Question

Hey there!

having a question which is from ( JEE Advanced 2014 )

$\huge\color{black}\boxed{\colorbox{pink}{Question♡}}$

A rocket is moving in a gravity free space with a constant acceleration of 2 −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 −1 relative to the rocket. At the same time, another ball is thrown in –x direction with a speed of 0.2 −1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is?

proper explanation needed~
thankyou~​

Answer 1

The time taken for the two balls to hit each other is 8 s.

The given parameters:

• Acceleration of the rocket, a = 2 m/s²
• Length of the chamber, s = 4 m
• Speed of the first ball, = V1 = 0.3 m/s
• Speed of the second ball, V2 = 0.2 m/s

The time taken for the two balls to hit each other is calculated by applying relative velocity formula as shown below;

$(V_1 - (-V_2) )t = s\\\\(V_1 + V_2) t = s\\\\(0.3 + 0.2) t = 4\\\\0.5t = 4\\\\t = \frac{4}{0.5} \\\\t = 8 \ s$

Thus, the time taken for the two balls to hit each other is 8 s.

Learn more about relative velocity here: https://brainly.com/question/17228388