Question

Two particles of equal mass $m$ have respective initial velocities $\overrightarrow{u_1}=u\hat{i}$ and $\overrightarrow{u_2}=\frac{u}{2}\hat{i}+\frac{u}{2}\hat{j}$. They collide completely inelastically. Find the loss in kinetic energy.

• A. $\frac{3m{u}^2}{4}$
• B. $\frac{\sqrt{2}m{u}^2}{\sqrt{3}}$
• C. $\frac{m{u}^2}{3}$
• D. $\frac{m{u}^2}{8}$

Answer 1

The correct option is D

$\frac{m{u}^2}{8}$

Step 1: Given

The mass of the two particles is the same and equal to $m$.

Initial velocities are $\overrightarrow{u_1}=u\hat{i}$ and $\overrightarrow{u_2}=\frac{u}{2}\hat{i}+\frac{u}{2}\hat{j}$

The collision is completely inelastic after the collision, the balls stick together

Let $\overrightarrow{u_3}=v_1\hat{i}+v_2\hat{j}$ be the final velocity

Step 2: Formulas used

From inelastic collision formula,
$m_1{v}_1+m_2{v}_2=\left(m_1+m_2\right)v$ [Since the collision is completely inelastic]

where, $m_1$ and $m_2$ are the masses of the two objects, and $v_1$ and $v_2$ are their respective velocities and $v$ is the final velocity

For a vector, $\overrightarrow{a}=x\hat{i}+y\hat{j}+z\hat{k}$, the magnitude is given as,
$\left|\overrightarrow{a}\right|=\sqrt{x^2+y^2+z^2}$

Step 3: Determine the expression for final vector

From the equation of inelastic colision,
$\begin{array}{cc}& m\overrightarrow{u_1}+m\overrightarrow{u_2}=\left(m+m\right)\overrightarrow{u_3}\\ \Rightarrow & m\left(\overrightarrow{u_1}+\overrightarrow{u_2}\right)=2m\overrightarrow{u_3}\\ \Rightarrow & \overrightarrow{u_3}=\frac{\overrightarrow{u_1}+\overrightarrow{u_2}}{2}\\ \Rightarrow & v_1\hat{i}+v_2\hat{j}=\frac{u\hat{i}+{\displaystyle \frac{u}{2}}\hat{i}+{\displaystyle \frac{u}{2}}\hat{j}}{2}\end{array}$

Comparing the magnitudes in each direction,
$$\begin{gathered} \begin{array}{ccc}\Rightarrow & v_1=\frac{u+{\displaystyle \frac{u}{2}}}{2}=\frac{3u}{4}& v_2=\frac{u}{4}\end{array} \\ \Rightarrow u_3=\frac{3u}{4}\hat{i}+\frac{u}{4}\hat{j} \end{gathered}$$

Step 4: Calculate magnitudes of the vectors

We have,
$\begin{array}{rcl}\left|\overrightarrow{u_1}\right|& =& \sqrt{u^2}=u\\ \left|\overrightarrow{u_2}\right|& =& \sqrt{{\left(\frac{u}{2}\right)}^2+{\left(\frac{u}{2}\right)}^2}\\ & =& \frac{u}{\sqrt{2}}\\ \left|\overrightarrow{u_3}\right|& =& \sqrt{{\left(\frac{3u}{4}\right)}^2+{\left(\frac{u}{4}\right)}^2}\\ & =& \sqrt{\frac{10u^2}{16}}\\ & =& \frac{\sqrt{10}u}{4}\end{array}$

Step 4: Calculate loss in kinetic energy

The loss in kinetic energy is the difference between initial and final kinetic energies.
If $∆E$ is the loss in kinetic energy are $K{E}_i$ and $K{E}_f$ are initial and final kinetic energies, then
$\begin{array}{rcl}∆E& =& K{E}_i-K{E}_f\\ & =& \left(\frac{1}{2}m{u_1}^2+\frac{1}{2}m{u_2}^2\right)-\frac{1}{2}2m{u_3}^2\\ & =& \frac{m}{2}\left(u^2+\frac{u^2}{2}-\frac{20u^2}{16}\right)\\ & =& \frac{m}{2}\left(\frac{16u^2+8u^2-20u^2}{16}\right)\\ & =& \frac{m}{2}\left(\frac{u^2}{4}\right)\\ \therefore ∆E& =& \frac{m{u}^2}{8}\end{array}$

Therefore, the loss in in kinetic energy is $\frac{m{u}^2}{8}$.

Hence, option D is correct.