Q.A ball is bouncing elastically with a speed 1 m/s between walls of a railway compartment of size 10 m in a direction perpendicular to walls. The train is moving at a constant velocity of 10 m/s parallel to the direction of motion of the ball. As seen from the ground, (Note: more than one of the given options may be correct.)

Concept and Intuition

When two objects are thrown from the same point at the same time, their relative motion is easiest to analyze by considering components. Stone A moves purely vertically; stone B has both horizontal and vertical components. Crucially, both experience the same gravitational acceleration downward. That means their vertical motions are identical if they start with the same vertical velocity component. Here, A’s vertical velocity is 10 m/s upward; B’s vertical velocity is $10 \sin 30^\circ = 5$ m/s upward — wait, that’s not the same. So they will have different vertical positions. But the problem asks for the separation between them, which is the straight-line distance. We can compute each stone’s position vector after 2 seconds and then find the distance.

Step-by-step solution

1. Set up coordinates

   Let the point of throw be the origin. Take upward as positive $y$ and horizontal (in the direction of B’s throw) as positive $x$. For stone A (vertical throw):

   • Initial velocity: $u_A = 10 \, \text{m/s}$ upward, so $u_{Ax}=0$, $u_{Ay}=10$. For stone B (projectile at $30^\circ$):
   • $u_B = 10 \, \text{m/s}$ at $30^\circ$ above horizontal.
   • Components: $u_{Bx} = 10 \cos 30^\circ = 10 \cdot \frac{\sqrt{3}}{2} = 5\sqrt{3} \, \text{m/s}$, $u_{By} = 10 \sin 30^\circ = 10 \cdot \frac12 = 5 \, \text{m/s}$.

2. Find positions after $t = 2$ s

   Use $y = u_y t - \frac12 g t^2$ and $x = u_x t$ (no horizontal acceleration). Take $g = 10 \, \text{m/s}^2$ for simplicity (common in such problems).

   • Stone A: $x_A = 0$, $y_A = 10 \cdot 2 - \frac12 \cdot 10 \cdot (2)^2 = 20 - 20 = 0$. So A is back at the starting point after 2 seconds.
   • Stone B: $x_B = (5\sqrt{3}) \cdot 2 = 10\sqrt{3} \, \text{m}$, $y_B = 5 \cdot 2 - \frac12 \cdot 10 \cdot 4 = 10 - 20 = -10 \, \text{m}$. …

1. The governing principle

For two vectors $\vec{a}, \vec{b}$ with angle $\theta$ between them,

Since $\cos\theta$ ranges over $[-1, +1]$:

• $\theta = 180^\circ$ (anti-parallel) $\Rightarrow R_{\min} = |a - b|$
• $\theta = 0^\circ$ (parallel) $\Rightarrow R_{\max} = a + b$

and $R$ takes every value in between continuously. So a resultant of 3 is achievable iff $|a-b| \le 3 \le a+b$.

2. Test each pair

Concept & Intuition

When two objects move along a straight line, the relative speed tells us how fast the distance between them changes.

• If they move in the same direction, the relative speed is the difference of their speeds: $|v_1 - v_2|$.
• If they move towards each other (opposite directions), the relative speed is the sum: $v_1 + v_2$.

The problem gives how much the distance changes per second, which is exactly the relative speed. So we set up two simple equations and solve.

Step-by-step solution

1. Define variables

   Let the constant speeds of the two objects be $u$ and $v$ (in $\mathrm{m/s}$), with $u \ge v$ without loss of generality.

2. Same direction case

   When moving in the same direction, the distance increases by $6\ \mathrm{m}$ each second. That means the relative speed is $6\ \mathrm{m/s}$ and equals the difference of their speeds:

3. Towards each other case When moving towards each other, the distance decreases by $8\ \mathrm{m}$ each second. The relative speed is $8\ \mathrm{m/s}$ and equals the sum of their speeds:

4. Solve the system Add equations (1) and (2):

Substitute $u = 7$ into (2):

$7 + v = 8 \implies v = 1$ …

The drone flies due west at $10\ \text{m/s}$ while the train moves due east at $20\ \text{m/s}$ — they travel in opposite directions, so their speeds add:

$v_{rel} = 10 + 20 = 30\ \text{m/s}$ …

The key here is that a collision problem with a given coefficient of restitution $e$ is solved by two equations: one from momentum conservation (always true for an isolated system) and one from the definition of $e$, which relates the relative speed of separation to the relative speed of approach.

For a head-on collision, the coefficient of restitution is defined as:

The relative speed of approach is the speed with which the two bodies come toward each other before collision. The relative speed of separation is the speed with which they move apart after collision. The sign convention matters: we take one direction as positive and stick to it.

Let’s set the initial direction of the $1~\mathrm{kg}$ ball as positive. So:

• Mass $m_1 = 1~\mathrm{kg}$, initial velocity $u_1 = +12~\mathrm{m/s}$
• Mass $m_2 = 2~\mathrm{kg}$, initial velocity $u_2 = -24~\mathrm{m/s}$ (opposite direction)

We need final velocities $v_1$ and $v_2$.

1. Apply conservation of linear momentum

   Total momentum before = total momentum after:

Substitute:

2. Apply the coefficient of restitution

   Relative speed of approach = $u_1 - u_2$ (taking $u_1$ as positive direction, so $u_2$ is negative — this automatically gives the correct magnitude):

Relative speed of separation = $v_2 - v_1$ (since after collision, if both move in the same direction, the faster one is ahead; the formula $v_2 - v_1$ gives the speed of separation regardless of sign — we just need the magnitude relation with $e$).

The definition:

Substitute $e = \frac{2}{3}$ and $u_1 - u_2 = 36$:

$v_2 - v_1 = 24 \quad \text{(Equation 2)}$ …

The key idea is relative velocity. Rain falls vertically downward at 12 m/s. The woman moves horizontally (east to west) at 12 m/s. To her, the rain does not appear to fall straight down — it has an apparent horizontal component opposite to her motion. She must tilt her umbrella into the direction from which the rain appears to come, so that the umbrella’s surface is perpendicular to the relative velocity of the rain.

Think of it this way: if you stand still in vertical rain, you hold the umbrella straight up. If you walk forward, the rain seems to hit you from the front, so you tilt the umbrella forward. Here, the woman is moving west, so the rain appears to come from the west — she tilts the umbrella towards the west.

1. Define the velocities as vectors.

   Let downward be positive $y$ and east be positive $x$.

   Rain velocity: $\vec{v}_r = 0\,\hat{i} - 12\,\hat{j}$ (vertical downward).

   Woman’s velocity (west = negative $x$): $\vec{v}_w = -12\,\hat{i} + 0\,\hat{j}$.

2. Find the relative velocity of rain with respect to the woman.

   The relative velocity is what the woman observes:

So the rain appears to have a horizontal component of 12 m/s towards east (positive $x$) and a vertical component of 12 m/s downward.

3. Interpret the direction. …