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

Step 1: Identify the given parameters.

The initial values relative to the rocket and the chamber length are specified.

$$a = 2\,\text{m/s}^{2}$$ $$L = 4\,\text{m}$$ $$u_1 = 0.3\,\text{m/s}$$ $$u_2 = -0.2\,\text{m/s}$$

Step 2: Determine the relative acceleration of the two balls.

In the accelerating frame of the rocket, both balls experience the same pseudo-acceleration.

$$a_1 = -2\,\text{m/s}^{2}$$ $$a_2 = -2\,\text{m/s}^{2}$$ $$a_{\text{rel}} = a_2 - a_1 = 0$$

Step 3: Calculate the relative initial velocity.

The relative velocity is the difference between the velocities of the two balls.

$$u_{\text{rel}} = u_1 - u_2$$ $$u_{\text{rel}} = 0.3 - (-0.2) = 0.5\,\text{m/s}$$

Step 4: Calculate the time of collision.

Since the relative acceleration is zero, the relative motion is uniform.

$$t = \frac{L}{u_{\text{rel}}}$$ $$t = \frac{4}{0.5}$$ $$t = 8\,\text{s}$$

Step 5: Verify using the ground frame of reference.

Let $V$ be the velocity of the rocket at the instant the balls are thrown.

$$x_1 = x_{10} + (V + 0.3)t$$ $$x_2 = x_{10} + 4 + (V - 0.2)t$$

Set $x_1 = x_2$ to find the collision time.

$$0.3t = 4 - 0.2t$$ $$0.5t = 4$$ $$t = 8\,\text{s}$$

Answer: $8$