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Magazine Chapter 6: Problem 5
Show that in an elastic collision between a neutron of mass \(m\) and a nucleus of mass \(M\) (with the nucleus initially at rest), \(\mathcal{M}\) will recoil with a velocity given by $$w=\frac{2 m}{m+M} v$$ where \(v\) is the initial velocity of the incoming neutron.
Short Answer
Expert verified
The nucleus will recoil with a velocity \( w = \frac{2m}{m+M} v \).
Step by step solution
01
Understand the Conservation Laws
In an elastic collision, both momentum and kinetic energy are conserved. We'll start by applying these conservation laws.
02
Write the Conservation of Momentum Equation
Before the collision, the total momentum is the momentum of the moving neutron, which is given by: \[ p_{\text{initial}} = m v \] After the collision, the total momentum is the sum of the momentum of the neutron and the nucleus: \[ p_{\text{final}} = m v' + M w \] Here, \( v' \) is the velocity of the neutron after the collision and \( w \) is the velocity of the nucleus. The conservation of momentum requires: \[ m v = m v' + M w \]
03
Write the Conservation of Kinetic Energy Equation
The kinetic energy before the collision is: \[ KE_{\text{initial}} = \frac{1}{2} m v^2 \] After the collision, the total kinetic energy is: \[ KE_{\text{final}} = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \] The conservation of kinetic energy requires: \[ \frac{1}{2} m v^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \]
04
Solve the System of Equations
We now have two equations:1. \( m v = m v' + M w \)2. \( \frac{1}{2} m v^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \)From equation 1, solve for \( v' \): \[ v' = \frac{m v - M w}{m} \]Substitute \( v' \) into equation 2 and solve for \( w \).
05
Simplify and Solve for \( w \)
Substituting \( v' \) from Step 4 into the kinetic energy equation:\[ \frac{1}{2} m v^2 = \frac{1}{2} m \left( \frac{m v - M w}{m} \right)^2 + \frac{1}{2} M w^2 \]Expanding and simplifying:\[ m v^2 = \left( m v - M w \right)^2 + m M w^2 \]\[ m v^2 = m^2 v^2 - 2 m M w v + M^2 w^2 + m M w^2 \]\[ m v^2 = m^2 v^2 - 2 m M w v + M w^2 \left( M + m \right) \]Now collect like terms and solve for \( w \):\[ 2 m M w v = M (m + M) w^2 \]Divide both sides by \( M (m + M) \):\[ w = \frac{2 m v}{m + M} \]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
conservation of momentum
In an elastic collision, the total momentum of the system before the collision is equal to the total momentum after the collision. Momentum is calculated by multiplying the mass of an object by its velocity. Let's consider the neutron and nucleus in this problem. Initially, only the neutron is moving, so the momentum before the collision is:
- \[ p_{\text{initial}} = m v \]
- \[ p_{\text{final}} = m v' + M w \]
- \[ m v = m v' + M w \]
conservation of kinetic energy
In elastic collisions, not only is momentum conserved, but kinetic energy is also conserved. Kinetic energy is given by the expression \( \frac{1}{2} m v^2 \). Before the collision, the kinetic energy of the system is simply the kinetic energy of the moving neutron:
- \[ KE_{\text{initial}} = \frac{1}{2} m v^2 \]
- \[ KE_{\text{final}} = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \]
- \[ \frac{1}{2} m v^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \]
recoil velocity
The recoil velocity of the nucleus \( w \) is what we're trying to find in this problem. By using the conservation laws of momentum and kinetic energy, we can derive this recoil velocity. From the momentum equation:
- \[ m v = m v' + M w \]
- \[ v' = \frac{m v - M w}{m} \]
- \[ \frac{1}{2} m v^2 = \frac{1}{2} m \bigg( \frac{m v - M w }{m} \bigg)^2 + \frac{1}{2} M w^2 \]
- \[ 2 m M w v = M (m + M) w^2 \]
- \[ w = \frac{2 m v}{m + M} \]
elastic collisions
Elastic collisions are a type of collision where both momentum and kinetic energy are conserved. This means that there is no loss of kinetic energy in the system as a result of the collision. Instead, the kinetic energy is redistributed among the colliding bodies. Elastic collisions often occur at the atomic or subatomic level, such as in the case of a neutron colliding with a nucleus.
- Such collisions can be analyzed using the laws of physics and some algebra to solve for unknown quantities.
- The key properties of an elastic collision involve the conservation of momentum and conservation of kinetic energy.
- This contrasts with inelastic collisions, where some of the kinetic energy is converted into other forms of energy such as heat or sound.
momentum equations
In physics, momentum is defined as the product of mass and velocity, i.e., \( p = mv \). During collisions, the conservation laws can be expressed using momentum equations. For our problem involving a neutron and a nucleus, we used two main equations:
- Conservation of Momentum:\[ m v = m v' + M w \]
- Conservation of Kinetic Energy:\[ \frac{1}{2} m v^2 = \frac{1}{2} m v'^2 + \frac{1}{2} M w^2 \]
