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 rebound at an angle x = 43° from the neutron's initial direction. The neutron's initial speed is 6.4*10^5 m/s. Determine the angle at which the neutron rebounds, measured from its initial direction.

I have been trying to answer this question for days. Below are the answers I have tried plugging in, but it was incorrect!

a) Determine the angle at which the neutron rebounds, measured from its initial direction. Incorrect: 79.94°

b) What is the speed of the neutron after the collision? Incorrect: 187226.4m/s

c) What is the speed of the helium nucleus after the collision? Incorrect: 545854.07 m/s

Answer 1

Form the conservation of momentum, P_x = M_n v_n = m_n v_n' cos theta_n' + M_HC V He' cos theta' He Rightarrow V_n - 4 VHe' cos theta 'He = Vn' cos theta_n' P_y: 0 = m_n vn' sin theta n' - M_He VHe' sin theta' He 4v' He sin theta' He = vn^' sin theta n' the collision is elastic, & so KE is also conserved KE = 1/2 m_n V_n^2 = 1/2 M_n V_n^'2 + 1/2 M_he V_he^'2 v_n^= v_n^'2 + 4 V_HE^'2 V_N^'2 = V_n^2 - 4V_He^'2 (v_n - 4v_he^' cos theta_he^')^2 = (v_n^' cos theta_n^')^2: (4 ve^' sin theta_he')^2 = (v_n' sin theta_) v_n^2 - 8 v_n V_he^' cos theta_he^' + 16 v_he^'2 cos^2 theta_he^' + 16 v_he^'2 sin^2 theta^'he = v_n^'2 cos^2 theta_n^' + v_n^'2 sin^2 theta n^' v_n^2 - 8 v_n V_he^' cos theta^'he + 16 V_he^'2 = v_n^'2 = v_n^2 - 4v_he^'2 V_he^' = 0.4 V_n cos^theta_he = 0.4 (6..4 times 10^5) cos 43 degree = 1.872 times 10^5 m/s \r\n V_n^'2 = v^n^2 - 4_he^'2 RIghtarrow v_n' = squareroot v_n^2 - 4v' He^2