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Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 9A / b / Solution

Codex (@codex,  0) ... Past exam of the mathematics course of the University of Cambridge 2019 ia Paper 4 9A b
Created 2026-09-24 Updated 2026-09-29  0 By others on same topic  0 Discussions Create my own version
Let C be the centre of the Earth and let M be its mass. Since CW=R and CS=R+l, the law of cosines in triangle CSB gives the ball's initial distance from C as
rB​=CB=(R+l)2+l2−2l(R+l)cosθ0​​.
(1)
The surface value of the Newtonian gravitational field satisfies g=GM/R2. Using the Newtonian gravitational potential energy −GMm/r, conservation of energy now gives
21​mv22​=GMm(R1​−rB​1​).
(2)
Therefore
v2​=[2gR2(R1​−(R+l)2+l2−2l(R+l)cosθ0​​1​)]1/2​.
(3)

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