Question

A proton (mass 1.67 × 10 −27kg) is moving at 1.30 ×10 −6m/s directly toward a stationary helium nucleus (mass 6.64 × 10 −27kg).

a)After a head-on elastic collision, what is the proton's velocity?.

b)After a head-on elastic collision, what is the helium's velocity?

Answer 1

Solution-
Let us assume that particle 1 be proton and helium nucleus is particle 2
proton's initial velocity = v1
proton's final velocity = v1'
helium nucleus final velocity = v2'
m1 is the mass of the proton and m2 is the mass of the helium nucleus.

Initial momentum = final momentum (By applying the conservation of momentum)
m1v1 = m1v1' + m2v2' (eq 1)

Initial KE = Final KE (because this is an elastic collision)
½m1v1² = ½m1v1'² + ½m2v2'² (eq 2)

v1' = [(m1-m2)/(m1+m2)] v1
=> [(1.67*10^-27 - 6.64*10^-27)/(1.6*10^-27 + 6.64*10^-27)] *1.30*10^-6
=> [(1.67 - 6.64)/(1.67 + 6.64)] * 1.30*10^-6
= -7.77*10^-7 m/s

v2' = [2m1/(m1+m2)] v1
= [ 2 x 1.67*10^-27 /(1.67*10^-27 + 6.64*10^-27)] x 1.30*10^-6
= [ 2 x 1.67 /(1.67 + 6.64)] x 1.30*10^-6
=5.22x10^-7 m/s