A rocket is moving in a gravity free space with a constant acceleration of 2 ms^-2 along +x directio

Q: Why do we use a pseudo force here?

Because the rocket is an accelerating (non-inertial) frame of reference. To apply Newton's laws inside the rocket as if it were at rest, we must apply a pseudo force in the direction opposite to the rocket's acceleration.

Q: Why is the relative acceleration zero?

Both balls are inside the same rocket, so they both experience the exact same pseudo acceleration of $-2 \text{ m/s}^2$. When you subtract them ($a_A - a_B$), they cancel out perfectly.

Q: Why did we get two different answers (2 and 8)?

The answer 8 seconds assumes the chamber is open (no walls), so the balls can freely pass the boundaries. The answer 2 seconds assumes a closed chamber where Ball A hits the left wall and stops, forcing Ball B to cover the entire distance.

Q: How do we know ball A hits the left wall?

By calculating its maximum displacement to the right. It only goes $0.0225 \text{ m}$ forward before the pseudo acceleration pulls it back. It returns to the starting point (the left wall) in just $0.3 \text{ s}$.

Q: What happens to ball A after it hits the wall?

In the closed chamber assumption, we assume the collision with the wall is inelastic or it just stays pinned against the wall due to the continuous pseudo force pushing it leftward. Thus, its position remains $x=0$ while Ball B approaches.

Question

A rocket is moving in a gravity free space with a constant acceleration of $2 	ext{ ms}^{-2}$ along +x direction (see figure). The length of a chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in +x direction with a speed of $0.3 	ext{ ms}^{-1}$ relative to the rocket. At the same time, another ball is thrown in -x direction with a speed of $0.2 	ext{ ms}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other is

Sigma Percentile · Accepted Answer

### The Illusion of Simplicity: A Tale of Two Frames

Imagine you are standing inside a rocket that is accelerating forward through deep space. According to Newton's laws, if you are in an accelerating frame of reference, you must account for a **pseudo force**. In this problem, the rocket accelerates at $2 	ext{ m/s}^2$ in the $+x$ direction. Therefore, every object inside the rocket experiences a pseudo acceleration of $2 	ext{ m/s}^2$ in the $-x$ direction.

This is the crucial first step. By shifting our perspective to the rocket's frame, we freeze the rocket and only deal with the motion of the two balls under this pseudo acceleration.

### The First Instinct (The Trap)

Let's look at the two balls. Ball A is thrown to the right with $u_A = 0.3 	ext{ m/s}$, and Ball B is thrown to the left with $u_B = -0.2 	ext{ m/s}$. Because both balls are in the same accelerating rocket, they both experience the exact same pseudo acceleration: $a_A = -2 	ext{ m/s}^2$ and $a_B = -2 	ext{ m/s}^2$.

When we calculate the relative acceleration, something beautiful happens:

$$a_{	ext{rel}} = a_A - a_B = -2 - (-2) = 0 	ext{ m/s}^2$$

The relative acceleration is zero! This means the relative motion between the two balls is perfectly uniform. The relative velocity is simply $u_{	ext{rel}} = 0.3 - (-0.2) = 0.5 	ext{ m/s}$.

If the relative velocity is constant, the time to collide should just be the distance divided by the speed:

$$t = \frac{4}{0.5} = 8 	ext{ seconds}$$

This seems like a perfectly elegant solution. But in physics, math without physical intuition can lead you off a cliff.

### The Reality Check

Let's ask a simple question: Where exactly is Ball A after 8 seconds?

Using the second equation of motion in the rocket's frame:

$$x_A = u_A t + \frac{1}{2} a_A t^2$$

$$x_A = 0.3(8) - \frac{1}{2}(2)(8)^2 = 2.4 - 64 = -61.6 	ext{ m}$$

Wait a minute. The chamber is only 4 meters long! If the chamber is a closed room, Ball A cannot possibly be at $-61.6 	ext{ m}$. It would have smashed into the left wall long before 8 seconds passed. The 8-second answer is only valid if the chamber is an open tube with no walls.

### The True Path of Ball A

Let's analyze what actually happens in a closed chamber. Ball A is thrown to the right, but it is being pulled to the left by the pseudo acceleration. It will slow down, stop momentarily, and then reverse direction.

How far forward does it go?

$$v_A^2 = u_A^2 + 2 a_A s \implies 0 = (0.3)^2 - 2(2)s$$

$$s = \frac{0.09}{4} = 0.0225 	ext{ m}$$

Ball A only travels $2.25 	ext{ cm}$ forward before turning back! It hits the left wall at $t = 0.3 	ext{ s}$. For the rest of the time, it is stuck at the left wall ($x = 0$).

### The Final Collision

Since Ball A is pinned against the left wall, Ball B has to travel the entire 4-meter length of the chamber to hit it.

Let's set up the equation for Ball B's displacement:

$$|s_B| = u_B t + \frac{1}{2} |a_B| t^2$$

$$4 = 0.2 t + \frac{1}{2} (2) t^2$$

This simplifies to a quadratic equation:

$$t^2 + 0.2t - 4 = 0$$

Using the quadratic formula:

$$t = \frac{-0.2 + \sqrt{0.04 - 4(1)(-4)}}{2} = \frac{-0.2 + \sqrt{16.04}}{2}$$

$$t \approx \frac{-0.2 + 4.005}{2} \approx 1.9 	ext{ seconds}$$

Rounding to the nearest integer, we get **2 seconds**.

This problem is a masterpiece of JEE Advanced. It tests your ability to apply relative motion, but more importantly, it tests your situational awareness. Always check your boundary conditions!

Visualized Solution

Frame of Reference

Relative Motion Setup

Case 1: Open Chamber

Checking Absolute Displacements

Case 2: Closed Chamber

Collision at Left Wall

Solving for Time

The Sigma Insight: Relative Velocity

The Illusion of Simplicity

The First Instinct (The Trap)

The Reality Check

The True Path of Ball A

The Final Collision

Similar Questions

Comprehension Passage

When and for how long can the particle C fulfil the given condition?

What is speed of the particle C at the instant $t = 5 s$ ?

What is modulus of acceleration of the particle C at the instant $t = 5 s$ ?

At the instant, when the line joining locations of A and B is perpendicular to the line joining locations of B and C, what are the magnitudes of velocities of C relative to A and B respectively?

List I describes four systems, each with two particles A and B in relative motion as shown in figure. List II gives possible magnitudes of their relative velocities (in $ms^{- 1}$ ) at time $t = \frac{π}{3} s$ .

List-I

List-II