High School Does acceleration affect impact energy vs constant velocity?

[Ray033]

I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question.

Ex. 1 - A moving object impacts a static object at a constant velocity.

Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact.

Assuming the mass of the objects is the same and the velocity at the moment of impact is the same in both examples.

Question - Does the acceleration component in example 2 increase the energy of the impact compared to the constant velocity used in example 1?

Any help in answering this question would be appreciated.

 

[PeroK]

Question - Does the acceleration component in example 2 increase the energy of the impact compared to the constant velocity used in example 1?

Any help in answering this question would be appreciated.

It depends whether the force producing the acceleration is maintained throughout the collision. For example, if you have a vertical pile driver, then the force of gravity continues to act while the object is being pile driven. And, indeed, you have to take this into account in your energy equations.

You might find a case where the accelerating force switches off at the moment of impact - and, in that case, the prior acceleration is irrelevant. In that case, only the speed at impact matters.

 

[Dale]

You might find a case where the accelerating force switches off at the moment of impact - and, in that case, the prior acceleration is irrelevant. In that case, only the speed at impact matters.

Or if the distance moved at the point of application of the external force is small, then the additional energy would be negligible.

 

[Ray033]

I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question.

Ex. 1 - A moving object impacts a static object at a constant velocity.

Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact.

Assuming the mass of the objects is the same and the velocity at the moment of impact is the same in both examples.

Question - Does the acceleration component in example 2 increase the energy of the impact compared to the constant velocity used in example 1?

Any help in answering this question would be appreciated.

I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question.

Ex. 1 - A moving object impacts a static object at a constant velocity.

Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact.

Assuming the mass of the objects is the same and the velocity at the moment of impact is the same in both examples.

Question - Does the acceleration component in example 2 increase the energy of the impact compared to the constant velocity used in example 1?

Any help in answering this question would be appreciated.

It depends whether the force producing the acceleration is maintained throughout the collision. For example, if you have a vertical pile driver, then the force of gravity continues to act while the object is being pile driven. And, indeed, you have to take this into account in your energy equations.

You might find a case where the accelerating force switches off at the moment of impact - and, in that case, the prior acceleration is irrelevant. In that case, only the speed at impact matters.

Thanks for the reply. So, acceleration can

I do not have a good working knowledge of physics yet. I tried to piece this together but after researching this, I couldn’t figure out the correct laws of physics to combine to develop a formula to answer this question.

Ex. 1 - A moving object impacts a static object at a constant velocity.

Ex. 2 - A moving object impacts a static object at the same velocity but is accelerating at the moment of impact.

Assuming the mass of the objects is the same and the velocity at the moment of impact is the same in both examples.

Question - Does the acceleration component in example 2 increase the energy of the impact compared to the constant velocity used in example 1?

Any help in answering this question would be appreciated.

It depends whether the force producing the acceleration is maintained throughout the collision. For example, if you have a vertical pile driver, then the force of gravity continues to act while the object is being pile driven. And, indeed, you have to take this into account in your energy equations.

You might find a case where the accelerating force switches off at the moment of impact - and, in that case, the prior acceleration is irrelevant. In that case, only the speed at impact matters.

Thanks for the reply. So, acceleration can increase the energy at the moment of impact vs constant velocity assuming that the force producing acceleration is maintained at impact. So, would I have to use the formula for kinetic energy and combine it with a formula from Newtons Laws of Motion to incorporate the acceleration into the calculation?

 

[PeroK]

Thanks for the reply. So, acceleration can


Thanks for the reply. So, acceleration can increase the energy at the moment of impact vs constant velocity assuming that the force producing acceleration is maintained at impact. So, would I have to use the formula for kinetic energy and combine it with a formula from Newtons Laws of Motion to incorporate the acceleration into the calculation?

That's a specific calculation for each problem. The impact in a particle collision is usually assumed to be of a sufficiently short time that the external forces are neglected during the collision itself. This allows you to use conservation of momentum and (for an elastic collision) conservation of kinetic energy to study the collision itself. The external accelerating force is applied to the particles before and after the collision, but not during it.

That said, you have to be careful that this is a reasonable assumption. And, you also have to keep an eye on how an external force like gravity may affect the solution to a collision problem.

 

[Ray033]

I really appreciate the reply. At this point, I was trying to test this in its simplest form and figure out how to calculate if acceleration could increase impact energy compared to constant velocity at impact. Just to give you an idea, I've changed my examples by adding basic values so I can try to eliminate any complicated variables when trying to figure out the calculations.

Ex. 1 - A 5 lb steel ball impacts a steel plate at a constant velocity of 1000 mph.

Ex. 2 – A 5 lb steel ball impacts a steel plate at 1000 mph but is accelerating 50 mph/sec at the moment of impact and the force producing the acceleration is maintained at the moment of impact.

Assuming the mass of the objects is the same and the velocity at the moment of impact is the same in both examples.

From what you told me earlier acceleration can increase impact energy compared to a constant velocity impact. Now I have to figure out the calculations to prove it.

I think your help has me pointed in the right direction.

Thanks again.

 

[jbriggs444]

From what you told me earlier acceleration can increase impact energy compared to a constant velocity impact. Now I have to figure out the calculations to prove it.

You have had opinions offered both ways. Both opinions are accurate. It depends on the details. Primarily on the duration of the collision. Or, almost equivalently, on the depth of penetration.

Say that you have this 5 pound steel ball impacting a steel plate at 1000 mph.

That is a bit more than 2 kg of steel at a density of 8 grams per cubic centimeter. I make it about 280 cubic centimeters. The volume of a spherical ball is $V = \frac{4}{3}\pi r^3$. If we solve for $r$ in terms of $V$ we get $r = \sqrt[3]{\frac{3V}{4\pi}} \ = \ \sqrt[3]{\frac{284 \times 3}{4\pi}} \approx 4$ cm radius. About a three inch diameter sphere.

At 1000 mph or about 1500 feet per second that ball can traverse its own diameter 6000 times in one second. If we imagine bringing this to rest with a constant rate of deceleration within its own diameter (so it winds up nestled in its own crater) that would take 1/3000 of a second.

[If it is slowing down steadily to a stop, its average speed during the collision is half of its speed before the collision]

That is a deceleration rate of 1500 feet per second in 1/3000 of a second. Or about 4.5 million feet per second squared.

Meanwhile, you specified an external force producing an acceleration rate of about 73 feet per second squared. Compared to the 4.5 million feet per second squared from the collision, the added force is negligible.

 

[Ray033]

Thanks for explaining all of that to me and showing me the math. It’ll take a little while to absorb it.

With the random values I chose the result you got was negligible. However, if I understand the general principle that you showed me, by increasing only the rate of acceleration, the result would still be an increase in the impact energy even if it’s a very small amount.

When I get this figured out, I’ll try to find values that will result in a more measurable increase in impact energy compared to constant velocity.

Again, thank you for all of the information.

 

[jbriggs444]

With the random values I chose the result you got was negligible. However, if I understand the general principle that you showed me, by increasing only the rate of acceleration, the result would still be an increase in the impact energy even if it’s a very small amount.

Yes.

It is sometimes helpful to consider what happens when one takes the scenario to extremes.

Suppose that we have this same 5 pound ball. But instead of travelling at 1000 miles per hour, it is travelling at one mile per hour. It is being pushed three inches into the steel plate by a hydraulic press under a force of twenty tons.

This time the impact energy is negligible. It is the hydraulic press that contributes essentially all of the energy to the ensuing collision. We can do the calculations for energy.

The kinetic energy of the ball is given by $KE = \frac{1}{2} mv^2$. We want to shift to metric units. English engineering units could be used but we do not want to be dealing with slugs, poundals or missing factors of 32.17. So we have a 2 kg ball moving at about 0.5 meters per second. $KE = \frac{1}{2}mv^2 \approx \frac{1 \times 2 \times {0.5}^2}{2} = 0.25$ Joules.

So about a quarter of a Joule of impact energy.

We are pushing this ball 6 centimeters into a steel plate encountering a hypothetical resistance of twenty tons. [I have no idea if this is the right order of magnitude force for pressing a steel ball into a steel plate]

20 tons (imperial units) is tolerably close to 20 metric tons. Or 200,000 Newtons. Work is force multiplied by distance. Multiply 200,000 Newtons by 6 cm (0.06 meters) and we get about 12,000 Joules.

As expected, this time the impact energy from the ball's pre-existing motion is dwarfed by the work done by the hydraulic press over the duration of the interaction.

 

[Ray033]

This explanation was very helpful. It gave me a completely different way to look at the problem and made it a bit easier to understand overall. I'll be careful with choosing extreme values to test a result thinking that larger numbers will produce larger results or discrepancies for comparison and I'll stick to the metric system for everything, even a simple example from now on.

I'm going to think about the info you gave me for a bit and then start trying to tackle the calculations needed to see what the results are when comparing the 2 different scenarios

I can't thank you enough. The information that you shared with me was great and I appreciate all of the time that you spent on this.