A particle of mass 2 kg moving with a speed of 6 m/s collides elastically with another particle of mass 4 kg travelling in same direction with a speed of 2m/s. The maximum possible deflection of the 2 kg particle is

Options

A. `37^(@)`
B. `45^(@)`
C. `53^(@)`
D. `60^(@)`

Answer

To solve the problem of finding the maximum possible deflection of a 2 kg particle after it collides elastically with a 4 kg particle, we will follow these steps:

### Step 1: Understand the Problem
We have two particles:
- Particle 1: Mass \( m_1 = 2 \, \text{kg} \), Initial speed \( u_1 = 6 \, \text{m/s} \)
- Particle 2: Mass \( m_2 = 4 \, \text{kg} \), Initial speed \( u_2 = 2 \, \text{m/s} \)

The collision is elastic, meaning both momentum and kinetic energy are conserved.

### Step 2: Apply Conservation of Momentum
The total momentum before the collision is equal to the total momentum after the collision.

\[ m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \]

Where \( v_1 \) and \( v_2 \) are the final velocities of the 2 kg and 4 kg particles, respectively.

Substituting the values:

\[ (2 \times 6) + (4 \times 2) = 2v_1 + 4v_2 \]

Calculating the left side:

\[ 12 + 8 = 2v_1 + 4v_2 \implies 20 = 2v_1 + 4v_2 \tag{1} \]

### Step 3: Apply Conservation of Kinetic Energy
The total kinetic energy before the collision is equal to the total kinetic energy after the collision.

\[ \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \]

Substituting the values:

\[ \frac{1}{2} (2)(6^2) + \frac{1}{2} (4)(2^2) = \frac{1}{2} (2)v_1^2 + \frac{1}{2} (4)v_2^2 \]

Calculating the left side:

\[ \frac{1}{2} (2)(36) + \frac{1}{2} (4)(4) = 36 + 8 = 44 \]

So we have:

\[ 44 = v_1^2 + 2v_2^2 \tag{2} \]

### Step 4: Solve the Equations
Now we have two equations (1) and (2):

1. \( 20 = 2v_1 + 4v_2 \)
2. \( 44 = v_1^2 + 2v_2^2 \)

From equation (1), we can express \( v_1 \) in terms of \( v_2 \):

\[ 2v_1 = 20 - 4v_2 \implies v_1 = 10 - 2v_2 \tag{3} \]

Substituting equation (3) into equation (2):

\[ 44 = (10 - 2v_2)^2 + 2v_2^2 \]

Expanding:

\[ 44 = 100 - 40v_2 + 4v_2^2 + 2v_2^2 \]
\[ 44 = 100 - 40v_2 + 6v_2^2 \]
\[ 6v_2^2 - 40v_2 + 56 = 0 \]

### Step 5: Solve the Quadratic Equation
Dividing the entire equation by 2:

\[ 3v_2^2 - 20v_2 + 28 = 0 \]

Using the quadratic formula \( v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ v_2 = \frac{20 \pm \sqrt{(-20)^2 - 4 \cdot 3 \cdot 28}}{2 \cdot 3} \]
\[ v_2 = \frac{20 \pm \sqrt{400 - 336}}{6} \]
\[ v_2 = \frac{20 \pm \sqrt{64}}{6} \]
\[ v_2 = \frac{20 \pm 8}{6} \]

Calculating the two possible values:

1. \( v_2 = \frac{28}{6} = \frac{14}{3} \)
2. \( v_2 = \frac{12}{6} = 2 \)

### Step 6: Find \( v_1 \)
Using equation (3) to find \( v_1 \):

1. For \( v_2 = \frac{14}{3} \):

\[ v_1 = 10 - 2 \cdot \frac{14}{3} = 10 - \frac{28}{3} = \frac{30 - 28}{3} = \frac{2}{3} \]

2. For \( v_2 = 2 \):

\[ v_1 = 10 - 2 \cdot 2 = 10 - 4 = 6 \]

### Step 7: Calculate the Maximum Deflection
The maximum deflection occurs when we consider the ratio of the velocities:

\[ \tan \theta = \frac{v_1}{v_2} \]

For \( v_2 = \frac{14}{3} \) and \( v_1 = \frac{2}{3} \):

\[ \tan \theta = \frac{\frac{2}{3}}{\frac{14}{3}} = \frac{2}{14} = \frac{1}{7} \]

For \( v_2 = 2 \) and \( v_1 = 6 \):

\[ \tan \theta = \frac{6}{2} = 3 \]

### Step 8: Find the Angle
To find the angle \( \theta \):

1. For \( \tan \theta = \frac{1}{7} \):

\[ \theta = \tan^{-1}\left(\frac{1}{7}\right) \approx 8.13^\circ \]

2. For \( \tan \theta = 3 \):

\[ \theta = \tan^{-1}(3) \approx 71.57^\circ \]

### Conclusion
The maximum possible deflection of the 2 kg particle after the elastic collision is approximately \( 71.57^\circ \).