.611v = sqrt(1-v^2/c^2)

43.714v^2 = (c^2-v^2)/c^2

v^2 = c^2/(43.714c^2 +1)

v= 0.023c (apprx) if wrong please check the calculation plzz

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