A particle of mass $$m$$ moving in the $$x$$ direction with speed $$2v$$ is hit by another particle of mass $$2m$$ moving in the $$y$$ direction with speed $$v$$. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to

We have two particles approaching each other at right angles. The first particle has mass $$m$$ and moves along the $$x$$-axis with speed $$2v$$, therefore its velocity vector is $$\vec u_1=(2v,\,0)$$. The second particle has mass $$2m$$ and moves along the $$y$$-axis with speed $$v$$, so its velocity vector is $$\vec u_2=(0,\,v)$$.

Because the collision is perfectly inelastic, the two particles stick together after impact and move as a single composite body of mass $$m_{\text{final}} = m + 2m = 3m.$$

According to the law of conservation of linear momentum, the vector sum of momenta before collision equals the momentum after collision:

$$m\vec u_1 + 2m\vec u_2 = (3m)\vec V,$$

where $$\vec V=(V_x,\,V_y)$$ is the common velocity of the combined mass.

Substituting the given velocities, we obtain

$$m(2v,\,0) + 2m(0,\,v) = (2mv,\,2mv) = 3m\,(V_x,\,V_y).$$

Dividing both components by $$3m$$ gives

$$V_x = \frac{2mv}{3m} = \frac{2v}{3}, \qquad V_y = \frac{2mv}{3m} = \frac{2v}{3}.$$

Hence the magnitude of the final velocity is

$$|\vec V| = \sqrt{V_x^2 + V_y^2} = \sqrt{\left(\frac{2v}{3}\right)^2 + \left(\frac{2v}{3}\right)^2} = \frac{2v}{3}\sqrt{2}.$$

Now we compare kinetic energies. The formula for kinetic energy is $$K=\tfrac{1}{2}mv^{2}.$$ First, the initial kinetic energy of each particle:

For mass $$m$$: $$K_1 = \frac{1}{2}m(2v)^2 = \frac{1}{2}m\cdot4v^{2} = 2mv^{2}.$$

For mass $$2m$$: $$K_2 = \frac{1}{2}(2m)(v)^2 = m v^{2}.$$

So the total initial kinetic energy is

$$K_{\text{initial}} = K_1 + K_2 = 2mv^{2} + mv^{2} = 3mv^{2}.$$

Next, the final kinetic energy of the composite mass:

$$K_{\text{final}} = \frac{1}{2}(3m)\,|\vec V|^{2} = \frac{1}{2}(3m)\left(\frac{2v\sqrt{2}}{3}\right)^{2} = \frac{1}{2}(3m)\left(\frac{8v^{2}}{9}\right) = \frac{24}{18}mv^{2} = \frac{4}{3}mv^{2}.$$

The loss in kinetic energy is therefore

$$\Delta K = K_{\text{initial}} - K_{\text{final}} = 3mv^{2} - \frac{4}{3}mv^{2} = \frac{9}{3}mv^{2} - \frac{4}{3}mv^{2} = \frac{5}{3}mv^{2}.$$

Finally, the percentage loss is

$$\text{Percentage loss} = \left(\frac{\Delta K}{K_{\text{initial}}}\right)\times100 = \left(\frac{\tfrac{5}{3}mv^{2}}{3mv^{2}}\right)\times100 = \left(\frac{5}{9}\right)\times100 \approx 55.6\%.$$

This value is closest to $$56\%$$.

Hence, the correct answer is Option D.