OK, this one comes out of the mechanical conservation of energy section.(adsbygoogle = window.adsbygoogle || []).push({});

A spring loaded cannon is compressed 1.1cm and shot off a the edge of a horizontal table. The marble lands 1.93m from the table. How far would you have to compress the spring to make the marble land 2.20m from the table?

Ok, so I tried doing this with potential energy of a spring = kinetic energy of marble. Now, I can find the inital velocity with v2^2 = v1^2 + 2ax [a=-9.8; v2=0; x=1.93] and get that the 1.1cm compression gave an inital velocity of 6.15m/s. Same method, for a 2.20m distance it would need to give an inital velocity of 6.57m/s. So, Uspring = Ke. Since both have a factor of 1/2: kx^2 = mv^2. Now, I don't know m or k. So, I just wrote k = m[va]^2/[xa]^2, and used numbers for v and x (6.15 and 0.11), yielding k = 312500m . Then I wrote the same thing for the second firing, substituting the "solved" value of k, well, for k. So, m*v^2 = 312500*m*x^2. Now, v^2 = 312500*x^2. 6.57^2/312500 = x^2. x = sqrt(6.57^2/312500) which is .01175 = 1.18cm.

Now, the book says the answer is 1.25 cm. Just to see, I tried this:

[xa]/[xb] = [da]/[db] where da=1.93m, db=2.20m, and xa=0.11m.

Solving for xb, the compression of the spring in the second case, I got 1.25cm.

Now, why did that work? I was kind of hoping, since it was in the conservation of energy section, that I could use that but apparently I did something wrong in there. I also tried having mv^2=kx^2 and just dividing the two, leaving [va]^2/[vb]^2 = [xa]^2/[xb]^2 but that gave the 1.18cm answer again!

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# Homework Help: Spring fired marble distance needed

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