Force of impact

[budullewraagh]

if there is an object with a mass of, say, 80 tons traveling at a rate of 236.9312m/s, hypothetically, and it hypothetically crashes into a wall, would you happen to know, offhand, the force of impact? or whatever it is that knocks the wall down.

[Radagast]

Wouldn't one have to use J = (Mass)(Change in Velocity) ?

 

Then use J = (Force) (Change in Time) ?

 

I can't quite remember, but thats what I thought one would do.

[Geode]

Force is generally given as either a function of mass for a given velocity or mass for a given acceleration, usually expressed as:

 

F=mv

 

or

 

F=ma

 

 

Given the parameters of your question, the former equation would be chosen, F=mv (force=mass times velocity).

 

Using the values, F= 80 tons * 236.9312m/s. The result is expressed in terms of newtons (kilograms/meters/sec).

 

 

Geode

[Radagast]

The result is expressed in terms of newtons (kilograms/meters/sec).

 

Isn't Newtons [ (Kilograms) x (Meters) / Seconds] ?

[swansont]

Force is generally given as either a function of mass for a given velocity or mass for a given acceleration' date=' usually expressed as:

 

F=mv

 

or

 

F=ma

[/quote']

 

mv is momentum, not force.

[mak10]

Isn't Newtons [ (Kilograms) x (Meters) / Seconds] ?

 

F = m x a

Therefore... Newton = Kg m/s^2

[mak10]

if there is an object with a mass of, say, 80 tons traveling at a rate of 236.9312m/s, hypothetically, and it hypothetically crashes into a wall, would you happen to know, offhand, the force of impact? or whatever it is that knocks the wall down.

 

given the values, we can only know the force of impact when the object comes in contact with the wall.... by first determining its momentum and then dividing it by the time it stays in contact with the wall. atleast, thats what i think!

 

-mak10

[YT2095]

wouldn`t it be 80 tons divided by 100g (we`ll call the answer `x`)

 

and then X times the velocity in m/s (in this case nearly 237) and that should give you the Wattage.

 

or something like that, I hate maths

 

either way, the wall is Toast!

[mak10]

why divide 80t by 100g??

[YT2095]

because 100g moving 1m/s = 1 watt (or there abouts).

 

I`m no mathematician, so I break everything down into nice little parcels that I can use

[mak10]

but isn't mass supposed to be in SI units (kg) in order to us it in equations of this sort (P = F X V) ?? And i think its more like 1N application producing a velocity of 1 m/s = 1 W. don't know if its right though...

 

-mak10

[blike]

if there is an object with a mass of, say, 80 tons traveling at a rate of 236.9312m/s, hypothetically, and it hypothetically crashes into a wall, would you happen to know, offhand, the force of impact? or whatever it is that knocks the wall down.

Go with what Mak10 said. You have to know the time period it stays in contact with the wall.

 

Then you can use the momentum divided by the time to find the average force.

 

f=ma

 

a= dv/dt (change in velocity over change in time)

 

thus f = m (v/t)

 

So it varies based on impact time and the change in velocity.

[swansont]

because 100g moving 1m/s = 1 watt (or there abouts).

 

I`m no mathematician' date=' so I break everything down into nice little parcels that I can use [/quote']

 

Wrong units. You can't deduce power solely from knowing a mass and a speed.

[[Tycho?]]

Well you can figure it out using conservation of momentum, but then you'd have to know the mass of the wall. mv=mv, and you can assume the mass comes to a complete stop, then find the velocity of the wall. This equation doesn't look familiar, although I can't think of a reason it wont work for rudamentery purposes.

 

if you want the energy released in such a reaction you can find the kenetic energy. That is E=1/2mv^2. Where m is mass in kg and v is velocity in m/s, and E is energy in Joules.

[swansont]

']Well you can figure it out using conservation of momentum' date=' but then you'd have to know the mass of the wall. mv=mv, and you can assume the mass comes to a complete stop, then find the velocity of the wall. This equation doesn't look familiar, although I can't think of a reason it wont work for rudamentery purposes.

 

if you want the energy released in such a reaction you can find the kenetic energy. That is E=1/2mv^2. Where m is mass in kg and v is velocity in m/s, and E is energy in Joules.[/quote']

 

You'd need to know the coefficient of restitution. The energy dissipated is maximized if the object comes to a stop, and is less if it bounces off.

 

Too many unknowns.

[Ophiolite]

Hmm. I'm an empiricalist. What you do Bud is get yourself a wall, a sensor array and an 80 ton train. ..............

[[Tycho?]]

You'd need to know the coefficient of restitution. The energy dissipated is maximized if the object comes to a stop' date=' and is less if it bounces off.

 

Too many unknowns.[/quote']

 

Yeah I knew it was just a basic example.

 

What is the coefficient of resitution however? I dont recall ever hearing about this.

[jgerlica]

I believe it would be 1,502,714,931 foot-pounds. Of course that is the total energy potential of the mass, not the energy dissipated into the target wall.

[[Tycho?]]

Man, imperial system is rough. Metric is so much easier.

[jgerlica]

I know, we Americans can be a bit As# backwards.

[Geode]

Clarification; newtons= kilogram*meters/seconds

 

 

Geode

[mak10]

once again... the unit of acceleration is meters/seconds^2... and since 1N is equal to the product of 1kg and 1 m/s^2 ..... newton = kg m/s^2.

 

-mak10

[swansont]

'']What is the coefficient of resitution however? I dont recall ever hearing about this.

 

It's a measure of the elasticity (or inelasticity, depending on how you look at it) of an object in a collision.

[cyeokpeng]

If this is a totally elastic collision, since you assume this is hypothetical, then applying the conservsation of momentum, also assuming that no friction is present, then the massive object should bounce back from the wall with exactly the same velocity, unless energy is lost by other means like heat of sound. Think in terms of billiard ball bouncing off the wall if it approach at right angles. Then applying F = d(mv)/dt and since mass is constant,

average impact force = m*dv/dt = m*(vf-vi)/t

Is this correct?

[mak10]

If this is a totally elastic collision, since you assume this is hypothetical, then applying the conservsation of momentum, also assuming that no friction is present, then the massive object should bounce back from the wall with exactly the same velocity, unless energy is lost by other means like heat of sound.

 

energy isn't lost in totally elastic collisions.

 

-mak10