A problem on Relative acceleration.

[Satvik Pandey]

Homework Statement

A rocket is moving in a gravity free space with a constant acceleration of 2m/s² along +x direction. The length of chamber inside the rocket is 4m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3m/s relative to rocket. At the same time, another ball is thrown in the -x direction with the speed of 0.2m/s from the right end of the rocket relative to the rocket. Find the tims in seconds when the two balls hit each other.

Homework Equations

The Attempt at a Solution

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According to my book the answer is 2 seconds. I am pretty much convinced by the solution of this question given but I am unable to find mistakes in my approach.

I tried to find $a_{ba}$
$a_{r}=2$ ($a_{r}$ is the acceleration of the rocket)
So $a_{br}=-2$
and $a_{ar}=-2$

Also $a_{br}=-2=a_{b}-a_{r}$...(1) (a_{br} is the acceleration of B wrt rocket)

and $a_{ar}=-2=a_{a}-a_{r}$...(2)

On subtracting eq(2) from eq(1) we get

$a_b-a_a=0$ So $a_{ba} =0$

As the acceleration is constant so

$s_{relative}=u_{relative}t+0.5a_{relative}t^{2}$

Here $u_{relative}=0.5$

As acceleration is zero so I got
$t=\frac{4}{0.5}=8 sec.$

What am I doing wrong?

 

[Orodruin]

Assuming the balls do not hit the back of the rocket as the rocket is accelerating and the balls not, you are correct. The difference in velocities is 0.5 m/s and the gap to close is 4 m.

 

[Satvik Pandey]

Assuming the balls do not hit the back of the rocket as the rocket is accelerating and the balls not, you are correct. The difference in velocities is 0.5 m/s and the gap to close is 4 m.

Thanks for the help.
If we don't make such assumption then the ball A will collide with the back of rocket and during the whole motion it's acceleration relative to rocket will not always remain -2 m/s². As acceleration would not be constant so we can not use the formula mentioned in #post1. Am I correct?

 

[Orodruin]

Correct, but you would also need to know what happens in the bounce (ie, elastic/inelastic).

 

[Satvik Pandey]

Correct, but you would also need to know what happens in the bounce (ie, elastic/inelastic).

But it is not mentioned in the question.
In my book's solution the distance traveled by the ball B is neglected. And the answer is found by finding the time for A to travel 4m relative to the rocket. This question appeared in the engineering entrance of India and one of the option was 8 seconds also.

 

[Orodruin]

Well, with the problem statement as presented, 8 s would be the correct answer in my opinion.

 

[Satvik Pandey]

Well, with the problem statement as presented, 8 s would be the correct answer in my opinion.

But the acceleration is not constant so we can not use the formula $s=ut+0.5t^{2}$. Then,why is it correct?

 

[Orodruin]

Why would acceleration not be constant? The acceleration of the rocket was 2 m/s², right? Anyway, this does not matter at all, just study the problem from the instantaneous rest frame of the rocket at the time when the balls are thrown and you do not need to care one bit about the acceleration.

 

[Satvik Pandey]

Why would acceleration not be constant? The acceleration of the rocket was 2 m/s², right? Anyway, this does not matter at all, just study the problem from the instantaneous rest frame of the rocket at the time when the balls are thrown and you do not need to care one bit about the acceleration.

I am talking about the acceleration of of ball B wrt A. According to given value, the ball A will collide with the back of the rocket before colliding with the ball B. When it will collide with the back of the rocket then it's velocity would be change from -0.3m/s to $+xm/s$ (relative to rocket). As the coefficient of restitution is not given so we can not find the final velocity of the ball after collision. And this would give some +ve acceleration. So, it is clear that the acceleration of the ball A is not constant wrt rocket. So acceleration of B relative A would not be constant either.

 

[Chestermiller]

The ball on the left is going to have to bounce. Otherwise, it is going to go right through the left wall (as will the ball on the right). You probably need to assume that the collisions are elastic, or the problem can't be solved. I solved the problem this way, and got an answer very close to 2 seconds.

Chet

 

[ehild]

The problem is the same when you throw a ball up with 3 m/s velocity from the ground, an an other one with 2 m/s with velocity downward from 4 m height, along the same vertical. At what height they meet? As the gravitational acceleration is the same, it will cancel. Only the relative velocity counts, which is 5 m/s.

Edit: sorry, the velocities are 0.3 m/s and 0.2 m/s, and the relative velocity 0.5 m/s.

 

[Orodruin]

Well, if the ball is thrown inside the rocket, yes. If the collision is assumed completely inelastic (or the ball will bounce several times, it is a mess), then you would have the question when B collides with the back of the rocket. The result 2 s can only be obtained if you compute the time it would take a ball released at B with relative velocity 0 to reach the back of the rocket.

 

[Satvik Pandey]

Thanks for the help. I got that.

The problem is the same when you throw a ball up with 3 m/s velocity from the ground, an an other one with 2 m/s with velocity downward from 4 m height, along the same vertical. At what height they meet? As the gravitational acceleration is the same, it will cancel. Only the relative velocity counts, which is 5 m/s.

I also thought like that. But initially, while solving the problem I didn't notice that A would collide with base before colliding with the ball B.

 

[Orodruin]

But initially, while solving the problem I didn't notice that A would collide with base before colliding with the ball B.

I still think the problem is not very well defined. There is nothing about the coefficient of restitution or anything hinting to whether or not the ball is actually hitting the back (or if there is a hole in the rocket to let it out, etc).

 

[Satvik Pandey]

I still think the problem is not very well defined. There is nothing about the coefficient of restitution or anything hinting to whether or not the ball is actually hitting the back (or if there is a hole in the rocket to let it out, etc).

I too think that. The main problem is that this appeared in one of the most competitive exams of India. Now I wonder how many people have marked the wrong option.

 

[Orodruin]

Now I wonder how many people have marked the wrong option.

Well ... "wrong" would seem to be ambiguous here. Perhaps someone should fire the people responsible for creating the exam - they do not seem overly competitive ...

 

[Chestermiller]

I got ~1.9 seconds after 6 elastic bounces. If one assumes inelastic bounces, the time is not much different from this. This is because the left ball never gets more than about 0.03 m from the left wall, whether or not it collides elastically.

Chet

 

[Orodruin]

I got ~1.97 seconds after 6 elastic bounces. If one assumes inelastic bounces, the time is not much different from this. This is because the left ball never gets more than about 0.03 m from the left wall, whether or not it collides elastically.

Chet

The point is that you can never get 2 s regardless of how the ball bounces. This is the time it takes for the ball from B to reach the end of the rocket if it is thrown with 0 velocity.

 

[ehild]

I didn't notice that A would collide with base before colliding with the ball B.

Neither did I