Question

a proton moving with a speed of 10^6m/s collides with a second proton initial at rest.one proton emerge from the collision at the angle 30 and other 60 degree from the incident direction.calculate the speed of proton.

Answer 1

Let us assume the mass of proton as 'm'

As collision takes place so momentum will be conserved here

$p_i=p_f\\p_i=m\times10^6+m\times0$

Let us see the final case one proton is at 30^o and other one is at 60^o

in y direction final momentum will be zero

$mv_1\sin30^o=mv_2\sin60^o\\v_1=v_2\sqrt3$ ..............................(1)

now let's see the momentum in the x direction

$mv_1\cos30^o+mv_2\cos60^o=m\times10^6\\$

substituting the value of $v_1$ in the above equation

$v_2\times\frac{3}2+v_2\times\frac12=10^6\\v_2=5\times10^5m/s$

from eq. (1)

we get $v_1=5\sqrt3\times10^5m/s$