How hard of a Hit force does it take to make a mass jump?

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  • #1
MIKES279
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I am looking for a formula that I can use to find out how high a steel rod or ball will jump up.
If I hit mass M with force F how high will it jump up. If there is a program that would be great.
every thing will be in pounds
So if ball or rod of 10 lbs is hit on the bottem with 14 lbs of force how high will it go up.
These are NOT THE NUMBERS they just to give you an idea of what I am looking for.
 

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  • #2
phinds
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What research have you done? What have you found so far?
 
  • #3
jbriggs444
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I am looking for a formula that I can use to find out how high a steel rod or ball will jump up.
If I hit mass M with force F how high will it jump up. If there is a program that would be great.
every thing will be in pounds
So if ball or rod of 10 lbs is hit on the bottem with 14 lbs of force how high will it go up.
These are NOT THE NUMBERS they just to give you an idea of what I am looking for.
A single number for force does not tell you much about a collision.

Is that a number for peak force? Or average force? Averaged over what interval? Averaged over time?

For a collision, one would normally want to know the mass of your hammer, the mass of your target, the speed of your hammer and the elasticity of the collision. "Force" would not be an important value. This is just as well, since "Force" is more difficult to predict, measure or control than those other things.
 
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  • #4
Lnewqban
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I am looking for a formula that I can use to find out how high a steel rod or ball will jump up.
If I hit mass M with force F how high will it jump up. If there is a program that would be great.
The direction of the force, the relative velocities, the shape of the coliding bodies, as well as the nature of the surface is important to know.

Hitting the rod sideways with a wood block over a flat wood surface is very different than applying a force vertically with a falling foam block over mud.
 
  • #5
MIKES279
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What research have you done? What have you found so far?
I want to get some numbers before I make a test unit.
 
  • #6
MIKES279
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A single number for force does not tell you much about a collision.

Is that a number for peak force? Or average force? Averaged over what interval? Averaged over time?

For a collision, one would normally want to know the mass of your hammer, the mass of your target, the speed of your hammer and the elasticity of the collision. "Force" would not be an important value. This is just as well, since "Force" is more difficult to predict, measure or control than those other things.
it will be peak time 100th of a sec or so.
 
  • #7
phinds
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What research have you done? What have you found so far?

I want to get some numbers before I make a test unit.
Sounds like you don't understand the difference between research and experiments. I didn't ask anything about experiments.
 
  • #8
MIKES279
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A single number for force does not tell you much about a collision.

Is that a number for peak force? Or average force? Averaged over what interval? Averaged over time?

For a collision, one would normally want to know the mass of your hammer, the mass of your target, the speed of your hammer and the elasticity of the collision. "Force" would not be an important value. This is just as well, since "Force" is more difficult to predict, measure or control than those other things.
I will change all of them to get the high's I am looking for. I want the formula so I can work out what I need and not waste money. That and after all everything needs to be fine tuned.
 
  • #9
MIKES279
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Sounds like you don't understand the difference between research and experiments. I didn't ask anything about experiments.
This part of the research and none of the books I have deal with this. There was a old dos game with a canonball that did this but that is long ago and far away. LOL
 
  • #11
MIKES279
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The direction of the force, the relative velocities, the shape of the coliding bodies, as well as the nature of the surface is important to know.

Hitting the rod sideways with a wood block over a flat wood surface is very different than applying a force vertically with a falling foam block over mud.
Steel on steel and it is on rails so up and down. Force and mass will change how high is what matters.
 
  • #12
Lnewqban
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Steel on steel and it is on rails so up and down. Force and mass will change how high is what matters.
Do you know about how high the velocity of the impact wold be?
 
  • #14
MIKES279
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Do you know about how high the velocity of the impact wold be?
No i will have to change that to make it work 100th of a sec if that helps. It will be fine tuned i realy want the formula to work it out.
 
  • #15
jbriggs444
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No i will have to change that to make it work 100th of a sec if that helps. It will be fine tuned i realy want the formula to work it out.
Where does the 1/100th of a second number come from? It sounds made up.

A steel on steel collision at 1 meter per second lasting 1/100th of a second will have 1/200th of a second of collision and 1/200th of a second of rebound. Assuming constant decelleration, that is an average speed of 1/2 meter per second so about 2.5 millimeters of deformation shared between the two objects.

At 10 meters per second, you'd expect 2.5 centimeters (about one inch) of deformation.

Offhand, that sounds high. But I do not know the scale of your objects.
 
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  • #16
Drakkith
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I am looking for a formula that I can use to find out how high a steel rod or ball will jump up.
If I hit mass M with force F how high will it jump up. If there is a program that would be great.
every thing will be in pounds
I recommend working in terms of mass, momentum, and kinetic energy, not force and mass. Energy is usually much easier to work with than force. Then you can calculate ideal elastic and/or inelastic collisions to quickly get a rough idea of what you'll be working with.

Khan Academy page about elastic and inelastic collisions: Link here.
 
  • #17
Lnewqban
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No i will have to change that to make it work 100th of a sec if that helps. It will be fine tuned i realy want the formula to work it out.
If I understand the situation correctly, we have a steel ball or rod of mass M that is suspended in mid air and between two vertical rails.
That mass M receives a vertical impact from another mass of steel.

That collision produces a force F acting upwards during a very short period of time.
Inmediately after that, the mass M slides directly upwards between the rails until reaching certain high, which you are trying to find a formula for, and then stops and perhaps falls back down.

Is the above assumption correct?
 
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  • #18
Lnewqban
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That situation remains me of this fair game:

 
  • #19
russ_watters
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I was not able to find it and I gave more data now
Find what? Look, it's been explained a bunch of time already that the approach you are trying to take doesn't work. If you want to solve this problem you need to change your approach. We could help you figure this out in minutes if you want. Or you can go additional weeks without getting anywhere. Up to you.

[Edit] And also, since you say you want to actually do this.- you can't. There's no easy way to control "impact force". It's just not a useful thing.
 
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  • #20
MIKES279
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If I understand the situation correctly, we have a steel ball or rod of mass M that is suspended in mid air and between two vertical rails.
That mass M receives a vertical impact from another mass of steel.

That collision produces a force F acting upwards during a very short period of time.
Inmediately after that, the mass M slides directly upwards between the rails until reaching certain high, which you are trying to find a formula for, and then stops and perhaps falls back down.

Is the above assumption correct?
Yes just about it.There will be more later.
 
  • #21
berkeman
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Yes just about it.There will be more later.

1670194817121.png

https://magisterrex.files.wordpress.com/2010/07/blogmousetrap1963.jpg
 
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  • #22
MIKES279
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I am looking for a formula that I can use to find out how high a steel rod or ball will jump up.
If I hit mass M with force F how high will it jump up. If there is a program that would be great.
every thing will be in pounds
So if ball or rod of 10 lbs is hit on the bottem with 14 lbs of force how high will it go up.
These are NOT THE NUMBERS they just to give you an idea of what I am looking for.
Ok the rod at point a is moving up at 2.18 feet ^2 at point a how high will it go till it starts to fall back to point a and how long till it starts to fall.
 
  • #23
Drakkith
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Ok the rod at point a is moving up at 2.18 feet ^2 at point a how high will it go till it starts to fall back to point a and how long till it starts to fall.
What does 2.18 square feet mean? That's an area, not a distance, velocity, or acceleration.
 
  • #24
MIKES279
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What does 2.18 square feet mean? That's an area, not a distance, velocity, or acceleration.
2.18 feet per sec
 
  • #27
Lnewqban
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If that is the initial velocity, this is a projectile problem, considering zero friction among the rails a guides.
 
  • #28
berkeman
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Well thank goodness we got that cleared up! :wink:

I didn't read the whole thread, but are you familiar with the Kinematic Equations of Motion for an object that is under the influence of a constant force (the force of gravity down) with no other influences? If you want to calculate the projectile motion of that object given an initial upward velocity, those are the equations you will use. Check out the Wikipedia article for more information...
 
  • #29
MIKES279
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If that is the initial velocity, this is a projectile problem, considering zero friction among the rails a guides.
I will adjust for that
 
  • #30
jbriggs444
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2.18 feet ^2
Guessing that you meant to write that a block is moving upward at 2.18 feet / sec^2, unimpeded by anything other than gravity.

The easy way to approach this from first principles is to start by computing how long it will take for the object to slow to a stop. That result can be obtained by $$t = \frac{v}{a}$$where ##t## is the time it takes, ##v## is the initial upward velocity and ##a## is the downward acceleration of gravity. In units of feet and seconds, the acceleration of gravity is approximately 32.2 feet per second squared.

Since the acceleration is uniform, the average velocity is the mean of the initial velocity and the final velocity. The final velocity is zero, so this simplifies to$$v_\text{avg}=\frac{v}{2}$$where ##v_\text{avg}## is our sought-after average velocity and ##v## is the given 2.18 feet per second^2.

The upward distance travelled will be the product of the two:$$h = v_\text{avg} t = \frac{v}{a} \frac{v}{2} = \frac{v^2}{2a}$$If we use a calculator we can calculate:

Elapsed time: 67 milliseconds.
Average velocity: 1.09 feet per second
Height gained: 0.737 feet = 0.885 inches
 
  • #31
MIKES279
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Guessing that you meant to write that a block is moving upward at 2.18 feet / sec^2, unimpeded by anything other than gravity.

The easy way to approach this from first principles is to start by computing how long it will take for the object to slow to a stop. That result can be obtained by $$t = \frac{v}{a}$$where ##t## is the time it takes, ##v## is the initial upward velocity and ##a## is the downward acceleration of gravity. In units of feet and seconds, the acceleration of gravity is approximately 32.2 feet per second squared.

Since the acceleration is uniform, the average velocity is the mean of the initial velocity and the final velocity. The final velocity is zero, so this simplifies to$$v_\text{avg}=\frac{v}{2}$$where ##v_\text{avg}## is our sought-after average velocity and ##v## is the given 2.18 feet per second^2.

The upward distance travelled will be the product of the two:$$h = v_\text{avg} t = \frac{v}{a} \frac{v}{2} = \frac{v^2}{2a}$$If we use a calculator we can calculate:

Elapsed time: 67 milliseconds.
Average velocity: 1.09 feet per second
Height gained: 0.737 feet = 0.885 inches
thats not in C or C++?
 
  • #32
jbriggs444
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thats not in C or C++?
Ideally, it would be in "number 2 pencil and a sheet of paper". However, all I have handy is this forum software that uses ##\LaTeX## and a Windows calculator.
 
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  • #33
Drakkith
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thats not in C or C++?
Nothing here needs to be in a computer program. It's all simple formulas that can be done quickly with pen and paper and a calculator. If anything a spreadsheet program like Excel would be just fine if you're looking for some barebones automation. You certainly don't need to go with C or C++ unless you plan on writing a serious program or you're practicing your coding.
 

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