Special relativity - mass energy and momentum I've just started studying relativity and I think I have just about gotten my head around time dilation and length contraction and I have now been able to derive the two equations 
$$t=\frac{t_0}{\sqrt {1-\frac {v^2}{ c^2}}}$$
$$l=l_0 \sqrt{1- \frac{v^2}{c^2}}$$
I've now started to look at energy and mass, which made very little sense. Would someone be able to explain the principle of energy, mass and momentum and then derive the equation
$$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}$$
Everything I've found online has gone way to fast and it makes no sense to me. I am only an A-level student so you'll have to take it slow. Any help would be great and really appreciated. 
 A: The purpose of relativistic mass is to make the Newtonian formula:
$$ \vec p = m\vec v $$
valid in special relativity, when if fact the correct formula is:
$$ \vec p = \gamma m_0 \vec v $$
This is only part of the story, since it is a 3-vector equation. The manifestly covariant equation from which it comes is:
$$ p_{\mu} = m_0 u_{\mu} $$
where the 4-momentum is:
$$ p_{\mu} =(E/c, \vec p) = ((\gamma m_0c^2)/c,\gamma m_0 \vec v) $$
and the 4-velocity is:
$$ u_{\mu} =\gamma(c, \vec v) $$
4-vectors are things that are transformed by Lorentz transformation.
One problem with relativistic mass is that it is property of the observer, not the observed. For instance, in the reference frame of a high energy cosmic ray you have a very, very high mass right now. Does that mean anything to you? No. 
A: Special relativity came with the idea that there is a maximum speed that nothing can exceed, which is $c$ the speed of light. To understand why the concept of mass has to change due to special relativity we must first study how linear momentum $\mathbf{p}$ changes when seen by a relativistic observer and how to know if in, let's say, an elastic collision it is still conserved. 
However, a more intuitive way of thinking why the mass has to have a dependence on the velocity of the moving object $\mathbf{u}$ is to go back and analyze the linear momentum considering the case where $u \rightarrow \infty$ in classical physics. We know that $p = m u$, and thus if $u \rightarrow \infty$, then $p \rightarrow \infty$ as well, independently of the mass. But when one tries to achieve such momenta in a relativistic regime, supposing the classical equation still holds $p = m u$, then it would be impossible to have $p \rightarrow \infty$, since $u \leq c $. So, one way of achieving the previous result and reproduce the classical version of infinite momentum in special relativity, we make $m = m(u)$, with $m(c) \rightarrow \infty$.
Another way of seeing why $m$ must be a function of the velocity $u$ is to analyze elastic collisions. To do so, first consider the following thought 2D experiment. Imagine you have two balls of equal mass $m$ (suppose it's not a function of $u$) colliding seen by two reference frames $S$ and $S'$. In $S$, one ball has a initial velocity $\mathbf{u}_{A_i} = -u_{Ay}\hat{y} +u_{Ax}\hat{x}$ (let's call this ball A), whereas the other ball has velocity $\mathbf{u_{B_i}} = u\hat{y}$ (and call this ball B). Let $S'$ be the reference frame where ball A has velocity $u_{A_i}' =-u\hat{y}$, while B has $u_{B_i}' = u_{Bx}'\hat{x} +u_{By}'\hat{y}$. Let $S'$ be moving in the $\hat{x}$ direction with speed $V$ with respect to $S$ and let us suppose the collision is elastic and there is no loss or dissipation of either momentum and energy. Then:
$$u_{Ax_i} = \frac{dx_A}{dt} = V,$$
$$u_{Ay_i} = \frac{dy_A}{dt} = \frac{-u}{\gamma\left(1-\frac{u_{Ax_i}' V}{c^2} \right)}=-\frac{u}{\gamma},$$
where $\gamma \equiv (1-V^2/c^2)^{-1/2}$. I've used the Lorentz transformations to conclude the above equations. It is important to emphasize that these are the speeds before the collision. After the collision, we must have $u_{By_f} = - u \hat{y}$, $u_{Ax_f} = V$ and $u_{Ay_f} = u/\gamma$, by symmetry. Now let's take a look in the conservation of momentum in the $x$ and $y$ directions:
In $x$:
$$p_{Ax_i} +p_{Bx_i} = mV = p_{Ax_f} + p_{Bx_f},$$
which proves momentum in the $x$ direction is conserved.
In $y$:
$$p_{Ay_i}+p_{By_i}=-m\frac{u}{\gamma} + mu,$$
while after the collision
$$p_{Ay_f}+p_{By_f} = m \frac{u}{\gamma} - m u.$$
This gives us:
$$2mu = \frac{2mu}{\gamma},$$
which would only be valid if $\gamma = 1$. Since we want to reproduce the momentum conservation law in the theory of special relativity, we should then allow $m$ to depend on $u$. Suppose now that $m_A(u_A)=\gamma(u_A) m_0$ and $m_B(u) = \gamma(u) m_0$ where $\gamma(v)$ is a function that correlates $m_0$ (later called rest mass) to the mass measurement of a moving object with speed $v$. So, by conservation of momentum, we now get
$$\gamma(u) = \frac{\gamma(u_A)}{\gamma} \Rightarrow m_B = \frac{m_A}{\gamma}.$$
This shows how to relate the mass measurement for the ball A with the mass measurement for ball B. We conventionally define the function $\gamma(v)$ as being the $\gamma$ factor so that the relativistic mass of a moving object with speed $u$ and velocity $\mathbf{u}$ becomes
$$m(u) = \gamma(u) m_0 = \frac{m_0}{\sqrt{1-u^2/c^2}}, $$
and we can easily check this definition would satisfy $m(c) \rightarrow \infty$, and the momentum conservation law as well.
