# What is the angle at which the neutron rebounds?

• PhyzicsOfHockey
In summary, a neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus rebounds at an angle of 45° from the neutron's initial direction. The neutron's initial speed is 6.6x 10^5 m/s. The angle at which the neutron rebounds, measured from its initial direction, is approximately 75.964 degrees. The speed of the neutron after the collision is 544249.9425 m/s, and the speed of the helium nucleus after the collision is 186676.1902 m/s.
PhyzicsOfHockey

## 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.

Last edited:
maybe to simplify, let's 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 that's 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.

Last edited:
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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## What is an elastic non head on collision?

An elastic non head on collision is a type of collision where two objects collide with each other, but not directly head on. This means that the objects do not have the same initial velocity, and they also do not travel in the same direction after the collision.

## How is momentum conserved in an elastic non head on collision?

In an elastic non head on collision, the total momentum of the system before and after the collision remains the same. This means that the sum of the individual momentums of the two objects remains constant, even though the objects may change direction and speed after the collision.

## What is the difference between elastic and inelastic collisions?

In an elastic collision, the total kinetic energy of the system is conserved, meaning that no energy is lost during the collision. In an inelastic collision, some of the kinetic energy is lost and converted into other forms of energy, such as heat or sound.

## How is the coefficient of restitution related to elastic non head on collisions?

The coefficient of restitution is a measure of the elasticity of a collision. In an elastic non head on collision, the coefficient of restitution is equal to the ratio of the final velocity of separation to the initial velocity of approach. This value is always between 0 and 1, where 1 represents a perfectly elastic collision and 0 represents a perfectly inelastic collision.

## What are some real-world examples of elastic non head on collisions?

Some examples of elastic non head on collisions include a tennis ball hitting a racket, two cars colliding at an angle, or a billiard ball hitting another ball at an angle. These collisions can be observed in sports, transportation, and everyday life.

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