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

Answer

8 or 2

Explanation

Assuming open chamber

`V_("relative")=0.5 m//s`
`S_("relative")=4m`
time `=4/0.5=8 m//s`
Alternate
Assuming closed chamber
In the frame of of chamber:

Maximum displacement of ball A from its left end is
`u_(A)^(2)/(2a)=((0.3)^(2))/(2(2))=0.0225 m`
This is negligible with respect to the length of chamber i.e. 4m. 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+1/2 at^(2)`
`4=(0.2)(t)+1/2 (2)(t)^(2)`
Solving this, we get
`t~~2s`