January 23, 2017

Homework Help: Physics

Posted by micole on Wednesday, May 30, 2007 at 1:28pm.

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 = 41° from the neutron's initial direction. The neutron's initial speed is 5.6 105 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?
I have been working with a tutor on this problem for an hour and a half. We can not figure it out. I started with the kinetic energy equation, then tried to find the x and y components, but we keep getting stuck. Please help!!!!

The conservation of momentum may be written in two directions, x, y at 90 deg to each other. That gives you two equations.

The KE is only one, so you have three equations, enough. You may not need the energy equation.

This is an exercise in algebra. Other than checking the initial equations, there is not much I can do for you.

There is a simple systematic way to solve these sort of problems. You first transform to the zero momentum (= center of mass) frame. In that frame you write down how the particles move after the collision. Then you transform back to the original frame.

In the zero momentum frame the total momentum of the particles is zero. So, there you have an equation like:

p1 + p2 = q1 + q2 = 0 (1)

where p and q are the momenta before and after the collision.

The kinetic energy for a particle of mass m and momentum p can be written as p^2/(2m). Conservation of energy for the particles thus implies:

p1^2/(2m1) + p2^2/(2m2) =
q1^2/(2m1) + q2^2/(2m2) (2)

From (1) you see that p2 = -p1 and q2 = -q1. If you substitute that in (2) you get:

p1^2 [1/(2m1) + 1/(2m2)] =
q1^2 [1/(2m1) + 1/(2m2)] --->

p1^2 = q1^2

and thus also:

p2^2 = q2^2

So, the magnitude of the two momenta stay the same before and after the collision! So, in the zero momentum, frame before the collision, the two particles approach each other. Then they collide and after the collision boith paticles move away from each other each. The speeds don't change, but the directions can change.

Of course, if one particle moves in a certain direction after the collision, the other particle must move in exactly the opposite direction, because the total momentum is zero.

Let's put:

Velocity of the neutron = V_n

mass of the neutron = m

angle at which the helium nucleus rebounds = beta

The initial velocity of the helium nucleus is zero. Suppose you move with velocity W in the same direction as the neuron. Then in your frame the neutron is moving with velocity:

V_n' = V_n - W

The helium nucleus is now moving with velocity:

V_h' = -W

The total momentum in your frame is thus:

m V_n' + 4m V_h' = m V_n - 5 m W

The total momentum is zero if:

W = 1/5 V_n

In this zero momentum frame we have:

V_n' = V_n - W = 4/5 V_n

V_h' = -W = -1/5 V_n

Let's denote the velocity of the neutron after the collision by U_n' and the velocity of the helium nucleus by U_h' (in the zero momentum frame and in the original frame without the prime). We know that the magnitude of the velocities stay the same only the direction changes.

Now, angles in the zero momentum frame are not the same as angles in the original frame, so, let's just assume that after the collision the helium nucleus moves at an angle alpha relative to the positive x-direction in the center of mass frame. Then:

U_h' = 1/5 |V_n| cos(alpha)x-hat +
1/5 |V_n| sin(alpha)y-hat

x-hat and y-hat are unit vectors in the x and y direction, respectively.

The velocity of the neutron after the collision is:

U_n' = -4/5 |V_n| cos(alpha)x-hat -
4/5 |V_n| sin(alpha)y-hat

Now transform back to the original frame. All you have to do is add W to the above velocities. Note that W is oriented in the x-direction, i.e.

W = |W| x-hat = 1/5 |V_n| x-hat

We thus have:

U_h = U_h' + W =

1/5 |V_n| [cos(alpha) + 1]x-hat +
1/5 |V_n| sin(alpha)y-hat (3)


U_n = U_n' + W =

1/5 |V_n| [1 - 4 cos(alpha)] x-hat -
4/5 |V_n| sin(alpha)y-hat (4)

All we have to do now is to express alpha in terms of the given angle beta:

The ratio of the y and x components of U_h is tan(beta). Using Eq. (3) you find:

sin(alpha)/[cos(alpha) + 1] = tan(beta)


sin(alpha) = (cos(alpha) + 1)
sin(beta)/cos(beta) ----->

sin(alpha)cos(beta) - cos(alpha)*
sin(beta) = sin(beta) ---->

sin(alpha - beta) = sin(beta) --->

alpha - beta = beta (or alpha - beta = pi - beta but this correponds to the velocity of the helium nucleus being zero). So we have:

alpha = 2 beta

and the problem is solved!

Thank you sooooooo much!!!! My tutor and I and 4 other students worked on this for over an hour and a half and got nowhere. I can't thank you enough!

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