# Mechanics: Chapter 16 – Collisions

If there is no external force acting on a system, the total momentum of a system is always conserved. In addition, if a collision is elastic, the total kinetic energy before a collision is equal to the total kinetic energy after the collision i.e kinetic energy is also conserved. In other words, in an elastic collision between two particles of mass $m_{1}$ and $m_{2}$,

• momentum is conserved: $m_{1}v_{1} + m_{2}v_{2} = m_{1}v_{1}^{'} + m_{2}v_{2}^{'}$
• kinetic energy is conserved: $\frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2} = \frac{1}{2}m_{1}v_{1}^{'2} + \frac{1}{2}m_{2}v_{2}^{'2}$

If the particle with mass $m_{2}$ is stationary, i.e $v_{2} = 0$, we can derive the following two expressions,

• $v_{1}^{'} = \frac{m_{1} - m_{2}}{m_{1} + m_{2}}v_{1}$
• $v_{2}^{'} = \frac{2m_{1}}{m_{1} + m_{2}}v_{1}$

If $m_{2} >> m_{1}$, we can neglect $m_{1}$ in our expressions and we get

• $v_{1}^{'} = - v_{1}$
• $v_{2}^{'} = 0$

Which makes sense. For example, imagine a tennis ball colliding with a wall.

If $m_{1} = m_{2}$, we get

• $v_{1}^{'} = 0$
• $v_{2}^{'} = v_{1}$

which also makes sense. Imagine two snooker balls colliding.

If a collision is inelastic its kinetic energy is not conserved (but the momentum is still conserved). However, if a collision is completely inelastic i.e the colliding particles get stuck together after the collision, we can still derive an easy expression to figure out their combined velocity simply from conservation of momentum:

• $v_{1}^{'} = v_{2}^{'} = v^{'} = \frac{m_{1}v_{1} + m_{2}v_{2}}{m_{1} + m_{2}}$

# Mechanics: Chapter 15 – Power

Power is the rate of work done.

• $P = \frac{W}{\Delta t} = \frac{dW}{dt}$

Since a infinitesimal amount of work done $dW = F \cdot ds$,

• $P = \frac{dW}{dt} = \frac{F \cdot ds}{dt} = F \cdot \frac{ds}{dt} = \overrightarrow{F} \cdot \overrightarrow{v}$

Hence, power is also the dot product of the Force and the velocity of the object. In other words,

• $P = Fvcos(\theta)$

Similarly, because of the fundamental theorem of calculus,

• $W = \int_{t_{2}}^{t_{1}} P dt$

# Mechanics: Chapter 14 – Potential Energy Diagrams and Equilibrium

Since the Work done,

• $W = \int F \cdot dr$

And, for a conservative force,

• $W = - \Delta U$

One can deduce that, for a conservative force,

• $\Delta U = - \int F \cdot dr$

In other words, the change in the potential energy of an object under the influence of a conservative force as it moves from position $r_{1}$ to $r_{2}$ is

• $\Delta U = - \int_{r_{1}}^{r_{2}} F \cdot dr = \int_{r_{2}}^{r_{1}} F \cdot dr$

Similarly, because of the Fundamental Theorem of Calculus, we can deduce that,

• $F = -\frac{dU}{dr}$

potential energy diagram is a diagram with the potential energy on one axis and the position on the other. The local maxima on a potential energy diagram are said to be points of unstable equilibrium, while the local minima are points of stable equilibrium.

Here is a graph of what is called the Lennard-Jones potential for the interaction of a pair of atoms:

Now the local minimum on this graph (which also happens to be the absolute minimum) is the point where the system reaches stable equilibrium. If given an equation for the potential energy $U$, we can use elementary calculus to find the minimum point which would be the point at which stable equilibrium is reached.

In the above potential energy diagram there are two points of unstable equilibrium (two local maxima) and three points of stable equilibrium (three local minima).

# Mechanics: Chapter 13 – Conservation of Mechanical Energy

When a force is conservative, the work done by the force on an object equals the change in the potential energy of the object. If the work done is positive, there is a decrease in the potential energy of the object. Similarly, if the work done is negative, there is an increase in the potential energy of the object. Hence,

• $W = - \Delta U$

Energy is always conserved. Essentially the sum of the kinetic energy $K$ and the potential energy $U$ of a system is constant.

• $K + U = c$ where $c$ is a constant

This also implies that if no external energy is added or removed from the system, the initial energy $E_{i}$ of a system is equal to the final energy $E_{f}$ of a system, and hence

• $K_{i} + U_{i} = K_{f} + U_{f}$

This is easy to prove in the case of a conservative force like gravity. Let’s say an object is dropped from rest in free fall, now the work done, $W$ by gravity on the object is, as I’ve already established,

• $W = - \Delta U$

and since the force of gravity accelerates the object from rest to a speed $v$, the work done by the force of gravity on the object increases the Kinetic energy of the object. In other words,

• $W = \Delta K$

Now

• $\Delta K = - \Delta U \Rightarrow \Delta K + \Delta U = 0$

If the initial Kinetic energy and potential energy of the system were $K_{i}$ and $U_{i}$ and

• $K_{i} + U_{i} = c$

Then the final kinetic and potential energies, $K_{f}$ and $U_{f}$ of the system can be written as the sum of the initial energies and the change in energies. Hence,

• $K_{f} + U_{f} = (K_{i} + \Delta K) + (U_{i} + \Delta U) = (K_{i} + U_{i}) + (\Delta K + \Delta U)$

We know that $\Delta K + \Delta U = 0$,  hence

• $K_{i} + U_{i} = K_{f} + U_{f}$

Quod erat demonstrandum.

# Mechanics: Chapter 12 – Conservative Forces

A force is conservative if its curl is a zero vector. In other words, a force $F$ is conservative if

• $\bigtriangledown \times \overrightarrow{F} = \overrightarrow{0}$

Two common conservative forces are the gravitational force and the spring force. In vector form, the gravitational force is

• $\overrightarrow{F} = \frac{GMm}{|r|^2}|\hat{r}|$
• $\Rightarrow \overrightarrow{F} = \frac{GMm}{x^2 + y^2 + z^2} \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2+y^2+z^2}}$
• $\Rightarrow \overrightarrow{F} = \frac{GMm}{x^2 + y^2 + z^2} \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2+y^2+z^2}}$
• $\Rightarrow \overrightarrow{F} = (GMm)\frac{x\hat{i} + y\hat{j} + z\hat{k}}{(x^2+y^2+z^2)^{\frac{3}{2}}}$

Taking the curl of this, one gets

• $\bigtriangledown \times \overrightarrow{F} = (GMm)\frac{2yz\hat{i} -2yz\hat{i}}{(x^2+y^2+z^2)^{\frac{5}{2}}} + (GMm)\frac{2xz\hat{j} -2xz\hat{j}}{(x^2+y^2+z^2)^{\frac{5}{2}}} + (GMm)\frac{2xy\hat{k} -2xy\hat{k}}{(x^2+y^2+z^2)^{\frac{5}{2}}} = \overrightarrow{0}$

Similarly, in vector form the spring force can be written as

• $\overrightarrow{F} = -k \overrightarrow{s} = -k (x\hat{i} + y\hat{j} + z\hat{k})$
• $\bigtriangledown \times \overrightarrow{F} = -k (0\hat{i} + 0\hat{j} + 0\hat{k}) = \overrightarrow{0}$

Another (equivalent) way of thinking about a conservative force is that if a force does zero net work when a particle does a round trip (i.e its final position is the same as its initial position). Forces like friction are not conservative the net work done on a round trio by friction is not zero (and hence its curl is not a zero vector. These conditions are equivalent).

# Mechanics: Chapter 11 – Relative Velocity and Acceleration

If two reference frames A and B are different from each other only because they are moving at different constant velocities relative to each other, we can convert the displacements, velocities and accelerations with respect to A to the displacements, velocities and accelerations with respect to B using Galilean Transformation.

If at time $t = 0$, the origins of both A and B coincide, and B is moving at a constant velocity $v_{0}$ with respect to A, then the displacement with respect to B, $r_{B}$ can be calculated from the displacement with respect to A, $r_{A}$ at time $t$ using:

• $r_{B} = r_{A} - v_{0}t$

Differentiating this with respect to time, we get,

• $v_{B} = v_{A} - v_{0}$

and taking the second derivative, we get

• $a_{B} = a_{A}$

Interestingly, acceleration remains the same in reference frames that can be transformed using Galilean transformation.

# Mechanics: Chapter 10 – Work Done by Forces

There are some easy derivations for the work done by common forces. The work done by a constant gravitational force is pretty simple. The force exerted by a constant gravitational force is just $mg$ and the displacement in the direction of the force is just the change in height $\Delta h$, hence

• Work done by constant gravitational force: $W = mg \Delta h$

The force exerted by a spring is given by Hooke’s Law, $F = -kx$, now this force varies with extension $x$ (which is essentially the displacement). Hence, we can’t simply take the product, we have to take the integral of $F$ with respect to $dx$, i.e

• Work done by spring: $W = \int F \cdot dx = - \int kx \cdot dx = -\frac{1}{2}kx^2$

The work done to accelerate an object from rest to a velocity $v$ is called the Kinetic Energy of the object. Now the force needed to accelerate the object is, by Newton’s second law:

• $F = ma$
• $F = m \frac{dv}{dt}$

Since the work done by $F$ is

• $W = \int F \cdot ds$
• $\Rightarrow W = \int m \frac{dv}{dt} \cdot ds$
• $\Rightarrow W = \int m \frac{ds}{dt} \cdot dv$
• $\Rightarrow W = \int mv \cdot dv$
• $\Rightarrow W = \frac{1}{2}mv^2$

Hence,

• Kinetic Energy of an object: $E_{K} = \frac{1}{2}mv^2$

# Mechanics: Chapter 9 – Introduction to Work

Work is the dot product of force and displacement. Force and displacement are vectors. Since the dot product returns a scalar, Work is a scalar quantity.

• $W = F \cdot s = Fscos(\theta)$

Where $W$ is the Work done, $F$ is the Force, $s$ is the displacement and $\theta$ is the angle between the direction of the force and the direction of the displacement. You can also calculate the dot product by expressing the the Force and displacement in terms of perpendicular unit vectors $\hat{i} \hat{j} \hat{k}$ etc. In which case, the dot project is just

• $(A_{x}\hat{i} + A_{y}\hat{j}) \cdot (B_{x}\hat{i} + B_{y}\hat{j}) = A_{x}B_{x} + A_{y}B_{y}$

When the force varies with displacement s, the work done is the integral of Force with respect to ds.

• $W = \int F \cdot ds$

# Mechanics: Chapter 8 – Free Body Diagrams

Drawing a free body diagram is an efficient way to understand and evaluate the forces acting on an object. The simplest way to model an object in a free-body diagram is to consider it to be a particle. This is a highly simplified model and is useful only when the rotational motion of the object is negligible. A particle (or point particle) is an ideal object that has mass but no size. In a free-body diagram of a point particle, essentially all one has to do is sketch the particle and arrows showing the directions of different forces acting on the particle. Here are some free body diagram examples I made,

# Mechanics: Chapter 7 – Hooke’s Law and Centripetal Force

The magnitude and direction of the force of a spring that restores its shape can be predicted by what is called Hooke’s Law.

• $F = -kx$

The direction of the force is always towards the relaxed position of the spring i.e against the motion of spring (hence the minus sign). $k$ is the spring constant. $x$ is the extension of the spring.

The centripetal force on an object in circular motion is always towards the center of the circle and can be calculated by

• $F_{c} = ma_{c}$

Where $a_{c}$ is the centripetal acceleration which I discussed in Chapter 4.

Note: I am not sure whether the force due to air resistance should be included at this point. I might append it to this chapter later.