Energy
The energy of a body is defined as its capacity or ability to do work. It cannot be created or destroyed, only transformed from one form to another. Like work, energy is a scalar quantity having magnitude only and no direction. The dimensions of energy are the same as the dimensions of work, i.e. \(\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\). It is measured in the same unit as work, i.e. joule in SI and erg in CGS system.
Energy is broadly classified into two main types:
Potential Energy:
Potential energy is stored energy that an object has because of its position, condition, or arrangement, which has the capacity to do work when released. This stored energy is typically classified by the force field acting on the object.
Examples of Potential Energy
Kinetic Energy:
Kinetic energy is the energy an object possesses due to its motion. Any moving object has kinetic energy.
Examples of Kinetic Energy
Energy can exist in various forms such as mechanical energy (potential energy and kinetic energy), sound energy, heat energy, light energy, etc.
Kinetic Energy
The energy possessed by a body by virtue of its motion is called kinetic energy. Kinetic energy of a body can be calculated by the amount of work done in stopping the moving body or from the amount of work done in giving it same velocity from state of rest.
If an object of mass \(m\) has velocity \({v}\), then its kinetic energy is given by
\(
\mathrm{KE}=\frac{1}{2} m v^2
\)
The kinetic energy of a system having \(n\) particles is equal to the sum of the kinetic energies of all its constituent particles.
i.e.
\(
\begin{gathered}
K=\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2+\frac{1}{2} m_3 v_3^2+\ldots \ldots+\frac{1}{2} m_n v_n^2 \\
K=\sum_{i=1}^n \frac{1}{2} m_i v_i^2
\end{gathered}
\)
Regarding the kinetic energy, the following two points are important to note
Example 1: In a ballistics demonstration a police officer fires a bullet of mass \(50.0 \mathrm{~g}\) with speed \(200 \mathrm{~m} \mathrm{~s}^{-1}\) on soft plywood of thickness \(2.00 \mathrm{~cm}\). The bullet emerges with only \(10 \%\) of its initial kinetic energy. What is the emergent speed of the bullet?
Solution: The initial kinetic energy of the bullet is \(m v^{2} / 2=1000 \mathrm{~J}\). It has a final kinetic energy of \(0.1 \times 1000=100 \mathrm{~J}\). If \(v_{f}\) is the emergent speed of the bullet,
\(
\begin{array}{l}
\frac{1}{2} m v_{f}^{2}=100 \mathrm{~J} \\
v_{f}=\sqrt{\frac{2 \times 100 \mathrm{~J}}{0.05 \mathrm{~kg}}} \\
=63.2 \mathrm{~m} \mathrm{~s}^{-1}
\end{array}
\)
Example 2: When a man increases his speed by \(2 \mathrm{~ms}^{-1}\), he finds that his kinetic energy is doubled. Find the original speed of the man.
Solution: Man possesses kinetic energy because of his velocity (\(v\)). If \(m\) is mass of man, then \(K=\frac{1}{2} m v^2\)
Given,
\(
v_1=v, m_1=m_2=m
\)
When
\(
v_2=(v+2) \mathrm{ms}^{-1} \text {, then }
\)
\(
K_2=2 K_1
\)
\(
\begin{aligned}
& \because \quad \frac{K_1}{K_2}=\frac{v_1^2}{v_2^2} \Rightarrow \frac{K_1}{2 K_1}=\frac{v^2}{(v+2)^2} \\
& \Rightarrow \quad v^2-4 v-4=0 \\
& \text { This gives, } \quad v_1=\frac{4+\sqrt{16+16}}{2}=\frac{4+\sqrt{32}}{2} \\
& \Rightarrow \quad v_1=2(\sqrt{2}+1) \mathrm{ms}^{-1}
\end{aligned}
\)
Example 3: A body of mass 0.8 kg has initial velocity \((3 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}) \mathrm{ms}^{-1}\) and final velocity \((-6 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathrm{ms}^{-1}\). Find change in kinetic energy of the body.
Solution: Change in kinetic energy,
where,
\(
\Delta \mathrm{KE}=\frac{1}{2} m v_f^2-\frac{1}{2} m v_i^2
\)
and
\(
v_f=\sqrt{6^2+2^2}=\sqrt{40} \mathrm{~ms}^{-1}
\)
and \(\quad v_i=\sqrt{3^2+4^2}=\sqrt{25} \mathrm{~ms}^{-1}\)
\(
\begin{aligned}
\therefore \quad \Delta \mathrm{KE} & =\frac{1}{2} \times 0.8[\sqrt{40})^2-(\sqrt{25})^2] \\
& =0.4[40-25]=0.4(15)=6 \mathrm{~J}
\end{aligned}
\)
Relation between kinetic energy and linear momentum
The linear momentum of a body is given by \(p=m v\), where \(m\) is the mass and \(v\) is the velocity of a body. Then, kinetic energy of the body,
\(
\begin{array}{r}
\mathrm{KE}=\frac{1}{2} m v^2=\frac{1}{2 m}\left(m^2 v^2\right) \\
\mathrm{KE}=\frac{p^2}{2 m} \text { or } p^2=2 m \mathrm{KE} \\
\Rightarrow \quad \text { Linear momentum, } p=\sqrt{2 m \mathrm{KE}}
\end{array}
\)
Example 4: Two bodies \(A\) and \(B\) having masses in the ratio of \(3: 1\) possess the same kinetic energy. Obtain the ratio of linear momentum of \(B\) to that of \(A\).
Solution: Kinetic energy of the body is given by
\(
E_K=\frac{1}{2} m v^2 \dots(i)
\)
and linear momentum, \(p=m v \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
E_K=\frac{m^2 v^2}{2 m}=\frac{p^2}{2 m}
\)
Now,
\(
E_{K_1}=E_{K_2}
\)
\(\Rightarrow \quad \frac{p_1^2}{2 m_1}=\frac{p_2^2}{2 m_2}\) or \(\frac{p_1}{p_2}=\sqrt{\frac{m_1}{m_2}}\)
\(
\frac{p_1}{p_2}=\sqrt{\frac{3}{1}} \text { or } \frac{p_2}{p_1}=\frac{1}{\sqrt{3}}
\)
Example 5: Kinetic energy of a particle is increased by \(300 \%\). Find the percentage increase in momentum.
Solution: Kinetic energy, \(E=\frac{1}{2} m v^2\)
and momentum, \(p=m v\)
When \(E\) is increased by \(300 \%\),
\(
\begin{aligned}
E^{\prime} & =E+3 E=4 E \\
& =4\left(\frac{1}{2} m v^2\right)=2 m v^2
\end{aligned}
\)
If \(v^{\prime}\) is velocity of body, then
\(
\begin{aligned}
& & \frac{1}{2} m\left(v^{\prime}\right)^2 & =2 m v^2 \\
\Rightarrow & & v^{\prime} & =2 v \\
& \text { So, } & p^{\prime} & =m v^{\prime}=2 m v
\end{aligned}
\)
Hence, percentage change in momentum
\(
=\frac{2 m v-m v}{m v} \times 100=100 \%
\)
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