Two small particles of equal masses start moving in opposite directions from a point A in a horizont

Q: Wait, why do they exchange velocities?

In a perfectly elastic head-on collision between two objects of equal mass, the conservation of momentum and kinetic energy mathematically dictates that they simply swap their velocity vectors. It's like billiard balls transferring their state perfectly!

Q: Why can we treat motion on a circle like 1D motion?

Because the particles are constrained to move along the 1D curve of the circle. Their tangential speeds act just like linear speeds on a straight track, and the distance is just the arc length $R\theta$.

Q: Does the faster particle always cover $240^\circ$ between collisions?

Yes! Because the relative speed is always $3v$, the time between any two consecutive collisions is constant. In that time, the particle moving at $2v$ will always cover twice the distance of the particle moving at $v$, which translates to $240^\circ$ and $120^\circ$ respectively.

Q: What if the masses were different?

If the masses were different, they wouldn't simply exchange velocities. We would have to use the full 1D elastic collision formulas to find their new speeds and directions after every single bounce, making the problem much more complex!

Q: Why is the answer 2 and not 3?

The particles do collide for a third time exactly at the starting point $A$. However, the question specifically asks for the number of collisions *other than that at A*. Therefore, we only count the first two collisions.

Question

Two small particles of equal masses start moving in opposite directions from a point $A$ in a horizontal circular orbit. Their tangential velocities are $v$ and $2v$ respectively, as shown in the figure. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at $A$, these two particles will again reach the point $A$?

Options:

(A) 4

(B) 3

(C) 2

(D) 1

Sigma Percentile · Accepted Answer

### The Setup: A Circular Chase

Imagine a perfectly smooth circular track of radius $R$. Two identical particles start at the very top, at point $A$.

They shoot off in opposite directions. Particle 1 moves counter-clockwise with a speed of $v$, while Particle 2 zooms clockwise at twice the speed, $2v$. Because they are confined to this circular path, they are destined to collide.

### The First Encounter

To find out where they meet, we can use the concept of relative motion. Since they are moving towards each other, their relative speed is the sum of their individual speeds:

$$v_{	ext{rel}} = v + 2v = 3v$$

The time it takes for them to cover the entire circumference of the circle and crash into each other is:

$$t = \frac{2\pi R}{3v}$$

In this time $t$, how far does each particle travel? Let's calculate the angular distance for Particle 1:

$$	heta_1 = \frac{v}{R} t = \frac{v}{R} \left( \frac{2\pi R}{3v} ight) = \frac{2\pi}{3} = 120^\circ$$

Particle 1 covers exactly one-third of the circle. Naturally, Particle 2, moving twice as fast, covers the remaining two-thirds, or $240^\circ$. They meet at a point $120^\circ$ counter-clockwise from $A$.

### The Magic of Equal Masses

Here is where the physics gets beautiful. The problem states that the collision is perfectly elastic and that the particles have equal masses.

There is a golden rule in mechanics for this exact scenario: **in a 1D perfectly elastic collision between two identical masses, the objects simply exchange their velocities.**

So, Particle 1, which was moving counter-clockwise at $v$, suddenly bounces back and moves clockwise at $2v$. Particle 2, which was moving clockwise at $2v$, now moves counter-clockwise at $v$. It is a complete role reversal!

### The Second Encounter

The chase begins anew. Because their speeds are still $v$ and $2v$, their relative speed remains $3v$. The time to the next collision is exactly the same.

In this second leg, Particle 1 (now moving clockwise at $2v$) covers $240^\circ$. Particle 2 (now moving counter-clockwise at $v$) covers $120^\circ$.

They meet again, advancing another $120^\circ$ counter-clockwise around the circle. This places the second collision at $240^\circ$ from the starting point $A$.

### The Grand Finale

At the second collision, they swap velocities once more. Particle 1 regains its original state, moving counter-clockwise at $v$. Particle 2 also resets, moving clockwise at $2v$.

In the third leg of the journey, Particle 1 covers another $120^\circ$ counter-clockwise. Particle 2 covers $240^\circ$ clockwise.

If we add up the angles, $240^\circ + 120^\circ = 360^\circ$. They both arrive exactly back at the starting point, $A$!

The question asks for the number of collisions **other than that at A**. Since they collided at $120^\circ$ and $240^\circ$ before returning to $A$, there are exactly **2** collisions.

Select Answer:

Visualized Solution

Initial Setup

Time to First Collision

Position of First Collision

The Elastic Exchange

Velocities After 1st Collision

Position of Second Collision

Velocities After 2nd Collision

The Third Collision

Final Conclusion

The Sigma Insight: Head-on Collision

The Setup

The First Encounter

The Magic of Equal Masses

The Second Encounter

The Grand Finale

Similar Questions

In a collinear collision, a particle with an initial speed $v_{0}$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50%$ greater than the original kinetic energy, the magnitude of the relative velocity between the two particles after collision, is

A body of mass 2 kg makes an elastic collision with a second body at rest and continues to move in the original direction but with one-fourth of its original speed. What is the mass of the second body?

A point mass of $1 kg$ collides elastically with a stationary point mass of $5 kg$ . After their collision, the $1 kg$ mass reverses its direction and moves with a speed of $2 ms^{- 1}$ . Which of the following statement(s) is/are correct for the system of these two masses?

An object of mass $m_{1}$ collides with another object of mass $m_{2}$ , which is at rest. After the collision, the objects move with equal speeds in opposite direction. The ratio of the masses $m_{2} : m_{1}$ is

A body of mass $2 kg$ moving with a speed of $4 m/s$ makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its initial speed. The speed of the two body centre of mass is $\frac{x}{10} m/s$ . Then, the value of $x$ is ......... .