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Conservation of energy and momentum

Conservation of energy

In the problem set of lecture 1 we saw that the work \(W\) done on an object by a conservative force \(F\) along a (one-dimensional) path is given by

\[ W = \int_{x_0}^{x_1} F \, dx = - \int_{x_0}^{x_1} \frac{dV}{dx} \, dx = V(x_0) - V(x_1)\, . \]

Using Newton’s second law, \(F = dp/dt\) and equation (\ref{eq:bkjngerhwoejs}), we can relate the potential change between two positions to the change in momentum:

\[\begin{split} \begin{aligned} V(x) - V(x_0) &= - \int_{x_0}^x \frac{dp}{dt} \, dx = - \int_{p(x_0)}^{p(x)} v \, dp = - \int_{p(x_0)}^{p(x)} \frac{p}{m}\, dp = - \left[ \frac{p^2}{2m}\right]_{p(x_0)}^{p(x)} \\ &= -\left(\frac{p(x)^2}{2m} - \frac{p(x_0)^2}{2m}\right)\, . \end{aligned} \end{split}\]

You will recognise the right-hand side in the last line as the change of kinetic energy \(T\) of the object, and this means that without any further effects on the object, the potential energy is transformed into kinetic energy. As the potential energy decreases (\(V(x) < V(x_0)\)), the kinetic energy increases and vice versa.

Bringing everything to the left-hand side of the equation, we end up with an expression for the conservation of energy:

\[ \Delta V + \Delta T = 0 \, , \]

where \(\Delta\) (delta) denotes a change. Note that this is a statement about changes. We can roll a marble down a ramp in the basement or in the penthouse (assuming constant \(g\)), and we will get the same results for the change in kinetic energy even though the potential energy of the marble in the penthouse is much greater than in the basement. Conservation of energy applies to the changes in energy in a process.

This is a special case of a broader conservation principle: the total energy in a closed system is conserved. It doesn’t mean that energy is always in a useful form. For example, when a friction force slows down a block sliding down a ramp, the final velocity is lower than if all potential energy was converted into kinetic energy. The “missing” energy has become the heat generated by the friction between the block and the ramp. It is typically not possible to recover this form of energy, but when we take it into account the total energy in the system is conserved.

When working with energy conservation in practical problems, we often compare the initial state of our physical system to its final state and do some simple bookkeeping.

• Initial state (block at rest atop a ramp at height \(h\)):

\[ E_{\rm before} = mgh \, , \]

where \(m\) is the mass of the block.

• Final state (at the bottom of the ramp):

\[ E_{\rm after} = \tfrac12 m v^2 + Q \, , \]

where \(v\) is the final speed of the block and \(Q\) is the heat generated by friction.

When there is no friction (\(Q=0\)), we can calculate \(v\) from \(h\) and \(m\). If there is friction (\(Q\neq0\)), we can determine the heat it generates by measuring \(v\) and using energy conservation, \(E_{\rm after} = E_{\rm before}\):

\[ Q = mgh - \tfrac12 m v^2 \, . \]

We will use this method when we discuss collisions. It will be critical to determine whether there is any heat generated in a collision, so you can tell whether you can use energy conservation for the particles that are colliding.

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Momentum conservation

Another important law of physics is the conservation of momentum: for any system that does not experience a net external force, its momentum is conserved.

For a single particle this is simply Newton’s second law:

\[ \mathbf{F} = 0 \quad \to \quad \frac{d\mathbf{p}}{dt} = 0 \, . \]

However, we saw that this law also holds for composite bodies consisting of \(N\) components with mass \(m_j\), centre of mass coordinate \(\mathbf{R}\), and centre of mass momentum \(\mathbf{P}\):

\[ \mathbf{F}_{\rm ext} = 0 \quad \to \quad \frac{d\mathbf{P}}{dt} = 0 \, . \]

Example: A supernova (ignoring the external gravitational force of the galaxy). In the rest frame of the star before the explosion,

\[ \mathbf{P}_{\rm before} = \mathbf{p}_1 + \mathbf{p}_2 + \cdots + \mathbf{p}_N = \sum_{j=0}^N \mathbf{p}_j = 0 \, , \]

since the velocity of the centre of mass \(\mathbf{V}=0\), even though individual parts are moving.

After the explosion,

\[ \mathbf{P}_{\rm after} = \mathbf{p}_1' + \mathbf{p}_2' + \cdots + \mathbf{p}_M' = \sum_{j=0}^M \mathbf{p}_j' = 0 \, . \]

We may no longer have the same number of parts (\(M \neq N\)), but the total momentum is conserved. The increased velocities of the constituents come from the release of nuclear potential energy.

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Momentum is not always conserved. For example, a ball bouncing off a wall changes momentum:

\[ \mathbf{p}_{\rm before} = m \mathbf{v}, \qquad \mathbf{p}_{\rm after} = -m \mathbf{v} \, . \]

The wall exerts a force on the ball. The impulse \(\mathbf{S}\) is

\[ \mathbf{S} = \mathbf{p}_{\rm after} - \mathbf{p}_{\rm before} = \mathbf{F}_{\rm wall} \, \Delta t \, , \]

where \(\Delta t\) is the contact time.

When considering the ball + wall system, momentum is conserved (Newton’s third law applies).

• Before:

\[ \mathbf{p}_{\rm before} = m\mathbf{v} + 0 \]

• After:

\[ \mathbf{p}_{\rm after} = -m\mathbf{v} + M \mathbf{v}_{\rm w} \, , \]

where \(M\) is the mass of the wall and \(\mathbf{v}_{\rm w}\) its velocity after the bounce.
Momentum conservation \(\mathbf{p}_{\rm after} = \mathbf{p}_{\rm before}\) gives:

\[ -m\mathbf{v} + M \mathbf{v}_{\rm w} = m\mathbf{v} \quad \to \quad \mathbf{v}_{\rm w} = \frac{2m}{M} \mathbf{v} \, , \]

which is tiny when \(M \gg m\).

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Collisions

The ball bouncing against a wall is a simple example of a collision problem. More generally, we distinguish between:

• Elastic collisions: both energy and momentum conserved.

• Inelastic collisions: momentum conserved, but kinetic energy changes:

\[ T_{\rm before} = T_{\rm after} + Q \, , \]

where \(Q\) is heat or other energy.

• Super-elastic collisions: \(Q<0\), e.g. when stored potential energy is released (supernovae, mines).

• Perfectly elastic: \(Q=0\).

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Example: Elastic collision in 2D

Consider particle \(m_1\) with velocity \(\mathbf{v}_1\) colliding with particle \(m_2\) at rest at the origin.

• Before:

\[ \mathbf{p}_{\rm before} = m_1 \mathbf{v}_1, \qquad T_{\rm before} = \tfrac12 m_1 v_1^2 \]

• After:

\[ \mathbf{p}_{\rm after} = m_1 \mathbf{u}_1 + m_2 \mathbf{u}_2, \qquad T_{\rm after} = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 \]

From momentum conservation:

\[ m_2 \mathbf{u}_2 = m_1 (\mathbf{v}_1 - \mathbf{u}_1) \, . \]

Squaring:

\[ m_2^2 u_2^2 = m_1^2 (v_1^2 + u_1^2) - 2 m_1^2 v_1 u_1 \cos\theta \, , \]

where \(\theta\) is the deflection angle of \(m_1\).

Thus,

\[ m_2 u_2^2 = \frac{m_1^2}{m_2}(v_1^2 + u_1^2) - 2 \frac{m_1^2}{m_2} v_1 u_1 \cos\theta \, . \]

Energy conservation:

\[ \tfrac12 m_1 v_1^2 = \tfrac12 m_1 u_1^2 + \tfrac12 m_2 u_2^2 \, . \]

Substituting gives:

\[ u_1^2 - v_1^2 + \frac{m_1}{m_2}(v_1^2 + u_1^2) - 2 \frac{m_1}{m_2} v_1 u_1 \cos\theta = 0 \, . \]

Special case: \(m_1=m_2=m\)

\[ \cos\theta = \frac{u_1}{v_1} \, . \]

Also,

\[ m\mathbf{v}_1 = m\mathbf{u}_1 + m\mathbf{u}_2 \quad \to \quad m^2 v_1^2 = m^2 u_1^2 + m^2 u_2^2 + 2m^2 \mathbf{u}_1\cdot\mathbf{u}_2 \, . \]

Using energy conservation,

\[ \mathbf{u}_1 \cdot \mathbf{u}_2 = 0 \, . \]

Thus, after the collision, the velocities of the two equal masses are perpendicular.
(Fig. \ref{fig;vngihreuwuiojesd})

Fig. 5 Left: two equal-mass snooker balls make an angle of \(90^\circ\) after collision. Right: a collision of two alpha particles make an angle of \(90^\circ\) between their outgoing velocities.

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Centre-of-mass frame

Collisions are often easiest in the centre-of-mass (COM) frame.

Here,

\[ \mathbf{p}_1 = -\mathbf{p}_2, \qquad \mathbf{p}_1' = -\mathbf{p}_2' \, . \]

Energy conservation (allowing for changing masses, as in particle physics):

\[ \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} = \frac{{p_1'}^2}{2m_1'} + \frac{{p_2'}^2}{2m_2'} + Q \, . \]

Using momentum conservation,

\[ \tfrac12 \Big(\tfrac{1}{m_1} + \tfrac{1}{m_2}\Big) p_1^2 = \tfrac12 \Big(\tfrac{1}{m_1'} + \tfrac{1}{m_2'}\Big) {p_1'}^2 + Q \, . \]

The bracketed terms are the reduced masses \(\mu, \mu'\):

\[ \frac{p_1^2}{2\mu} = \frac{{p_1'}^2}{2\mu'} + Q \, . \]

• If the collision is elastic (\(Q=0\)) and \(\mu=\mu'\), then \(p_1 = p_1'\), i.e. momentum magnitude in COM is unchanged (though direction changes).