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 rebound at an angle Theta'2 = 41 degrees from the neutron's initial direction. The neutron's initial speed is 5.8 X 10^5 m/s. Determine the angle at which the neutron rebounds,Theta'1, measured from its initial direction. What is the speed of the neutron after the collision? What is the speed of the helium nucleus after the collision? 7. -9 points My Notes 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 #2 = 41° from the neutron's initial direction. The neutron's initial speed is 5.8 105 m/s. Determine the angle at which the neutron rebounds, e', measured from its initial direction What is the speed of the neutron after the collision? m/s What is the speed of the helium nucleus after the collision? m/s Submit Answer Save Progress

Explanation / Answer

Given,

theta2' = 41 deg ; u = 5.8 x 10^5 m/s

from conservation of momentum along X direction:

mv = 4m (Va) cos45 + m (Vn) cos(theta)

v = 2.83 Va + Vn cos(theta)

Along vertical:

0 = 4m (Va) sin45 - m Vn sin(theta)

Vn sin(theta) = 2.83 Va (1)

squaring this we get

Vn^2 sin^2(theta) = 8 Va^2

1/2 Vn^2 sin^(theta) = 4 Va^2 (2)

enegy is conserved in elastic collision

1/2 m v^2 = 1/2 4m Va^2 + 1/2 m Vn^2

v^2 = 4 Va^2 + Vn^2

putting the value of 4 Va^2 from 2

v^2 = 1/2 Vn^2 sin^(theta) + Vn^2

v = Vn sqrt (1/2 sin^2(theta)) + 1

using 1,

v = Vn sin(theta) + Vn cos(theta)

v = Vn (sin(theta) + cos(theta))

eqaying the two values of v

Vn sqrt (1/2 sin^2(theta)) + 1 = Vn (sin(theta) + cos(theta))

1/2 sin^2(theta) = 2 sin(theta) cos(theta)

tan(theta) = 4 => theta = 75.96 deg

Hence, theta = 75.96 deg

putting theta in

v = Vn [sin(theta) + cos(theta)]

5.8 x 10^5 = Vn (sin75.96 + cos75.96) = 1.213 Vn

Vn = 4.78 x 10^5 m/s

Putting in

v sin(theta) = 2.83 Va

Va = 5.8 x 10^5 x sin75.96/2.83 = 1.99 x 10^5 m/s

Hence, Vn = 4.78 x 10^5 m/s and Va = 1.99 x 10^5 m/s

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