Year 1 Sem 1/LC Classical Mechanics and Relativity 1/Lecture 16

Lecture 16: Mechanics XII: End of CoLM
Thu 20 Nov 2025 15:00

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In this lecture:

Examples of problems using the conservation of linear momentum.
The start of gravitation.

Conservation of Linear Momentum

As a recap, if we have a single particle:

$$\pmb{p} = m \pmb{v}$$

And for $N$ separate particles in a system:

$$\pmb{p} = \sum_{k=1}^{N} \pmb{p}_k = \sum_{k=1}^{N} m_k \pmb{v}_k$$

And the force given on a particle, or collection thereof, is given by:

$$\pmb{F} = \frac{d}{dt} \pmb{p}$$

Example

Consider two particles. $m_1$ initially travels along the positive x-axis with velocity $u$. It elastically collides with a stationary particle $m_2$.

Being totally elastic, both total kinetic energy and total momentum are conserved. $m_1$ and $m_2$ scatter off in two different directions, each making angle $\alpha$ and $\beta$ respectively from the x-axis in a top down view (i.e. snooker).

Say we know $m_1, m_2, u, |v_1|$. What are unknowns $\alpha, \beta$ and $|v_2|$?

In order to find the three unknowns, we need to have three constraint equations.

Conservation of Energy

Before the collision, the total kinetic energy is given by:

$$\sum E_k = \frac{1}{2}m_1 u^2$$

And after the collision:

$$\sum E_k = \frac{1}{2} \left(m_1 v_1^2 + m_2 v_2^2\right)$$

As we are conserving energy, we can equate these:

$$\frac{1}{2}m_1 u^2 = \frac{1}{2} \left(m_1 v_1^2 + m_2 v_2^2\right)$$

$$m_1 u^2 = m_1 v_1^2 + m_2 v_2^2 \tag{1}$$

Conservation of Linear Momentum

In our 2D top-down plane, we can separately conserve left/right $x$ momentum and upwards/downwards $y$ momentum.

$$\text{x: } \quad m_1 u = m_1 v_1 \cos \alpha + m_2 v_2 \cos \beta \tag{2}$$

$$\text{y: } \quad 0 = m_1 v_1 \sin \alpha - m_2 v_2 \sin \alpha \tag{3}$$

This gives us three equations and three unknowns which are fairly simple to solve for.

Impulse and Work

Again, we have:

$$\pmb{F} = \frac{d \pmb{p}}{dt} \implies d \pmb{p} = \pmb{F} dt = dJ$$

And that the work done by a force is:

$$dW = \pmb{F} \cdot d \pmb{s} = dE_k$$

The first equation gives the infinitesimal change in momentum, the “impulse” while the second gives an infinitesimal change in kinetic energy. It’s important not to confuse them because they seem similar.

Example

Consider crashing a car into a wall. The car of mass $m$ starts a velocity $v$ and, after crashing, ends up as a wreck of mass $m$ and no velocity.

What is the work done by the force stopping the car and the impulse exerted by the wall?

The work done is simply $W = |\Delta E_k| = E_\text{final} - E_\text{initial} = \frac{1}{2}mv^2$.

The impulse exerted is $|\Delta p| = p_\text{final} - p_\text{initial} = mv$

Which is simple! What if we crash a car of half the mass, travelling at twice the speed. How do the impulse and work required to stop the car change?

$$|\Delta p| = \frac{m}{2} \times 2v = mv$$

So the impulse remains the same! The change in kinetic energy is:

$$|\Delta E_k| = \frac{1}{2} \frac{1}{2} \left(\frac{m}{2}\right) \times \left(2v\right)^2 = mv^2$$

So, although the impulse is the same, the required energy is twice as much.

Manoeuvring Spaceships

Consider a rocket of mass $m$ travelling at $v$. In order to manoeuvre the rocket, we need to use a thruster to eject material. Everything is taking place linearly along the $x$-axis.

The effect of expelling material (for example from the back, in the negative x-axis, for a rocket travelling in the +x) is an increase in the rocket speed, and decrease in rocket mass.

After an infinitesimal expulsion of material backwards, the new forwards velocity is $v + dv$ and new mass $m + dm$ (where $dm$ is some negative quantity representing the amount of fuel expelled.)

The expelled material (mass of $-dm$) is expelled with velocity $u$. As total momentum is conserved:

$$p_\text{before} = mv$$

$$p_\text{after} = (v+dv)(m+dm) + (-dm)(v-u)$$

$$\implies mv = (v+dv)(m+dm) + (-dm)(v-u)$$

$$mv = mv + vdm + mdv + dvdm -vdm + udm$$

The $dv dm$ term is an infinitesimal change squared, so can be neglected.

$$\implies 0 = m dv + u dm$$

$$mdv = - udm$$

$$dv = -u \frac{dm}{m}$$

$$\int_{v_0}^{v_1} dv = -u \int_{m_0}^{m_1} \frac{dm}{m}$$

$$v_1 - v_0 = -u \left[\ln m\right]^{m_1}_{m_0}$$

$$v_1 = v_0 - u \ln \left(\frac{m_1}{m_0}\right)$$

Since $m_1 < m_0$, the log term is negative, so cancelling out the existing negative and $v_1 > v_0$. The change in velocity is proportional to the ejection speed, hence why thruster efficiency is very important.

Say you’ve been able to eject $99\%$ of the rocket mass, so $m_1 = 10^{-12}m_0$:

$$v_1 = v_0 - u \ln(10^{-2})$$

$$v_1 = v_0 + 2u(\ln 10) \approx v_0 + 2u(2.3)$$

So space manoeuvres are very costly for mass, and a large drop in mass represents a small increase in velocity. This is unlike ejection velocity, which has a much larger (linear) effect.

Centres of Mass

Consider a collection of particles scattered in an $(x, y, z)$ plane. Each particle has a mass $m_k$ and position $\pmb{r}_k$.

The centre of mass of this system (positionally denoted $\pmb{R}$) is defined as:

$$0 = \sum_k m_k \left(\pmb{r}_k - \pmb{R}\right)$$

$$\sum_k \left(m_k \pmb{r}_k - m_k \pmb{R}\right)$$

$$= \sum_k m_k \pmb{r}_k - \sum_k m_k \pmb{R}$$

And letting total system mass being $M$:

$$= \sum_k m_k \pmb{r}_k - M \pmb{R}$$

$$\implies \pmb{R} = \sum_k \left(\frac{m_k \pmb{r}_k}{M}\right)$$

If one mass is much larger than any other mass, we can approximate $M$ as being equal to this mass, and hence this object is the COM.

Total System Momentum

We can also consider the total momentum of this system:

$$\frac{d}{dt} \left[= \sum_k m_k \pmb{r}_k - M \pmb{R}\right] = 0$$

$$\sum_k m_k \pmb{v}_k - M \frac{d \pmb{R}}{dt} = 0$$

$$M \pmb{V} = \sum_k m \pmb{v}_k$$

Where $\pmb{V}$ is the velocity of the COM and the second term is the total system momentum.

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