Question

A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to scatter at an angle θ₂ = 52° from the neutron's initial direction. The neutron's initial speed is 52000·m/s.

Determine the angle at which the neutron scatters, θ₁, measured from its initial direction:

What is the speed of the neutron after the collision?

What is the speed of the helium nucleus after the collision?

Answer 1

Momentum is conserved.
horizontal components:
mv = 4m(vα)cos52° + m(vn)cosθ
v = 2.4626(vα) + (vn)cosθ ---- [1]

vertical components:
0 = 4m(vα)sin52° - m(vn)sinθ
(vn)sinθ = 3.152(vα) ----[2]

For elastic collisions, kinetic energy is conserved.
(1/2)mv² = (1/2)(4m)(vα)² + (1/2)(m)(vn)²
v² = 4(vα)² + (vn)² ----- [3]

[2]² :
(vn)²sin²θ = 9.935(vα)²
(1/2.484)(vn)²sin²θ = 4(vα)² -----[4]

Substitute 4(vα)² from [4] into [3]:
v² = (1/2.484)(vn)²sin²θ + (vn)²
v = (vn)sqrt[(1/2.484)sin²θ + 1] ------ [5]

Substitute 2sqrt(2)(vα) from [2] into [1]:
v = 0.781(vn)sinθ + (vn)cosθ

v = (vn)[0.781sinθ + cosθ] ---- [6]

[5] = [6]:
(vn)sqrt[(1/2.484)sin²θ + 1] = (vn)[0.781sinθ + cosθ]
sqrt[(1/2.484)sin²θ + 1] = 0.781sinθ + cosθ
(1/2.484)sin²θ + 1 = 0.610sin²θ + 1.562sinθcosθ + cos²θ
(1/2.484)sin²θ + 1 = 1.562sinθcosθ + 0.610sin^2θ + cos^2θ

On solving

Sinθ = 3.88cosθ (eliminate sinθ as a possible solution)
tanθ = 3.88
θ = 75.54°

Substitute θ into [6] to solve for vn:
52000 m/s = (vn)[0.781sin75.54° + cos75.54°]
vn = 51691.7m/s

Substitute θ and vn into [2] to solve for vα
(51691.7 m/s)sin75.54° = 3.152(vα)
vα = 15880.15 m/s