Oxford PAT 2011, Question 22

2011_page11

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12 thoughts on “Oxford PAT 2011, Question 22

  1. Hello, I was just wondering why for part c) we no longer need to consider the EPE. Also for part b) are we taking GPE=0 when the catapult is fully extended?
    Many thanks!

    1. Hi Diya, thanks for your questions. Yes (to answer your second question) for part (b) when the catapult is fully extended we take GPE as zero because we’re not really interested in anything that happens below this point. The answer we get from part (b) gives us the KE (and hence velocity) at the point of release. At this point the mass leaves the cup and so it no longer has anything to do with the cup or the elastic — this is the reason why we don’t need to consider EPE (as you asked in your first question).

  2. If you plug in real values for the final equation on PART 2 of the question, you get root a negative number in any situation, except compression. Can someone explain this?

    1. If you plug in real values for the final equation on PART 2 of the question, you get root a negative number in any situation, except compression.

      No you don’t. Remember, x is the total length of the spring when it is pulled back, so the expression inside the square root is positive when x – l > 2gm/k.

  3. Hi,

    It took me a while to understand the solution to the problem. I’ve tried different combinations of events, i.e. if the spring was compressed instead. Would the equation, assuming the natural length is ‘L’ for the spring, and this is taken as the reference point, of E initial = 1/2 K (L-X)^2 + Mg(L-X) and E final = 1/2 MV^2, giving you 1/2MV^2 = 1/2 K(L-X)^2 + Mg(L-X) be correct?

    I think I need some clarification on why GPE is negative in the solution to the problem and does this have anything to do with the conservation of mechanical energy (0 = KE + GPE). If I use the work-energy principle and conservation of mechanical energy for the energy equations of compression, does that make the equation above incorrect since it now becomes 1/2 MV^2 – Mg(L-x)?

    Sorry for the lengthy question! Thanks!

  4. In part 2, the KE equals the elastic potential energy minus the gravitational potential energy. But in the GPE, why is the height given as x-l ? Is it not possible for the mass to exceed this height?

    1. Hi Brook — thanks for your question.

      In part 2, we’re talking about the point where the ball has its maximum velocity. (Writing capitals for X and L to make the text clearer) when the spring is extended to total length X, the ball is at its lowest point, and isn’t moving. The ball reaches its highest velocity at the point where the spring just becomes slack, i.e. when it has length L. So at this point the ball has been raised a total distance of X – L. So the GPE gained at the point of maximum velocity is mg(X – L), and the total KE at this point is the energy that was stored in the spring minus this quantity of GPE gained.

      Is that OK? Add another comment if you think there’s a mistake or you need more clarification.

      1. I think the max velocity occurs when the spring is extended by an amount, y, such that ky = mg. For extensions less than this the weight exceeds the spring force and so the ball is decelerating. This complicates the maths.

        1. Oh yes of course you’re absolutely right, thanks veru much for pointing that out. During the very last bit of the spring contraction, when ky < mg, the mass is decelerating, so the maximum velocity is when ky = mg.

          But looking back over the answer, I think it is still correct for the height that the mass reaches. It’s not correct for the maximum velocity but, as you say, the formula for the max velocity is a bit complicated and I suspect that the question was just asking for the velocity when exiting the catapult (i.e. when the spring went slack). If you regard the point of release as the point where the spring goes slack, and replace “point of max velocity” with “point of release” then I think the explanation is OK.

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