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: Relative velocity composition — the ball's bounce direction is parallel to the train's motion, so this is a 1-D superposition, not a 2-D vector sum.

Step 1. Each one-way trip between walls takes $t = \dfrac{10\ \text{m}}{1\ \text{m/s}} = 10\ \text{s}$.

Step 2. In the ground frame, the ball's velocity is the train's velocity plus the ball's velocity relative to the train, and both act along the same line:

Step 3 — Option check.

(A) The velocity is always positive (along $+x$); the direction never reverses as seen from the ground. False.

(B) Speed alternates $11 \to 9 \to 11 \to 9\ \text{m/s}$ every $10\ \text{s}$ (at each bounce). True. …

Setup

The walls are 10 m apart and the ball moves at 1 m/s relative to the train, so each one-way trip takes $t = \dfrac{10}{1} = 10$ s. Elastic bounces off the (uniformly moving) walls only reverse the ball's velocity in the train frame: it stays $\pm 1$ m/s. The train adds $+10$ m/s.

Ground-frame velocity:

Checking each option

(A) Direction changes every 10 s. The velocity is $+11$ m/s, then $+9$ m/s, then $+11$ m/s - always pointing along $+x$. The direction of motion never changes as seen from the ground. (A) is false.

(B) Speed changes every 10 s. Speed alternates $11 \to 9 \to 11 \to 9$ m/s at each bounce, i.e. every 10 s. (B) is true. …

Concept: Ground-Frame Motion via Periodicity and Vertical Graph-Shifting

Method: Shift the Whole $v$-$t$ Graph, Then Use the Ball's Own Periodicity (not a per-bounce velocity computation)

The stored answer computes the two ground-frame speeds ($10\pm1 = 11$ and $9$ m/s) directly and then checks each option against those two numbers. This method instead treats the entire ground-frame $v$-$t$ graph as a vertical shift of the simpler train-frame graph, and answers option (C) — the hardest one — using a periodicity argument that never needs to add up $11\times10 + 9\times10$ by hand.

Setting up the shift

In the train's frame, the ball's motion is the simplest possible periodic motion: it bounces between two fixed walls $10\ \text{m}$ apart at a constant $1\ \text{m/s}$, so its velocity is a square wave alternating $+1, -1, +1, -1,\ldots$ every $10\ \text{s}$ (period $20\ \text{s}$).

Galilean transformation to the ground frame is nothing more than adding the train's velocity, $+10\ \text{m/s}$, as a constant, to every point of this graph — i.e. shifting the whole square wave vertically upward by $10$. A square wave between $-1$ and $+1$, shifted up by $10$, becomes a square wave between $\mathbf{9}$ and $\mathbf{11}\ \text{m/s}$ — read directly off the shifted picture, with no case-by-case bounce arithmetic.

Reading each option off the shifted graph

1. (A) Direction reversal. The shifted graph's lower bound is $9>0$ — it never dips to zero or below, so the ball's ground-frame velocity is always positive; a vertical shift can never make a strictly-positive-after-shifting graph cross zero unless the shift is smaller than the original amplitude, which it isn't here ($10 \gg 1$). (A) is false — direction never reverses as seen from the ground.

2. (B) Speed changes every 10 s. The square wave (shifted or not) still jumps at every bounce, i.e. every $10\ \text{s}$ — a vertical shift changes the graph's height, never where it jumps. (B) is true. …