A new take on a relative velocity problem

In summary: If simplicity equates to a lack of rigour, then perhaps that's the case.In summary, the two methods of solving this question - relative Velocity and 2 kinematics equations - produce different results. The second method, which substitutes the given values in the 4 equations, gives the right answer.
  • #1
Shreya
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Homework Statement
A Balloon moves up at 25m/s, 5 seconds later a bullet is fired in the same direction. Find the minimum velocity of the bullet such that it pops the balloon.
Relevant Equations
$$v=u+at$$
$$s=ut+\frac {1}{2} at²$$
$$s=\frac{1}{2} (u+v)t$$
$$v²=u²+2as$$
Where:
$$s=25t + 125$$
$$v = 25m/s$$
$$a=10$$
I can solve this question using relative Velocity or using 2 kinematics equations. But a peer of mine, tried to do it in a different way. He substituted the above said values in the 4 equations and solved for u in terms of t and differentiated them and equated to to 0 to find a maxima. He found that for the 1, 3 and 4th equations, there was no value of t for which u was minimum but the 2nd equation gave the right answer.
Here's a graph of the 4 equations.

The question is - why did the second function have a minima while others didn't.
Please be kind to help!
 

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  • #2
Equations don't have minima, functions have minima. Your question is unclear because you are not asking about minima of functions.

I would solve the problem by equating the right-hand sides of the 2nd and 5th equations, putting a=-10 in the 2nd equation. That reflects that we must have a time at which both balloon and bullet are at the same height s.

That new equation is a quadratic in t. The quadratic will only have a solution if its discriminant is non-negative. That discriminant is a function of u. So we need to find the smallest value of u that keeps the discriminant non-negative. Find that value of u, then check the result by inserting it in the quadratic and solving to find t, then calculate the height s at which the bullet and balloon meet.

It's a silly question though, because the bullet won't pop the balloon in that case because when it finally touches the balloon, its relative velocity will be zero. So unless the balloon has a needle-sharp tip, it won't pop it!
 
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  • #3
andrewkirk said:
Equations don't have minima, functions have minima. Your question is unclear because you are not asking about minima of functions.
Sorry, it was wrong paraphrasing. I shall edit it.
What I meant was if we take the maxima of the four functions of u(t), which are:
$$u=10t+25$$
$$u=25+5t+\frac{250}{t}$$
$$u=\frac{25t+250}{t}$$
$$u=\sqrt{500t+3125}$$
andrewkirk said:
It's a silly question though, because the bullet won't pop the balloon in that case because when it finally touches the balloon, its relative velocity will be zero. So unless the balloon has a needle-sharp tip, it won't pop it!
That's true.
 
  • #4
Note that the balloon is at a height of ##h = 125 \ m## and moving upwards at ##25 \ m/s## when the bullet is fired. If we then consider a frame moving upwards at ##25 \ m/s##, then the bullet must have an initial upwards speed of ##u' = \sqrt{2gh}## in this frame in order to reach its maximum height ##h##. The balloon is stationary at height ##h## in this frame.

To get the required initial speed of the bullet in the ground frame we add ##25 \ m/s## to ##u'##.
 
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  • #5
Sorry someone explain me the basics here, if I understand well the balloon does uniform linear motion (the buoyancy force of the balloon eliminates its weight) while the bullet does uniform decelerating motion with deceleration g=10m/s^2?
 
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  • #6
Ok I see now, basically the *intuitive* claim here is that the minimum speed is such that when the bullet reaches the balloon it has the same speed as the balloon. How do we justify this claim more formally?
 
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  • #7
Delta2 said:
Ok I see now, basically the *intuitive* claim here is that the minimum speed is such that when the bullet reaches the balloon it has the same speed as the balloon. How do we justify this claim more formally?
Change frames so that the balloon is stationary?
 
  • #8
PeroK said:
Change frames so that the balloon is stationary?
Hm yes that seems to do the job, so we look for the minimum speed of bullet so that it reaches a height of 125m under the influence of gravity. Hm but something doesn't look 100% right with this, let me think it abit more.
 
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  • #9
Delta2 said:
Hm yes that seems to do the job, so we look for the minimum speed of bullet so that it reaches a height of 125m under the influence of gravity. Hm but something doesn't look 100% right with this, let me think it abit more.
How much speed a bullet needs to have in order to pop a balloon which is moving at a rate of 90 kilometers per hour?
 
  • #10
Lnewqban said:
How much speed a bullet needs to have in order to pop a balloon which is moving at a rate of 90 kilometers per hour?
at least 90km/h but I don't know something doesn't look 100% rigorous with this, I found the approach described in post #2 more rigorous than the approach of post #4.

However post #4 is what we call a "clever shortcut".
 
  • #11
Delta2 said:
I found the approach described in post #2 more rigorous than the approach of post #4.
If simplicity equates to a lack of rigour, then perhaps that's true!
 
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  • #12
PeroK said:
If simplicity equates to a lack of rigour, then perhaps that's true!
Not exactly this, but yeah I too find your method more elegant but for some reason less rigorous.
 
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  • #13
Delta2 said:
90km/h but I don't know something doesn't look 100% rigorous with this,...
Far from rigorous.
I believe that is what the designer of the problem inplied, but who knows what differential speed is needed for that pop to happen?
 
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  • #14
Lnewqban said:
Far from rigorous.
I believe that is what the designer of the problem inplied, but who knows what differential speed is needed for that pop to happen?
The bullet has a razor sharp tip such that balloon popping is ensured with the faintest of possible touches.
 
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  • #15
I think the question just asks for a lower limit of the speed.
But, when can't we take the minima of the 1st, 3rd or 4th function and get the right answer while the minima of the 2nd works?
 
  • #16
Shreya said:
I think the question just asks for a lower limit of the speed.
But, when can't we take the minima of the 1st, 3rd or 4th function and get the right answer while the minima of the 2nd works?
Sorry the 4 functions of u(t) are different and I guess they represent different things, are all represent the velocity of the bullet?
 
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  • #17
Shreya said:
I think the question just asks for a lower limit of the speed.
But, when can't we take the minima of the 1st, 3rd or 4th function and get the right answer while the minima of the 2nd works?
The second equation calculates the time, ##t##, for the bullet to hit the balloon:
$$ut -\frac 1 2gt^2 = (25 \ m/s)t + 125 \ m$$Which reduces to:
$$u = (25 \ m/s)t + \frac 1 2 gt + \frac{ 125 \ m}{t}$$You can look at this as ##u(t)## being function of the time taken, ##t##. And, by differentiating ##u(t)## you can actually find the minimum possible ##u##.

The other equations, I assume, are not calculating the time ##t## to hit the balloon. In any case, the other equations make no sense to me.
 
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  • #18
Shreya said:
I think the question just asks for a lower limit of the speed.
Technically it is a threshold speed. Below this speed, the bullet does not hit the balloon. And, above this speed it does. Technically what happens precisely at the threshold speed is not really the issue.
 
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  • #19
Yes they all represent the initial velocity of the bullet
This is how I got them.
 

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  • #20
The issue is that on finding the minima of u(t) - I get the right answer for only the 2nd equation. The other have no minima, as can be seen from the graph in OP
 
  • #21
Shreya said:
Yes they all represent the initial velocity of the bullet
... but only the second equation takes the motion of the balloon into account.
 
  • #22
Please explain that more @PeroK. I'm stuck there
 
  • #23
Perhaps the problem could have been reworded to ask for the minimum initial speed of the bullet to reach the balloon, the implication being that any speed higher than that would pop it. Setting that aside, I had some difficulty understanding this approach. Presumably, ##t## in the four equations in post #3 is not the usual independent variable but the catch-up time ##t_c## when the bullet reaches the balloon. That's OK.

The four equations in #3 were rearranged from the four equations in #1. What is not OK is the use of all four equations to find the minimum. That's because equations 3 and 4 in post #1 do not bring to the table anything new. They can be readily derived from the first two equations by eliminating one variable or another and all 4 equations are applicable to any problem where the acceleration is constant. In this particular problem substitutions of functional form for ##s## and ##v## and a number for ##a## adapt the equations to this particular problem, nevertheless equations 3 and 4 are still linearly dependent on 1 and 2. All one has to do, given post #3, is solve the first two equations as a system of 2 equations and two unknowns and ignore equations 3 and 4. This, by the way, is the traditional approach.
 
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  • #24
Shreya said:
The issue is that on finding the minima of u(t) - I get the right answer for only the 2nd equation. The other have no minima, as can be seen from the graph in OP
One of your problems here is a sloppiness in notation that many books follow. We shouldn't really use ##t## for both the continuous variable of time and the fixed time of the collision. For this we should use ##t_0## or ##t_1## or ##T##.

An equation like ##v = u + at## is just a general equation for the velocity of the bullet. It carries no specific information about this problem.

The second equation, however, is an equation for ##T## the time to hit the balloon:
$$u = (25 \ m/s)T + \frac 1 2 gT + \frac{ 125 \ m}{T}$$That equation contains all the information about the problem: in particular the initial height and speed of the balloon. We would normally think of ##T## being a function of ##u##: choose an initial speed, ##u##, and (assuming it's large enough) we get some time ##T## of impact.

Turning this round, we can look at choosing an impact time ##T## and, assuming its short enough, we can find the initial speed that gives that impact time.

Then we notice that minimising the speed with respect to possible impact times solves the problem.

Your problem is that you applied some mathematics (function minimisation) without understanding how the function related to your physical problem. And I had to piece it all together for you!
 
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  • #25
kuruman said:
. In this particular problem substitutions of functional form for s and v and a number for a adapt the equations to this particular problem, nevertheless equations 3 and 4 are still linearly dependent on 1 and 2
I appreciate the traditional approach. I actually didn't use all 4 together - I tried to solve in 4 independent ways.
kuruman said:
Presumably, t in the four equations in post #3 is not the usual independent variable but the catch-up time tc when the bullet reaches the balloon
I am sorry I didn't mention that
 
  • #26
PeroK said:
One of your problems here is a sloppiness in notation that many books follow. We shouldn't really use t for both the continuous variable of time and the fixed time of the collision. For this we should use t0 or t1 or T.
I realize that - I shan't do it again.
PeroK said:
That equation contains all the information about the problem: in particular the initial height and speed of the balloon. We would normally think of T being a function of u: choose an initial speed, u, and (assuming it's large enough) we get some time T of impact.

Turning this round, we can look at choosing an impact time T and, assuming its short enough, we can find the initial speed that gives that impact time.

Then we notice that minimising the speed with respect to possible impact times solves the problem.
That makes much more sense now. I am sorry, but could you also explain why the 3rd & 4th equation lacks the essential information. (The question is actually one of many peers'; & I've got to answer them).

I appreciate how you guys can make a problem so much more intuitive.
 
  • #27
Shreya said:
I appreciate the traditional approach. I actually didn't use all 4 together - I tried to solve in 4 independent ways.
Let's analyse your last attempt. From ##v^2 - u^2 = 2as## you got:
$$u^2 = v^2 +2g(h_0 + v_b t)$$But, then you plugged in ##v = 25 \ m/s##, which destroys the minimisation approach - as this ##v## is a specific solution. Instead, you need to use ##v = u -gt## to get:
$$u^2 = u^2 -2ugt + g^2t^2 +2g(h_0 + v_bt)$$Hence:
$$2ugt = g^2t^2 + 2gv_bt + 2gh_0$$And:
$$u = \frac 1 2 gt + v_b + \frac{h_0}{t}$$Which is the same equation we got above. Again, it might be better to use ##T## to indicate the time of impact.
 
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  • #28
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  • #29
I did not contribute much other than trying to ensure balloon popping! But thanks for including me in the "credits" list all the same.!
 
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  • #30
I didnt contribute much either, I think 95% of the thanks should go to Perok, Andrewkirk and Kuruman
 
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  • #31
PeroK said:
One of your problems here is a sloppiness in notation that many books follow. We shouldn't really use ##t## for both the continuous variable of time and the fixed time of the collision. For this we should use ##t_0## or ##t_1## or ##T##.

An equation like ##v = u + at## is just a general equation for the velocity of the bullet. It carries no specific information about this problem.

The second equation, however, is an equation for ##T## the time to hit the balloon:
$$u = (25 \ m/s)T + \frac 1 2 gT + \frac{ 125 \ m}{T}$$That equation contains all the information about the problem: in particular the initial height and speed of the balloon. We would normally think of ##T## being a function of ##u##: choose an initial speed, ##u##, and (assuming it's large enough) we get some time ##T## of impact.

Turning this round, we can look at choosing an impact time ##T## and, assuming its short enough, we can find the initial speed that gives that impact time.

Then we notice that minimising the speed with respect to possible impact times solves the problem.

Your problem is that you applied some mathematics (function minimisation) without understanding how the function related to your physical problem. And I had to piece it all together for you!
But t here(v=u+at) is still the same time taken for the bullet to reach the balloon,right? Also could you mention what information this equation is missing?
 
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  • #32
PeroK said:
Let's analyse your last attempt. From ##v^2 - u^2 = 2as## you got:
$$u^2 = v^2 +2g(h_0 + v_b t)$$But, then you plugged in ##v = 25 \ m/s##, which destroys the minimisation approach - as this ##v## is a specific solution. Instead, you need to use ##v = u -gt## to get:
$$u^2 = u^2 -2ugt + g^2t^2 +2g(h_0 + v_bt)$$Hence:
$$2ugt = g^2t^2 + 2gv_bt + 2gh_0$$And:
$$u = \frac 1 2 gt + v_b + \frac{h_0}{t}$$Which is the same equation we got above. Again, it might be better to use ##T## to indicate the time of impact.
Why does it work only when we keep v as u - gt ?
 
  • #33
Differentiate it said:
But t here(v=u+at) is still the same time taken for the bullet to reach the balloon,right? Also could you mention what information this equation is missing?
That equation applies to the bullet regardless of what the balloon is doing.
 
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  • #34
PeroK said:
That equation applies to the bullet regardless of what the balloon is doing.
Whoops, my bad, that was a silly question 😅
What about the other question though?
 
  • #35
Differentiate it said:
Why does it work only when we keep v as u - gt ?
Because ##v = 25 \ m/s## implies the specific solution and determines ##u## and ##t##. These are then no longer continuous variables. If ##v = 25 m/s##, then ##u = 75 \ m/s## and ##t = 5 \ s## and there is no function to differentiate.
 
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