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A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times...

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.

a) Incorrect: Your answer is incorrect. 73.96°

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

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

Solutions

Expert Solution

The x and y axis are chosen as shown in the diagram. The positive x axis is along the direction of v2 (final velocity of He nucleus). The y axis is the axis perpendicular to it.

The components of momentum along these x and y axis are calculated.

Conservation of momentum equations are set up for these components as follows:

Horizontal direction:

Total initial momentum along x axis = total final momentum along x axis.

So, mXu1XCos43 = 4XmXv2 + mXv1x

where u1 = 6.4X105 m/s

So, v1x = 468066.3690 - 4 X v2 ------

(1)

Total initial momentum along y axis = total final momentum along y axis.

So, mXu1XSin47 = mXv1y

So, v1y = 468066.3690 m/s -------

(2)

By conservation of kinetic energy in elastic collisions,

total initial kinetic energy = total final kinetic energy.

-------------(3)

---------------(4)

By using all above four equations, we get

v2 = 187226.4 m/s

substitute v2 in eq (5) then we get

v1x = -280839.231 m/s

v1 = -280839.231 i + 468066.3690 j

= 468066.3690 /280839.231

= 59.030

angle at which neutron rebounds, measured from its initial direction, θ1 = 430 + 900 - = 73.970

magnitude of v1 = 545854.1925 m/s

genius_generous answered 2 years ago
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