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 4.2×10⁵ m/s .

Determine the speeds of the two particles, v′n and v′He, after the collision. Express your answers using two significant figures. Enter your answers numerically separated by a comma.

Answer 1

Given,

m2 = 4 m1; theta'2 = 45 deg ; vi = 4.2 x 10^5 m/s

momentum is conserved in all the direction. So along x

mv = 4 m v'He cos45 + m v'n cos(theta)

v = 2 sqrt (2) v'He + v'n cos(theta)

along y

0 = 4m v'He sin45 - m v'n sin(theta)

v'n sin(theta) = 2 sqrt(2) v'He

v'v^2 sin^2(theta) = 8 v'He

from conservation of energy

1/2 m v^2 = 1/2 4m + v'He^2 + 1/2 m v'n^2

v^2 = 4 v'He^2 + v'n^2

putting v'v^2 sin^2(theta) = 8 v'He

v = v'n sqrt (1/2 sin^2(theta) + 1)

putting in first eqn

v = v' n(sin(theta) + cos(theta))

v'n sqrt (1/2 sin^2(theta) + 1) = n'n sin(theta) + cos(theta)

sin(theta)/cos(theta) = 4

theta = 76 deg

4.5 x 10^5 = v'n (sin76 + cos76)

v'n = 3.71 x 10^5 m/s

v'n sin(theta) = 2 sqrt(2) v'He

3.71 x 10^5 sin76 = 2 x 1.414 v'He

v'He = 1.27 x 10^5 m/s