A particle of mass m moving in the x direction with speed 2v is hit by 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. 0.44
B. 0.5
C. 0.56
D. 0.62

Explanation

To solve the problem, we need to calculate the percentage loss in kinetic energy during a perfectly inelastic collision between two particles. Let's break down the solution step by step:

### Step 1: Identify the masses and velocities of the particles
- Particle 1 has mass \( m \) and is moving in the x-direction with speed \( 2v \).
- Particle 2 has mass \( 2m \) and is moving in the y-direction with speed \( v \).

### Step 2: Calculate the initial kinetic energy of both particles
The kinetic energy (KE) of an object is given by the formula:
\[ KE = \frac{1}{2} mv^2 \]

- For Particle 1:
\[ KE_1 = \frac{1}{2} m (2v)^2 = \frac{1}{2} m \cdot 4v^2 = 2mv^2 \]

- For Particle 2:
\[ KE_2 = \frac{1}{2} (2m) v^2 = mv^2 \]

- Total initial kinetic energy \( KE_{initial} \):
\[ KE_{initial} = KE_1 + KE_2 = 2mv^2 + mv^2 = 3mv^2 \]

### Step 3: Determine the final velocity after the collision
In a perfectly inelastic collision, the two particles stick together and move with a common velocity. We can use the conservation of momentum to find this velocity.

- Initial momentum in the x-direction:
\[ p_{x, initial} = m \cdot (2v) + 0 = 2mv \]

- Initial momentum in the y-direction:
\[ p_{y, initial} = 0 + (2m) \cdot v = 2mv \]

- Total initial momentum vector:
\[ \vec{p}_{initial} = (2mv) \hat{i} + (2mv) \hat{j} \]

The total mass after the collision is:
\[ m_{total} = m + 2m = 3m \]

Using conservation of momentum:
\[ \vec{p}_{initial} = \vec{p}_{final} \]
\[ (2mv) \hat{i} + (2mv) \hat{j} = (3m) \vec{v}_{final} \]

Dividing by \( 3m \):
\[ \vec{v}_{final} = \left( \frac{2v}{3} \hat{i} + \frac{2v}{3} \hat{j} \right) \]

### Step 4: Calculate the magnitude of the final velocity
The magnitude of the final velocity \( v_{final} \) is:
\[ v_{final} = \sqrt{\left(\frac{2v}{3}\right)^2 + \left(\frac{2v}{3}\right)^2} = \sqrt{\frac{4v^2}{9} + \frac{4v^2}{9}} = \sqrt{\frac{8v^2}{9}} = \frac{2\sqrt{2}v}{3} \]

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

### Step 6: Calculate the change in kinetic energy
The change in kinetic energy \( \Delta KE \) is:
\[ \Delta KE = KE_{final} - KE_{initial} = \frac{4mv^2}{3} - 3mv^2 \]
\[ = \frac{4mv^2}{3} - \frac{9mv^2}{3} = -\frac{5mv^2}{3} \]

### Step 7: Calculate the percentage loss in kinetic energy
The percentage loss in kinetic energy is given by:
\[ \text{Percentage Loss} = \left(\frac{\Delta KE}{KE_{initial}}\right) \times 100 \]
\[ = \left(\frac{-\frac{5mv^2}{3}}{3mv^2}\right) \times 100 = \left(-\frac{5}{9}\right) \times 100 = -55.56\% \]

Since we are interested in the magnitude of the loss:
\[ \text{Percentage Loss} \approx 55.56\% \]

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

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