A neutron collides elastically with a helium nucleus (at rest initially) whose m
ID: 1652652 • Letter: A
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 degree from the neutron's initial direction. The neutron's initial speed is 5.8 times 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? m/s What is the speed of the helium nucleus after the collision? m/sExplanation / Answer
Momentum is conserved.
horizontal components:
mv = 4m(v)cos41° + m(vn)cos
v = 3.02(v) + (vn)cos [1]
vertical components:
0 = 4m(v)sin41° - m(vn)sin
(vn)sin = 2.62(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]
Lets break variables and equations into x,y,z for simplicity.
x = 3.02(y) + (z)cos; --1
(z)sin = 2.62(y); --2
x² = 4(y)² + (z)²; --3
y = zsin/2.62 from 2,
x = 3.02(zsin/2.62) + zcos ---5
x^2 = 4(z^2*sin^2/2.62^2) + z^2 --- 6
Squaring 1,
x^2 = (3.02(zsin/2.62))^2 + z^2cos^2 + 2((3.02(zsin/2.62)*zcos) --7
Equating 6 and 7,
4(z^2*sin^2/2.62^2) + z^2 = (3.02(zsin/2.62))^2 + z^2cos^2 + 2((3.02(zsin/2.62)*zcos)
0.582717 z^2 sin^2() + z^2 = 1.32865 z^2 sin^2() + z^2 cos^2() + 2.30534 z^2 sin() cos()
-0.745933 z^2 sin^2() + z^2 = z^2 cos^2() + 2.30534 z^2 sin() cos()
-0.745933sin^2() + 1 = cos^2() + 2.30534sin()cos()
-0.745933sin^2() + sin^2() = + 2.30534sin()cos()
-0.745933sin() = + 2.30534cos()
=> tan() = 2.30534/-0.745933
=> = arctan(2.30534/-0.745933)
= 72.07 degree
= 72.07°
Substitute into [6] to solve for vn:
5.8e5 m/s = (vn)[sin72.07° + cos72.07°]
vn = 4.6e5 m/s
Substitute and vn into [2] to solve for v
(4.6e5 m/s)sin72.07° = 2sqrt(2)(v)
v = 1.54e5 m/s
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