Tutor profile: Wiktor W.

Using the Snell's law of refraction $$n \sin(\theta) = \rm{const}$$ find the angle of total internal reflection.

Let us consider an interface between the two media with refractive indices given by $$n_1$$ and $$n_2$$, respectively. The light propagates from medium 1 towards medium 2. The incident angle is denoted a $$\theta_1$$. For all incident angles $$\theta_1$$ yielding the angle of refraction $$\theta_2$$ smaller than 90 degrees, refraction in possible. The limiting refraction angle is therefore $$\theta_2 = 90^o$$. Writing the Snell's law for this case: $$n_1 \sin(\theta_1) = n_2 \sin(90^o) = n_2$$ we conclude that the minimum angle for which the total internal reflection occurs is: $$\theta_1 = \arcsin(n_2/n_1)$$. Based on this result, we conclude also that the total internal reflection can only occur in the case of light propagating form the medium with higher refractive index to the lower refractive index ($$n_1>n_2$$), as the argument of the $$\arcsin(x)$$ function is smaller than 1 (to be exact $$|x|<1$$).

There is a man standing at the end of a stationary boat floating on the water. How far the boat moves, if the man walks to the other end of the boat? The mass of the man is M and the mass of the boat is B. The length of the boat is L. Please neglect the friction of the water.

We will use here the first principle of dynamics and it's application to the system described by the center of mass. The boat and the man consist of a closed system, and no external force is acting on this system. In the initial state, the system as a whole was stationary. Therefore, from the first principle of dynamics. the system, in particular understood as its center of mass, will remain stationary. Knowing that we have to compute the location of the center of mass in the initial state. From the fact that it will be the same in the final state, we will be able to compute the distance the boat traveled. Let us assume that in the initial state the boat was centered at x = 0 and that the man stands at the left hand-side of the boat (x<0). The initial position of the man is $$-L/2$$. Then the center of mass of the system is: $$C_i = -ML/2 + B\cdot 0 = -ML/2$$. When mans walks the distance L to the right (in the frame of the boat), the boat moves by an unknown distance D to the left (in the frame of a stationary external observer) (third principle of dynamics). As a result the man moves by the distance L-D towards the right, in the external reference frame. The new location of the man is $$-L/2 + (L-D) = L/2-D$$ Therefore, the new center of mass (expressed in the external reference frame) is given by $$G_f = M(L/2-D) - BD$$. As stated above, the center of mass of the system did not move: $$C_i = C_f$$. So $$-ML/2 = M(L/2-D) - BD$$. Solving for $$D$$, we get: $$D = \frac{ML}{M+B}$$.

While cooling down from boiling to room temperature, a glass of water emits the energy of about 67200J. What is the height to which car with a mass of 2000 kg can be lifted using this energy?

Let's define: Available energy: $$E = 67200$$ J Mass of the car: $$m = 2000$$ km Earth's gravitational acceleration $$9.8$$ m/s$$^2$$ Height: $$h = $$ ? Let us use the expression for the potential energy $$E_p = m g h$$. It can be transformed to the expression for the height: $$h = \frac{E_p}{mg}$$. Using the principle of conservation of energy, the entire thermal energy $$E$$ will be transformed into the potential energy $$E_p$$. Inserting these values into the expression gives: $$h = \frac{67200}{2000 \cdot 9.8} \approx 3.43$$ m. Therefore, using the thermal energy from a single glass of water we could lift a car to 3.43 m height.