That boy should have been careful 2

We define the following variables: $v_{ice} = \text{ horizontal velocity of the ice mound with respect to the ground} \\ v_x = \text{ horizontal velocity of the boy with respect to the ground} \\ v_y = \text{ vertical velocity of the boy with respect to the ground} \\ h = \text{ height of the boy above ground}$ and from these, we define the following "helper" variables $v_{x, ice} = \text{ horizontal velocity of the boy with respect to the ice mound } = v_x - v_{ice} \\ \rho = \frac{h}{R}$ Here, I find it pertinent to note that we assume all relativity effects are completely Newtonian. This is a fairly harmless assumption for the numerical value of $a/b$ , but it probably would change $a+b$ (if such a value even exists in Einstein Relativity). Also, we should note that, with these variables, we are actually looking for the value of $\rho$ when the boy loses contact.

We now consider a few equations:

• By conservation of linear momentum, we have $Mv_{ice} + mv_x = 0$
• All interactions are frictionless, so we have a conservation of energy in terms of only potential and kinetic energy of the boy and ice mound system: $mgR = mgh + \frac{1}{2}m\left(v_x^2 + v_y^2\right) + \frac{1}{2}Mv_{ice}^2$
• For $h \in \left[\frac{a}{b}R, R\right]$ , the boy's velocity relative to the ice mound must by tangent to the ice mound. Since the ice mound is circular, this gives us $\frac{v_y}{v_{x, ice}} = -\frac{\sqrt{R^2 - h^2}}{h}$
• Lastly, when $h=\frac{a}{b}R$ , the boy loses contact with the ice mound, so he is no longer applying a force to it, which means at this moment the ice mound reference frame is inertial. Therefore, in this instant we can apply the formulas of centripetal acceleration to the ice mound frame. This gives $\frac{v_{x, ice}^2 + v_y^2}{R} = g\frac{h}{R}$ since at the moment the boy loses contact, the radial component of the force of gravity is exactly equal to the centripetal force.

If we solve those equations for $\rho$ (if people really want to see this, I'll show it below, but it really is just shuffling around symbols), we find the equation $0 = \rho^3 - \left(1+\frac{m}{M}\right)\left(\rho^3 - 3\rho + 2\right)$

Setting $\frac{m}{M} = \frac{25}{7}$ and solving for the only solution $\rho \in [0,1]$ yields $\rho = \frac{4}{5}$ . It follows that $a=4$ , $b=5$ , and $\boxed{a+b = 9}$

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Note: When we let $\frac{m}{M}$ tend to $0$ , then the conservation of linear momentum equation would give a constant $v_{ice}=0$ , as in the problem that inspired this one . You might also note that in this case, the equation I have given for $\rho$ becomes $0 = 3\rho - 2$ , giving the correct answer for that problem.