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 4.1×10⁵ 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

mn = m

mHe = 4*mn = 4 m

vn = 4.1*10^5 m/s

total momentum is conserved

along the horizontal ( x axis)

mn*vnx + mHe*vHex = mn*v'n*cosalpha + mHe*v'He*costheta

vn + 0 = v'n*cosalpha + 4*v'He*cos45

v'n*cosalpha = vn - 4*v'He*cos45 - .........(1)

along the vertical ( y axis)

mn*vny + mHe*vHey = mn*v'n*sinalpha - mHe*v'He*sintheta

0 + 0 = v'n*sinalpha - 4*v'He*sin45

v'n*sinalpha = 4*v'He*sin45 ................(2)

(1)^2 + (2)^2

(v'n)^2 = (vn)^2 - 2*vn*4*v'He*cos45 + 16*(v'He)^2 ..............(3)

from energy conservation

total kinetic energy is constant

(1/2)*mn*vn^2 = (1/2)*mn*v'n^2 + (1/2)*mHe*v'He^2

(vn)^2 = (v'n)^2 + 4*(v'He)^2.............(4)

substituting 3 and 4

(v'n)^2 = (v'n)^2 + 4*(v'He)^2 - 2*vn*4*v'He*cos45 + 16*(v'He)^2

- 2*vn*4*v'He*cos45 + 20*(v'He)^2 = 0

v'He = (8/17)*vn*cos45 = (8/20)*4.1*10^5*cos45

v'He = 1.2*10^5 m/s

using v'He in eq 4

v'n = 3.4*10^5 m/s