A helpful yet elementary answer may do the trick, If you are familiar with the Euler-Lagrange equation then it will be straight forward and you can skip ahead a little. If not then you have to accept that there is an equation in physics that generalises classical mechanics called the Euler-Lagrange equation. For a particle moving in one dimension under a conservative force it is written,
\begin{equation}
\frac{d}{dt}\bigg(\frac{\partial T}{\partial \dot x}\bigg)+\frac{\partial V}{\partial x}=0
\end{equation}
Where $T$ is the kinetic energy of the system and $V$ is the potential energy, $x$ is the particles position and $\dot x=\frac {d}{dt}x$ is the velocity of the particle. We define the momentum of the particle to be,
\begin{equation}
p:=\frac{\partial T}{\partial \dot x}
\end{equation}
And you will note that we can now write the Euler-Lagrange equation as,
\begin{equation}
\frac{d}{dt}(p)+\frac{\partial V}{\partial x}=0
\end{equation}
This is Newton's second law of motion. The momentum is changed by the action of a force on the particle, if there are no forces then the time derivative of the momentum is zero. If the time derivative is zero then the momentum does NOT change as time evolves and will have the same value at the end of the experiment as it did at the beginning.
In this way the Euler-Lagrange equation has given us a conservation law for $p$ only when $\partial V/\partial x=0$. The invariance of the potential with respect to $x$ leads to a conservation law.
In general we do not write the Euler-Lagrange equation for a one dimensional particle. The general form is written,
\begin{equation}
\frac{d}{dt}\bigg(\frac{\partial L}{\partial \dot x}\bigg)-\frac{\partial L}{\partial x}=0
\end{equation}
Where $L(x,\dot x)=T(\dot x)-V(x)$ is the Lagrangian of the system. Check that this will give the above stated equation. In general if the Lagrangian for a particular system is not a function of $x$ then you can clearly see that,
\begin{equation}
\frac{\partial L}{\partial \dot x}=constant
\end{equation}
Since the time derivative vanishes. When the Lagrangian is not a function of $x$ we say that the Lagrangian has a symmetry. When the Lagrangian has a symmetry, there is a conservation law.
