A particle of mass m moving in the x-direction with speed 2v is hit by another particle of mass 2m m

Q: Why isn't kinetic energy conserved in this collision?

In a perfectly inelastic collision, the maximum possible kinetic energy is lost. This energy is dissipated as heat, sound, and work done in permanently deforming the bodies so they stick together.

Q: Can we just conserve momentum using vector addition instead of components?

Yes! You can write $\vec{P}_i = m(2v\hat{i}) + 2m(v\hat{j})$ and equate it to $\vec{P}_f = 3m\vec{v}_f$. This directly gives $\vec{v}_f = \frac{2v}{3}\hat{i} + \frac{2v}{3}\hat{j}$, which is mathematically identical to the component method.

Q: What does 'perfectly inelastic' mean?

It means the coefficient of restitution $e = 0$. The relative velocity of separation after the collision is zero, meaning the objects stick together and move as a single unit.

Q: Why do we add the masses for the final state?

Because the collision is perfectly inelastic, the two particles fuse into a single object. By the law of conservation of mass, the new object's mass is simply the sum of the individual masses ($m + 2m = 3m$).

Q: Would the percentage loss change if the masses were different?

Yes, the percentage loss depends heavily on the mass ratio and the initial velocity ratio. If the masses or velocities were different, the final velocity and thus the final kinetic energy would change, leading to a different percentage loss.

Question

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

Options:

(A) 44%

(B) 50%

(C) 56%

(D) 62%

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are observing a microscopic traffic intersection. Two particles are hurtling towards the origin from perpendicular directions. Particle 1, with mass $m$, is speeding along the x-axis at a brisk velocity of $2v$. Meanwhile, Particle 2, which is twice as heavy with mass $2m$, is moving along the y-axis at a velocity of $v$.

They are on a collision course. But this isn't just any collision; the problem states it is a **perfectly inelastic collision**. What does that mean physically? It means that upon impact, the two particles don't bounce off each other. Instead, they crumple and stick together, forming a single, heavier combined mass that moves off in a new direction.

### The Master Equation: Conservation of Momentum

In the chaotic moment of any collision, one fundamental law of the universe stands firm: the **Conservation of Linear Momentum**. Because there are no external forces acting on our two-particle system, the total momentum before the crash must exactly equal the total momentum after the crash.

Since momentum is a vector, we must conserve it independently along both the x and y axes. Let's define the final velocity of our new combined mass ($3m$) as $\vec{v}_f = v_x \hat{i} + v_y \hat{j}$.

First, let's look at the x-direction. Before the collision, only Particle 1 has momentum along the x-axis.
$p_{ix} = m(2v)$

After the collision, the combined mass $3m$ moves with velocity $v_x$.
$p_{fx} = (3m)v_x$

Equating them:
$m(2v) = 3m v_x \implies v_x = \frac{2v}{3}$

Now, let's apply the same logic to the y-direction. Before the collision, only Particle 2 has momentum along the y-axis.
$p_{iy} = (2m)v$

After the collision, the combined mass $3m$ moves with velocity $v_y$.
$p_{fy} = (3m)v_y$

Equating them:
$2mv = 3m v_y \implies v_y = \frac{2v}{3}$

### Calculating the Energy Toll

While momentum is strictly conserved, kinetic energy is a different story. In a perfectly inelastic collision, a significant amount of kinetic energy is lost—transformed into heat, sound, and the deformation of the particles as they fuse together. To find out exactly how much is lost, we need to calculate the kinetic energy before and after the event.

Let's calculate the total initial kinetic energy ($E_i$). It is simply the sum of the kinetic energies of the two individual particles:
$E_i = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)(v)^2$

$E_i = \frac{1}{2}m(4v^2) + mv^2$

$E_i = 2mv^2 + mv^2 = 3mv^2$

Next, we calculate the final kinetic energy ($E_f$) of the combined mass $3m$ moving with velocity components $v_x$ and $v_y$:
$E_f = \frac{1}{2}(3m)(v_x^2 + v_y^2)$

Substitute the values we found earlier:
$E_f = \frac{1}{2}(3m)\left[\left(\frac{2v}{3}ight)^2 + \left(\frac{2v}{3}ight)^2ight]$

$E_f = \frac{1}{2}(3m)\left[\frac{4v^2}{9} + \frac{4v^2}{9}ight]$

$E_f = \frac{1}{2}(3m)\left[\frac{8v^2}{9}ight] = \frac{4}{3}mv^2$

### Final Calculation: The Percentage Loss

Now we can clearly see the energy toll. The loss in kinetic energy ($\Delta E$) is the difference between the initial and final states:
$\Delta E = E_i - E_f$

$\Delta E = 3mv^2 - \frac{4}{3}mv^2$

$\Delta E = \frac{9mv^2 - 4mv^2}{3} = \frac{5}{3}mv^2$

The question asks for the **percentage loss** in energy. This is the ratio of the lost energy to the initial energy, multiplied by 100:
$	ext{Percentage Loss} = \frac{\Delta E}{E_i} 	imes 100$

$	ext{Percentage Loss} = \frac{\frac{5}{3}mv^2}{3mv^2} 	imes 100$

$	ext{Percentage Loss} = \frac{5}{9} 	imes 100 \approx 55.56\%$

Rounding to the nearest integer, we get **$56\%$**. The collision was violent enough to dissipate more than half of the system's initial kinetic energy!

Select Answer:

Visualized Solution

Visualizing the Initial State

Perfectly Inelastic Collision

Conservation of Momentum (X-axis)

Conservation of Momentum (Y-axis)

Initial Kinetic Energy

Final Kinetic Energy

Loss in Kinetic Energy

Percentage Loss in Energy

The Way Forward

The Sigma Insight: Oblique Collision

Analyzing the Setup

The Master Equation

Calculating the Energy Toll

Final Calculation

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