Question

A particle (𝑝₁) of mass 𝑚 moving with speed 𝑣 collides with a stationary identical particle (𝑝₂). The particles bounce off each other elastically with 𝑝₁ getting deflected by an angle 𝜃 = 30° from its original direction. Then, which of the following statement(s) is/are true after the collision?

• A. Speed of 𝑝₁ is \(\frac{\sqrt3}{2}\) V
• B. Kinetic energy of 𝑝₂ is 25% of the total energy
• C. Angle between the directions of motion of the two particles is 90°
• D. The kinetic energy of the centre of mass of 𝑝₁ and 𝑝₂ decreases

Answer 1

The Correct Option is A, B, C

To solve this problem, we need to analyze an elastic collision between two identical particles. Here's a detailed step-by-step solution to evaluate each given statement:

Step 1: Analyze the Collision

The collision is elastic, which means both momentum and kinetic energy are conserved. Initially, particle \( p_1 \) has mass \( m \) and velocity \( v \), while particle \( p_2 \) is stationary.

Step 2: Conservation of Momentum

The vector components of momentum are conserved both in the x-direction and y-direction.

Step 3: Calculate Speed of \( p_1 \) after Collision

After the collision, particle \( p_1 \) is deflected at an angle \( \theta = 30^\circ \). The final speed of \( p_1 \) can be determined using trigonometric components:

\(v_{1f} = v \cos(30^\circ) = v \times \frac{\sqrt{3}}{2}\) 

This matches the statement given as one of the options:

Speed of \( p_1 \) is \( \frac{\sqrt{3}}{2} V \)

Hence, this statement is true.

Step 4: Calculate Kinetic Energy of \( p_2 \)

Using the conservation of kinetic energy:

\(\frac{1}{2}mv^2 = \frac{1}{2}m(v_{1f})^2 + \frac{1}{2}m(v_{2f})^2\)

Substituting \(v_{1f} = \frac{\sqrt{3}}{2}v\), we find:

\(\frac{1}{2}mv_{2f}^2 = \frac{1}{2}mv^2 - \frac{1}{2}m\left(\left(\frac{\sqrt{3}}{2}v\right)^2\right)\)

\( \Rightarrow \frac{1}{2}mv_{2f}^2 = \frac{1}{4}mv^2 \)

Thus, \( v_{2f}^2 = \frac{1}{2}v^2 \), meaning:

The kinetic energy of \( p_2 \) is 25% of the initial total energy, confirming:

Kinetic energy of \( p_2 \) is 25% of the total energy

Hence, this statement is true too.

Step 5: Direction of the Particles

Due to conservation of momentum in perpendicular directions, and since the initial momentum vector of \( p_2 \) was zero, the two particles move perpendicularly after the collision. Hence:

Angle between the directions of motion of the two particles is 90°

This statement is true as well.

Step 6: Kinetic Energy of the Centre of Mass

The kinetic energy of the center of mass will not change because it depends on the system's total initial and final velocities, which remain the same for an isolated system.

This disproves the last statement:

The kinetic energy of the centre of mass of \( p_1 \) and \( p_2 \) decreases

This is false.

Conclusion:

The correctly validated statements are:

1. Speed of \( p_1 \) is \( \frac{\sqrt{3}}{2} V \)
2. Kinetic energy of \( p_2 \) is 25% of the total energy
3. Angle between the directions of motion of the two particles is 90°