A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving in they 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. `56%`
B. `62%`
C. `44%`
D. `50%`

Answer

To solve the problem of the perfectly inelastic collision between two particles, we will follow these steps:

### Step 1: Determine Initial Momentum
We have two particles:
- Particle 1: mass \( m \), moving in the x-direction with speed \( 2v \).
- Particle 2: mass \( 2m \), moving in the y-direction with speed \( v \).

The initial momentum in the x-direction (\( P_{initial,x} \)) is:
\[ P_{initial,x} = m \cdot 2v = 2mv \]

The initial momentum in the y-direction (\( P_{initial,y} \)) is:
\[ P_{initial,y} = 2m \cdot v = 2mv \]

### Step 2: Apply Conservation of Momentum
Since the collision is perfectly inelastic, the two particles will stick together after the collision. Let \( V_x \) and \( V_y \) be the final velocities in the x and y directions, respectively.

Using conservation of momentum in the x-direction:
\[ P_{initial,x} = P_{final,x} \]
\[ 2mv = (3m) V_x \quad \text{(total mass after collision is } 3m\text{)} \]
\[ V_x = \frac{2mv}{3m} = \frac{2}{3}v \]

Using conservation of momentum in the y-direction:
\[ P_{initial,y} = P_{final,y} \]
\[ 2mv = (3m) V_y \]
\[ V_y = \frac{2mv}{3m} = \frac{2}{3}v \]

### Step 3: Calculate Initial Kinetic Energy
The initial kinetic energy (\( KE_{initial} \)) is the sum of the kinetic energies of both particles:
\[ KE_{initial} = \frac{1}{2} m (2v)^2 + \frac{1}{2} (2m) v^2 \]
\[ = \frac{1}{2} m \cdot 4v^2 + \frac{1}{2} \cdot 2m \cdot v^2 \]
\[ = 2mv^2 + mv^2 = 3mv^2 \]

### Step 4: Calculate Final Kinetic Energy
The final kinetic energy (\( KE_{final} \)) after the collision is:
\[ KE_{final} = \frac{1}{2} (3m) \left( V_x^2 + V_y^2 \right) \]
Substituting \( V_x \) and \( V_y \):
\[ = \frac{1}{2} (3m) \left( \left( \frac{2}{3}v \right)^2 + \left( \frac{2}{3}v \right)^2 \right) \]
\[ = \frac{1}{2} (3m) \left( \frac{4}{9}v^2 + \frac{4}{9}v^2 \right) \]
\[ = \frac{1}{2} (3m) \left( \frac{8}{9}v^2 \right) \]
\[ = \frac{12mv^2}{18} = \frac{2mv^2}{3} \]

### Step 5: Calculate Percentage Loss in Kinetic Energy
The percentage loss in kinetic energy is given by:
\[ \text{Percentage Loss} = \frac{KE_{initial} - KE_{final}}{KE_{initial}} \times 100 \]
\[ = \frac{3mv^2 - \frac{2mv^2}{3}}{3mv^2} \times 100 \]
\[ = \frac{9mv^2 - 2mv^2}{9mv^2} \times 100 \]
\[ = \frac{7mv^2}{9mv^2} \times 100 = \frac{700}{9} \approx 77.78\% \]

Thus, the percentage loss in energy during the collision is approximately **77.78%**.