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. `50%`
B. `56%`
C. `62%`
D. `44%`

Answer

To solve the problem, we need to calculate the percentage loss in kinetic energy during a perfectly inelastic collision between two particles. Here are the steps to arrive at the solution:

### Step 1: Identify the initial kinetic energies of both particles.
- The first particle has mass \( m \) and speed \( 2v \):
\[ KE_1 = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \cdot 4v^2 = 2mv^2 \]

- The second particle has mass \( 2m \) and speed \( v \):
\[ KE_2 = \frac{1}{2} (2m) v^2 = mv^2 \]

### Step 2: Calculate the total initial kinetic energy.
\[ KE_{\text{initial}} = KE_1 + KE_2 = 2mv^2 + mv^2 = 3mv^2 \]

### Step 3: Apply conservation of momentum to find the final velocity after the collision.
- The momentum in the x-direction before the collision:
\[ P_{x,\text{initial}} = m \cdot 2v + 0 = 2mv \]
- The momentum in the y-direction before the collision:
\[ P_{y,\text{initial}} = 0 + 2m \cdot v = 2mv \]

After the collision, the two particles stick together, so the total mass is \( 3m \). Let \( V_f \) be the final velocity vector. By conservation of momentum:
- In the x-direction:
\[ 2mv = 3m V_{fx} \implies V_{fx} = \frac{2}{3}v \]
- In the y-direction:
\[ 2mv = 3m V_{fy} \implies V_{fy} = \frac{2}{3}v \]

### Step 4: Calculate the magnitude of the final velocity.
Using the Pythagorean theorem:
\[ V_f = \sqrt{V_{fx}^2 + V_{fy}^2} = \sqrt{\left(\frac{2}{3}v\right)^2 + \left(\frac{2}{3}v\right)^2} = \sqrt{2 \cdot \left(\frac{2}{3}v\right)^2} = \frac{2v}{3} \sqrt{2} \]

### Step 5: Calculate the final kinetic energy.
\[ KE_{\text{final}} = \frac{1}{2} (3m) V_f^2 = \frac{1}{2} (3m) \left(\frac{2v}{3} \sqrt{2}\right)^2 = \frac{1}{2} (3m) \cdot \frac{8v^2}{9} = \frac{4mv^2}{3} \]

### Step 6: Calculate the percentage loss in kinetic energy.
\[ \text{Loss in KE} = KE_{\text{initial}} - KE_{\text{final}} = 3mv^2 - \frac{4mv^2}{3} = \frac{9mv^2}{3} - \frac{4mv^2}{3} = \frac{5mv^2}{3} \]

Now, the percentage loss in kinetic energy is given by:
\[ \text{Percentage loss} = \frac{\text{Loss in KE}}{KE_{\text{initial}}} \times 100 = \frac{\frac{5mv^2}{3}}{3mv^2} \times 100 = \frac{5}{9} \times 100 \approx 55.56\% \]

### Final Answer
The percentage loss in kinetic energy during the collision is approximately **56%**.

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