A neutron of mass 1.67 x 10^-27 collides head on with a deuterium at rest. If the velocity before the collision of the neutron is 8.8km/s?

A)
The neutron: m₁ = 1.67 × 10⁻²⁷ kg, u₁ = 8.8 km/s, v₁ = ? km/s
The deuterium nucleus: m₂ = 3.34 × 10⁻²⁷ kg = 2m₁, u₂ = 0 km/s, v₂ = ? km/s

Conservation of momentum:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ + (2m₁)u₂ = m₁v₁ + (2m₁)v₂
u₁ + 2u₂ = v₁ + 2v₂
8.8 + 2(0) = v₁ + 2v₂
v₁ = 8.8 - 2v₂ …… [1]

Elastic collision (conservation of kinetic energy):
(1/2)m₁u₁² + (1/2)m₂u₂² = (1/2)m₁v₁² + (1/2)m₂v₂²
(1/2)m₁u₁² + (1/2)(2m₁)u₂² = (1/2)m₁v₁² + (1/2)(2m₁)v₂²
u₁² + 2u₂² = v₁² + 2v₂²
(8.8)² + 2(0) = v₁² + 2v₂²
v₁² + 2v₂² = (8.8)² …… [2]

Substitute [1] into [2] :
(8.8 - 2v₂)² + 2v₂² = (8.8)²
(8.8)² - 35.2v₂ + 4v₂² + 2v₂² = (8.8)²
6v₂² - 35.2v₂ = 0
v₂(6v₂ - 35.2) = 0
v₂ = 0 (rejected) or v₂ = 35.2/6 = 5.87 km/s

Substitute v₂ = 35.2/6 km/s into [1]:
v₁ = 8.8 - 2(35.2/6)
v₁ = -2.93 km/s

Speed of the deuterium nucleus, |v₂| = 5.87 km/s
Speed of the neutron, |v₁| = 2.93 km/s


B)
Method 1 :
Kinetic energy after the collision
= (1/2)m₁v₁² + (1/2)m₂v₂²
= (1/2)(1.67 × 10⁻²⁷ kg)(-2.93 × 1000 m/s)² + (1/2)(3.34 × 10⁻²⁷ kg)(5.87 × 1000 m/s)²
= 6.47 × 10⁻²⁰ J

Method 2 :
As the collision is elastic, kinetic energy is conserved.
Kinetic energy after the collision
= Kinetic energy before the collision
= (1/2)m₁u₁² + (1/2)m₂u₂²
= (1/2)(1.67 × 10⁻²⁷ kg)(8.8 × 1000 m/s)² + 0 J
= 6.47 × 10⁻²⁰ J