All parts, please and thanks! 3. 0). The initial A moving neutron collides elast
ID: 1731060 • Letter: A
Question
All parts, please and thanks!
3. 0). The initial A moving neutron collides elastically with a helium nucleus which is at rest (UHe speed of the neutron (vn) is 6.2 x 10 m s and it is moving horizontally to the right. The mass of the helium nucleus is 4 times the mass o the neutron (mHe -4mn) After the collision the helium nucleus and neutron are observed to move off at angles, @we-45°, and ??-??, from the direction of motion of the incoming neutron. The speeds of the neutron (vh) and helium nucleus (vHe) after the collision are also unknown Draw sketches showing the motion of the neutron and helium nucleus before and after the collision, marking the information given above (including knowns) onto your diagram a. [1 mark] b. Write down the energy and momentum conservation laws for this collision using components resolved horizontally and vertically where appropriate 3 marks] c. Using your equation(s) for momentum show that, for this collision [2 marks] d. Using the expression from c. and your equation for energy, show that the speed of the helium nucleus after the collision is about 1.75 x 105 ms. 3 marks] e. Hence, calculate the speed of the neutron after the collision (v) and the deflection angle for the neutron (en) [2 marks]Explanation / Answer
3. given moving neutron, collides with helium at rest, Vhe = 0
Vn = 6.2*10^5 m/s
moving horizontally to right
Mhe = 4*Mn
after collision
thetahe = 45 deg
thetan = ?
V'n, V'he = ?
a. the sketch is as under
b. energy conservation
0.5Mn*Vn^2 = 0.5Mn*Vn'^2 + 0.5*Mhe*Vhe'^2
momentum conseration
horizontal
MnVn = MnVn'*cos(tehtan) + Mhe*Vhe'*cos(thetahe)
vertical
MnVn'*sin(thetan) = Mhe*Vhe'*sin(thetahe)
c. Mn*Vn^2 = Mn*Vn'^2 + 4Mn*Vhe'^2
Vn^2 = Vn'^2 + 4Vhe'^2
also
Vn'*sin(thetan) = 4*(Vhe')*sin(thetahe)
Vn = Vn'*cos(thetan) + 4*Vhe'*cos(thetahe)
squaring and adding
Vn'^2 = 16Vhe'^2 + Vn^2 - 8Vn*Vhe'*cos(thetahe)
d. Vn^2 = Vn'^2 + 4Vhe'^2
hence
20Vhe' = 8Vn*cos(thetahe)
Vhe' = 175362.481734263 m/s
e. Vn' = 511265.09757658 m/s
hence
thetan = 75.963 deg