Mathematics - IB Physics

1. An electron is initially at rest in a uniform electric field of strength $E = 1.5 \times 10^3 \text{ N C}^{-1}$. What is the speed of the electron after it has travelled a distance of $5.0 \text{ cm}$ in the field?
   Answer: $5.1 \times 10^6 \text{ m s}^{-1}$ The work done by the electric field on the electron provides it with kinetic energy. Work done $W = Fd = (qE)d$. The kinetic energy gained is $E_k = \frac{1}{2}mv^2$. Setting $W = E_k$ gives $qEd = \frac{1}{2}mv^2$. Rearranging for $v$: $v = \sqrt{\frac{2qEd}{m}}$. Using values for an electron ($q=e$, $m=m_e$) from the data booklet: $v = \sqrt{\frac{2(1.60 \times 10^{-19})(1.5 \times 10^3)(0.050)}{9.11 \times 10^{-31}}} = \sqrt{\frac{2.4 \times 10^{-17}}{9.11 \times 10^{-31}}} \approx 5.1 \times 10^6 \text{ m s}^{-1}$.
2. Car A starts from rest and accelerates uniformly at $3.0 \text{ m s}^{-2}$. At the same instant, Car B, which is 100 m ahead of Car A, is moving in the same direction at a constant velocity of $5.0 \text{ m s}^{-1}$. At what time does Car A overtake Car B?
   Answer: 10 s Let the starting position of Car A be $s=0$. The position of Car A at time $t$ is $s_A = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2}(3.0)t^2 = 1.5t^2$. The initial position of Car B is $s=100$ m. Its position at time $t$ is $s_B = s_0 + vt = 100 + 5.0t$. Car A overtakes Car B when $s_A = s_B$. So, $1.5t^2 = 100 + 5.0t$. This gives the quadratic equation $1.5t^2 - 5.0t - 100 = 0$. Using the quadratic formula, $t = \frac{-(-5.0) \pm \sqrt{(-5.0)^2 - 4(1.5)(-100)}}{2(1.5)} = \frac{5 \pm \sqrt{25 + 600}}{3} = \frac{5 \pm \sqrt{625}}{3} = \frac{5 \pm 25}{3}$. Since time must be positive, $t = \frac{30}{3} = 10 \text{ s}$.
3. A 60 W incandescent light bulb is about 5% efficient at producing visible light. Estimate the order of magnitude for the number of visible-light photons it emits per second. (Assume the average wavelength of visible light is 500 nm).
   Answer: $10^{19}$ First, calculate the power of the visible light emitted: $P_{vis} = 60 \text{ W} \times 0.05 = 3.0 \text{ W}$. Next, calculate the energy of a single photon: $E_{photon} = hf = hc/\lambda$. Using approximations: $E \approx \frac{(6.6 \times 10^{-34})(3 \times 10^8)}{500 \times 10^{-9}} \approx \frac{20 \times 10^{-26}}{5 \times 10^{-7}} = 4 \times 10^{-19} \text{ J}$. The number of photons per second is $N = P_{vis} / E_{photon} = 3.0 / (4 \times 10^{-19}) = 0.75 \times 10^{19}$. The order of magnitude is $10^{19}$.
4. A student plots a graph to determine the acceleration due to gravity, $g$, using the simple pendulum equation $T = 2\pi\sqrt{\frac{l}{g}}$. The graph produces a straight line that passes through the origin. Which of the following correctly identifies the quantities on the axes and the expression for the gradient of the graph?
   Answer: y-axis: $T^2$, x-axis: $l$, gradient: $\frac{4\pi^2}{g}$ To obtain a linear relationship, the equation $T = 2\pi\sqrt{\frac{l}{g}}$ should be rearranged into the form $y = mx + c$. Squaring both sides gives $T^2 = 4\pi^2 \frac{l}{g}$, which can be written as $T^2 = \left(\frac{4\pi^2}{g}\right)l$. If $T^2$ is plotted on the y-axis and $l$ is plotted on the x-axis, the graph will be a straight line through the origin with a gradient $m = \frac{4\pi^2}{g}$.
5. What is the order of magnitude of the ratio of the electrostatic force to the gravitational force between two protons?
   Answer: $10^{36}$ The ratio is $\frac{F_E}{F_G} = \frac{k e^2 / r^2}{G m_p^2 / r^2} = \frac{k e^2}{G m_p^2}$. Using approximate values from the data booklet: $k \approx 9 \times 10^9$, $e \approx 1.6 \times 10^{-19}$, $G \approx 6.7 \times 10^{-11}$, $m_p \approx 1.7 \times 10^{-27}$. The ratio is approximately $\frac{(9 \times 10^9)(1.6 \times 10^{-19})^2}{(6.7 \times 10^{-11})(1.7 \times 10^{-27})^2} \approx \frac{10^{10} \cdot (10^{-19})^2}{10^{-10} \cdot (10^{-27})^2} = \frac{10^{-28}}{10^{-64}} = 10^{36}$. This shows the immense relative strength of the electrostatic force.
6. The unit of pressure, the pascal (Pa), can be expressed in terms of the fundamental SI units for mass (kg), length (m), and time (s). Which of the following is the correct expression for the pascal?
   Answer: $\text{kg m}^{-1} \text{s}^{-2}$ Pressure is defined as force per unit area ($P = F/A$). The SI unit of force is the newton (N) and the unit of area is square meters ($\text{m}^2$), so $1 \text{ Pa} = 1 \text{ N m}^{-2}$. Force is defined by Newton's second law ($F=ma$), so $1 \text{ N} = 1 \text{ kg} \cdot \text{m s}^{-2}$. Substituting this into the expression for the pascal gives $1 \text{ Pa} = (1 \text{ kg m s}^{-2}) \text{m}^{-2} = 1 \text{ kg m}^{-1} \text{s}^{-2}$.
7. A beam of light enters a block of glass with a refractive index of 1.50 from air (refractive index ≈ 1.00). The angle of incidence is $60^\circ$. What is the angle of refraction inside the glass?
   Answer: $35^\circ$ According to Snell's Law, $n_1 \sin{\theta_1} = n_2 \sin{\theta_2}$. Here, $n_1 = 1.00$ (air), $\theta_1 = 60^\circ$, and $n_2 = 1.50$ (glass). We need to find $\theta_2$. $1.00 \times \sin(60^\circ) = 1.50 \times \sin{\theta_2}$. $\sin{\theta_2} = \frac{\sin(60^\circ)}{1.50} = \frac{0.866}{1.50} \approx 0.577$. The angle of refraction is $\theta_2 = \arcsin(0.577) \approx 35.3^\circ$. The closest answer is $35^\circ$.
8. In an experiment to determine the resistivity $\rho$ of a wire, the resistance $R$ of different lengths $L$ of the wire is measured. The wire has a uniform cross-sectional area $A$. If a graph of $R$ is plotted on the y-axis against $L$ on the x-axis, what does the gradient of the resulting straight-line graph represent?
   Answer: $\rho/A$ The formula for resistance in terms of resistivity is $R = \frac{\rho L}{A}$. To analyse this relationship with a linear graph, we can write it as $R = \left(\frac{\rho}{A}\right)L$. This equation is in the form $y = mx + c$, where $y=R$, $x=L$, the y-intercept $c=0$, and the gradient $m = \frac{\rho}{A}$.
9. Two long, parallel wires carry currents $I_1$ and $I_2$ and are separated by a distance $r$. They experience a magnetic force per unit length $f$. If the current in both wires is doubled and their separation distance is halved, what is the new force per unit length?
   Answer: $8f$ The force per unit length $f$ between two parallel wires is given by $f = \frac{\mu_0 I_1 I_2}{2\pi r}$. This shows that $f$ is directly proportional to the product of the currents ($I_1 I_2$) and inversely proportional to the distance ($r$). If both currents are doubled ($I_1 \rightarrow 2I_1, I_2 \rightarrow 2I_2$) and the distance is halved ($r \rightarrow r/2$), the new force $f_{new}$ will be proportional to $\frac{(2I_1)(2I_2)}{(r/2)} = \frac{4 I_1 I_2}{r/2} = 8 \frac{I_1 I_2}{r}$. Thus, the new force per unit length is $8f$.
10. A black body has a thermodynamic temperature $T$ and radiates power $P$. If the thermodynamic temperature of the black body is doubled to $2T$, what is the new power radiated?
    Answer: $16P$ According to the Stefan-Boltzmann law, the power $P$ radiated by a black body is proportional to the fourth power of its thermodynamic temperature $T$ ($P \propto T^4$). If the temperature is doubled ($T \rightarrow 2T$), the new power $P_{new}$ will be proportional to $(2T)^4 = 2^4 T^4 = 16T^4$. Therefore, the new power is 16 times the original power, so $P_{new} = 16P$.
11. Two point charges experience an electrostatic force $F$. The distance between the charges is then tripled, and the magnitude of one of the charges is doubled. What is the new electrostatic force in terms of $F$?
    Answer: $\frac{2}{9}F$ According to Coulomb's Law, the electrostatic force $F$ is proportional to the product of the charges ($q_1q_2$) and inversely proportional to the square of the distance ($r^2$) between them: $F \propto \frac{q_1q_2}{r^2}$. If one charge is doubled ($q_1 \rightarrow 2q_1$) and the distance is tripled ($r \rightarrow 3r$), the new force $F_{new}$ will be proportional to $\frac{(2q_1)q_2}{(3r)^2} = \frac{2q_1q_2}{9r^2} = \frac{2}{9} \frac{q_1q_2}{r^2}$. Therefore, $F_{new} = \frac{2}{9}F$.
12. An experimenter plots data points with error bars on a graph to find a physical constant from the gradient. To determine the uncertainty in the gradient, two additional lines are drawn. Which statement best describes these two lines?
    Answer: A line of maximum possible slope and a line of minimum possible slope that pass through the error bars of all data points. The uncertainty in the gradient is found by considering the range of possible straight lines that can be reasonably drawn through the data, given the uncertainties represented by the error bars. This range is defined by the line of maximum possible slope (the 'steepest' line) and the line of minimum possible slope (the 'shallowest' line) that still pass through or touch the error bars of all data points. The uncertainty in the gradient is typically taken as half the difference between these maximum and minimum gradients.
13. A block of weight $W = mg$ is placed on a frictionless plane inclined at an angle $\theta$ to the horizontal. What is the magnitude of the component of the block's weight that acts parallel to the surface of the plane?
    Answer: $mg \sin{\theta}$ The weight $mg$ acts vertically downwards. This vector can be resolved into two components: one perpendicular to the inclined plane and one parallel to it. Using trigonometry, the component of weight parallel to the plane is $mg \sin{\theta}$, and the component perpendicular to the plane is $mg \cos{\theta}$. The parallel component is the one that would cause the block to accelerate down the plane.
14. A radioactive sample has a half-life of $T_{1/2}$. What is the time required for the activity of the sample to decrease to $\frac{1}{10}$ of its initial value?
    Answer: $T_{1/2} \frac{\ln(10)}{\ln(2)}$ The activity $A$ follows the decay law $A = A_0 e^{-\lambda t}$, where $\lambda$ is the decay constant. We are given $A/A_0 = 1/10$. So, $1/10 = e^{-\lambda t}$. Taking the natural logarithm of both sides gives $\ln(1/10) = -\lambda t$, which simplifies to $t = \frac{\ln(10)}{\lambda}$. The decay constant is related to the half-life by $\lambda = \frac{\ln(2)}{T_{1/2}}$. Substituting for $\lambda$ gives $t = \frac{\ln(10)}{\ln(2)/T_{1/2}} = T_{1/2} \frac{\ln(10)}{\ln(2)}$.
15. A graph shows the variation of the resultant force $F$ acting on an object as a function of time $t$. What physical quantity is represented by the area under the force-time graph?
    Answer: The change in momentum of the object. By definition, impulse is the product of the average force and the time interval over which it acts ($J = F \Delta t$). For a variable force, the impulse is the integral of the force with respect to time, which corresponds to the area under the force-time graph. The impulse-momentum theorem states that the impulse on an object is equal to the change in its momentum ($J = \Delta p$).
16. A swimmer can swim at a speed of $1.5 \text{ m s}^{-1}$ in still water. She swims across a river that has a current of $2.0 \text{ m s}^{-1}$ flowing at a right angle to her intended direction of travel. What is the magnitude of her resultant velocity relative to the river bank?
    Answer: $2.5 \text{ m s}^{-1}$ The swimmer's velocity relative to the water and the velocity of the water current are perpendicular vectors. The resultant velocity is the vector sum of these two velocities. The magnitude of the resultant velocity can be found using the Pythagorean theorem: $v_{resultant} = \sqrt{v_{swimmer}^2 + v_{current}^2} = \sqrt{(1.5)^2 + (2.0)^2} = \sqrt{2.25 + 4.00} = \sqrt{6.25} = 2.5 \text{ m s}^{-1}$.
17. A satellite of mass $m$ orbits a planet of mass $M$ in a stable circular orbit of radius $r$ with an orbital period $T$. Which expression correctly gives the mass of the planet $M$? ($G$ is the gravitational constant.)
    Answer: $M = \frac{4\pi^2 r^3}{GT^2}$ For a circular orbit, the gravitational force provides the centripetal force. So, $F_g = F_c \Rightarrow \frac{GMm}{r^2} = \frac{mv^2}{r}$. The orbital speed is $v = \frac{2\pi r}{T}$. Substituting for $v$: $\frac{GMm}{r^2} = \frac{m}{r} \left(\frac{2\pi r}{T}\right)^2 = \frac{m}{r} \frac{4\pi^2 r^2}{T^2}$. Simplifying gives $\frac{GM}{r^2} = \frac{4\pi^2 r}{T^2}$. Rearranging for $M$ gives Kepler's third law in this form: $M = \frac{4\pi^2 r^3}{GT^2}$.

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