JEE Advanced · Physics · 4. Motion in 2D
A rocket is moving in a gravity free space with a constant acceleration of 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 relative to the rocket. At the same time, another ball is thrown in - x direction with a speed of from its right end relative to the rocket. The time in seconds when the two balls hit each other for the first time is

- A 1
- B 2
- C 3
- D 4
Answer & Solution
Correct Answer
(B) 2
Step-by-step Solution
Detailed explanation
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 \(=2 \mathrm{~s}\)
\(\frac{u^2}{2 a}=\frac{0.3 \times 0.3}{2 \times 2}=\frac{0.09}{4}=0.02 m\)
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 \(=2 \mathrm{~s}\)
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