How do I do this problem with an angle involved?

Question

I can do it without the angle, so can someone show me the steps to do it with the angle?

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 = 42° from the neutron's initial direction. The neutron's initial speed is 5.0 105 m/s. Determine the angle at which the neutron rebounds, '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?

ok, I think I have the answer to the thrid one: 1.486289651E5

Answer 1

Elastic collision
____/
.........\

okay - so the initial speed of the neutron is 5.0105 m/sec

OOOOOHH - I see 5.0 x 10^5
that's different!

Mass of neutron is m

He (mass 4m) rebounds at 42 degrees from the neutron's initial direction

You need to use trig to find this all out.

Conservation of momentum
p = mv
= 5.0 x 10^5m (direction due east across my piece of paper) kgm/sec

angle of neutron rebound is unknown angle theta (th)
Velocity of the helium atom is VH
Velocity of the neutron is VN

5.0 x 10^5 = 4m.VH.cos(42) + m.VN.cos(th)

Also - the vertical components of the neutron and helium atom momentum cancel each other out

4m.VH.sin(42) - 4m.VN.sin(th) = 0

VHsin(42) = VNsin(th)

And now we have Helium atom speed VH = 1.486289651x10^5 m/sec I hope

Good

So we have VN.sin(th) = 1.486...x10^5.sin(42)

AND

m.VN.cos(th) = 5.0x10^5 - 4m.cos(42).1.486...x10^5

Are you given the mass m?