A skier weighing 600 Ngoes over a frictionless circular hill of radius R = 20 m (Fig. 8-62). Assume that the effects of air resistance on the skier are negligible. As she comes up the hill, her speed is 0.8 m/s at point B, at angle $\mathit{\theta }\mathbf{=}\mathbf{20}\mathbf{°}$. (a) What is her speed at the hilltop (point A) if she coasts without using her poles? (b) What minimum speed can she have at B and still coast to the hilltop? (c) Do the answers to these two questions increase, decrease, or remain the same if the skier weighs 700 N instead of 600 N ?

1. Weight of the skier w = 600 N
2. Radius of the circular hill R =20 m
3. Angular displacement$\theta =20°$
4. Velocity of skier at point B${\mathrm{v}}_{\mathrm{b}}=8.0\frac{\mathrm{m}}{\mathrm{s}}$
5. Frictional force and drag force are zero

If the system is isolated, only conservative forces will act on it. For this condition, we can write the sum of K and U for any state of a system = Sum of K and U for any other state of the system. Using this we can solve for unknown velocities of the system.

Formula:

$K_a+U_a=K_b+U_b$

$K=\frac{1}{2}m{v}^2$

U = m g h

As there is no frictional and drag force, the skier and circular hill forms the isolated system. Hence, only conservative forces are acting on the system. Therefore

$K_a+U_a=K_b+U_b$ (i)

At point A, $K_a=\frac{1}{2}m{v}_a^2$ (ii)

$U_a=mgh$ (iii)

From figure,

$$\begin{gathered} \cos \theta =\frac{R-h}{R} \\ \Rightarrow \cos \left(20\right)=1-\frac{h}{R} \\ \Rightarrow 0.9396=1-\frac{h}{20} \end{gathered}$$

By solving for h, we get

h = 1.208 m (iv)

And using it in equation (ii), we get

$$\begin{gathered} U_{a}mgh \\ \Rightarrow U_a=2\times 9.8\times 1.208 \\ \Rightarrow U_a=23.6768J \\ {K}_b=\frac{1}{2}m{v}_b^2 \\ {U}_b=0asBisthelowestpoint,h=0 \end{gathered}$$ (v)

Using equation (ii), (iii), (iv), (v) in equation (i)

$\frac{1}{2}m{v}_a^2+mgh=\frac{1}{2}m{v}_b^2$.

Simplifying the equation

$v_a^2=v_b^2-2gh$ (vi)

$$\begin{gathered} \Rightarrow {\mathrm{v}}_{\mathrm{a}}^2=40.3232 \\ {\mathrm{v}}_{\mathrm{a}}=6.35\frac{\mathrm{m}}{\mathrm{s}} \\ \approx 6.4\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

If the skier reaches at point A with minimum velocity at point B, without using her poles, then velocity at point A will be. Using this in equation (vi), we get

$$\begin{gathered} v_a^2=v_b^2-2gh \\ \Rightarrow 0=v_b^2-2gh \\ \Rightarrow v_b^2-2gh \\ \Rightarrow v_b^2=2\times 9.8\times 1.208 \\ \Rightarrow v_b^2=23.6768 \end{gathered}$$

Solve further as:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{b}}=\sqrt{23.6768} \\ {\mathrm{v}}_{\mathrm{b}}=4.86\frac{\mathrm{m}}{\mathrm{s}} \\ \approx 4.9\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

As both equation (vi) and (vii) are independent of mass, answer for$v_{a}and{v}_b$ will not change even if the weight of the skier changes.