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Step-by-Step Solution
Step 1: Understand the Problem
We have two particles, each of mass $m$, moving initially with velocities:
Particle 1: $\vec{u}_1 = u\,\hat{i}$
Particle 2: $\vec{u}_2 = u\,\dfrac{\hat{i} + \hat{j}}{2}$
They collide completely inelastically (meaning they stick together), and we need to find the loss in kinetic energy during this collision.
Step 2: Apply Conservation of Momentum
In a completely inelastic collision, total momentum before collision equals total momentum after collision. Let $\vec{v}$ be the common velocity after collision. Then,
$\vec{P}_\text{i} = \vec{P}_\text{f}$
So,
$m\,\bigl(u\,\hat{i}\bigr) + m\Bigl(u\,\frac{\hat{i} + \hat{j}}{2}\Bigr) \;=\; (2m)\,\vec{v}.$
Simplifying,
$\vec{v} \;=\; \frac{1}{2m} \left[ m\,u\,\hat{i} + m\left(u\,\frac{\hat{i} + \hat{j}}{2}\right) \right]
\;=\; \frac{1}{2}\left[u\,\hat{i} + \left(u\,\frac{\hat{i} + \hat{j}}{2}\right)\right]
\;=\; \frac{1}{2}\left[u\,\hat{i} + \frac{u}{2}\,\hat{i} + \frac{u}{2}\,\hat{j}\right].
$
Combining terms carefully, we get:
$\vec{v} \;=\; \frac{3u}{4}\,\hat{i} \;+\; \frac{u}{4}\,\hat{j}.$
Step 3: Find the Magnitude of the Common Velocity
The squared magnitude of $\vec{v}$ is
$\bigl|\vec{v}\bigr|^2 \;=\; \left(\frac{3u}{4}\right)^2 + \left(\frac{u}{4}\right)^2
\;=\; \frac{9u^2}{16} \;+\; \frac{u^2}{16}
\;=\; \frac{10u^2}{16}
\;=\; \frac{5u^2}{8}.$
So,
$\bigl|\vec{v}\bigr| \;=\; \sqrt{\frac{5u^2}{8}} \;=\; u\,\sqrt{\frac{5}{8}}.$
Step 4: Compute the Final Kinetic Energy
After collision, the two masses move together with velocity $\vec{v}$. The total mass is $2m$, so the final kinetic energy is:
$\text{KE}_\text{f} \;=\; \frac{1}{2}\,(2m)\,\bigl|\vec{v}\bigr|^2
\;=\; m\,\left(\frac{5u^2}{8}\right)
\;=\; \frac{5}{8}\,m\,u^2.$
Step 5: Compute the Initial Kinetic Energy
Initial kinetic energy is the sum of kinetic energies of both particles:
Particle 1: $\frac{1}{2}\,m\,u^2$
Particle 2:
$\frac{1}{2}\,m\,\left(\frac{u}{\sqrt{2}}\right)^2
= \frac{1}{2}\,m\,\left(\frac{u^2}{2}\right)
= \frac{1}{4}\,m\,u^2.$
Therefore,
$\text{KE}_\text{i} \;=\; \frac{1}{2}\,m\,u^2 \;+\; \frac{1}{4}\,m\,u^2
\;=\; \frac{3}{4}\,m\,u^2.$
Step 6: Calculate the Energy Lost
The energy lost $\Delta \text{KE}$ during the collision is:
$\Delta \text{KE}
\;=\; \text{KE}_\text{f} \;-\; \text{KE}_\text{i}
\;=\; \frac{5}{8}\,m\,u^2 \;-\; \frac{3}{4}\,m\,u^2.
$
Converting fractions to a common denominator:
$\frac{5}{8}\,m\,u^2 \;=\; \frac{5}{8}\,m\,u^2,\quad
\frac{3}{4}\,m\,u^2 \;=\; \frac{6}{8}\,m\,u^2.$
So,
$\Delta \text{KE}
\;=\; \frac{5}{8}\,m\,u^2 - \frac{6}{8}\,m\,u^2
\;=\; -\frac{1}{8}\,m\,u^2.
$
A negative value indicates the kinetic energy is lost. Hence, the magnitude of the lost energy is:
$\boxed{\frac{1}{8}\,m\,u^2}.
