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Step-by-Step Solution
Step 1: Identify the given information
• Mass of first particle = $m$, moving along the x-axis with speed $2v$.
• Mass of second particle = $2m$, moving along the y-axis with speed $v$.
• The collision is perfectly inelastic, meaning the two particles stick together after collision.
Step 2: Write down initial momenta
Since there is no motion in the other axes initially, we conserve momentum in each direction:
• In the x-direction initially:
$$p_{ix} = m \cdot 2v = 2 m v$$
• In the y-direction initially:
$$p_{iy} = (2m) \cdot v = 2 m v$$
Step 3: Final momentum and final velocity
After collision, the combined mass is $(m + 2m) = 3m$. Let the final velocity of the combined mass be $\vec{v}_f = (v_{fx}, v_{fy})$.
• From momentum conservation along x-axis:
$$3m \cdot v_{fx} = 2 m v \quad \Rightarrow \quad v_{fx} = \frac{2v}{3}$$
• From momentum conservation along y-axis:
$$3m \cdot v_{fy} = 2 m v \quad \Rightarrow \quad v_{fy} = \frac{2v}{3}$$
Hence,
$$\vec{v}_f = \left(\frac{2v}{3}, \frac{2v}{3}\right).$$
The magnitude of the final velocity is
$$|\vec{v}_f| = \sqrt{\left(\frac{2v}{3}\right)^2 + \left(\frac{2v}{3}\right)^2} = \frac{2v}{3} \sqrt{2}.$$
Step 4: Calculate initial kinetic energy
• Kinetic energy of the first particle (mass $m$, speed $2v$):
$$KE_{1} = \frac{1}{2} \cdot m \cdot (2v)^2 = 2 m v^2.$$
• Kinetic energy of the second particle (mass $2m$, speed $v$):
$$KE_{2} = \frac{1}{2} \cdot (2m) \cdot v^2 = m v^2.$$
Thus, the total initial kinetic energy is
$$KE_{\text{initial}} = 2 m v^2 + m v^2 = 3 m v^2.$$
Step 5: Calculate final kinetic energy
After collision, the combined mass is $3m$, moving with velocity $\frac{2v}{3}\sqrt{2}$:
$$
KE_{\text{final}} = \frac{1}{2} \cdot (3m) \cdot \left(\frac{2v}{3}\sqrt{2}\right)^2
= \frac{1}{2} \cdot 3m \cdot \frac{8v^2}{9}
= \frac{3m \times 8v^2}{18}
= \frac{24 m v^2}{18}
= \frac{4}{3} m v^2.
$$
Step 6: Calculate the percentage loss in kinetic energy
• Loss in kinetic energy:
$$
\Delta KE = KE_{\text{initial}} - KE_{\text{final}}
= 3m v^2 - \frac{4}{3}m v^2
= \frac{9}{3}m v^2 - \frac{4}{3}m v^2
= \frac{5}{3} m v^2.
$$
• Percentage loss in kinetic energy:
$$
\text{Percentage loss}
= \frac{\Delta KE}{KE_{\text{initial}}} \times 100\%
= \frac{\frac{5}{3} m v^2}{3 m v^2} \times 100\%
= \frac{5}{9} \times 100\%
= 55.55\% \approx 56\%.
$$
Final Answer
The percentage loss in kinetic energy in the perfectly inelastic collision is approximately $56\%$.
