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MythBusters and double the speed

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In the last episode of MythBusters, Adam and Jamie wanted to test something that Jamie had said earlier:

“Two cars crashing head on at 50 mph is the same as one car crashing into a wall at 100 mph”

Jamie was wrong, but that is not what I am going to talk about. Instead, I am going to talk about Adam’s small scale test of this situation. Really, it was a nice set up. Basically, he wanted to collide something into a wall at one speed and then double that speed. Then he was going to collide two things together at the lower speed. He had a cool way of measuring the collision. He put a piece of clay between two masses. When the object collided with something, the clay would get smashed and he could measure how smashed it got. Here is a simple diagram of his apparatus.

i-db3c95c8705c707c80d64d2e62340909-2010-05-06_untitled_1.jpg

Where would you have to put the first object so that the second position is twice as fast?

How fast as a function of angle?

Since these objects are not moving in straight lines, they actually don’t have constant accelerations. I know the distance they travel though, so that hints that I should use the work-energy principle. In short, this says that if I look at the mass plus the Earth as my system, there will be no change in total energy. (Gravity won’t do any work because it is an internal force to my system). This means:

i-16335f60d4519f53c8211c9c42a901a6-2010-05-06_la_te_xi_t_1_1.jpg

Going from Position 1 to the bottom (which I will call position 0), I can write this change in energy equation. For simplicity, I will call the bottom of the motion the zero gravitational potential energy. This gives:

i-bd34e6f0544da7fa8d978219206726a1-2010-05-06_la_te_xi_t_1_2.jpg

Notice that the mass cancels – that is a good thing. Also, in order to double the velocity at the bottom, y2 would have to be 4 times higher than y1. Or, if you want to work backwards. Suppose you put the one mass at the horizontal position where position 2 is. What angle would position 1 be at? Here is the expression for the velocity from position 2:

i-7db2e47029365b3b45c3980657458b7c-2010-05-06_la_te_xi_t_1_3.jpg

Now, how high would y1 be so that v0-1 is half of that value?

i-609228aeb9c12200430a5c93b7854f0b-2010-05-06_la_te_xi_t_1_4.jpg

If height is determined by an angle measured from the vertical axis, then (and the radius of the circle is R):

i-ab2d94620b67c82ec186005b3d885477-2010-05-06_la_te_xi_t_1_5.jpg

Ok, I admit. I first looked at the video and thought Adam had put the position 1 at 45 degrees (which wouldn’t even be half the height). I was wrong. Here is a screen shot of the set up.

i-169632fe24e24666dfa1b1af889c2dd8-2010-05-06_vid00924mp4.jpg

See the mark that says “49″? That must be the angle as measured from the top. Hats off to you, MythBusters.


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