Classical Mechanics - Linear Momentum

A helium atom, mass 4u travels with non relativistic speed v normal to the surface of a certain material, makes an elastic collision with an (essentially free) surface atom, and leaves in the opposite direction with speed 0.6v. The atom on the surface must be an atom of

A. Hydrogen, mass 1u
B. Helium, mass 4u
C. Carbon, mass 12u
D. Oxygen, mass 16u
E. Silicon, mass 28u

(GR9677 #20)

Solution:

m_a = 4u
v_a  = v
v_b  = 0
v_a' = − 0.6v

Conservation of momentum of the system:

m_av_a + m_bv_b = m_av_a' + m_bv_b' 
4uv = 4u(− 0.6v) + m_bv_b' 
4uv = − 2.4uv  + m_bv_b' 
m_bv_b' = 6.4uv (Eq.1)

Conservation of kinetic energy of the system:

½ m_av_a² + ½ m_bv_b² = ½ m_av_a'² + ½ m_bv_b'²
4uv² = 4u(− 0.6v)² + m_bv_b'²
4uv² = 4u(0.36v²) + m_bv_b'²
m_bv_b'² =  4uv²  − 1.44uv²
m_bv_b'² =  2.56uv² (Eq.2)

(Eq.1) → (Eq.2)
6.4 (v_b') = 2.56v
v_b' =  (2.56/6.4)v = 0.4v  (Eq.3)

(Eq.3) → (Eq.1)
m_b = 6.4uv / 0.4v = 16u

Answer: D