A particle of mass `m` moving with a speed `v` hits elastically another staionary particle of mass `2m` on a smooth horizontal circular tube of radius `r`. Find the time when the next collision will take place?

Options

A. `(2pir)/(v)`
B. `(4pir)/(v)`
C. `(3pir)/(2v)`
D. `(pir)/(v)`

Explanation

To solve the problem, we need to analyze the elastic collision between two particles and determine the time until the next collision occurs. Here’s a step-by-step breakdown of the solution:

### Step 1: Conservation of Momentum
In an elastic collision, both momentum and kinetic energy are conserved. Initially, only the particle of mass `m` is moving with speed `v`, while the particle of mass `2m` is stationary.

The initial momentum \( p_i \) is given by:
\[ p_i = mv \]

After the collision, let \( V_1 \) be the final velocity of mass \( m \) and \( V_2 \) be the final velocity of mass \( 2m \). The final momentum \( p_f \) is:
\[ p_f = mV_1 + 2mV_2 \]

Setting initial momentum equal to final momentum:
\[ mv = mV_1 + 2mV_2 \]

Dividing through by \( m \):
\[ v = V_1 + 2V_2 \quad \text{(Equation 1)} \]

### Step 2: Conservation of Kinetic Energy
The initial kinetic energy \( KE_i \) is:
\[ KE_i = \frac{1}{2} mv^2 \]

The final kinetic energy \( KE_f \) is:
\[ KE_f = \frac{1}{2} mV_1^2 + \frac{1}{2} (2m)V_2^2 = \frac{1}{2} mV_1^2 + mV_2^2 \]

Setting initial kinetic energy equal to final kinetic energy:
\[ \frac{1}{2} mv^2 = \frac{1}{2} mV_1^2 + mV_2^2 \]

Dividing through by \( \frac{1}{2} m \):
\[ v^2 = V_1^2 + 2V_2^2 \quad \text{(Equation 2)} \]

### Step 3: Solving the Equations
We have two equations:
1. \( v = V_1 + 2V_2 \)
2. \( v^2 = V_1^2 + 2V_2^2 \)

From Equation 1, we can express \( V_1 \):
\[ V_1 = v - 2V_2 \]

Substituting \( V_1 \) into Equation 2:
\[ v^2 = (v - 2V_2)^2 + 2V_2^2 \]

Expanding the square:
\[ v^2 = v^2 - 4vV_2 + 4V_2^2 + 2V_2^2 \]
\[ v^2 = v^2 - 4vV_2 + 6V_2^2 \]

Cancelling \( v^2 \) from both sides:
\[ 0 = -4vV_2 + 6V_2^2 \]

Factoring out \( V_2 \):
\[ V_2(6V_2 - 4v) = 0 \]

Thus, either \( V_2 = 0 \) (not possible since it’s a collision) or:
\[ 6V_2 = 4v \implies V_2 = \frac{2v}{3} \]

Substituting \( V_2 \) back into Equation 1 to find \( V_1 \):
\[ V_1 = v - 2\left(\frac{2v}{3}\right) = v - \frac{4v}{3} = -\frac{v}{3} \]

### Step 4: Finding the Time for Next Collision
The distance traveled by both particles until the next collision is the circumference of the circular tube:
\[ \text{Distance} = 2\pi r \]

The relative speed between the two particles is:
\[ V = V_2 - V_1 = \frac{2v}{3} - \left(-\frac{v}{3}\right) = \frac{2v}{3} + \frac{v}{3} = v \]

The time \( T \) until the next collision is given by:
\[ T = \frac{\text{Distance}}{\text{Relative Speed}} = \frac{2\pi r}{v} \]

### Final Answer
Thus, the time until the next collision will take place is:
\[ T = \frac{2\pi r}{v} \]