Question

A rocket is moving in a gravity-free space with a constant acceleration of $2{\text{ms}}^{-2}$ 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 $0.3{\text{ms}}^{-1}$ relative to the rocket. At the same time. Another ball is thrown in $-x$ direction with a speed of $0.2{\text{ms}}^{-1}$ from its right end relative to the rocket. The time in $\text{seconds}$ when the two balls hit each other is (nearest integer)

Answer 1

$\text{Assuming open chamber}$

For ball $\text{A}$,
$u_{A,r}=0.3\text{m/s};a_{A,r}=-2{\text{m/s}}^2$
Using the equation of motion,
$S_{A,r}=u_{A,r}t+\frac{1}{2}{a}_{A,r}{t}^2$
$\Rightarrow x_A=0.3t+\frac{1}{2}(-2)t^2$
$x_A=0.3t-t^2$ ....$\left(\text{i}\right)$

For ball $\text{B}$,
$u_{B,r}=0.2\text{m/s};a_{B,r}=2{\text{m/s}}^2$
Using the equation of motion,
$S_{B,r}=u_{B,r}t+\frac{1}{2}{a}_{B,r}{t}^2$
$\Rightarrow x_B=0.2t+\frac{1}{2}\left(2\right)t^2=0.2t+t^2$ ....$\left(\text{ii}\right)$
and $x_A+x_B=4\text{m}$ ...$\left(\text{iii}\right)$

from Eqn. $\text{(i),(ii) \& (iii)}$
$t=8\text{sec}$

$\text{Assuming closed chamber}$

The maximum displacement of ball $A$ from its left end is
$v_{A,r}^2-u_{A,r}^2=2a{S}_{A,r}$
$⟹0-(0.3{)}^2=2\times (2)\times S_A$
$⟹S_A=0.0225\text{m}$

This is negligible with respect to the length of the chamber i.e. $4\text{m}$. So, the collision will be very close to the left end.

Hence, the time taken by ball $B$ to reach the left end will be given by
$S=u_{B}t+\frac{1}{2}a{t}^2$
$4=0.2\times t+\frac{1}{2}\times 2\times t^2$
$\therefore t=2\text{s}\left(\text{Nearest integer}\right)$