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 the elastic collision between two particles, we will follow these steps:

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

Both particles are moving in the same direction, and we need to find the maximum possible deflection of the 2 kg particle after the elastic collision.

### 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 known values:

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

Calculating the left side:

\[ 12 + 8 = 2 v_1 + 4 v_2 \]
\[ 20 = 2 v_1 + 4 v_2 \quad \text{(Equation 1)} \]

### Step 3: Apply conservation of kinetic energy
Since the collision is elastic, 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 known values:

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

Calculating the left side:

\[ \frac{1}{2} \times 2 \times 36 + \frac{1}{2} \times 4 \times 4 = v_1^2 + 2 v_2^2 \]
\[ 36 + 8 = v_1^2 + 2 v_2^2 \]
\[ 44 = v_1^2 + 2 v_2^2 \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 1, we can express \( v_1 \) in terms of \( v_2 \):

\[ 20 = 2 v_1 + 4 v_2 \implies v_1 = 10 - 2v_2 \]

Substituting \( v_1 \) into Equation 2:

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

Expanding the equation:

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

### Step 5: Solve the quadratic equation
Dividing the entire equation by 2:

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

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

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

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 \( v_2 = \frac{14}{3} \):

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

### Step 7: Calculate the deflection angle
The deflection angle \( \theta \) can be calculated using the tangent function:

\[ \tan \theta = \frac{v_2}{v_1} = \frac{\frac{14}{3}}{\frac{2}{3}} = 7 \]

Thus,

\[ \theta = \tan^{-1}(7) \]

Calculating \( \theta \):

\[ \theta \approx 81.87^\circ \]

### Conclusion
The maximum possible deflection of the 2 kg particle is approximately \( 81.87^\circ \).