a) C. 9 m/s
First of all, let's calculate the difference in height between the starting point of the motion and the end point of the first section. It is given by:
[tex]\Delta h = L sin \theta[/tex]
where
L = 6.0 m
[tex]\theta=45^{\circ}[/tex]
Substituting,
[tex]\Delta h = (6.0)(sin 45^{\circ})=4.2 m[/tex]
Assuming the rider starts from rest, its initial speed is zero. For the law of conservation of energy, the decrease in gravitational potential energy of the rider will be equal to its gain in kinetic energy, so we can write:
[tex]mg\Delta h = \frac{1}{2}mv^2[/tex]
where m is the rider's mass, g = 9.8 m/s^2 is the acceleration of gravity, and v is the speed at the end of the first section. Solving for v, we find:
[tex]v=\sqrt{2g\Delta h}=\sqrt{2(9.8)(4.2)}=9.1 m/s \sim 9 m/s[/tex]
b) B. Increase the radius of the circular segment
In fact, the acceleration during the second section of the motion (circular motion) is given by the formula for the centripetal acceleration:
[tex]a=\frac{v^2}{r}[/tex]
where
v is the speed at the end of the first section
r is the radius of the circle
We notice that the acceleration is
- Proportional to the square of the speed
- Inversely proportional to the radius
So, we immediately see that if we increase the radius of the circle (choice B), the acceleration will decrease.
c) B. 3.4 m/s
When the rider exits the second section of the motion, he has a speed (completely horizontal) of 9.1 m/s (calculated in part (a); it didn't change, because the speed during the second section does not change).
The vertical component of his velocity is instead zero, since his motion is completely horizontal. Therefore, we can use the following SUVAT equation along the vertical direction:
[tex]v_y^2 - u_y^2 = 2g\Delta h[/tex]
where
[tex]v_y[/tex] is the vertical component of the velocity as the rider hits the water
[tex]u_y=0[/tex] is the vertical component of the velocity as the rider starts the 3rd section
[tex]\Delta h = 0.60 m[/tex] is the difference in height
Solving for [tex]v_y[/tex],
[tex]v_y = \sqrt{2g\Delta h}=\sqrt{2(9.8)(0.60)}=3.4 m/s[/tex]
d) C. 2.4 m
We want here the rider to land twice as far compared to before.
The horizontal distance travelled by the rider in section 3 is entirely determined by his horizontal motion. The horizontal component of the velocity, which is constant, is
[tex]v_x = 9.1 m/s[/tex]
calculated at part (a) and remained unchanged during section 2. The horizontal distance travelled during section 3 is
[tex]d=v_x t[/tex] (1)
where t is the time of the fall. This can be rewritten as
[tex]t=\frac{d}{v_x}[/tex]
We also know that the vertical displacement is:
[tex]h=\frac{1}{2}gt^2[/tex]
Substituting t from (1) into this equation, we find
[tex]h=\frac{gd^2}{2v_x^2}[/tex]
So we see that the height needed is proportional to the square of the distance: [tex]h \propto d^2[/tex]. Therefore, in order to land at twice the previous distance, the height must be 4 times the previous one, so:
[tex]h=4 (0.6 m)=2.4 m[/tex]
e) B. The second
We need to calculate the acceleration in each section.
In section 1 (motion along the slope), it is:
[tex]a=g sin \theta = (9.8)(sin 45^{\circ})=6.9 m/s^2[/tex]
In section 2 (circular motion), the acceleration is the centripetal acceleration:
[tex]a=\frac{v^2}{r}=\frac{9.1^2}{1.5}=55.2 m/s^2[/tex]
In section 3, the motion is free fall, so the acceleration is equal to the acceleration of gravity:
[tex]a=g=9.8 m/s^2[/tex]
Therefore, the rider experiences the largest acceleration in section 2.
