Oh, sure, you’ve seen a watermelon dropped from a balcony onto a trampoline. But what happens when you drop a car from a high tower onto a trampoline? That’s a whole new level of physics fun, and it’s exactly what happens in this video from Mark Rober and the How Ridiculous guys. First they built their own monster trampoline with overlapping sheets of bulletproof kevlar for the pad, supported by a thick steel frame and 144 big old garage door springs.
Then they tested it with a bunch of other things, dropping a whole sack of watermelons, 20 bowling balls and a 66-pound Atlas stone onto a bed of water balloons. The car drop happens near the end of the video, starting at 9:20. Even if you don't think this is awesome (c’mon, it’s empirically proven to be awesome), it's still a great source for some physics problems you can work out at home, while we’re all doing this social distancing thing. I’m going to solve some of these for you—and I’ll pretend I’m doing them as examples.
The truth? I can’t help myself; I just love physics. Can you tell from the video how far the car falls before hitting the trampoline? This is the best question, and I'm going to spoil it by giving you the answer.
So pause here if you want to try it on your own first. Ready? If you know your physics, you realized that to find the distance, all you need to do is measure the free-fall time. Let's start with the basics.
Once an object leaves a person's hand, the only force acting on it is the downward gravitational force. The magnitude of this force is the product of its mass (m) and the gravitational field (g = 9. 8 N/kg). Since the acceleration of an object also depends on the mass, all free falling objects have the same downward acceleration of 9.
8 m/s2. But what’s the connection between fall time and height? I'm going to derive this—and no, I won't just say "Use a kinematic equation. " The definition of acceleration in one dimension is the change in velocity (Δv) divided by a change in time (Δt).
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If I know the elapsed time (I can get that from the video), and I know the acceleration (because this is on Earth), then I can solve for the change in velocity. Note, I'm using negative g for the acceleration, since it's moving downward. In this expression, v1 is the starting velocity of the object, which in this case is zero, and v2 is the final velocity. Now for another definition—average velocity (in one dimension) looks like this, where (Δy) is the change in vertical position: For an object with a constant acceleration (like we have here), the average velocity is just the sum of the initial and final velocity divided by two—it's literally the average of the velocities.
And since the initial velocity is zero, the average velocity is just half of the final velocity. I can use this to find the change in position, i. e. , the distance it falls: Yes, the change in the y position is negative, since the object is moving down.
All that's left is the time. I looked at the part of the video with the dropped watermelons. Some of the shots are in slow motion, but some appear to be in regular time. I can get the fall time from those shots.
You could try to use the time stamp on YouTube to do this, but it's not detailed enough. I like to use the Tracker video analysis tool—it's my go-to for this kind of thing (and it's free). From that, I get a fall time of 2. 749 seconds.
Plugging that into the equation above, I get a fall height of 37. 0 meters (121. 5 feet). Boom, that's one question solved.
If you drop an object from rest (i. e. , zero initial velocity), how fast will it be traveling right before it hits the trampoline? Oh, you thought I was going to answer this question too?
Nope. Actually, this one's not too difficult. You can use the time and the definition of acceleration to find this answer. You can do it.
I believe in you. Let's walk through this whole motion. The car drops. While falling, the gravitational force pulls on it, causing it to speed up, more and more, till it contacts the trampoline.
At this point, the springs on the trampoline stretch and create an upward pushing force on the car. The farther the springs stretch, the greater the upward pushing force. Remember that in order for an object to slow down, there needs to be a net force pushing in the opposite direction as the motion. When the car first hits the trampoline, the backwards pushing force is LESS than gravity, so the net force is still downward, and the car keeps speeding up.
This is something that students tend not to have a good intuition for. Remember, it's the net force that determines acceleration. It’s not until the spring force becomes greater than the downward pushing gravity force that the car starts slowing down. Of course, it's still moving down, so the springs stretch even more, and this increases the spring force.
Eventually the car stops falling and starts moving back up. Now, how can we quantify that? One way to model the force from a spring is with Hooke's law. This says that the spring force (Fs) is proportional to the distance (s) that the spring stretches or compresses.
This proportionality constant is called the spring constant, k. You can think of k as the stiffness of the spring. Actually, we can’t apply this model directly to our trampoline, because it assumes that the springs are in line with the motion of the car. In fact, if the car moves down 10 cm, the springs stretch even more than that, because of the geometry of the situation.
But don't worry, we can just pretend everything is in one dimension, and that will give us an overall effective spring constant. That makes the problem look like this: Now we can find an expression for the spring constant k by using the work-energy principle. This says that the work done on a system is equal to the change in energy in that system. So if we define our system to consist of Earth, the car, and the spring, there are no external interactions in the system and thus no work done.
That means the total energy must be constant—energy is conserved. For this system, there are really just three kinds of energy involved. Here are the equations for these energies along with explanations below: Kinetic energy (K): This is the energy an object has when it’s moving. The kinetic energy depends on both the mass of the object and its velocity.
Gravitational potential energy (Ug): When two objects are gravitationally interacting (like the car and the Earth), there’s a potential energy associated with their position. On Earth’s surface we can approximate this as being proportional to the car's mass and some arbitrary vertical position. (Don't worry about this position; it's only the change in position that really matters. ) Elastic potential energy (Us): Also called spring potential energy.
This depends on both the amount the spring is compressed or stretched and the spring constant. Boom—this is how we will get an expression for the spring stiffness. You know what's so great about using the work-energy principle?
“I can just look at changes from one state to another and ignore all the stuff in between.”
This means I can start with the car at rest (at the top of the drop) and end with the car at the bottom of the spring (again at rest). I don't need to know how fast the car is moving at points in the middle—that just doesn't matter. Putting all of this together, I get the following. Just some notes.
I'm using the 1 subscript for the position and velocity at the top of the drop, and the 3 subscript for the bottom. (Stage 2 is when it hits the spring). At both of these positions, the kinetic energy is zero. That means the change in kinetic energy is also zero.
The change in height (y3 – y1) is just –h (from the diagram above). For the stretch at the beginning of the drop (s1), this is just zero, since the spring hasn't yet been compressed. Now I can use this (along with my notation from the diagram) to solve for the spring constant, k. That’s making some progress.
All we need now is the stretch distance s (how far the trampoline moves down) and the mass of the car. The stretch distance shouldn't be too hard to estimate—it looks to be roughly 1. 5 meters. But what about the mass?
Mark said he adjusted the mass of the car, but he didn't say what the resulting mass was. Oh, I could just ask him? No. Where’s the fun in that?
Try to come up with a good guess for the mass to finish the question. OK, we assumed above that the springs are in line with the car’s motion, but that’s clearly not the case. The cool thing about a trampoline is that the springs stretch a distance that is different than the distance the trampoline moves down. Let's make a very simplified trampoline so we can see what's going on.
This version has a horizontal bar supported by two horizontal springs. When a mass is on top of the bar, it moves down and stretches the springs. Here's a diagram: A few things we need to consider: First, if the trampoline moves down a distance of y, how much does a spring (with an unstretched length of L0) stretch? That's not too hard to figure out from the diagram.
Second, what component of this spring force is in the upward direction? The spring on the left exerts a force pulling up and to the left, while the one on the right pulls up and to the right. If the springs are equal, the horizontal components of these spring forces cancel, and we’re left with just the upward component. But how much that is depends on the angle of the spring with respect to the horizontal (θ in my diagram).
Here's what you can do next: Just pick some values for the spring constant and the unstretched length. Now plot the net vertical spring force as a function of vertical position. Is this plot linear? That's what you would expect for a single Hooke's law spring.
Honestly, I'm not sure what you will get—that's why this is a great homework question. Although I derived an expression for the effective spring constant of the trampoline, I didn’t get a numerical value. If you want to get a rough estimate of this value, you could start with 144 garage door springs. You can estimate the unstretched length (maybe about 75 centimeters).
I'm not sure about the garage door spring constant. They say these are "450-pound" springs, but it's not clear what that means. Just guess. Once you have the effective spring constant (or a force as a function of distance), you can then go back to the previous problem and solve for the mass of the car.
This would be great. Don't cheat and ask Mark. I have no idea what kind of car they dropped. Maybe it’s some Australian model?
But I do know they changed the mass, and my suspicion is that they did so by removing the engine. Doing that might make this stunt easier to pull off—without an engine it might be more likely to fall in a “wheels up” position without rotating. Why do I think that? Because of the center of mass.
An object’s center of mass is the point at which you can pretend there’s a single gravitational force acting on it. Of course the car is made of a bunch of small pieces, and each one interacts gravitationally with the Earth. But it's simpler to treat all of these forces as just one force. And once you have a single force, you need a single location for that force—that's the center of mass.
Most cars have a center of mass that is not in the center. That's because of this very massive auto part called the engine, which shifts the center of mass toward the front. But what if you hang a car from a cable? To keep it from rotating, both the tension force from the cable and the gravitational force have to pass through the same point, so that they don't exert a torque.
That means you can draw a line from the cable extending through the car and it will pass through the center of mass. Here's a shot of that hanging car: If you use three attachment points (as seen in the photo), the car can still rotate a little bit to have the center of mass in line with the main cable, but it won’t swing too much. Now for the homework. Estimate the location of the center of mass and see how much it would move toward the front if you put the engine back in.
Oh, you don't want any more homework questions? Too bad. When the car is falling, my earlier analysis assumed that the only force acting on it was gravity. Is that legit?
Clearly it's not completely true, but it might be OK. As the car falls, it moves through the air. Since it has to push the air out of the way, the air pushes back on the car. This is the essence of the air drag force.
It's a force in the opposite direction from the velocity, and it can usually be modeled with the following equation: In this model, ρ is the density of the air, A is the cross-sectional area, C is a drag coefficient that depends on the shape, and of course v is velocity. If you want to really model the motion of a falling object with air drag, things can get spicy. Since the car will change velocity, and the air drag force depends on the velocity, you can't use simple assumptions like we did before. Really, the best way to solve for the motion of something with air drag is to break it into small time steps and use a numerical calculation.
Here is an example of that. But I'm pretty sure we can ignore the air drag force here. Here’s why: The listed height of the tower is 45 meters. Since the air drag force is in the opposite direction from the gravitational force, a significant air drag would increase the falling time.
Using a longer time (while ignoring the air drag like I did before) would give a calculated tower height greater than 45 meters. I didn't find that, so I don't think the air resistance matters. But you should still model it. Ha!
This should keep you busy for a while. Actually, this isn't a homework question, but it's probably the best part of the video. Here’s what Mark Rober says: "It's this loop of designing something in CAD and then analyzing it to see if it's good enough, and then you test it to check your answers. Using computers to analyze designs allows us to make much more complicated systems than before, when computers weren't as powerful.
" "This idea that we can understand and predict the world around us using math and equations is what first made me fall in love with science, when I took high school physics. " Yes. It's all about models.
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