# Physics with Applications, PHYS*1130 - Sample Exam 1

- A standing wave on a guitar string is described by the equation

\(\mathrm{y = 3.0 \cos(2760t)\sin(195x)}\)

with the amplitude measured in \(\mathrm{mm,}\) \(\mathrm{t}\) in seconds and \(\mathrm{x}\) in metres. Determine the displacement of the string when \(\mathrm{t = 0.28\; T}\) seconds and \(\mathrm{x = 0.04\; \lambda}\) metres where \(\mathrm{T}\) is the period and \(\mathrm{\lambda}\) is the wavelength

(A) \(\mathrm{3.0\; mm}\)

(B) \(\mathrm{-3.0\; mm}\)

(C) \(\mathrm{1.6\; mm}\)

(D) \(\mathrm{0.14\; mm}\)

(E) \(\mathrm{-0.14 \;mm}\)

**Answer is E**\(\mathrm{y = A \cos (\omega t) \sin (kx)}\)

\(\mathrm{y = A \cos(2\pi /T)t \; \sin(2\pi /T)x}\)at \(\mathrm{t = 0.28T \;and\; x = 0.04\; \lambda}\)

\(\mathrm{y = A\cos[(2\pi /T)0.28T] \sin[(2\pi /\lambda )0.04\lambda ] \\ = 3.0 \cos(0.28\times 2\pi ) \sin(0.04\times 2\pi ) \\ = 3.0\; \cos100.8^\circ \sin14.4^\circ \\ = 3.0(-0.1784)(0.2487) \\ = -0.14 \;mm}\)

Note: The numbers 2760 and 195 given in the problem are irrelevant. If they are used in the solution they will just cancel out again.

- A student's reading problems are corrected with lenses of power \(\mathrm{+3.0\; diopters}\) to give a corrected near point of \(\mathrm{0.25\; m.}\) The near point without the glasses is:

(A) \(\mathrm{1.0\; m}\)

(B) \(\mathrm{0.5\; m}\)

(C) \(\mathrm{2.0\; m}\)

(D) \(\mathrm{1.5\; m}\)

(E) \(\mathrm{2.5 \;m}\)

**Answer is A**If an object is placed at \(\mathrm{0.25 \;m}\) (the normal near point), where will the image be? This is the place where the object must be placed in the absence of the correcting lens.

\(\mathrm{1/p + 1/q = P \\ 1/0.25 + 1/q = 3 \\ -1/p - 1/q = 3 - 1/0.25 = 3 - 4 = -1 \\ q = -1 \;m}\)

- An object is placed 0.05 m in front of a lens made of glass of index of refraction 1.5. The magnification is found to be 2.5. The radius of curvature of the front (first surface) of the lens is 0.6 m. Determine the radius of curvature of the back surface.

(A) \(\mathrm{-0.14 \;m}\)

(B) \(\mathrm{+0.38 \;m}\)

(C) \(\mathrm{-0.71 \;m}\)

(D) \(\mathrm{+0.04 \;m}\)

(E) \(\mathrm{-0.38\; m}\)

**Answer is D**

\(\mathrm{m = -q/p = 2.5 \\ q = -2.5p = -2.5\times 0.05 = -0.125 \;m}\)Lensmaker's formula:

\(\mathrm{1/p + 1/q = (n-1)(1/R_1 - 1/R_2) \\ 1/0.05 - 1/0.125 = 20 - 8 = 12 = (1.5 - 1)(1/0.6 - 1/R_2) = (1/2)(1/0.6 - 1/R_2) \\ 1/2\;R_2 = 0.8333 - 12 = -11.166 \\ R_2 = -0.0447 = -0.045\; m}\)

- Cloud-to-cloud lightning produces thunder which can be heard by observers on the ground below. One particularly brilliant flash produced a thunder clap which reached the ground observer 4.0 seconds after the flash and registered 105 db on an intensity level meter held by the observer. What was the acoustical power produced by this particular lightning flash?

(A) \(\mathrm{6.1 \times 10^{-2} \;watts}\)

(B) \(\mathrm{5.8 \times 10^2\; watts}\)

(C) \(\mathrm{3.4 \times 10^7\; watts}\)

(D) \(\mathrm{7.3 \times 10^5\; watts}\)

(E) \(\mathrm{1.9 \times 10^3\; watts}\)

**Answer is D**

\(\mathrm{D = vt = 340\times 4 = 1360\;m \\ IL = 10 \log(I/10^{-12}) \\ 105 = 10 \log(I/10^{-12}) \\ \log(I/10^{-12}) = 10.5 \\ I = 3.16\times 10^{10}\times 10^{-12} = 3.16 \times10^{-2}\; W/m^2 \\ P = 4\pi (1360)^2\times 3.16\times 10^{-2} = 7.35\times 10^5 \;W}\)

- A bat is flying directly toward a telephone pole at \(\mathrm{25\; m/s}\) while emitting a constant frequency acoustical signal at \(\mathrm{60\; kHz}\). What echo frequency will the bat hear?

(A) \(\mathrm{40 \;kHz}\)

(B) \(\mathrm{50\; kHz}\)

(C) \(\mathrm{60\; kHz}\)

(D) \(\mathrm{70\; kHz}\)

(E) \(\mathrm{80\; kHz}\)

**Answer is D**\(\mathrm{\Delta f/f_0 = v/c \\ \Delta f = (25/340)60\;kHz = 4.41\;kHz}\)

Frequency incident on the pole and reflected is \(\mathrm{f_1 = 60 + 4.41 = 64.41\; kHz}\)

Heard by the bat (moving receiver): \(\mathrm{f_2 = 64.41 + 4.41 = 68.8 = 70\; kHz}\)

- A nucleus of \(\mathrm{^{218}{_{84}} \;Po}\) decays first by alpha emission and then by beta minus emission. The final nucleus is:

(A) \(\mathrm{^{214}{_{84}} \;Po}\)

(B) \(\mathrm{^{214}{_{82}} \; Po}\)

(C) \(\mathrm{^{214}{_{83}} \; Po}\)

(D) \(\mathrm{^{216}{_{82}} \; Po}\)

(E) \(\mathrm{^{210}{_{82}} \; Po}\)

**Answer is B**For alpha emission:

mass number decreases by \(\mathrm{4; 218 - 4 = 214}\)

atomic number decreases by \(\mathrm{2; 84 - 2 = 82}\)Isotope is \(\mathrm{_{82}{^{214}} \;Pb}\)

For Beta minus decay:

mass number is unchanged

atomic number increases by \(\mathrm{1: 82+1=83}\)

Isotope is\(\mathrm{_{83}{^{214}} \; Pb}\)

- A mass of \(\mathrm{0.85 \;kg}\) is attached to the end of a spring and placed on a frictionless horizontal table. The other end of the spring is attached to a wall. When the spring and mass is displaced \(\mathrm{0.15\; m}\) from rest position and released, the system exhibits simple harmonic motion with a frequency of \(\mathrm{6.4\; Hz}\) Assuming the spring's mass is negligible, determine the spring constant.

(A) \(\mathrm{1.98 \;kg \;s^{-2} (or\; N/m)}\)

(B) \(\mathrm{1370\; kg\; s^{-2} (or \;N/m)}\)

(C) \(\mathrm{1525\; kg\; s^{-2} (or\; N/m)}\)

(D) \(\mathrm{0.82\; kg\; s^{-2} (or \;N/m)}\)

(E) \(\mathrm{4.6 \times 10^{-2} \;kg\; s^{-2} (or\; N/m)}\)

**Answer is B**\(\mathrm{F = 6.4\; Hz}\)

\(\mathrm{\omega = 2\pi \; f = 40.2\; rad/s}\)\(\mathrm{\omega = (k/m)^{1/2}}\)

\(\mathrm{\omega^ 2 = k/m}\)

\(\mathrm{k = m\omega ^ 2 = 0.85\times 40.2^2 = 1374\; N/m \;or\; kg/s^2}\)

- Which of the following are proper dimensions for intensity?

(A) \(\mathrm{ML^3/T}\)

(B) \(\mathrm{M^2L^2/T}\)

(C) \(\mathrm{L^2/T}\)

(D) \(\mathrm{M/T^3}\)

(E) \(\mathrm{T^2/L}\)

**Answer is D**\(\mathrm{I = P/A = E/tm^2 = Fd/tm^2 \\ [I] = [F][d]/[t][m^2] = MLT^{-2} L/TL^2 = MT^{-3}}\)

- A radar technician is testing different reflectors for use on pleasure boats such as small sailboats. One model, a spherical shape of diameter \(\mathrm{0.5\;m}\) was especially good at reflecting radar waves in all directions. The receiver which the technician used to measure the "echo" has a reception area of 1 square metre and receives a signal which is \(\mathrm{1 \times 10^{-10}}\) times the power of the transmitted signal. What is the distance between the reflector and the echo receiver? Assume both the transmitted signal and the echo are radiated uniformly in all directions.

(A) \(\mathrm{6 \;m}\)

(B) \(\mathrm{60 \;m}\)

(C) \(\mathrm{600\; m}\)

(D) \(\mathrm{6000\; m}\)

(E) \(\mathrm{60000\; m}\)

**Answer is B**\(\mathrm{P = power\; of\; transmitter \\ Intensity \; at \; reflector = I_1 = P/4\pi \;R^2}\)

\(\mathrm{Power \;reflected = P1 = P= I_1A_1,}\) \(\mathrm{where\; A_1 \;is\; the\; area\; of\; the \;reflector.}\)

\(\mathrm{P_1 = PA_1/4\pi \; R^2}\)

The intensity at the receiver (same distance) \(\mathrm{= I_2 = P_1/4\pi \; R^2 = PA_1/16\pi \; ^2R^4}\)

If \(\mathrm{A_2}\) is the area of the receiver, then the power detected \(\mathrm{= I_2A_2 = PA_1A_2/16\pi ^2R^4 = 1\times 10^{-10}P}\)

\(\mathrm{R^4 = A_1A_2/16\pi ^210^{-10} = (\pi /4)(0.5)^2(1)/(16\pi ^210^{-10}) = 1.243 \times10^7}\)

\(\mathrm{R = 60 \;m}\)

- Three acoustical speakers produce sounds of intensity 0.1, 0.3, and 1.2 watts per square metre respectively at a detector placed ten metres away. If all three speakers produce the sounds simultaneously, what is the expected intensity level at the detector?

(A) \(\mathrm{92\; db}\)

(B) \(\mathrm{320\; db}\)

(C) \(\mathrm{122\; db}\)

(D) \(\mathrm{355 \;db}\)

(E) \(\mathrm{68 \;db}\)

**Answer is C**Intensities add

\(\mathrm{I = 0.1 + 0.3 + 1.2 = 1.6 \;W \\ IL = 10 \log(1.6/10^{-12}) = 10 \log 1.6\times 10^{12} \\ = 122\;dB}\)

- Three resistances (20, 40 and 80 ohm) are connected in parallel and the resulting set is connected in series to a 100 ohm resistance. What is the equivalent resistance of this set of resistances?

(A) \(\mathrm{124\; \Omega}\)

(B) \(\mathrm{96 \; \Omega}\)

(C) \(\mathrm{89 \; \Omega}\)

(D) \(\mathrm{240\; \Omega}\)

(E) \(\mathrm{111\; \Omega}\)

**Answer is E**

Parallel equivalent:\(\mathrm{1/R_p = 1/R_1 + 1/R_2 = 1/20 + 1/40 + 1/80 = (4 + 2 + 1)/80 = 7/80}\)

\(\mathrm{R_p = 80/7\; \Omega}\)

\(\mathrm{R_p\; in\; series\; with\; 100\; \Omega}\)

\(\mathrm{R_{eq}= 80/7 + 100 = 780/7 = 111\; \Omega}\)

- What is the electrical potential at the point p shown on the diagram below? The value of \(\mathrm{q_1}\) is 5.0 microcoulombs and the value of \(\mathrm{q_2}\) is minus 15.0 microcoulombs.

(A) \(\mathrm{-1.6 \times 10^6\; V}\)

(B) \(\mathrm{+1.6 \times 10^6\; V}\)

(C) \(\mathrm{+0.36 \times 10^6\; V}\)

(D) \(\mathrm{0\; V}\)

(E) \(\mathrm{-.36 \times 10^6\; V}\)

**Answer is D**

\(\mathrm{V = kq/r \\ V_1 = k5\times 10^{-6}/0.25 = 20\times 10^{-6}k \;volts \\ V_2 = k(-15\times 10^{-6})/(1 - 0.25) = -20\times 10^{-6} \;volts \\ V = V_1 + V_2 = 0}\)

- BONUS QUESTION

Rays from naturally radioactive materials were initially called

(A) Becquerel rays

(B) Curie rays

(C) X-rays

(D) Heisenberg rays

(E) Rutherford rays

**Answer is A**

- The isotope \(\mathrm{^{214}{_{84}} \;Po}\) might have two possible modes of decay. It could decay by alpha emission or by beta(-) emission. The half-life for the alpha decay is 3.1 minutes.

(A) Write the reaction for both decay processes and determine, using the data below, if both modes are possible. Find the total kinetic energy of the products.

\(\mathrm{^{214}{_{84}} \;Po --> ^{214}{_{85}}At + ^0{_{-1}}\beta}\)

Reaction Decay \(\mathrm{^{214}{_{84}} \;Po}\) \(\mathrm{^{214}{_{85}}At + ^0{_{-1}}\beta}\) 213.99519 213.99631 - 000.00055 - 213.99686 = total Not enough mass in the original Po, so the prosess is not possible.

\(\mathrm{^{214}{_{84}} \;Po -->^{210}{_{82}} \;Pb + 4_2He}\)Reaction Decay \(\mathrm{^{214}{_{84}} \;Po}\) \(\mathrm{^{210}{_{82}}Pb + 4_2He}\) 213.99519 209.98413 - 004.002603 - 213.98673 = total Po has more mass than the products so the process is possible.

Mass difference = 213.99519 - 213.98673 = 0.00846 atomic mass units

Since 1 amu = 931 MeV, the kinetic energy of the products is 931 X 0.00849 = 7.9 MeV

(B) Find the decay constant for alpha decay.

\(\mathrm{\lambda = 0.693/3.1 = 0.2235\; min^{-1}}\)

(C) A \(\mathrm{0.001\; \mu g}\) sample of the Po isotope is prepared by neutron irradiation of some smaller nucleus. How many atoms of Po remain after 2 minutes?

\(\mathrm{N = N_{0e}{^{-\lambda t}} = 0.001e^{-0.2235}\times 2 =0.001e^{-0.447} = 6.395\times 10^{-4} gm \\ N = (6.395\times 10^{-4})(6.02\times 1023)/214 = 1.8\times 10^{18} \;atoms}\)

Nuclide Mass \(\mathrm{^{214}{_{84}}\; Po}\) 213.99519 amu \(\mathrm{^{214}{_{85}} \;At}\) 213.99631 amu \(\mathrm{^{210}{_{82}} \; Pb}\) 209.98413 amu \(\mathrm{4_2\; He}\) 4.002603 amu \(\mathrm{e}\) 0.000550 amu

- Find the current in each branch of the circuit below, and find the equivalent resistance of the circuit.

**Loop 1**

\(\mathrm{14 - 2(I_1 - I_3) - 1(I_1 - I_2) = 0}\)

therefore

\(\mathrm{3I_1 - I_2 -2I_3 = 14[Eqn. 1]}\)**Loop 2**

\(\mathrm{1(I_2 - I_1) + 1(I_2 - I_3) + 2I_2 = 0}\)

therefore

\(\mathrm{I_1 -4I_2 + I_3 = 0 [Eqn. 2]}\)**Loop 3**

\(\mathrm{1(I_3 - I_2) + 2(I_3 - I_1) + I_3 = 0}\)

therefore

\(\mathrm{2I_1 + I_2 - 4I_3 = 0 [Eqn. 3]}\)

Solve Eqns. 1, 2, 3 for the currents.

From \(\mathrm{[2] I_1 = 4I_2 - I_3,}\) substitute this in \(\mathrm{[1]}\) and \(\mathrm{[2]}\) and get

\(\mathrm{11I_2 - 5I_3 = 14 [Eqn. 4]}\)

and

\(\mathrm{I_2 = (2/3)I_3\; [Eqn. 5]}\)

Solve [4] and [5], to get

\(\mathrm{I_3 = 6\; amp \\ I_2 = 4 \;amp \\ I_1 = 10\; amp}\)

Equivalent \(\mathrm{R = V/I}\) where \(\mathrm{I}\) is the total current out of the battery, i.e., \(\mathrm{I_1}\)\(\mathrm{R = 14/10 = 1.4\; \Omega}\)

- An object has a height of 2 cm. Its image is formed by a converging lens on a screen that is 20 cm from the lens. When the lens is moved 10 cm closer to the screen (with the object fixed), another image is formed on the screen. Determine the focal length of the lens and the distance from the screen to the object.

First case \(\mathrm{1/p + 1/20 = P}\)Second case \(\mathrm{1/(p + 10) + 1/10 = P}\)

Equating:

\(\mathrm{1/p + 1/20 = 1/(p + 10) + 1/10 \\ p^2 + 10p - 200 = 0}\)Solving the quadratic;

\(\mathrm{d = [-10 +/-(10^2- 4(1)(200)^{1/2}]/2 = (1/2)(-10 +/- 30)}\)

Only the positive solution is relevant

\(\mathrm{d = (1/2)(-10 + 30) = 10 \;cm}\)

- A baby's "jolly jumper" consists essentially of a (massless) seat attached to the end of an ideal spring, with the other end of the spring securely fastened to the ceiling. When a 10 kg baby is put into the seat, the spring lengthens by 30 cm. You then pull the jolly jumper down 45 cm below its equilibrium position and release.

(A) How fast is the baby travelling as he reaches the equilibrium position?

\(\mathrm{F = -kx}\)

\(\mathrm{-10(9.8) = -k(0.3)}\)\(\mathrm{k = 327\; N/m}\)

\(\mathrm{\omega = (k/m)^{1/2} = (327/10)^{1/2} = 5.716\; rad/s}\)

When released from rest at \(\mathrm{-0.45\;m}\) the graph of \(\mathrm{y}\) vs \(\mathrm{t}\) is a minus cosine function

\(\mathrm{y = -y_0\cos \omega \; t}\)\(\mathrm{v = dx/dt = y_0\omega\; \sin \omega\; t}\)

at \(\mathrm{t = T/4 = 2\pi /4\omega}\)

\(\mathrm{v - 0.45(5.716) \sin [\omega (2\pi /4\omega )] = 0.45(5.716) \sin \pi /2 = 2.57\; m/s}\)

(B) How long after the baby is released does he reach a point 20 cm above the equilibrium position?

Time to go from \(\mathrm{0}\) to \(\mathrm{0.20\;m = t}\)

\(\mathrm{0.20 = y = - 0.45 \cos 5.716t \\ \cos \;5.716t = - 0.4444 \\ 5.716t = 2.031 \;rad \\ t = 2.031/5.716 = 0.255\; s}\)

- Two positive charges \(\mathrm{Q_A = 5\; \mu C}\) and \(\mathrm{Q_B = 15\; \mu C}\), are separated by a distance of \(\mathrm{600\; nm.}\) Point P lies on a line that bisects the line joining \(\mathrm{Q_A}\) and \(\mathrm{Q_B.}\) If the distance from \(\mathrm{P}\) to \(\mathrm{O}\) is \(\mathrm{400\; nm,}\) determine the electric field at \(\mathrm{P.}\)

\(\mathrm{r = (400^2 + 300^2)^{1/2} = 500\;nm \\ \theta = \tan^{-1} 400/300 = 53.1½ \\ E_A = kQ_A/r^2 \;and\; E_B = kQ_B/r^2 = 3kQ_A/r^2}\)

x comp:

\(\mathrm{E_{ax} + E_{bx} = (kQ_A/r^2)\cos \theta - (3kQ_A/r^2)\cos \theta = -(2kQ_A/r^2)\cos \theta = 2(kQ_A/r^2)(3/5) \\ = (6/5)(9\times 10^9)(5\times 10^{-6})/600\times 10^{-9} = 0.15\; N/C}\)

y comp:

\(\mathrm{E_{ay} + E_{by} = (kQ_A/r^2)\sin \theta + (3kQ_A/r^2)\sin \theta = (4kQ_A/r^2)\sin \theta = 4(kQ_A/r^2)(4/5) \\ = (16/5)(9\times 10^9)(5\times 10^{-6})/600\times 10^{-9} = 0.40 \;N/C}\)

\(\mathrm{E = (0.15^2 + 0.40^2)^{1/2} = 0.43\; N/C}\)at an angle \(\mathrm{ \phi}\) where

\(\mathrm{\phi = \tan-10.4/0.15 = 69½}\) up to the left.