A

To calculate the pressure exerted by the ball on the chute at point B, we need to consider the forces acting on the ball.
Step 1: Determine the potential energy (PE) at height h.
The potential energy is given by:
$$ PE = mgh $$
where $$ h = \frac{5}{2}R $$, therefore:
$$ PE = mg \left(\frac{5}{2}R\right) = \frac{5}{2}mgR $$
Step 2: Energy conservation to find speed at point B.
When the ball reaches point B, all potential energy converts to kinetic energy (KE) plus the potential energy at point B (where height = R cos(\alpha)).
So, we have:
$$ KE + PE_{B} = PE \Rightarrow \frac{1}{2} mv^2 + mgR \cos(\alpha) = \frac{5}{2} mgR $$
Rearranging gives:
$$ \frac{1}{2} mv^2 = \frac{5}{2} mgR - mgR \cos(\alpha) \Rightarrow mv^2 = 5mgR - 2mgR \cos(\alpha) = mg(5R - 2R \cos(\alpha)) $$
So,
$$ v^2 = g(5 - 2 \cos(\alpha)) $$
Step 3: Centripetal force at point B.
At point B, we have a centripetal force requirement to keep the ball moving in a circular path:
$$ \frac{mv^2}{R} = mg \cos(\alpha) + N $$
where N is the normal force exerted by the chute on the ball. Rearranging gives:
$$ N = \frac{mv^2}{R} - mg \cos(\alpha) $$
Substitute for v^2:
$$ N = \frac{m(g(5 - 2 \cos(\alpha)))}{R} - mg \cos(\alpha) $$
Combine like terms:
$$ N = mg \left(\frac{5}{R} - \frac{2 \cos(\alpha)}{R} - \cos(\alpha)\right) = mg \left(\frac{5}{R} - \cos(\alpha) \left(\frac{2}{R} + 1\right)\right) $$
Step 4: Pressure Calculation.
Pressure is defined as force per unit area. Assuming the area is negligible, we focus on the force. For a small area A,
$$ P = \frac{N}{A} $$
To express pressure in terms of known quantities:
$$ P = \frac{mg}{A} \left(\frac{5}{R} - \cos(\alpha) \left(\frac{2}{R} + 1\right)\right) $$
Thus, you can conclude with the pressure exerted by the ball on the chute depending on the angle \alpha.