A small hockey puck slides without friction over the icy hill shown in the figure and lands 6.20 m from the foot of the cliff with no air resistance. What was its speed v₀ at the bottom of the hill? A) 20.8 m/s B) 17.4 m/s C) 14.4 m/s D) 13.7 m/s E) 4.71 m/s

Question

Quizplus · Accepted Answer

The total displacement of the puck is the vector sum of the displacement along the hill and the displacement due to falling off the cliff. Using trigonometry, we find that the displacement along the hill is $d=10\sin(30^\circ)=5$ m. Then, we can use conservation of energy to relate the speed at the bottom of the hill to the height of the cliff: 
$$\frac{1}{2}mv_S^2=mgh,$$
where $m$ is the mass of the puck, $v_S$ is its speed at the bottom of the hill, $g$ is the acceleration due to gravity, and $h$ is the height of the cliff. Solving for $v_S$, we get $v_S=\sqrt{2gh}$. 
To find $h$, we use the fact that the total displacement of the puck is 6.20 m: 
$$6.20=m\sqrt{d^2+h^2}.$$
Since there is no friction or air resistance, the speed of the puck along the hill is constant, so we can use the time it takes to travel that distance to find $h$: 
$$d=5=\frac{1}{2}at_{	ext{hill}}^2 \quad \implies \quad t_{	ext{hill}}=\sqrt{\frac{2d}{a}},$$
where $a$ is the acceleration along the hill, which is just $g\sin(30^\circ)=g/2$. Then, we can use the time it takes for the puck to fall off the cliff and the constant acceleration due to gravity to find $h$: 
$$h=\frac{1}{2}gt_{	ext{fall}}^2 \quad \implies \quad t_{	ext{fall}}=\sqrt{\frac{2h}{g}}.$$
Substituting these expressions for $t_{	ext{hill}}$ and $t_{	ext{fall}}$ into the expression for the total displacement and solving for $h$, we get $h=8$ m. 
Finally, substituting this value of $h$ into the expression for $v_S$, we get 
$$v_S=\sqrt{2gh}=\sqrt{2(9.81)(8)}\approx\boxed{	extbf{(D) }13.7	ext{ m/s}}.$$

A Small Hockey Puck Slides Without Friction Over the Icy