# Elastic non head on Collision

## 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 Theta2 = 45° from the neutron's initial direction. The neutron's initial speed is 6.6x 10^5 m/s.

Determine the angle at which the neutron rebounds, O, 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?

## Homework Equations

Conservation of KE
Conservation of momentum
1/2M1V1^2= 1/2M1V'1^2 + 1/2M2V'2^2
M1V1=M1V'1*cosO + M2V2cos45
0=M1V'1*sinO + M2V'2sin45
M2=4M1
V1=V'1 + 2*V'2
V1=V'1*cosO + 4*V'2*cos45
V'1*sinO= 4*V'2*sin 45

## The Attempt at a Solution

I have 3 equations and 3 unknows (V'1, V'2, O) I tried solving for O and it becomes a huge mess. Is there a simple way to do this? Or do I really have to figure out all this mess? Any help and tips are appreciated.

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maybe to simplify, lets just pretend they are a couple of bowling balls, one 4 # the other 16.

You are right: in an elastic collision we conserve both momentum and energy.

But the '2 is confusing as hell,

Also, where I get lost is the "helium atom rebounds". Since its a neutron, there are no electrical forces, so how does anything thats hit hard in one direction come back at ya?

Entirely possible that the nuetron comes back at you and what is defined as 45 degrees is actually 135.

If so,

'2 was suppose to be theta 2

Sorry, in a dense state, is that 2(theta)=45?

no its just the angle the He nucleus makes with respect to the x axis.

oh, ok.

well the momentum needs to be conserved in this case in both the x and y directions.

scroll down on the following link to the description of 2D elastic and notice the vector summation, see if that simplifies things for you.
http://en.wikipedia.org/wiki/Elastic_collision

postscript: answers using that method where final velocities along the line of force were for the neutron using the 1D eqns were:
-3/5sin(45)*V0 and for the helium,
2/5sin(45)*V0 calculating the angle given that thee tangential component is unchanged gives 76 degrees below the horizontal.

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See the attachment for diagram and solution with proper superscripts and subscripts.

The x and y axis are chosen as shown in the diagram. The positive x axis is along the direction of v2 (final velocity of He nucleus). The y axis is the axis perpendicular to it.
The components of momentum along these x and y axis are calculated.
Conservation of momentum equations are set up for these components as follows:
Total initial momentum along x axis = total final momentum along x axis.
So, m*u1*Cos45 = 4*m*v2 + m*v1x
where u1 = 6.6*105 m/s
So, v1x = 466690.4756 - 4*v2 ------(1)

Total initial momentum along y axis = total final momentum along y axis.
So, m*u1*Sin45 = m*v1y
So, v1y = 466690.4756 m/s -------(2)

By conservation of kinetic energy in elastic collisions, total initial kinetic energy = total final kinetic energy.
So,
(1/2)*m*u12 = (1/2)*(4m)*v22 + (1/2)*m*v12
So, u12 = 4v22 +v12 -------(3)
v12 = v1x2 + v1y2 --------(4)
Using eqn(4), eqn(3), eqn(1) and eqn(2), we get
v2 = 186676.1902 m/s -----(5)

Plugging in eqn(5) in eqn(1) and solving, we get v1x = -280014.2852 m/s
Thus v1 = -280014.2852 i + 466690.4756 j m/s
So tanφ = 466690.4756/280014.2852
So φ = 59.036o
angle at which neutron rebounds, measured from its initial direction, θ1 = 45o + 90o - φ = 75.964o
magnitude of v1 = 544249.9425 m/s

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