A student skates up a ramp that makes an angl | vedclass

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(B) By the principle of conservation of mechanical energy between the bottom point and point $y$:$\frac{1}{2} mv_0^2 = mgh + \frac{1}{2} mv_1^2$$	her

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$A, D$

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(B) By the principle of conservation of mechanical energy between the bottom point and point $y$:$\frac{1}{2} mv_0^2 = mgh + \frac{1}{2} mv_1^2$$	herefore v_1^2 = v_0^2 - 2gh \quad \dots (i)$At point $y$,the semicircular path lies on the inclined plane. The component of gravity acting towards the center of the circular path is $mg \sin 30^{\circ}$. This component provides the necessary centripetal force:$mg \sin 30^{\circ} = \frac{mv_1^2}{R}$$\frac{1}{2} mg = \frac{mv_1^2}{R} \implies v_1^2 = \frac{gR}{2}$Substituting this into equation $(i)$:$v_0^2 - 2gh = \frac{gR}{2}$. Thus,statement $(A)$ is correct.At points $x$ and $z$,the skater is at a lower height than at point $y$. By energy conservation,the velocity $v$ at $x$ and $z$ is greater than $v_1$ at $y$. Since the centripetal force required is $F_c = \frac{mv^2}{R}$,and $v > v_1$,the required centripetal force at $x$ and $z$ is greater than that at $y$. Thus,statement $(D)$ is correct.

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