roblem .00 Part A A neutron collides elastically with a helium nucleus (at rest

Question

roblem .00 Part A 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 ,-45°. The neutron's initial speed is 4.8x105 m/s. Determine the angle of the neutron. , after the collision Express your answer using two significant figures A2p o below the initial direction of the neutron Submit My Answers Give Up Part B Determine the speeds of the two particles, vn and vHe, after the collision Express your answers using two significant figures. Enter your answers numerically sep by a comma m/s Submit My Answers Give Up

Explanation / Answer

A)

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)

eq(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

tan = 4

= 75.96°

B)

Substitute into (6) to solve for vn:

4.8 x 10^5 m/s = (vn)[sin75.96° + cos75.96°]

vn = 3.96 x 10^5 m/s

Substitute and vn into (2) to solve for v

(3.96 x 10^5 m/s)sin75.96° = 2sqrt(2)(v)

v = 1.358 x 10^5 m/s

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