mgh, terminal velocity and impact force

[casrip1@gmx.com]

hey guys, as far as i know the impact force depends on velocity just before impact (its the impulse of the crash right?). http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html this website uses mgh to calculate impact force. so if mgh can be used to calculate impact force, does that mean there is a maximum cap on the potential energy based on terminal velocity?

 

i'm sure i'm mixing up a lot of things here, so if someone here who knows the relation between the 3 concepts (mgh, TV and IF) please explain to me the relationship of these 3 forces?

[EdEarl]

F=ma See: http://en.wikipedia.org/wiki/Newtonian_mechanics#Forces.3B_Newton.27s_second_law

 

The force depends on how quickly the crash occurs. Crashing a steel ball into a thick steel plate creates more force than crashing a sponge (same mass as steel ball) into a steel plate, because the sponge doesn't stop instantly, rather a little at a time. Even steel has some elasticity and does not stop instantly.

Edited June 22, 2013 by EdEarl

[J.C.MacSwell]

hey guys, as far as i know the impact force depends on velocity just before impact (its the impulse of the crash right?). http://hyperphysics.phy-astr.gsu.edu/hbase/flobi.html this website uses mgh to calculate impact force. so if mgh can be used to calculate impact force, does that mean there is a maximum cap on the potential energy based on terminal velocity?

 

i'm sure i'm mixing up a lot of things here, so if someone here who knows the relation between the 3 concepts (mgh, TV and IF) please explain to me the relationship of these 3 forces?

The terminal velocity caps the velocity due to aerodynamic drag, therefore capping the kinetic energy of the falling mass not the potential energy, as the drag reduces the rest of the potential energy to other forms, ultimately to heat.

[EdEarl]

Force can me measured in Newtons: 1 N = 1 kg·m/s²

 

Assume a 1 g mass travels 1 m/s and is stopped in 10^-9s.

 

F=ma = 1g * 1m/s / 10⁹s = 10⁹g·m/s² = 10⁶kg·m/s² = 1,000,000 N (unless I made an arithmetic error)

 

Yes, the mass and velocity of the moving object make an important difference to the force created on impact. But the force can be mediated with crush structures as found in modern automobiles, because they make the crash occur over a longer period of time.

[casrip1@gmx.com]

correct or incorrect:

 

1. |impact force| = |impulse| [(p2-p1)/t] where p = momentum

2. mgh = impact force given terminal velocity is not reached

3. mgh =/= impact force given terminal velocity is achieved midway though the fall (as the potential energy after terminal velocity would be converted into heat instead of velocity)

[EdEarl]

1. is correct except t is the change in time between p1 and p2 and the word impulse is not relevant (does not affect the equation). It is basically the impulse momentum equation (i.e., [math]F\Delta t = \Delta p [/math])

Edited June 23, 2013 by EdEarl

[casrip1@gmx.com]

right, i meant change in t, forgot since in general we used t to represent (delta)t as we always set t1 = 0 (thats how its been throughout the whole high school)

 

also, could you confirm whether 2 and 3 are correct? if they are, then all my confusions have cleared for the moment, if not, well s**t...

[EdEarl]

2. and 3. are statements defining mgh, which is OK, one may define terms. I do not see anything illogical with your statements.

[John Cuthber]

Given the conventional use of the symbols, mgh is the gravitational potential energy and not a force.

The units are wrong so the equation is wrong.

[D H]

2. mgh = impact force given terminal velocity is not reached

 

False. Suppose a falling object is just shy of terminal velocity. Here mgh will yield far too high a value.

 

It's not an either/or situation. Air resistance will always reduce impact velocity (and hence impact force) to some extent. You can safely ignore air resistance only if the impact velocity is orders of magnitude smaller than terminal velocity.

[J.C.MacSwell]

correct or incorrect:

 

1. |impact force| = |impulse| [(p2-p1)/t] where p = momentum

2. mgh = impact force given terminal velocity is not reached

3. mgh =/= impact force given terminal velocity is achieved midway though the fall (as the potential energy after terminal velocity would be converted into heat instead of velocity)

This is true of the PE that remains after terminal velocity is reached. Prior to that, as per DH's post, as soon as any velocity is achieved some of the potential energy starts to be converted to heat, and the remainder accelerates the object to terminal velocity as kinetic energy.