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Lebombo

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Hi, I'd like to better understand the physics of an object's net force when another object collides with it, such as a pile driver applying a force to drive a pile into the ground.

Questions:

A pile driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. The speed of the driver just as it hits the pile is 100m/s.

If a pile driver strikes a pile into the dirt, a collision occurs. From the instant of impact, if the pile driver and pile remain in contact until the pile is driven into the ground some known distance, then does this mean that the pile must have (or could have) reached the speed of the pile driver at some point at the beginning of impact?

In other words, how does the acceleration of the pile occur? Does the pile instantly begin traveling at a rate of 100m/s and then slow down to 0m/s. Or does the pile accelerate up to 100m/s in some very small fraction of a second, and then slow down to 0m/s. Or does the pile never accelerate up to 100m/s?

Two scenarios.

Scenario 1)

10kg driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. Driver remains in contact with pile. Pile is driven .2 meters into the ground. (All energy is transferred into work)

Xf = ½(a)(t^2) + vt + r = ½(10)(t^2) , therefore time in free fall = 10 seconds

Vf = at + v = (10)(10) + 0 = 100m/s, velocity at time of impact

ΔKE = ½(m)(v^2) – ½(m)(v^2) = ½(10)(100^2) – ½(10)(0) = 50,000J

ΔKE = work = force(distance) => 50,000J = F(.2) => 50,000J/.2m = 250,000NCan you make sure my wording is correct as to what is being applied to what in the scenario:

The gravity is applying a force to the driver.

The driver is building kinetic energy as it falls.

The driver transfers all of its kinetic energy to the pile.

The driver applies force to the pile.

The driver applies a force to the pile over a distance of .2 meters.

The driver applies a force to the pile of 250,000N only at the instant upon impact.Scenario 2)

10 kg driver drops from 500 meters (falls for 10 seconds at 10m/s^2 constant acceleration) and hits pile. Driver bounces off pile, remaining in contact for only .01 seconds. Pile is driven .2 meters into the ground.Xf = ½(a)(t^2) + vt + r = ½(10)(t^2) , therefore time in free fall = 10 seconds

Vf = at + v = (10)(10) + 0 = 100m/s, velocity at time of impact

ΔKE = ½(m)(v^2) – ½(m)(v^2) = ½(10)(100^2) – ½(10)(0) = 50,000JI'm a bit curious as to what is happening here as well..

If a collision is elastic, such as the driver hitting the pile for .01 seconds, then is it at all possible to calculate the force that the driver applies to the pile when the distance that the pile is driven into the ground is given?

In other words, when the driver strikes the pile and transfers all of its energy to the pile, and the pile is driven .2 meters into the ground, will the force applied to the pile be 250,000N regardless of the duration of contact?Thank you.P.S. This is a question of my own. Not homework or any other kind of assigned work. I am not looking for any specific numerical answer, but more of a conceptual understanding. Thank you.