A rocket is moving in a gravity free space with a constant acceleration of  $2m{s}^{-2}$ along $+x$
Direction (see figure). The length of a chamber inside the rocket is 12cm. A ball is thrown from the left end of the chamber in +x direction at same time another ball is thrown -x direction with a speed of  $0.2m{s}^{-1}$ from its right end relative to the rocket. The time in seconds when the balls hit each other is 
 

Assuming open chamber   
           $V_{relative}=0.5m/s$
           $S_{relative}=4m$
Time =  $\frac{4}{0.5}$ = 8 m/s
Alterante  
Assuming closed chamder 
In the frame of  chamber :  
 
Maimum displacement of ball A from its left end is  $\frac{u_A^2}{2a}$ = $\frac{{\left(0.3\right)}^2}{2\left(2\right)}$  =0.0225m
This is negligible with respect to the length of chamber ie. 4 m . So, the collision will be very close to the left end.
Hence, time taken by ball B to reach left end will be  given by
$S=u_{B}t+\frac{1}{2}a{t}^2$
$4=\left(0.2\right)\left(t\right)+\frac{1}{2}\left(2\right){\left(t\right)}^2$
Solving this, We get  $t\approx 2s$