0
$\begingroup$

Question statement:

A rocket is moving in a gravity free space with a constant acceleration of 2ms−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 0.3ms−1 relative to the rocket. At the same time, another ball is thrown in -x direction with a speed of 0.2ms−1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is__________ ?

This is a question from jee advanced. I followed the approach of relative velocity. In the frame of rocket we can write:

  • $s_{rel}$=$u_{rel}t$ since acceleration of both the particles is 0 in this frame . This gives answer as 8seconds which is wrong and it is 1.9 seconds

So my question is why relative motion approach can't be used here - is there some exception to cases where relative motion can be used?

Improve this question
$\endgroup$
1
  • $\begingroup$ The acceleration of both particles is not zero in the frame. If you sit inside the rocket, you will have to apply the pseudo force concept. Then you can use the appropriate equations for constant acceleration. $\endgroup$
    – Polario
    Commented Oct 17, 2020 at 14:33

4 Answers 4

Reset to default
2
$\begingroup$

The problem in this situation is that the left side of the chamber is accelerating and the balls are not. In the fixed system (in which the acceleration of the rocket is measured), the left wall starts at x = 0, picks up the left side ball, and then collides with the right side ball at x = $(v_o)t + (1/2)(2)(t^2) = 4 +[(v_o) - (.2)]t$.

Improve this answer
$\endgroup$
1
$\begingroup$

This one is quite of a famous problem of past iit-jee paper. Actually what happens here is relative motion DIDN'T go wrong. But the arel is not constant. As you may notice in the beginning, in any frame, the relative accn is zero. Now lets bring you, the observer, inside the rocket. You would see, the left ball chases the right one with a constant velocity of approach, what you could name urel . But as they are subject to an accn individually, they (say at time t0 take a U-turn (just to say; they would retrace the same straight lined path). This acceleration is due to pseudo force in the rocket's frame. While retracing the path, the left ball will reach its initial position position, and then collide with the wall with a BANG! and wouldn't move any further. Now only the right ball is moving and hence it is subject to an accn. You see, at this moment, the relative acceleration changes. And that's the part where your mathematics went wrong. I hope you can picture this all. If not, I've posted a picture to make this picture pictured in your brain forever. ROCKETMANNN!!!

(Note: this visualization is made in the rocket's frame. You may use ground frame as per your convenience as explained by R.W. Bird.) also, I leave you with the Maths here ;)

Improve this answer
$\endgroup$
2
  • $\begingroup$ U literally gave me the vibe. $\endgroup$
    – imposter
    Commented Dec 8, 2020 at 20:37
  • $\begingroup$ Glad you liked it 😃 $\endgroup$
    – SteelCubes
    Commented Dec 8, 2020 at 20:40
1
$\begingroup$

You have to look at the parameters in the problem. You can solve it as 2 balls coming together at 0.5 m/s in any frame (internal, non inertial,...) you want, and get 8 seconds. The thing is, the rocket is accelerating at 2 m/s, so it should have changed velocity by 16 m/s, which is way more than the ball's velocities.

Or, you could calculate where the balls meet in the rocket frame ($x=-61.6$ m) and compare that with the 4 meter chamber.

However you solve it, one of the balls hits the rocket before it hits the other ball.

Improve this answer
$\endgroup$
0
$\begingroup$

The rocket's frame isn't an inertial frame.So in that frame,you wouldn’t get the same value(which is constant in lab frame)for $S_{rel}$

Improve this answer
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.