Question

A proton (p) of mass 2.36-u(unified atomic mass units) traveling with a speed 4.93 x 10⁴ m/s of has an elastic head-on collision with a helium (He) nucleus (m_He = 4.00 u) initially at rest.

What is the velocity of helium nucleus after the collision?
(As mentioned in Chapter 1, 1 u = 1.66 x 10^–27 kg but we won’t need this fact.)

Assume the collision takes place in nearly empty space.

Answer 1

1. m_pv_p + 0 = m_pv_p' + m_He v_He'

2.   m_pv_p² /2 + 0 = m_pv'_p² /2 + m_He v'²_He /2

3. From (1):   v_p' = v_p –( m_He/m_p)v_He'

Substitute (3) into (2)

                                               v'²_He -- v'_He [2m_pv_p /(m_p + m_He)]

This eqv. has 2 solution: 1)   v'_He =0,    v'_p = 4.93 x 10⁴ m/s (initial cond.)

                        and             2) v'_He = 2m_pv_p /(m_p + m_He) = 2.96*10^4m/s

                                          v'_p =   v_p - ( m_He/m_p)x[2m_pv_p /(m_p + m_He)] = -0.617v_p = -3x10⁴ m/s

i.e. proton has changed the direction of moving.