Finding the Velocity of an object due to an imparted force

  • #1
Hello folks, I was wondering how I could calculate/find the velocity of an object from an imparted force given its mass. My force is 26 MN, the object is 10 metric tons, or 10kKg, what will the objects velocity be if this force was applied to it rapidly, e.g. like a jab/punch. My goal is to find kinetic energy, however, the calculator I am using works with Force and Mass to find Acceleration, which tells me the object should have an acceleration of 2600 m/s, but of course, I need velocity, is there any way to work back from this to find velocity?

lets assume the object is a basic cube in shape at rest on a flat surface under regular Earth gravity.
 
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  • #2
Delta2
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You should either use the momentum-impulse theorem of which a simplified version is
$$F\cdot\Delta t=mv_f-mv_i$$
or the work-energy theorem of which a simplified version is
$$F\cdot \Delta s=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2$$.

In the first approach you need to know the duration ##\Delta t## of the applied force ##F##(for example duration of the time that the punch makes contact with the object and transfers momentum via the impulse), while in the second approach you need to know the distance ##\Delta s## that the object moves while the force ##F## is applied (for example a few cm).
In the above equations m is the mass of the object while ##v_f## and ##v_i## are its final and initial velocity respectively.
 
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  • #3
Thank you for the quick response, and straight to the point! :biggrin:

duration of the time that the punch makes contact with the object and transfers momentum via the impulse
I see, would a good estimate for this be in the millisecond range, e.g. 100 milliseconds or maybe even less? A punch does not take very long.

distance Δs that the object moves
I see, like above, can this be estimated? I mean, how far does a 10 ton object move if hit by 26 MN for say 100 milliseconds? I am not sure, maybe as you said, a few cm or a meter or so?

Is the result of acceleration (2600 m/s) from my calculator useful to me to work this out? e.g, is the velocity going to be "at least" that value upon point of impact?
 
  • #4
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I see, would a good estimate for this be in the millisecond range, e.g. 100 milliseconds or maybe even less? A punch does not take very long.
Yes i was about thinking the same, to generalize it abit a duration ##\Delta t## in the range of 50-100ms for a punch.
I see, like above, can this be estimated? I mean, how far does a 10 ton object move if hit by 26 MN for say 100 milliseconds? I am not sure, maybe as you said, a few cm or a meter or so?
If you know acceleration ##a## and duration of contact ##\Delta t##, you can calculate ##\Delta s## as $$\Delta s=v_i\cdot\Delta t+\frac{1}{2}a\Delta t^2$$. This last equation can be also used in "reverse", if you know the acceleration and the distance ##\Delta s## and you want to find the duration of contact ##\Delta t## but its a bit harder now cause one has to solve the respective quadratic equation.
Is the result of acceleration (2600 m/s) from my calculator useful to me to work this out? e.g, is the velocity going to be "at least" that value upon point of impact?
The acceleration is useful for what i said just above. Not sure what exactly you mean by this sentence, the velocity wont be equal to acceleration when the impact starts.
 
  • #5
Thanks, at least I know its possible.

The acceleration is useful for what i said just above. Not sure what exactly you mean by this sentence, the velocity wont be equal to acceleration when the impact starts.
I guess in my head I was thinking surely the object is going to be fastest immediately upon impact? I am curious, why is my calculator saying the acceleration is 2600 m/s, as in, without knowing distance what does that acceleration represent? Surely it will only accelerate at that rate for so long, is that the initial acceleration?

Anyway as a side note, do you mind giving me a basic example in practice of this calculation given my numbers? I don't often do math so some of your abbreviations and symbols like the little dot between the equations confuse me.

If we assume the contact time of the strike was 100 ms, force was 26 MN, weight of object is 10 metric tons (10kKg)what will the equation look like? I guess if I see the actual numbers and where they go, and I can try and work it out myself in the future.
 
  • #6
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The little dot is just multiplication, you are right i should have put that everywhere or omitted from where i put it ...
With the numbers you give and applying impulse-momentum theorem (additional data that the initial velocity ##v_i=0## that is the object is initially at rest) we will have the equation
$$26\cdot 10^6 N \cdot100 \cdot10^{-3}s=10^4Kg\cdot v_f-10^4kg\cdot 0$$ which if we solve for the final velocity ##v_f## we will have $$v_f=\frac{26\cdot 10^5 N\cdot s}{10^4Kg}=260\frac{m}{s}$$
 
  • #7
The little dot is just multiplication, you are right i should have put that everywhere or omitted from where i put it ...
With the numbers you give and applying impulse-momentum theorem (additional data that the initial velocity ##v_i=0## that is the object is initially at rest) we will have the equation
$$26\cdot 10^6 N \cdot100 \cdot10^{-3}s=10^4Kg\cdot v_f-10^4kg\cdot 0$$ which if we solve for the final velocity ##v_f## we will have $$v_f=\frac{26\cdot 10^5 N\cdot s}{10^4Kg}=260\frac{m}{s}$$
Cheers, I think I have a good idea with the equation, if I practice a bit, now I see where the numbers plug in. Also, 260 m/s is the acceleration over 1 second my calculator gave me divided by 10, or rather, 1/10th of a second, which makes sense to me.

If we assume the contact was for 50 ms, instead of 100, the result would be 130 m/s?
 
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  • #8
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I guess in my head I was thinking surely the object is going to be fastest immediately upon impact? I am curious, why is my calculator saying the acceleration is 2600 m/s, as in, without knowing distance what does that acceleration represent? Surely it will only accelerate at that rate for so long, is that the initial acceleration?
No the object wont be fastest immediately upon impact, there has to be a time interval ##\Delta t## during which the object accelerates and its velocity increases from an initial value to a final value. Acceleration represents the rate at which velocity increases (or decreases) as time passes. The units of acceleration are actually ##m/s^2## or to state it differently ##m/s for every second##. So an acceleration of ##2600m/s^2## means that the object increases its velocity by 2600m/s for every 1 second that passes. So if initially it starts from rest , that is initial velocity zero, after 1 second it will increase by 2600 m/s so its velocity will be 2600m/s, after the next 1 second (2seconds total time since impact) it will increase by another 2600m/s so the total velocity will be 5200m/s, after the next 1second(3 seconds total) it will be 5200+2600=7800m/s and so on. Note that the force must be applied for the total duration of the 3seconds in order for its velocity to reach 7800m/s. Just saying this because in your numbers the force is applied by a punch only for 100ms=0.1s.

And now that we have said all the above, which are not wrong, however if we know acceleration and the time duration we just multiply them to get the velocity. In your example we just multiply $$2600\cdot 0.1=260\frac{m}{s}$$ which is the same result as at post #6.
 
  • #10
Yes.
Cheers, I think I have got it. Thank you for the lesson sir! And have a good day :biggrin:
 
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  • #11
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@Physics quest I don't know what you're planning on doing with your answer. But please realize, that if you are going to guess at the duration of impact (your 100 millisec) then the velocity answer is no better than a guess. You can get any answer you want by adjusting the guess. Is 100 msec "reasonable?" Maybe, but maybe not. It depends entirely on the details (what are you hitting your object with, and what is your object made out of?).
 
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  • #12
I don't know what you're planning on doing with your answer
Nothing much really, I just wanted some idea of how to find velocity from a given amount of information.

Is 100 msec "reasonable?" Maybe, but maybe not. It depends entirely on the details
I mean, its important that its just a quick strike, like a punch, a little research (30 mins on google so not much) lead to a paper/pdf that talks about actually measuring a boxers punch and 20 msec was the contact time in that case so that may be more reasonable.

what is your object made out of?
Does it matter? I mean, lets say we have three objects, one iron, one granite and one bronze, surely the result will be same if they have the same mass and are all cubic blocks? Assuming their not destroyed/broken by the impact, but for the sake of argument in this problem their not.


This is a theoretical, not a practical you understand. Nothing directly relating to something I am doing in real life.
 
  • #13
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This is a theoretical, not a practical you understand. Nothing directly relating to something I am doing in real life.
Fair enough.

a little research (30 mins on google so not much) lead to a paper/pdf that talks about actually measuring a boxers punch and 20 msec was the contact time in that case
My force is 26 MN, the object is 10 metric tons
Clash of the Titans?
 
  • #14
Clash of the Titans?
Your along the right track, it is fictional. Superhumans :wink:
 
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  • #15
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In a real-life scenario (or not-too-soft science fiction scenario) the force will strongly depend on the impact duration and the thing that's more controllable is the total momentum.
"Superhuman who can punch something with 26 MN" is both unrealistic and oddly specific where there is no need for it. "Superhuman who can deal 106 kg m/s punches" is still oddly specific but at least a bit more realistic. Note that they'll need a wall or something else on the other side, or they will largely push themselves away from the object - unless they can violate conservation of momentum, but if you break that then why bother with numbers at all?
 
  • #16
ll need a wall or something else on the other side, or they will largely push themselves away from the object
Well its complicated, the force 26.8 MN is the base requirement we estimated for the superhuman in question to indent a relatively large hand sized area of metal, which we think is iron (The area of the hand I calculated to be 80.7 inches square and the Iron hardness MPA at 260) and create a hold, they use this hold to hold onto blocks which may weigh upwards of 40 tons.

The 10 tons is an estimate of a large cage which someone estimated at about 10 tons. Using the force, I wanted to figure out the velocity and kinetic energy the cage would have if hit by the first character.

You are right in that we are discussing a lot of rough elements, as is often the case with superhuman characters not everything adds up. We do make some allowances due to limitations (graphics engine for example in the game).
 
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