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**"neutron collides elastically with a helium nucleus"**

## Homework Statement

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 θ'2 = 40° from the neutron's initial direction. The neutron's initial speed is 5.0 e5 m/s. Determine the angle at which the neutron rebounds, θ'1, measured from its initial direction.

## Homework Equations

(1/2)mu1^2 = (1/2)(4m)(v2^2) + (1/2)(m)(v1)^2 where u1 = 5.0e5m/s

(1/2)u1^2 = (1/2)(4)(v2^2) + (1/2)(v1)^2

**u1^2 = (4)(v2^2) + (v1)^2**

m(u1)(cos40) = (4)(m)(v2) + (m)(v1x)

(u1)(cos40) = (4)(v2) + (v1x)

(u1)(.7660) = (4)(v2) + (v1x)

(5.0e5)(.7660)

383022.2216 = (4)(v2) + (v1x)

**v1x = 383022.2216 - (4)(v2)**

m(u1)(sin40) = (m)(v1y)

(u1)(sin40) = (v1y)

(5.0e5)(.6428) = (v1y)

**v1y = 321393.8048**

(v1^2) = (v1x)^2 + (v1y)^2

(v1^2) = { (383022.2216 - (4)(v2) }^2 + ( 321393.8048)^2

(v1^2) = (1.4671e11 - 16v2^2 + 1.0329e11)

(v1^2) = 2.50e11 - 16v2^2

**(v1) = 5.0e5 - 4v2)**

(2.50e11) = (4)(v2^2) + (2.5e11+16v2^2)

5.0e5 = 4v2 + 5.0e5+4v2

0 = 8v2 NOOOO!!!

*u1^2 = (4)(v2^2) + (v1)^2*(2.50e11) = (4)(v2^2) + (2.5e11+16v2^2)

5.0e5 = 4v2 + 5.0e5+4v2

0 = 8v2 NOOOO!!!

Thanks, any help is appreciated!!!