A neutron collides elastically with a helium nucleus (at rest initially) whose m

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  

'2 = 45 degrees. The neutron's initial speed is 4.5×10^5 m/s .

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

1 = ____ degrees below the initial direction of the neutron

Determine the speeds of the two particles, vn and vHe, after the collision. (Enter your answers numerically separated by a comma.)

v'n, v'He = ________ m,s

Explanation / Answer

Momentum is conserved.
horizontal components:
mv = 4m(v)cos45° + m(vn)cos
v = 2sqrt(2)(v) + (vn)cos [1]

vertical components:
0 = 4m(v)sin45° - m(vn)sin
(vn)sin = 2sqrt(2)(v) [2]

For elastic collisions, kinetic energy is conserved.
(1/2)mv² = (1/2)(4m)(v)² + (1/2)(m)(vn)²
v² = 4(v)² + (vn)² [3]

[2]² :
(vn)²sin² = 8(v)²
(1/2)(vn)²sin² = 4(v)² [4]

Substitute 4(v)² from [4] into [3]:
v² = (1/2)(vn)²sin² + (vn)²
v = (vn)sqrt[(1/2)sin² + 1] [5]

Substitute 2sqrt(2)(v) from [2] into [1]:
v = (vn)sin + (vn)cos
v = (vn)[sin + cos] [6]

[5] = [6]:
(vn)sqrt[(1/2)sin² + 1] = (vn)[sin + cos]
sqrt[(1/2)sin² + 1] = sin + cos
(1/2)sin² + 1 = sin² + 2sincos + cos²
(1/2)sin² + 1 = 2sincos + 1
(1/2)sin² = 2sincos
sin = 4cos (eliminate sin as a possible solution)
tan = 4
= 75.96°

Substitute into [6] to solve for vn:
6.5e5 m/s = (vn)[sin75.96° + cos75.96°]
vn = 5.3e5 m/s

Substitute and vn into [2] to solve for v
(5.3e5 m/s)sin75.96° = 2sqrt(2)(v)

= >5.1*e5= 2sqrt(2)(v)

=>v= 5.1*e5/2sqrt(2)
v = 1.8e5 m/s

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