Kinematics Rocket Question

  • #1
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Homework Statement



A rocket is moving in a gravity free space with a constant acceleration of 2 m/s2 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 from its right end relative to the rocket. The time in seconds when the two balls hit each other is ?

Image: http://postimg.org/image/fh9trs9r5/

The answer ranges from 0 to 9.

Homework Equations



http://en.wikipedia.org/wiki/Relative_velocity

The Attempt at a Solution



Here is what I did.

If rocket accelerates by 2 m/s2 in +x direction, then balls inside it also accelerates with it with same value and in same direction in ground frame. Thus when we switch to frame of the rocket there is no acceleration on balls. Inside that frame I impose the concept of relative velocity and convert this two body problem to one body.

Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.

This is the correct answer.

However exam in which this question was asked says that both 8 seconds and 2 seconds answers are acceptable.

One book offers solution as :

Maximum displacement of the left ball from the left wall of the chamber is 2.25 cm, so the right ball has to travel almost the whole length of the chamber (4m) to hit the left ball. So the time taken by the right ball is 1.9 sec (approximately 2 sec).

Now my doubt is as follows :

How is 2 seconds a correct answer ? Also how can there be "two" values of time of collision ?

Please help !!

Thanks in advance... :)
 
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Answers and Replies

  • #2
adjacent
Gold Member
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If rocket accelerates by 2 m/s2 in +x direction, then balls inside it also accelerates with it with same value and in same direction in ground frame.
How do you know that? If the ball is thrown from one end, then why should it accelerate along with the rocket?
 
  • #3
785
15
How do you know that? If the ball is thrown from one end, then why should it accelerate along with the rocket?

Observe from ground frame yourself.

See, if you are sitting in an accelerated train suppose, and you throw a ball vertically, it comes back to your hand right ? This is because it is accelerated with the train.

Waiting for someone else to reply though.
 
  • #4
adjacent
Gold Member
1,553
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See, if you are sitting in an accelerated train suppose, and you throw a ball vertically, it comes back to your hand right ? This is because it is accelerated with the train.
I know that. I actually forgot about the "relative" word in your question
 
  • #5
785
15
I know that. I actually forgot about the "relative" word in your question

No problem.

Just waiting for someone to reply and help me. xD

It's confusing question.. :confused:
 
  • #6
Observe from ground frame yourself.

See, if you are sitting in an accelerated train suppose, and you throw a ball vertically, it comes back to your hand right ? This is because it is accelerated with the train.

Waiting for someone else to reply though.

I think Im going to have to refute that. As long as the ball is in your hand while you are on the accelerated train, then yes it is accelerating, but as soon as you throw it what forces are acting on it? None. The train keeps accelerating due to some force, but the ball being in air is subject only to its initial velocity and g with respect to the ground, no? If the train is going at a constant velocity, then the relative velocity is zero, yet if the trian is accelerating with acceleration [itex] \hat{a} [/itex], then the balls acceleration relative to the train is [itex] -\hat{a}[/itex].

EDIT: Remember: [itex] \hat{a}_{\text{1,2}} = \hat{a}_{\text{1,g}} - \hat{a}_{\text{2,g}} [/itex]
 
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  • #7
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No problem.

Just waiting for someone to reply and help me. xD

It's confusing question.. :confused:
This problem can be solved for either the rocket frame of reference or for the fixed frame of reference. Both approaches will give the same answer. Which frame do you prefer?

Chet
 
  • #8
785
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This problem can be solved for either the rocket frame of reference or for the fixed frame of reference. Both approaches will give the same answer. Which frame do you prefer?

Chet

Hello Chet ! :)

I would like to solve the problem with both approaches. First I will do rocket frame approach. But do you not think I already solved it in my attempt at solution using rocket frame ? :confused:
 
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  • #9
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In the rocket frame of reference, the system behaves as if there is a gravitational acceleration of 2 m/s2 in the negative x-direction. So, for the ball thrown from the left end at x = 0, the ball's location at time t is xl = 0.3t - t2. For the ball thrown from the right end at x = 4, its location at time t is xr=4 -0.2t - t^2. The balls meet when xl=xr.

time left ball right ball

0 0 4
1 -0.7 2.8
2 -3.4 -0.4
3 -8.1 -5.6
4 -14.8 -12.8
etc.

These results make no sense. Are you sure about that acceleration? Could it be 0.2 m/s2? With this acceleration, the left ball falls back to the left wall in a very short time.

Chet
 
  • #10
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The left ball falls back to the left wall every 0.3 seconds, and it never gets further from the wall than 0.0225 m, even if it bounces elastically. So the question really boils down to how long it takes for the right ball to fall to ~ 0.

Chet
 
  • #11
1,540
135
Hi Sankalp

Nice question :)

Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.

This is the correct answer.

This approach is not correct because as a result of collision of the left ball with the left wall of the rocket ,the velocity of the left ball changes after every 0.3 seconds (assuming elastic collision) .

I think 8 seconds cannot be the correct answer .OTOH approx. 2 seconds looks alright.
 
  • #12
ehild
Homework Helper
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Observe from ground frame yourself.

See, if you are sitting in an accelerated train suppose, and you throw a ball vertically, it comes back to your hand right ? This is because it is accelerated with the train.

NO.

It is true if the train travels with constant velocity.

In an accelerating train, the released ball moves backward. Just like you if you stand and do not grab something fixed.

ehild
 
  • #13
ehild
Homework Helper
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Hi Sankalp

Nice question :)



This approach is not correct because as a result of collision of the left ball with the left wall of the rocket ,the velocity of the left ball changes after every 0.3 seconds (assuming elastic collision) .

I think 8 seconds cannot be the correct answer .OTOH approx. 2 seconds looks alright.

Very good!

ehild
 
  • #14
785
15
Thanks to all. :)

I think I figured it out.

From the rocket frame we have acceleration of 2 m/s2 in -x direction on each ball.

So maximum displacement traveled by left ball is u2/2a = 0.025 m.

Note that acceleration is still acting on left ball in -x direction.

Time taken :

0=0.3-2t
t=0.15 s.

In this time:

Distance by right ball :

-s=-0.2t-t2

s= 0.05 m.

Now 0.08 + 0.05 <<4

The right ball has to traverse a distance of 4-x, where x<0.02..

Hence approximating 4-x≈4 we have

-4=-0.2t-t2

This gives t=1.9 s ≈ 2 s...

Thanks to Chet.. His third post explains the reason for this theoretically.

Now why exam accepted 8 s ?

That's my reasoning :

If the balls are "kept on the floor of rocket cabin", then they have the same acceleration as rocket. Hence from rocket frame there is no acceleration on balls. Inside that frame I impose the concept of relative velocity and convert this two body problem to one body.

Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.


So both answers are correct.

Correct ?
 
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  • #16
1,540
135
I haven't been able to follow your reasoning.

Now why exam accepted 8 s ?

That's my reasoning :

If the balls are "kept on the floor of rocket cabin", then they have the same acceleration as rocket. Hence from rocket frame there is no acceleration on balls. Inside that frame I impose the concept of relative velocity and convert this two body problem to one body.

Velocity of left ball relative to right ball = 0.5 m/s in +x direction.

Thus time taken to collide = distance/Velocity of left ball relative to right ball = 4/0.5=8 seconds.


So both answers are correct.

Correct ?

That is still the same situation as in OP .Keeping the balls on the floor cabin do not give the balls same acceleration as that of rocket.

I think there is only one correct answer and that is 2 seconds.

I think I figured it out.

From the rocket frame we have acceleration of 2 m/s2 in -x direction on each ball.

So maximum displacement traveled by left ball is u2/2a = 0.025 m.

Note that acceleration is still acting on left ball in -x direction.

Time taken :

0.025 = 0.3t-t2

This gives imaginary roots of t.. That means .... I approximate 0.025 ≈ 0.02..

0.02 = 0.3t-t2

What is 0.02 on the LHS ?
 
  • #17
785
15
I haven't been able to follow your reasoning.



That is still the same situation as in OP .Keeping the balls on the floor cabin do not give the balls same acceleration as that of rocket.

Why ? When you are sitting in the train, you are accelerated with the train. When you throw the ball up, then that acceleration doesn't count with it.


I think there is only one correct answer and that is 2 seconds.



What is 0.02 on the LHS ?

I can't say that 8 s cannot be the answer.

0.02 is the distance traversed by the left ball when it velocity becomes zero due to deceleration.
 
  • #18
1,540
135
0.02 is the distance traversed by the left ball when it velocity becomes zero due to deceleration.

OK.

Why ? When you are sitting in the train, you are accelerated with the train. When you throw the ball up, then that acceleration doesn't count with it.

Assuming frictionless surface ,balls motion would be independent of the motion of the rocket .

In the train example ,when the person throws the ball up ,the acceleration of the ball will still be 9.8ms-2 . But if you are working from the train reference then surely you need to take into account the pseudo force .

But accounting for pseudo force is not the same thing as if the ball is having horizontal acceleration of the train in addition to its vertical gravitational acceleration.

I can't say that 8 s cannot be the answer.

If 2 sec is the correct answer ,what does 8 seconds signify ?
 
  • #19
785
15
OK.



Assuming frictionless surface ,balls motion would be independent of the motion of the rocket .

In the train example ,when the person throws the ball up ,the acceleration of the ball will still be 9.8ms-2 . But if you are working from the train reference then surely you need to take into account the pseudo force .

But accounting for pseudo force is not the same thing as if the ball is having horizontal acceleration of the train in addition to its vertical gravitational acceleration.



If 2 sec is the correct answer ,what does 8 seconds signify ?

More to the point.

8 s should signify that the right ball is taking approx. 6 seconds to move from left to right and then again left.

It has already taken 2 s to move from right to left initially.

Assume that left ball is fixed at x=0 for the sake of approximation.

Correct ?
 
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  • #20
1,540
135
More to the point.

8 s should signify that the right ball is taking approx. 6 seconds to move from left to right and then again left.

It has already taken 2 s to move from right to left initially.

Assume that left ball is fixed at x=0 for the sake of approximation.

Correct ?

Have you calculated the time it takes for the right ball to move from left to right end ,and left again ?

Does it come out to be 6 sec ?

I am curious to know what makes you so sure that 8 seconds is the correct answer :smile: .The answer key may be wrong .Which book are you referring to ?
 
  • #21
785
15
Have you calculated the time it takes for the right ball to move from left to right end ,and left again ?

Does it come out to be 6 sec ?

I am curious to know what makes you so sure that 8 seconds is the correct answer :smile: .The answer key may be wrong .Which book are you referring to ?

Nope :(

Actually it comes out to be 3 seconds and the final answer as 3+2=5 seconds :(

Now why is 5 s not the answer ?

BTW, this question was asked in IIT-JEE advanced 2014.
 
  • #22
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See there are two possibilities-

i)Collisions between the ball and the rocket don't occur (the rocket is assumed to be very long or whatever)

In this case the answer is found very easily to be 8 sec. (Just see the relative speeds of the balls and divide them by the distance between them)

ii)Collisions occur (seems more reasonable when we look at the figure)

This raises further questions. Specifically, what is the coefficient of restitution for the collision? Since this information isn't given, we cannot proceed further.

So the question is incorrect.
 
  • #23
1,540
135
Nope :(

Actually it comes out to be 3 seconds and the final answer as 3+2=5 seconds :(

Now why is 5 s not the answer ?

2 seconds is the correct answer . Wherever 8 seconds is published ,it is surely done hastily assuming that the left ball doesn't collide with the left wall ,which is an incorrect assumption.

It is implicit that the question asks for the time when the two balls collide for the first time and you agree that it is 2 seconds .Why should you be concerned about the time when they collide for the second or third or nth time ?
 
  • #24
785
15
2 seconds is the correct answer . Wherever 8 seconds is published ,it is surely done hastily assuming that the left ball doesn't collide with the left wall ,which is an incorrect assumption.

Actually it is published by IIT itself at www.jeeadv.iitd.ac.in. See it yourself. :)

It is implicit that the question asks for the time when the two balls collide for the first time and you agree that it is 2 seconds .Why should you be concerned about the time when they collide for the second or third or nth time ?

If they would have said minimum time taken for collision I would have agreed without questioning that it is 2 seconds and nothing more than that.

See there are two possibilities-

i)Collisions between the ball and the rocket don't occur (the rocket is assumed to be very long or whatever)

In this case the answer is found very easily to be 8 sec. (Just see the relative speeds of the balls and divide them by the distance between them)

Nope. It is a 3 body problem. If you see from rocket frame you are bringing rocket "virtually" to rest and this imposes acceleration of 2 m/s2 on each of the ball in -x direction.
You can do nothing more than that. You seem to be thinking that we find relative velocity of one ball w.r.t another. Albeit, that doesn't mean that you ignore acceleration of rocket.
You can't convert this 3 body problem to one body.

ii)Collisions occur (seems more reasonable when we look at the figure)

This raises further questions. Specifically, what is the coefficient of restitution for the collision? Since this information isn't given, we cannot proceed further.

So the question is incorrect.

Well you are right here but everyone will assume perfect elastic collision. I hope you have given this paper.

-----

Awaiting more replies. Please help. :(
 
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  • #25
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Probably the reason they accepted 8 seconds on the exam was that they felt that the students did not have time to reason it out in detail correctly. So they gave them a break. But, as Tanya pointed out, the only real correct answer is ≈ 2 sec.

Chet
 

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