Problem 93 - University Physics with Modern Physics

Question

A neutron with mass \(m\) makes a head-on, elastic collision with a nucleus of mass \(M,\) which is initially at rest. (a) Show that if the neutron's initial kinetic energy is \(K_{0}\) , the kinetic energy that it loses during the collision is 4\(m M K_{0} /(M+m)^{2}\) . (b) For what value of \(M\) does the incident neutron lose the most energy? (c) When \(M\) has the value calculated in part (b), what is the speed of the neutron after the collision?

Step-by-Step Solution

Answer

(a) The energy lost is \( \frac{4mMK_0}{(M+m)^2} \). (b) Maximum energy is lost when \( M = m \). (c) Neutron speed is 0 after the collision.

Step 1: Conservation of Momentum

In a head-on elastic collision, the conservation of momentum principle holds. The initial momentum of the system is the momentum of the neutron: \[ p_{i} = mv_{n,0} \]For the final momentum, both the neutron and the nucleus have velocities: \[ p_{f} = mv_{n} + MV \]This gives the equation: \[ mv_{n,0} = mv_{n} + MV \]

Step 2: Conservation of Kinetic Energy

In an elastic collision, both the kinetic energy and momentum are conserved. Initial kinetic energy is: \[ K_{i} = \frac{1}{2}mv_{n,0}^{2} = K_{0} \]Final kinetic energy is: \[ K_{f} = \frac{1}{2}mv_{n}^{2} + \frac{1}{2}MV^{2} \]Using conservation of kinetic energy, we have: \[ \frac{1}{2}mv_{n,0}^{2} = \frac{1}{2}mv_{n}^{2} + \frac{1}{2}MV^{2} \] This equation will be used in solving (a).

Step 3: Solving for Neutron's Final Velocity using Conservation Equations

From the conservation equations, solve for the final velocity of the neutron \( v_{n} \) after the collision. Using equations derived:\[ v_{n} = \frac{v_{n,0}(m-M)}{m+M} \]

Step 4: Solving for Nucleus's Final Velocity

We use the conservation of momentum to solve for the nucleus's final velocity \( V \):\[ V = \frac{2mv_{n,0}}{m+M} \]

Step 5: Calculating Kinetic Energy Loss

The kinetic energy loss, \( \Delta K \), is the difference between the initial and final kinetic energy of the neutron:\[ \Delta K = K_i - K_f = \frac{1}{2}m(v_{n,0}^{2} - v_{n}^2) \]Substitute the expression from Step 3 for final velocity \( v_{n} \).The loss of energy is:\[ \Delta K = K_0 \left(1 - \left( \frac{m-M}{m+M} \right)^2 \right) \]Simplifying, we find:\[ \Delta K = \frac{4mMK_0}{(M+m)^{2}} \]

Step 6: Maximizing Energy Loss

The energy loss expression is maximized by differentiating \( \Delta K = \frac{4mMK_0}{(M+m)^{2}} \) with respect to \( M \) and setting the derivative to zero.After differentiation and solving:\[ M = m \]This maximizes the energy loss.

Step 7: Calculating Neutron's Speed After Collision for Maximum Energy Loss

Substitute \( M = m \) into the expression for \( v_{n} \) from Step 3:\[ v_{n} = \frac{v_{n,0}(m-M)}{m+M} = 0 \] When the mass of the nucleus equals the neutron's mass, the neutron comes to rest after the collision.

Key Concepts

Conservation of Momentum

In the realm of physics, particularly when dealing with collisions, the conservation of momentum is a vital principle. Momentum, a product of mass and velocity, must be preserved in the system if no external forces act upon it. This means that the total momentum before the collision must equal the total momentum after.

In the original problem, a neutron collides with a nucleus that is initially at rest. The initial momentum is purely from the neutron:

• Initial momentum = \( mv_{n,0} \)

After the collision, both particles have momentum:

• Final momentum = \( mv_n + MV \)

Equating these, we have:

• \( mv_{n,0} = mv_{n} + MV \)

This equation is crucial as it links the initial movements with the final velocities of the neutron and nucleus, helping us to solve for unknown velocities after the collision.

Conservation of Kinetic Energy

In addition to momentum, elastic collisions conserve kinetic energy. This means that the total kinetic energy before the collision equals the total kinetic energy after the collision. For our problem, this principle applies because it is an elastic collision.

The initial kinetic energy of the neutron is given as:

• \( K_i = \frac{1}{2}mv_{n,0}^2 = K_0 \)

The final kinetic energy of the system, post-collision, is:

• \( K_f = \frac{1}{2}mv_{n}^2 + \frac{1}{2}MV^2 \)

Using the principle of kinetic energy conservation, we equate:

• \( K_i = K_f \)

This equation supports solving for the velocities of the neutron and nucleus after the collision, ensuring that energy remains constant.

Kinetic Energy Loss

Kinetic energy loss in collisions, especially elastic ones, is a crucial concept. In the stated problem, despite the collision being elastic where kinetic energy is generally conserved, the neutron still loses part of its original kinetic energy to the nucleus.

The initial problem requires calculating this kinetic energy loss, expressed as:

• \( \Delta K = \frac{4mMK_0}{(M+m)^2} \)

This formula shows how the initial kinetic energy \( K_0 \) of the neutron is spread between both particles after collision and gives insight into how masses affect the energy distribution.

To obtain this expression, we calculate the energy transferred to the nucleus and subtract it from the neutron's initial energy. The obtained formula indicates that maximum energy loss occurs when the nucleus's mass, \(M\), equals the neutron’s mass, \(m\). Understanding how energy loss is calculated is fundamental for analyzing momentum transfer in collisions.

Mechanics Problem Solving

Solving mechanics problems often involves applying fundamental physics principles to deduce unknown quantities. This problem is a great example, requiring systematic use of conservation laws in a step-by-step approach.

Step-by-Step Process:

1. **Identify the system and variables:** Here, the system includes a neutron and nucleus with given masses, and the task is to solve for kinetic energy loss and final velocities.

2. **Apply Conservation Laws:** Use the conservation of momentum and kinetic energy to set up equations that relate initial and final states.

3. **Solve Algebraically:** From these equations, solve for unknowns such as the final velocities of the particles. It may involve substitution and simplification of expressions.

4. **Check For Maximum or Special Conditions:** To find extreme or specific values, like the mass causing maximum energy loss, differentiate the energy equation with respect to mass \(M\) and set it to zero.

This approach not only yields accurate solutions but also reinforces understanding of physics fundamentals by connecting theory to practice through problem-solving.