A skier skis from rest from a vertical height h1 = 18 m over two successively lower hills of vertical heights h2 = 15 m and h3 = 7 m. The summit of the third hill fits a circle of... A skier skis from rest from a vertical height h1 = 18 m over two successively lower hills of vertical heights h2 = 15 m and h3 = 7 m. The summit of the third hill fits a circle of radius h3 centered at height 0 m. Friction with the snow and air resistance are negligible. (a) Find her speeds at x1, x2, and x3. (b) Does the skier leave the surface at x3? If not, what should h1 be so that she just leaves the surface at x3? Read more

Steps to Solve

1. Identify Relevant Energy Principles

Since friction and air resistance are negligible, we can apply the principle of conservation of mechanical energy. The total mechanical energy (potential + kinetic) remains constant.

2. Calculate Speed at x1

At point x1 (height $h_1 = 18 , \text{m}$), the skier starts from rest, so the initial kinetic energy is 0. The potential energy is given by:

$$ PE_1 = mgh_1 $$

At the bottom of the first hill, all potential energy converts to kinetic energy:

$$ PE_1 = KE_2 \implies mgh_1 = \frac{1}{2}mv_2^2 $$

Cancelling mass and solving for speed ( v_2 ):

$$ v_2 = \sqrt{2gh_1} $$

Substituting ( h_1 = 18 , \text{m} ):

$$ v_2 = \sqrt{2 \cdot 9.81 \cdot 18} $$

3. Calculate Speed at x2

At point x2 (height $h_2 = 15 , \text{m}$):

Using the conservation of energy from x1 to x2:

$$ KE_2 + PE_2 = KE_3 $$

We can use the kinetic energy at x1 and the potential energy at x2:

$$ \frac{1}{2}mv_2^2 + mgh_2 = \frac{1}{2}mv_3^2 $$

Cancelling mass and plugging in ( v_2 ):

$$ \frac{1}{2} \cdot 2gh_1 + gh_2 = \frac{1}{2}v_3^2 $$

Solve for ( v_3 ):

$$ v_3 = \sqrt{2g(h_1 - h_2)} $$

Substituting ( h_2 = 15 ) and ( h_1 = 18 ):

$$ v_3 = \sqrt{2 \cdot 9.81 \cdot (18 - 15)} $$

4. Calculate Speed at x3

At point x3 (height $h_3 = 7 , \text{m}$):

Using the conservation of energy from x2 to x3:

$$ KE_3 + PE_3 = KE_4 $$

Plugging values:

$$ \frac{1}{2}mv_3^2 + mgh_3 = \frac{1}{2}mv_4^2 $$

Cancelling mass and solving for ( v_4 ):

$$ v_4 = \sqrt{2g(h_2 - h_3)} $$

Substituting ( h_3 = 7 ) and ( h_2 = 15 ):

$$ v_4 = \sqrt{2 \cdot 9.81 \cdot (15 - 7)} $$

5. Determine if Skier Leaves Surface at x3

To determine if the skier leaves the surface at x3, we apply the centripetal force condition. The required centripetal acceleration is provided by gravity at height ( h_3 ):

$$ \frac{v^2}{r} = g $$

Substituting ( r = h_3 ):

$$ \frac{v_4^2}{h_3} = g $$

Check if ( v_4^2 < gh_3 ).

6. Finding Height h1 for Just Leaving the Surface

If the skier just leaves the surface at x3, we need:

$$ v_4^2 = gh_3 $$

Use conservation of energy to find the new height ( h'_1 ):

$$ mgh'_1 = \frac{1}{2}mv_4^2 + mgh_3 $$

Solving for ( h'_1 ):

$$ h'_1 = \frac{1}{2g}v_4^2 + h_3 $$

Substituting the condition ( v_4^2 = gh_3 ):

$$ h'_1 = \frac{1}{2}h_3 + h_3 $$