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 move off at an angle θ′2=45∘. The neutron's initial speed is 6.0×105 m/s .

Determine the angle of the neutron, θ′1, after the collision.

Determine the speeds of the two particles, v′n and v′He, after the collision

Answer 1

By conservation of momentum,

Horizontal component :

mv = 4m*(Va)*cos 45 degree + m*(Vn)* cos theta

v = 2*sqrt(2)*(Va) + Vncos theta [eq1]

Vertical component

0 = 4m*(Va) *sin 45 degree - m*(Vn) sin thet

Vn sin theta = 2sqrt(2)va [eq2]

For elastic collision KE is conserved,

(1/2)mv^2 = (1/2)(4m)(Va)^2 + (1/2)m(Vn)^2

v^2 = 4*Va^2 + (Vn) ^2 [eq3]

By Eq 2,

Vn^2 sin^2 theta = 8 Va^2 [eq4]

Solving eq4 and 3,

v^2 = (1/2)Vn^2 sin ^2 theta + Vn^2

v = Vnsqrt[(1/2)sin^2 theta + 1]. [Eq5]

Solving eq2 and eq1,

v = Vn sin theta + Vn cos theta

Substitute 2sqrt(2)(vα) from [2] into [1]:

v = (vn)sinθ + (vn)cosθ

v = Vn*(sin theta + cos theta) [eq6]

Equating eq5 and eq6,

Vnsqrt((1/2)sin^2 theta +1) = Vn( sin theta + cos theta)

Sqrt[(1/2)sin^2 theta + 1)] = (sin theta + cos theta)]

(1/2)sin^2 theta + 1 = sin^2 theta + 2sin theta cos theta + cos ^2 theta

(1/2)sin^2 theta + 1 = 1 + 2 sin theta cos theta

(1/2)sin^2 theta = 2sin theta cos theta

Sin theta = 4 cos theta

theta = arctan(4) = 75.96 degree.

Putting the value of theta in eq6,

6*10^5 = Vn*(sin 75.96 degree + cos 75.96 degree)

Vn = 4.94*10^5 m/s

Putting theta and Vn in eq2,

(4.94*10^5)*sin 75.96 degree = 2*sqrt(2)Va

Va = 1.69*10^5 m/s