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Impact Forces

40 mph doesn't seem like it's an incredibly high speed, does it? What happens if a vehicle travelling at this speed were to experience a direct, head-on impact, bringing it to rest? A passenger in a vehicle travelling at 40 mph would experience a force of over 300 00 newtons in less than a tenth of a second; that's the…

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Impact Forces

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Jetzt kostenlos anmelden40 mph doesn't seem like it's an incredibly high speed, does it? What happens if a vehicle travelling at this speed were to experience a direct, head-on impact, bringing it to rest? A passenger in a vehicle travelling at 40 mph would experience a force of over 300 00 newtons in less than a tenth of a second; that's the equivalent of the weight of five, fully-grown elephants! So 40 mph is a significant speed, especially if it were suddenly brought to zero. This type of force is called an impact force, and this article will teach you the basics of impact forces.

There are no quantitative (numerical) definitions of when something is classified as an impact, so there is also no quantitative definition of an impact force. Therefore, below is the best we can do.

An object experiences an **impact force** if the object experiences a force as a result of an impact. In practice, this means that an impact force causes a very large acceleration and has a very small duration.

So, if we were to plot the acceleration during an impact against time, the plot would look like a narrow high spike. See the graph below for an example.

If the force-time graph doesn't look like this, the graph does not describe an impact, according to the definition above.

In the event of an impact, there is a collision between two objects that causes these objects to change their velocities in a very short time. This changes their kinetic energy, and this change is caused by work being done on the objects (by each other).

If an object has a kinetic energy of$10\mathrm{J}$and then comes to an abrupt halt due to an impact, then the work done on that object is$10\mathrm{J}$because its kinetic energy after the impact is$0\mathrm{J}$(because its speed is zero).

In all situations, we assume that the acceleration of objects (or, equivalently, the impact force on objects) is always constant during the impact.

Remember that the equation for how work is calculated is multiplying the force by the distance over which the force is applied:$W=Fs$. Thus, to calculate the impact force on an object, we need to know its change in kinetic energy, and the distance over which this change happened. The following formula then gives us the impact force on an object:

$\mathrm{impact}\mathrm{force}=\frac{\mathrm{change}\mathrm{in}\mathrm{kinetic}\mathrm{energy}}{\mathrm{distance}\mathrm{of}\mathrm{impact}}$

Thus, written down using symbols, this formula becomes

$F=\frac{\Delta E}{s}$,

where$F$is the (average) impact force,$\Delta E$is the change in kinetic energy of the object, and$s$is the distance over which the change in kinetic energy happened.

Remember that the kinetic energy of an object with mass*$m$*and speed$v$is given by $\mathrm{kinetic}\mathrm{energy}=\frac{1}{2}\times \mathrm{mass}\times {\mathrm{speed}}^{2}$, or using symbols:$E=\frac{1}{2}m{v}^{2}$.

Note that we can now also calculate the acceleration$a$of an object according to the following formula:

$a=\frac{F}{m}=\frac{\left(\frac{\Delta E}{s}\right)}{m}=\frac{\frac{1}{2}m{\left(\Delta v\right)}^{2}}{sm}=\frac{{\left(\Delta v\right)}^{2}}{2s}$.

We see from this formula that the acceleration only depends on the speed difference and the distance over which this speed difference occurs!

Lastly, we are interested in the duration of the impact (because this should be short), which is equal to the distance over which the impact happens divided by the average speed during the impact. The average speed is half of the initial speed, due to our assumption of constant acceleration. Thus, we can calculate the duration$t$of the impact as follows:

$\mathrm{impact}\mathrm{duration}=\frac{\mathrm{impact}\mathrm{distance}}{\mathrm{average}\mathrm{speed}}=\frac{2\times \mathrm{impact}\mathrm{distance}}{\mathrm{initial}\mathrm{speed}}$.

Thus, in symbols, the calculation looks like this:

$t=\frac{s}{{v}_{\text{average}}}=\frac{2s}{{v}_{\text{initial}}}$.

Armed with the knowledge of these equations, we can now calculate impact forces in certain situations!

Let's say you are in a car going$50\mathrm{mph}$and you collide with an immovable wall. The crumple zone of your car has a length of$0.6\mathrm{m}$and you have a mass of$70\mathrm{kg}$. We can now calculate the impact force you will experience during the impact.

We first name and write down all the quantities we know:$m=70\mathrm{kg}$,$v=50\mathrm{mph}$, and the distance over which the force acts will be the length of our crumple zone, so$s=0.6\mathrm{m}$. We then calculate the change in kinetic energy:

$\Delta E=\frac{1}{2}m{\left(\Delta v\right)}^{2}=\frac{80\text{kg}\xb7{\left(50\text{mph}\right)}^{2}}{2}=40\text{kg}\xb7{\left(22.352\frac{\text{m}}{\text{s}}\right)}^{2}=20\text{kJ}$.

This means that the impact force on you will be

$F=\frac{\Delta E}{s}=\frac{20\text{kJ}}{0.6\text{m}}=33\text{kN}$.

This means that you will experience an acceleration of

$a=\frac{F}{m}=\frac{33\mathrm{kN}}{70\mathrm{kg}}=\frac{33000\frac{\mathrm{m}\overline{)\mathrm{kg}}}{{\mathrm{s}}^{2}}}{70\overline{)\mathrm{kg}}}=471\frac{\mathrm{m}}{{\mathrm{s}}^{2}}$.

This is more than 48 times the gravitational acceleration$g$. We can also calculate how long the impact took, and thus how long the impact force acted on us. The time it took to cover the crumple zone of the car is

$t=\frac{2s}{{v}_{\text{initial}}}=\frac{2\xb70.6\overline{)\text{m}}}{22.4\frac{\overline{)\mathrm{m}}}{\mathrm{s}}}=0.05\text{s}$.

The large acceleration and small duration of the collision certainly make this event a good example of an impact, and the accompanying force a good example of an impact force!

A less dark but slightly more difficult example is the bouncing of a bouncing ball. This ball experiences a sudden change of velocity when it makes contact with the ground or a wall. Let's calculate the impact force in a typical bouncing ball bounce scenario!

In this example, the bouncing ball has a mass of$50\mathrm{g}$, the depth of its squishy bit is$2\mathrm{mm}$, and we drop it from a height of$2\mathrm{m}$. For the sake of this example, the ball bounces back to a height of$2\mathrm{m}$(so no energy is lost) and all of the squishy bit will be squished during the bounce (see the image below for a schematic of the bounce).

The bounce of a bouncing ball with the squishy bit squished, Wikimedia Commons CC BY-SA 4.0.

We first name and write down all the quantities we know:$m=50\mathrm{g}$,$s=2.0\mathrm{mm}$, and$E=mgh=1.0\mathrm{J}$. The potential energy of the bouncing ball will be fully converted to kinetic energy once it hits the ground. However, the bouncing ball will always have the same amount of energy, because it bounces back to the same height, and so$\Delta E$will be zero over the whole bounce! We have to be careful to take into account that its velocity does change, so it will experience an acceleration, and we do this by splitting the problem into two parts. We first consider the first half of the bounce, where it is slowed down to a complete stop in an all-the-way-squished state. We calculate the impact force on the bouncing ball during the first half of the bounce as follows:

$F=\frac{\Delta E}{s}=\frac{1.0\text{J}}{2.0\text{mm}}=500\text{N}$.

This means that the acceleration of the bouncing ball during the first part of the bounce is

$a=\frac{F}{m}=\frac{500\text{N}}{50\text{g}}=\frac{500\frac{\mathrm{k}\overline{)\mathrm{g}}\mathrm{m}}{{\mathrm{s}}^{2}}}{50\overline{)\mathrm{g}}}=10\frac{\text{km}}{{\text{s}}^{2}}$.

The second part of the bounce will look exactly like the first part but reversed. The impact force and acceleration will be the same during the second part as during the first part of the bounce. If we wish to find out the duration of the bounce, we need to calculate the average speed during the bounce, which is half of the speed just before the bounce. This speed is calculated from the kinetic energy${E}_{\mathrm{kin}}$by

${v}_{\text{before}}=\sqrt{\frac{2{E}_{\text{kin}}}{m}}=\sqrt{2gh}=\sqrt{40}\frac{\text{m}}{\text{s}}=6.3\frac{\text{m}}{\text{s}}$.

Given that the total covered distance of the ball during the bounce is twice its squishy bit, we arrive at a bounce duration of

$t=\frac{2s}{{v}_{\text{average}}}=\frac{2\xb72.0\text{mm}}{3.2\frac{\mathrm{m}}{\mathrm{s}}}=1.3\text{ms}$.

- An impact force is a force that causes a very large acceleration and has a very small duration.
- The following formula gives us the impact force on an object:$F=\Delta E/s$, where$F$is the (average) impact force,$\Delta E$is the change in kinetic energy of the object, and$s$is the distance over which the change in kinetic energy happened.
- We can calculate the duration
*t*of an impact as follows:$t=2s/{v}_{\text{initial}}$, where${v}_{\text{initial}}$is the initial speed of the object.

Every force can be measured by a force meter.

*W=Fs*, this decreases the average impact force.

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