Sliding off

For the particle, let $v(x)$ and $v(y)$ be its horizontal and vertical velocities, with rightward and downward taken to be positive, respectively. Let $V$ be the velocity of the hemisphere, with leftward taken to be positive.

By conservation of momentum, $mv(x) = MV$

When the particle is at an angle of $θ$ , $(v(y)/(v(x) + V)$ = $\tan θ$

If $R$ is the radius of the hemisphere, the particle has fallen $R(1 - \cos θ)$ , so by conservation of energy, $((1/2)m((v(x)^2) + (v(y)^2))) + ((1/2)MV^2) = mgR(1 - \cos θ)$

Let $r = m/M$

Then $(v(x)^2) = (2gR(1 - \cos θ))/((1+ r)(1 + ((1+r)((\tan θ)^2)))$

Since $v(x)$ cannot decrease, the particle separates from the hemisphere when this function reaches a maximum. Taking the derivative and setting to zero yields:

0 = $(r * (\cos θ)^3) - (3(1+r) \cos θ) + 2(1+r)$

If $m = M$ , $r = 1$ , and this becomes:

$0 = (\cos θ)^3 - 6 \cos θ + 4$

This factors as $(\cos θ - 2)((\cos θ)^2 + 2 \cos θ - 2) = 0.$

since $\cos θ$ can never equal 2, solve the quadratic:

$\cos θ = \sqrt{3} - 1$

$θ = 42.9$ degrees