Determine the angle of the neutron

[Bones]

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 θ'2 = 45° from the neutron's initial direction. The neutron's initial speed is 7.0 × 105 m/s.

a) Determine the angle of the neutron, θ'1, measured from its initial direction.
_____°

b) What is the speed of the neutron, vn', after the collision?
_____m/s

c) What is the speed of the helium nucleus, vHe', after the collision?
_____m/s

x direction: m1*7.0x10^5+0=m1*v1'*Cosθn+4*m1*v2'*Cos45

y direction: 0=-m1*v1'*Sinθn+4*m1*v2'*Sin45

Change in KE: 1/2*m1*(7.0x10^5)^2=1/2*m1*v1'^2+1/2*4*m1*v2'^2

I know mass cancels and I need to do algebra to figure out the answer, but I am having trouble doing the algebra with so many unknowns.
Can anyone lend me a helping hand?

 

[Bones]

I figured it out ;)

 

[thomate1]

a) To determine the angle of the neutron, we can use the conservation of momentum and energy equations:

x direction: m1*7.0x10^5 = m1*v1'*cosθn + 4*m1*v2'*cos45

y direction: 0 = -m1*v1'*sinθn + 4*m1*v2'*sin45

Change in KE: 1/2*m1*(7.0x10^5)^2 = 1/2*m1*v1'^2 + 1/2*4*m1*v2'^2

We can use the second equation to solve for sinθn:

sinθn = (4*m1*v2'*sin45)/m1*v1'

sinθn = 4*v2'*sin45/v1'

sinθn = (4*v2'*0.707)/v1'

sinθn = (2.828*v2')/v1'

Next, we can use the first equation to solve for cosθn:

cosθn = (m1*7.0x10^5 - 4*m1*v2'*cos45)/m1*v1'

cosθn = (7.0x10^5 - 2*v2'*cos45)/v1'

Now, we can use the trigonometric identity cos^2θ + sin^2θ = 1 to solve for cosθn:

cos^2θn + sin^2θn = 1

(7.0x10^5 - 2*v2'*cos45)^2 + (2.828*v2')^2 = v1'^2

49x10^10 - 2*7.0x10^5*2*v2'*cos45 + (2*v2'*cos45)^2 + (2.828*v2')^2 = v1'^2

49x10^10 - 2*7.0x10^5*2*v2'*0.707 + (2*v2'*0.707)^2 + (2.828*v2')^2 = v1'^2

49x10^10 - 9.88x10^5*v2' + 2*v2'^2 + 2.828^2*v2'^2 = v1'^2

v1'^2 = 49x