Shooting bullets, work & power

  • #1
Taylan
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Homework Statement


When bullets are shot on Superman, they bounce off of his chest. 100 Bullets will be shot on his chest per minute, each with a mass of 9g and v= 1000m/s. Assume that the bullets will bounce off elastically back in x-direction.

a) What is the average power of all of the bullets?

b) What work should superman perform
i) when he stands still?
ii) when he moves towards the shooter with a speed of 20km/h?


Homework Equations


P=W/t
W=Fd/t
W=KE/t
KE=1/2mv^2

The Attempt at a Solution


a) P= [ (1/2mv^2)*100 ] / 60 = 7500watt

For a) I found the total kinetic energy of all the bullets and divided by the total time, 60s, to find the average power.

I found b) confusing
so in i) since he stands without moving ( unaffected by the bullets) I thought the work he is doing must be equal with the work all the bullets are doing and in opposite direction. So that he opposes all that work by the bullets. However as I realized, an object usually does move to do some work. so can he still be doing work standing still there? ( like if a ball hits a wall, I guess the ball is doing work but the wall is not? )

my suspected answer would be W= [ (1/2mv^2)*100 ] = 450,000watt

for ii) then for ii I would assume the work is the sum of work done in i) plus the work done for just moving which is m*v ( v is known but mass of the guy isn't). That would be my idea but seemingly it wouldn't work since I don't even know the guy's mass.

Can you please help me about this question?
 

Answers and Replies

  • #2
Andrew Mason
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Be careful...What is the definition of work? Is Superman doing work when he stands still?

For the secnd part of b, how would you measure the rate at which he does work? (Hint:what force does he apply to the bullets?)

AM
 
  • #3
haruspex
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can he still be doing work standing still there? ( like if a ball hits a wall, I guess the ball is doing work but the wall is not? )
They're both elastic collisions, so think about what happens during a bounce. At the halfway point nothing is moving, so where has all the work gone?
ii I would assume the work is the sum of work done in i) plus the work done for just moving which is m*v
mv is not work, so you can rule that out. And ifhe is moving at constant speed he requires no work to maintain that.
Think about the change the bullets undergo.
 
  • #4
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Since work = Pt, I don't see how it can be calculated without time information. Unless we mean work per second which - in any case - is power ?!
 
  • #5
Andrew Mason
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Since work = Pt, I don't see how it can be calculated without time information. Unless we mean work per second which - in any case - is power ?!
You are not applying the essential definition of work: ##W = \int \vec{F} \cdot d\vec{s}## - the application of a force over a distance. In part b i) does Superman exert a force through any distance? How about in b ii)?

AM
 
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  • #6
CWatters
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If any of this seems confusing consider how much KE the bullets had before and after their elastic collision with superman. Did it change?
 
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  • #7
haruspex
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You are not applying the essential definition of work: ##W = \int \vec{F} \cdot d\vec{s}## - the application of a force over a distance. In part b i) does Superman exert a force through any distance? How about in b ii)?

AM
You are missing the point being made. In bii) you can calculate the (nonzero) power, not any specific quantity of work.
 
  • #8
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If any of this seems confusing consider how much KE the bullets had before and after their elastic collision with superman. Did it change?
I would say it does indeed seem confusing. On the face of it, there is no change in KE yet the bullets are first 'stopped' by Superman and then sent back the way they came with equal and opposite velocity (if he is stationary). Perhaps in a case like this we need to employ a slightly different 'lexicon'. Eg 'What energy absorbance (per minute) capacity will Superman require' ? Where energy 'absorbance' reflects the ability to change the KE to some form of elastic energy and then release it again.
 
  • #9
Andrew Mason
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F = dp/dt. Think of the force exerted by the bullets in bouncing off Superman by the average change in bullet momentum per unit time. That is the force that Superman must exert on the bullets in order to remain stationary. Since Superman is not applying that force over a distance, does he perform work?

In b ii), Superman is now moving forward against the bullets. Is he now doing work? In order to determine the rate at which he is doing work, you must determine the average force he is applying to the bullets. Hint: it is more than the force applied in I) because the bullets are now traveling 20 km/hr faster relative to Superman so they rebound with the same relative speed.

AM
 
  • #10
haruspex
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F = dp/dt. Think of the force exerted by the bullets in bouncing off Superman by the average change in bullet momentum per unit time. That is the force that Superman must exert on the bullets in order to remain stationary. Since Superman is not applying that force over a distance, does he perform work?

In b ii), Superman is now moving forward against the bullets. Is he now doing work? In order to determine the rate at which he is doing work, you must determine the average force he is applying to the bullets. Hint: it is more than the force applied in I) because the bullets are now traveling 20 km/hr faster relative to Superman so they rebound with the same relative speed.

AM
I do not understand why you keep suggesting dealing with forces. Much simpler to consider work done on the bullets.
 
  • #11
Andrew Mason
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I do not understand why you keep suggesting dealing with forces. Much simpler to consider work done on the bullets.
The difficulty that the OP seemed to be having was in determining whether and at what rate work is being done by Superman. Now one can reason that since the bullets rebound with their incident energy, Superman does no work on the bullets. But the question asked whether and at what rate Superman does work. The OP understood that he did no work in i) on the bullets but was not sure that he did no work at all. In that case I find it helpful to go back to first principles.

This problem involves similar concepts used in the kinetic theory of gases. Superman is providing pressure to the 'bullet gas'. But no work is performed unless the pressure (force/area) acts through a volume (distance x area).

AM
 
  • #12
Taylan
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Thank you for all the answers!

So once again:
b)i) I will consider the conservation of momentum. P before collision= P after collision (P=mv)
Since superman doesn't experience any change in momentum after the collisions, I concluded that the bullets will move back in the x-axis with the same velocity. That means their kinetic energy is the same, thus supermen is not doing any work on the bullets (one of the definition of work is "energy you need to put into a substance to make it move" )

mv is not work, so you can rule that out. And ifhe is moving at constant speed he requires no work to maintain that.
Think about the change the bullets undergo.

for b)ii) I still have a bit of confusion here but what I concluded is:
since superman is moving with same speed before and after the collisions (being shot), his momentum is not changing. That is why the bullets should again be moving in the opposite direction with same velocity. So their kinetic energy is not changed. That means no additional energy ( work) is given to the bullets. So superman is again doing no work.

Does that sound good?
 
  • #13
Taylan
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Since work = Pt, I don't see how it can be calculated without time information. Unless we mean work per second which - in any case - is power ?!

but can ı calculate the power if I am asked to do so? like let's say if I am asked to find the work exerted by the guy on the bullets per minute? or do you mean the time of collision must be known?
 
  • #14
jbriggs444
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for b)ii) I still have a bit of confusion here but what I concluded is:
since superman is moving with same speed before and after the collisions (being shot), his momentum is not changing. That is why the bullets should again be moving in the opposite direction with same velocity.
Newton's third law says that if there is no acceleration there is no net force. Not that there is no force.
 
  • #15
Taylan
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Newton's third law says that if there is no acceleration there is no net force. Not that there is no force.

do you mean no net force exerted on the bullets by superman?
 
  • #16
haruspex
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That is why the bullets should again be moving in the opposite direction with same velocity.
How do you conclude that? What do you know about elastic collisions of two bodies that are both moving?
but can ı calculate the power if I am asked to do so?
Yes.
do you mean no net force exerted on the bullets by superman?
Are the bullets accelerating? Remember that velocity and acceleration are vectors; they have direction.
 
  • #17
Taylan
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How do you conclude that? What do you know about elastic collisions of two bodies that are both moving?

Momentum before collision = momentum after collision

since the question states that they bounce off and move towards the opposite direction I concluded they must move in the opposite direction after the collision and the guys mass or velocity is never changing so it must mean the momentum of bullets shouldn't be changing too since their mass aren't changing. So speed must be the same too. But when I calculated using the formula i can see the direction is actually somehow not changing.

m(20km/h *1000/3600) + 9x10^-3kg(-1000m/s) = m(20km/h*1000/3600) + 9x10^-3kg(v)

9x10^3kg(-1000m/s) = 9x10^-3kg(v)
v= -1000m/s

Yes.
is power not equal to work divided by time? how can I calculate the power if I can't calculate the work? Or can I calculate the work?

Are the bullets accelerating? Remember that velocity and acceleration are vectors; they have direction
They are not
 
  • #18
haruspex
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the guys mass or velocity is never changing
That works if you take more of a big picture view, looking at the changes in energy over a longish period of time (compared to the interval between bullets). But if you want to analyse it in terms of forces you need to get into finer detail. Mr S's speed will not actually be constant. Each bullet will necessarily slow him a little, and he will have to do some work to get back to his prior speed.
To pursue this line, introduce an unknown for his mass. It should disappear later.
the momentum of bullets shouldn't be changing
Momentum is a vector! If the direction changes, momentum changes.
how can I calculate the power if I can't calculate the work?
The reason you cannot calculate the work is that you are not given a time period over which to measure it. There is enough information to find the power.
E.g. if you know the radius of a car wheel and how fast it is rotating (without slipping) you can find the speed of the car, but you cannot say how far it went.
They are not
Again, acceleration is a vector. A change in direction of velocity constitutes an acceleration.
 
  • #19
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Let's try quantifying a little. For a perfectly elastic collision of two masses m1 and m2 in which m1 <<m2 , the equal and opposite collision impulse is given by the equation: $$Δp \approx 2m_1Δv$$ We note firstly that since - as stated - the impulse(s) (and hence force(s)) on the colliding bodies are equal and opposite, there is no net force it being assumed (naturally) that the colliding bodies are in contact for an equal time period Δt. Secondly that the collision impulse depends on Δv, their relative velocity. Hence the collision impulse applied to the bullets will increase if Superman moves in a direction opposite to their original direction. They will 'return to sender' at a velocity increased in magnitude by 2 x Superman's velocity.

This leaves us pondering the somewhat improbable physics of Superman absorbing bullets 'elastically'. Note that unless he has powers outside of the laws of physics, he will have to slow down however infinitesimally. Whatever kinetic energy is gained by the bullets must be lost by Superman otherwise the problem does not satisfy the assumptions on conservation of kinetic energy and of momentum for a 'perfectly elastic collision'.
 
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  • #20
Andrew Mason
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Let's try quantifying a little. For a perfectly elastic collision of two masses m1 and m2 in which m1 <<m2 , the equal and opposite collision impulse is given by the equation: $$Δp \approx 2m_1Δv$$ We note firstly that since - as stated - the impulse(s) (and hence force(s)) on the colliding bodies are equal and opposite, there is no net force it being assumed (naturally) that the colliding bodies are in contact for an equal time period Δt. Secondly that the collision impulse depends on Δv, their relative velocity. Hence the collision impulse applied to the bullets will increase if Superman moves in a direction opposite to their original direction. They will 'return to sender' at a velocity increased in magnitude by 2 x Superman's velocity.

This leaves us pondering the somewhat improbable physics of Superman absorbing bullets 'elastically'. Note that unless he has powers outside of the laws of physics, he will have to slow down however infinitesimally.
Well, he is Superman - faster than a speeding bullet and more powerful than a locomotive. He flies through the air without propellers and without jets or rockets. The bullets are obviously made out of something that allows them to deflect without permanently deforming and generating heat (no loss of energy). So I think we are to suspend the laws of physics for Superman and the bullets and just accept they do what the problem says they do - Superman deflects the bullets elastically while maintaining constant speed of 20 km/hr.
Whatever kinetic energy is gained by the bullets must be lost by Superman otherwise the problem does not satisfy the assumptions on conservation of kinetic energy and of momentum for a 'perfectly elastic collision'.
Whatever kinetic energy the bullets gain, Superman provides by doing work. You cannot determine the actual instantaneous forces on Superman. But you can determine the average force by F = Δp/Δt where Δp is the total change in bullet momentum of 100 bullets and Δt is 60 seconds. Keep in mind that in b ii) the change in momentum of the bullets in the lab frame is not from mv to -mv. The speed at which they bounce back off Superman is not the incoming speed of 1000 m/sec.

AM
 
  • #21
jbriggs444
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So I think we are to suspend the laws of physics for Superman and the bullets and just accept that what the problem says they do - Superman deflects the bullets elastically while maintaining constant speed of 20 km/hr.

Edit: On re-reading, I've mistaken the intent. We are to ignore Superman's mode of motive power, not his effect on the bullets -- which are treated as a plain old elastic collisions of small masses with a much larger one.


No can do. Invoking this particular flavor of magic then leaves the problem undefined. The problem is that energy is a frame-variant quantity.

Suppose that we do as you suggest and ignore all energy going into or out of Superman -- he's magic. We adopt the frame of the Earth and specify an elastic collision. That means that the bullets rebound at the same velocity they had at impact. Now shift to a frame in which Superman is at rest. An elastic collision now means that the bullets rebound at the same [superman-relative] velocity they had at impact. The two predictions for rebound velocity are different. That destroys the problem. And that is why such an interpretation of Superman cannot be permitted.

Suppose instead that you make superman very massive. He rebounds slightly from each bullet and powers his way forward through the hail of gunfire using whatever strength is required. That is a workable problem.


 
  • #22
Andrew Mason
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Just another way to look at this. Think of the bullets traveling due west on the equator and hitting a wall and bouncing back elastically. Then think of the wall actually moving from west to east at over 1600 km/hr or 465 m/sec with the rotating earth. What is the change in momentum of the bullets from the inertial frame of an observer in the Earth frame who is not rotating with the earth?

AM
 
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  • #23
haruspex
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I think we are to suspend the laws of physics
It is a physics problem. If we ignore the laws of physics there is no solution.
@neilparker62 's method works fine. It matters not whether S does work during each impact to maintain velocity or loses a little each time and then, by whatever means, regains it before the next. The latter is just a convenient method for obtaining the answer.
 
  • #24
Andrew Mason
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It is a physics problem. If we ignore the laws of physics there is no solution.
@neilparker62 's method works fine. It matters not whether S does work during each impact to maintain velocity or loses a little each time and then, by whatever means, regains it before the next. The latter is just a convenient method for obtaining the answer.
I wasn't suggesting that we suspend the laws of physics in solving the problem - just in acceptance of the problem as stated. S must do work during each impact and no work at all between impacts so unless he has infinite mass or is perfectly rigidly connected to something of infinite mass (neither of which are facts here) we just have to pretend that he somehow - ie. magically with his superhuman and supernatural propulsion system - maintains a constant speed throughout.

AM
 
  • #25
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... so unless he has infinite mass or is perfectly rigidly connected to something of infinite mass (neither of which are facts here) we just have to pretend that he somehow - ie. magically with his superhuman and supernatural propulsion system - maintains a constant speed throughout.

AM
The equation for collision impulse: ##Δp \approx 2m_1Δv## rests on the assumption that Superman's mass (or effective mass!) is indeed infinite or close enough to infinity to make the approximation sufficiently accurate. Strictly speaking the equation for collision impulse is ##Δp = 2μΔv## where μ is the reduced mass of the colliding objects: $$ μ=\frac{m_1m_2}{m_1+m_2} $$ It should be clear enough where the afore-mentioned approximation comes from. But the derivation of this formula itself rests on the principles of conservation of kinetic energy and of momentum in an elastic collision. I don't think we can consider a collision 'elastic' if there is an external injection of kinetic energy so I would rather limit the 'magic' to a source of energy which renders Superman's effective mass sufficiently close to infinity such that we can confidently apply the approximate formula. Superman's mass is not mentioned at all in the problem yet in principle it is an essential ingredient of the equation(s) needed to provide a numeric answer to part ii) of the question.
 
  • #26
Andrew Mason
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I don't think we can consider a collision 'elastic' if there is an external injection of kinetic energy so I would rather limit the 'magic' to a source of energy which renders Superman's effective mass sufficiently close to infinity such that we can confidently apply the approximate formula.
The collisions are "elastic" in the sense that they rebound with the same energy as their incident energy in Superman's frame of reference.

Superman's mass is not mentioned at all in the problem yet in principle it is an essential ingredient of the equation(s) needed to provide a numeric answer to part ii) of the question.
Unless we just accept, as the problem states, that he is moving toward the shooter at a constant 20 km/hr while the bullets are hitting him. That should not be problem. Indeed some may find difficult to accept that Superman, who normally travels faster than a speeding bullet, is moving at only 20 km/hr. (If they are kryptonite bullets that would explain it - and perhaps the bullets' perfectly elastic recoil as well).

AM
 
  • #27
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The collisions are "elastic" in the sense that they rebound with the same energy as their incident energy in Superman's frame of reference.

Unless we just accept, as the problem states, that he is moving toward the shooter at a constant 20 km/hr while the bullets are hitting him.

AM

I have no problem accepting that premise. It's very standard in collision problems for both colliding objects to have a velocity and that's why we calculate the relative velocity Δv in the formula ##Δp=2μΔv##. The fact remains that we still require the reduced mass of colliding objects for which we need knowledge of both masses or otherwise an assumption that one of them is (effectively) infinite.
 

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