Wednesday 16th September 2020
Today I had a go at solving question h) of BPhO 2018-19 Round 1. The question, as outlined on the photo above, was to find the minimum initial velocity of a round ball of mass m for it to lift the rectangular object with mass M from the ground.
Firstly, I approached the question by representing the tangential velocity of the round ball in terms of the initial velocity u, the vertical displacement h, and gravitational field strength g. Then I deduced the vertical component of the velocity vector by figuring out the gradient of the tangent by differentiating the equation of a circle. After that, my plan was to represent the vertical momentum of the ball in terms of u, g, h, and R (radius of circular path), re-organize it in terms of h, R, and time (t), and differentiate it with respect to time to figure out the vertical net force applied by the ball to the object with mass M. However, after doing so, I realized two problems. Firstly, the expression is too complicated to be rearranged in terms of t, and secondly, I have noticed that I have considered the vertical momentum of the ball exerting a force on M when it is moving from h = 0 to h = R; I should not have done this because during this interval the round ball has no effect in the vertical direction to object with mass M.
Realizing that my approach before was unnecessarily complicated and mostly wrong, I attempted approaching the question in a different way in terms of conservation of energy. Since total energy of the system is constant, the kinetic energy of the ball at the start (very bottom) would be equal to the sum of the kinetic energy of the ball at the very top and the gravitational potential energy of the ball at the same location. This enabled me to represent u (initial velocity) in terms of v (final velocity), g (gravitational field strength), and R (radius of trajectory path of the ball). Then, another insight into the problem for the second equation was realizing that the force exerted by the ball on object M would be equal to the centripetal force exerted by the ball into the path, since at the moment when the ball is at maximum height there is no acceleration to either up or down, meaning that the forces are balanced so upwards force is equal to downwards force which is centripetal force. Therefore the equation: (mv^2)/R (centripetal force) = Mg + mg (total weight of the system) can be established. Rearranging this equation in terms of v then substituting the resulting expression to the previous equation: u = sqrt(v^2 + 4gR) would give the final answer to the question: u = sqrt(Rg*((M+5m)/5))
Then, since my SAT is on the 26th of September, I had a go at solving a full set of reading questions, my weakest area in SAT. Luckily, I got only 4 questions out of 52 wrong, which was quite relieving. I hope I get the same or better score than this in my real SAT about 2 weeks later.
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