Mechanical energy of a system is the sum of its kinetic and potential energies. In the absence of non-conservative forces, the total mechanical energy of a system remains constant. The conservation of mechanical energy is a useful tool for solving problems.

**Problem (JEE Mains 2006):**
The potential energy of a 1 kg particle free to move along the $x$-axis is given by
\begin{align}
U(x)=\left(\frac{x^4}{4}-\frac{x^2}{2}\right)\mathrm{J}.\nonumber
\end{align}
The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is_______?

**Solution:**
The variation of potential energy $U(x)={x^4}/{4}-{x^2}/{2}$ with $x$ is shown in the figure. It has a local maxima at $x=0$ and two global minima at $x=-1$ and $x=1$. You can get these points by solving $\mathrm{d}U/\mathrm{d}x=0$. The minimum value of potential energy is,
\begin{align}
U_\text{min} & =U(x\!=\!-1) \\
&=U(x\!=\!1) \\
&=-1/4\;\mathrm{J}.\nonumber
\end{align}

The total energy of the particle is $E=2$ J. The kinetic energy, $K=E-U$, is maximum when the potential energy is minimum. The maximum kinetic energy of the particle is, \begin{align} K_\text{max} & =E-U_\text{min} \\ &=2-(-1/4)=9/4\;\mathrm{J}.\nonumber \end{align} The maximum velocity of the particle is, \begin{align} v_\text{max} & =\sqrt{2K_\text{max}/m} \\ &=\sqrt{2(9/4)/(1)} \\ &=3/\sqrt{2}\;\mathrm{J}.\nonumber \end{align}

This problem has beautiful hidden concepts. Let us explore some of them. Look at the figure. What is the particle's kinetic energy at $x=-2$ and $x=2$? It is zero. The particle comes to a momentary halt at these points and returns. It is traveling in a closed path between $x=-2$ and $x=2$. The particle is prohibited from entering the region $x<-2$ or $x>2$. The kinetic energy in these regions is negative, which is meaningless.

Let a particle of total energy $E=-1/2$ J is released near $x=1$. The particle will travel in a closed path between $x_\text{min}$ and $x_\text{max}$. Can you find the values of $x_\text{min}$ and $x_\text{max}$?

**Problem (IIT JEE 1997):**
An isolated particle of mass $m$ is moving in horizontal plane ($x\text{-}y$), along the $x$-axis, at a certain height above the ground. It suddenly explodes into two fragment of masses $m/4$ and $3m/4$. An instant later, the smaller fragment is at $y={15}\;\mathrm{cm}$. The larger fragment at this instant is at,

- $y={-5}\;\mathrm{cm}$
- $y={20}\;\mathrm{cm}$
- $y={5}\;\mathrm{cm}$
- $y={-20}\;\mathrm{cm}$

**Solution:**
There is no external force on the system in the $x\text{-}y$ plane. Hence, linear momentum of the system is conserved along $x$ and $y$ directions, separately. Initially, the particle was moving along $x$ axis with a speed $v_x$. Hence, coordinates of its centre of mass varies with time $t$ as,
\begin{align}
x_{c}=v_x t, \qquad y_{c}=0.\nonumber
\end{align}
By conservation of linear momentum, $y_{c}$ remains constant (here zero) after the collision i.e.,
\begin{align}
y_{c}=\frac{({m}/{4}) 15+({3m}/{4})y}{{m}/{4}+{3m}/{4}}=0.
\end{align}
Solve to get $y={-5}\;\mathrm{cm}$.