Oxford PAT 2015 example question 16



2 thoughts on “Oxford PAT 2015 example question 16

  1. Hello, for part b) when the catapult is hung vertically in equilibrium, mg=ke -> e=mg/k so the catapult is accelerated upwards through a distance of (x-L- mg/k) in stead of x-L in your equation. Is there anything wrong with my logic ?

    1. Yes you are absolutely right, in that the point of maximum speed is where you say it is.

      The question is a bit ambiguous: do they mean “to what velocity does the catapult now accelerate the ball” when it reaches its maximum velocity, or when it leaves the catapult. I think either answer is a good answer.

      Your answer gives us the point of maximum speed, from which you can get the maximum speed; my answer gives us the speed at the point of release, when the catapult stops pulling the ball upwards. In the final phase (when the ball moves from (x-L- mg/k) to (x-L)) the catapult is pulling the ball upwards, but gravity is pulling the ball downwards harder, so the ball is slowing down.

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