What is the velocity of a pendulum when impacted by another mass?

[TheFerruccio]

I'm computing an energy problem whereby a mass is being suspended from a pendulum. I think I have this right, but it would mean that I messed up with an earlier computation.

Two masses will impact, and they both hang on wires of 8.5 feet long. When both masses are hanging down, they are 1 inch apart. When I pull back one of the masses to impact the other, I need to exceed the 1 inch pullback, or the masses will not touch on the upswing. I need to compute the speed of the impact.

So, I pulled back the mass 3 inches, and it will swing through the 2 inch wide "dead zone" and then impact after exceeding 1 inch distance in the other direction. What is the velocity? I said that there is a transfer of potential energy to kinetic energy, and I used pythagorean's theorem to figure out the difference in heights.

Length of string (and hypotenuse) = always 8.5 feet.
Horizontal pullback distance = 7 inches
Energy difference = pullback at 7 inches - pullback at 1 inch

Thus, $v = \sqrt_{\left(\sqrt_{8.5^2-(1/12)^2} - \sqrt_{8.5^2-(7/12)^2}\right)2g}$

This gets me a velocity of approximately 1.12 feet per second. Is this right? Also, for some reason, the tex integration isn't working for me anymore. It just draws a box around what I wrote. I checked it with another tex generator, and it worked just fine.

 

[HomogenousCow]

I can't read the expression, but your logic sounds right, energy conservation n all that.
However, remember that collisions do not always conserve kinetic energy, whether this is the case or not depends on the information given in the problem. (I'm assuming this is school work)




Wait a minute, shouldn't all school work be posted in the other section?

 

[mfb]

It works without underscores:
$v = \sqrt{\left(\sqrt{8.5^2-(1/12)^2} - \sqrt{8.5^2-(7/12)^2}\right)2g}$
Looks fine.

So, I pulled back the mass 3 inches
[...]
Horizontal pullback distance = 7 inches

3? 7?

 

[TheFerruccio]

I can't read the expression, but your logic sounds right, energy conservation n all that.
However, remember that collisions do not always conserve kinetic energy, whether this is the case or not depends on the information given in the problem. (I'm assuming this is school work)




Wait a minute, shouldn't all school work be posted in the other section?

This isn't schoolwork. I'm solving for velocity, not so much the energy absorbed, as that is another problem in itself. I'm calling it an energy problem in the sense that the potential becomes kinetic when you change the height, regardless of the impact, and that other losses are minimal.

I think I just confused the 3 inches vs. 7 inches thing. Just assume it was always 7 inches. Anyway, thanks. I think I figured out the problem. Turns out, since I was doing everything in feet already, I made a mistake and converted from meters to feet, inflating my answer by a good factor of 3 or so. Oops! I redid the experiment with feet in mind, and, luckily, it worked out!