A rocket is moving in a gravity free space with a constant acceleration of $2\mathrm{ }\mathrm{m}{\mathrm{s}}^{-2}$ 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 $0.3\mathrm{ }\mathrm{m}{\mathrm{s}}^{-1}$ relative to the rocket. At the same time, another ball is thrown in - x direction with a speed of $0.2\mathrm{ }\mathrm{m}{\mathrm{s}}^{-1}$ from its right end relative to the rocket. The time in seconds when the two balls hit each other for the first time is

Solution

Consider motion of two balls with respect to rocket. \(\frac{u^2}{2 a}=\frac{0.3 \times 0.3}{2 \times 2}=\frac{0.09}{4}=0.02 m\) So, collision of two bals will take place very near to left wall. \(\begin{aligned} & \text { for } B S=u t+\frac{1}{2} a t^2 \\ & -4=-0.2 t\left(\frac{1}{2}\right) 2 t^2 \Rightarrow t^2+0.2 t \\ & -4=0 \\ & \Rightarrow f t=\frac{-0.2 \pm \sqrt{0.04+16}}{2}=1.9 \end{aligned}\) nearest integer \(=2 \mathrm{~s}\)

Hence, the correct answer is 2.