Oxford PAT 2014, Questions 11 and 12

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6 thoughts on “Oxford PAT 2014, Questions 11 and 12

  1. It means that we can assume that the ISS and the object that has detached from it are not moving relative to each other (i.e. when the object was detached it wasn’t shot out of a gun, it just drifted very slowly away).

    1. Hi Harry, thanks for your question.

      To understand the answer, think of a time T immediately after the object is detached from the ISS.

      At T the positions of the ISS and the object are the same, as they have just that instant become detached.

      The question tells us explicitly that the relative velocity of the ISS and the object is negligible after the object is detached, so the velocities are the same at T.

      We can work out the acceleration of the ISS and the object by looking at the forces involved. At T the Earth’s gravity is acting on the ISS with a force proportional to the mass of the ISS, and acting on the object with a force proportional to the mass of the object. There are no other forces involved. Because force is mass times acceleration, and the forces on the ISS and the object are proportional to their respective masses, the accelerations of the ISS and the object are the same at T.

      Since the position, velocity and acceleration are the same for the ISS and the object, they will follow the same path.

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