← Study with ScholarAB
Physics 30 Practice Set 1
Twenty diploma-style questions with diagrams, answers and worked solutions. Every question was written by ScholarAB. This is not an official Alberta Education examination.
Keep the official Physics 30 data sheets open while you work, since the real exam gives you the same sheets. Work each question on paper first, then open its answer.
- Momentum and impulse questions 1 to 3
- Forces and fields questions 4 to 10
- Electromagnetic radiation questions 11 to 15
- Atomic physics questions 16 to 20
Use the following information to answer the next two questions.
During a game of crokinole, a 6.00 g disc sliding east at 1.50 m/s strikes an identical disc that is at rest. Friction during the collision can be ignored.
Vector arrows are not drawn to scale.
1
The speed of Disc 2 immediately after the collision is blank m/s.
(Record your three-digit answer.)
Show the answer to question 1
The answer is 1.15 m/s
- The discs have equal mass, so momentum is conserved in the velocities alone.
- After the collision Disc 1 has components 0.900 cos 50.0° = 0.579 m/s east and 0.900 sin 50.0° = 0.689 m/s north.
- Disc 2 carries the rest, 1.50 − 0.579 = 0.921 m/s east and 0.689 m/s south.
- Its speed is √(0.921² + 0.689²) = 1.15 m/s, at 36.8° south of east.
2.
The collision is classified as i because ii.
The statement above is completed by the information in row
The statement above is completed by the information in row
| Row | i | ii |
|---|---|---|
| A. | elastic | the total momentum is the same before and after |
| B. | elastic | the total kinetic energy is the same before and after |
| C. | inelastic | the discs move apart at an angle |
| D. | inelastic | the total kinetic energy is less after the collision than before |
Show the answer to question 2
The answer is D
- Kinetic energy before is ½(0.00600 kg)(1.50 m/s)² = 6.75 × 10⁻³ J.
- After, it is ½(0.00600 kg)(0.900² + 1.15²) = 6.40 × 10⁻³ J, about 5% less, so the collision is inelastic.
- Row A is the trap. Momentum is conserved in every isolated collision, so it cannot tell elastic from inelastic.
Use the following information to answer the next question.
A volleyball and a beanbag of equal mass hit a gym wall head on at the same speed. The volleyball bounces straight back with about the same speed. The beanbag stops and drops.
3.
Compared with the impulse that the wall gives the beanbag, the impulse that the wall gives the volleyball is
A.smaller, because the volleyball is in contact with the wall for less timeB.the same, because both objects had the same momentum before impactC.greater, because the volleyball has a greater change in momentumD.the same, because both objects exert the same force on the wall
Show the answer to question 3
The answer is C
- Impulse equals the change in momentum.
- The beanbag goes from mv to zero, a change of mv. The volleyball reverses direction, a change of about 2mv.
- B is the trap. Equal momentum before impact says nothing about the change.
Use the following information to answer the next question.
Two point charges are fixed 0.400 m apart. Point P is halfway between them.
4
The magnitude of the electric field at point P, expressed in scientific notation, is a.bc × 10d N/C. The values of a, b, c, and d are blank, blank, blank, and blank.
(Record all four digits of your answer.)
Show the answer to question 4
The answer is 1126, for 1.12 × 10⁶ N/C
- Each charge is 0.200 m from P.
- The +3.00 μC charge gives (8.99 × 10⁹)(3.00 × 10⁻⁶) / 0.200² = 6.74 × 10⁵ N/C, pointing away from it, toward the negative charge.
- The −2.00 μC charge gives (8.99 × 10⁹)(2.00 × 10⁻⁶) / 0.200² = 4.50 × 10⁵ N/C, pointing toward it, which is the same direction.
- Fields in the same direction add, so the total is 1.12 × 10⁶ N/C. Subtracting them gives 2.25 × 10⁵, the most common wrong answer.
5
An electron moving at 3.00 × 106 m/s passes through a potential difference of 15.0 V that slows it down. The speed of the electron afterward is blank × 106 m/s.
(Record your three-digit answer.)
Show the answer to question 5
The answer is 1.93
- Kinetic energy before is ½(9.11 × 10⁻³¹ kg)(3.00 × 10⁶ m/s)² = 4.10 × 10⁻¹⁸ J.
- The field takes away qV = (1.60 × 10⁻¹⁹ C)(15.0 V) = 2.40 × 10⁻¹⁸ J, leaving 1.70 × 10⁻¹⁸ J.
- Then v = √(2 × 1.70 × 10⁻¹⁸ / 9.11 × 10⁻³¹) = 1.93 × 10⁶ m/s.
- Adding the energy instead of taking it away gives 3.78, a common slip when a field slows a charge down.
Use the following information to answer the next question.
An electron moves toward the right side of the page through a uniform magnetic field that is directed out of the page.
6.
Which of the following diagrams shows the direction of the magnetic force on the electron at the instant shown?
A.
B.
C.
D.
• Represents a magnetic field directed out of the page
Show the answer to question 6
The answer is A
- Use the left-hand rule for a negative charge, with the thumb along the velocity and the fingers along the field. The palm faces the top of the page.
- B is the answer for a positive charge.
- A magnetic force is always perpendicular to both the velocity and the field, which rules out C and D.
Use the following information to answer the next question.
A straight horizontal wire that has a mass of 12.0 g and a length of 0.250 m lies between the poles of a magnet. The uniform magnetic field between the poles is horizontal, perpendicular to the wire, and has a magnitude of 0.300 T.
⊗ Represents electron flow directed into the page
7
The current in the wire that makes the magnetic force on the wire exactly balance the gravitational force on it is blank A.
(Record your three-digit answer.)
Show the answer to question 7
The answer is 1.57 A
- The forces balance when IlB = mg.
- I = (0.0120 kg)(9.81 m/s²) / ((0.250 m)(0.300 T)) = 1.57 A.
- Leaving the mass as 12.0 instead of 0.0120 kg gives an answer a thousand times too big.
Use the following information to answer the next question.
An electronic air cleaner in a home furnace charges dust particles and then passes the air between two parallel metal plates that have a potential difference across them.
8.
The distance between the plates is halved and the potential difference does not change. Compared with before, the electric force on a charged dust particle between the plates is
A.half as greatB.the sameC.twice as greatD.four times as great
Show the answer to question 8
The answer is C
- Between parallel plates E = V/d. Halving d with V fixed doubles E, and F = qE, so the force doubles.
- D treats the plates like point charges with an inverse square law.
- B mixes up uniform with constant. The field is the same everywhere between the plates, but it still depends on V and d.
Use the following information to answer the next two questions.
Some silicon wafers used in electronics are treated by firing protons into them. In one machine, protons start from rest, are accelerated through a potential difference of 5.00 kV, and then enter a uniform 0.0800 T magnetic field perpendicular to their velocity. The field bends the beam onto the wafer.
× Represents a magnetic field directed into the page
9
The radius of the circular path of the protons in the magnetic field is blank cm.
(Record your three-digit answer.)
Show the answer to question 9
The answer is 12.8 cm
- Each proton gains qV = (1.60 × 10⁻¹⁹ C)(5.00 × 10³ V) = 8.00 × 10⁻¹⁶ J.
- So v = √(2 × 8.00 × 10⁻¹⁶ / 1.67 × 10⁻²⁷) = 9.79 × 10⁵ m/s.
- The magnetic force is the centripetal force, qvB = mv²/r, so r = mv/(qB) = (1.67 × 10⁻²⁷)(9.79 × 10⁵) / ((1.60 × 10⁻¹⁹)(0.0800)) = 0.128 m, or 12.8 cm.
- Reading 5.00 kV as 5.00 V gives 0.404 cm.
10
Two of the physics principles listed on the data sheet are needed to determine the radius in the previous question. The numbers of these two principles are blank and blank.
(Record your answer in either order.)
Show the answer to question 10
The answer is 25, or 52
- Principle 5, conservation of energy, turns qV into kinetic energy.
- Principle 2, uniform circular motion, sets the magnetic force equal to the centripetal force.
- Principle 3, the work energy theorem, describes the same first step, so 23 is a defensible answer too. Check how your teacher marks it.
Use the following information to answer the next two questions.
Violet light that has a wavelength of 414 nm shines on a metal surface in the apparatus shown below. The work function of the metal is 2.10 eV.
11
The stopping voltage for the emitted photoelectrons is blank V.
(Record your answer to two digits.)
Show the answer to question 11
The answer is 0.90 V
- Each photon has E = hc/λ = (4.14 × 10⁻¹⁵ eV·s)(3.00 × 10⁸ m/s) / (414 × 10⁻⁹ m) = 3.00 eV.
- The fastest photoelectrons leave with 3.00 − 2.10 = 0.90 eV, so the stopping voltage is 0.90 V.
12.
The violet light is made twice as bright and its wavelength does not change. Compared with before, the photoelectric current and the maximum kinetic energy of the photoelectrons will be, respectively,
A.greater, greaterB.greater, the sameC.the same, greaterD.the same, the same
Show the answer to question 12
The answer is B
- Brighter light means more photons each second, so more electrons each second and a greater current.
- Each photon still carries 3.00 eV, so the maximum kinetic energy of each electron does not change.
- A is what the wave model predicted, and the photoelectric effect is the experiment that disproved it.
Use the following information to answer the next question.
A laser shines on a diffraction grating that has 600 lines per millimetre. The first-order bright lines appear on a screen at an angle of 22.3° from the central bright line.
The diagram is not drawn to scale.
13
The wavelength of the laser light, expressed in scientific notation, is a.bc × 10−d m. The values of a, b, c, and d are blank, blank, blank, and blank.
(Record all four digits of your answer.)
Show the answer to question 13
The answer is 6327, for 6.32 × 10⁻⁷ m
- The slit spacing is d = 1 / (600 000 lines per metre) = 1.667 × 10⁻⁶ m.
- λ = d sin θ / n = (1.667 × 10⁻⁶ m)(sin 22.3°) / 1 = 6.32 × 10⁻⁷ m, red light.
- Forgetting to convert millimetres to metres gives an answer 1 000 times off.
Use the following information to answer the next question.
A student places a small candle 10.0 cm from a converging lens that has a focal length of 15.0 cm.
14.
Which row describes the image of the candle?
| Row | Type | Orientation | Size compared with the candle |
|---|---|---|---|
| A. | real | inverted | larger |
| B. | virtual | upright | larger |
| C. | virtual | upright | smaller |
| D. | real | inverted | smaller |
Show the answer to question 14
The answer is B
- 1/di = 1/15.0 − 1/10.0 = −1/30.0, so di = −30.0 cm. A negative image distance means a virtual image on the candle side of the lens.
- The magnification is −di/do = 30.0/10.0 = 3.00, positive so upright, and three times larger. The lens is working as a magnifying glass.
- A is what happens once the candle is beyond the focal point.
Use the following information to answer the next question.
A radio wave travels east across the prairie. At one instant, the electric field of the wave points straight up.
15.
At that instant, the magnetic field of the wave points
A.straight upB.east, along the direction of travelC.horizontally, along a north to south lineD.straight down
Show the answer to question 15
The answer is C
- In electromagnetic radiation the electric field, the magnetic field and the direction of travel are all perpendicular to each other.
- With travel east and the electric field vertical, the magnetic field lies along a north to south line.
- B would make it a longitudinal wave, and EMR is transverse.
Use the following information to answer the next two questions.
Energy Levels of Gas X
16.
An electron that has a kinetic energy of 5.00 eV collides with an atom of gas X in its ground state. Which of the following statements is correct?
A.The atom cannot absorb any energy, because 5.00 eV does not match a difference between levels.B.The atom can move to n = 2, and the electron leaves with 0.86 eV of kinetic energy.C.The atom can move to n = 3, and the electron leaves with no kinetic energy.D.The atom is ionized, because 5.00 eV is more than half of 9.80 eV.
Show the answer to question 16
The answer is B
- Moving from −9.80 eV to −5.66 eV takes 4.14 eV, which is less than 5.00 eV. The electron keeps the rest, 5.00 − 4.14 = 0.86 eV.
- Reaching n = 3 takes 6.60 eV, which is too much, and ionizing takes 9.80 eV.
- A is the trap. It is true for a photon, which must match a gap exactly. A colliding electron can give up just part of its energy.
17
When an atom of gas X returns from n = 2 to the ground state, it emits a photon that has a wavelength of blank × 10−7 m.
(Record your three-digit answer.)
Show the answer to question 17
The answer is 3.00
- The photon carries the 4.14 eV gap.
- λ = hc/E = (4.14 × 10⁻¹⁵ eV·s)(3.00 × 10⁸ m/s) / 4.14 eV = 3.00 × 10⁻⁷ m, in the ultraviolet.
18.
Some exit signs glow without electricity because they contain tritium, which undergoes beta-negative decay. Which of the following equations represents this decay?
A.31H → 32He + 0−1β + νB.31H → 21H + 10nC.31H → 32He + 0+1β + ν̄D.31H → 32He + 0−1β + ν̄
Show the answer to question 18
The answer is D
- In beta-negative decay a neutron becomes a proton, so the atomic number rises from 1 to 2 and the mass number stays 3.
- Charge balances, +1 = +2 − 1, and so do nucleons, 3 = 3. The electron is always paired with an antineutrino.
- A pairs the electron with a neutrino. B balances but is neutron emission. C does not conserve charge.
Use the following information to answer the next question.
Activity of a Medical Isotope as a Function of Time
19
The half-life of the isotope is blank days.
(Record your three-digit answer.)
Show the answer to question 19
The answer is 8.00 days (anything from 7.80 to 8.20 read from the graph)
- The activity starts at 640 kBq. Half of that, 320 kBq, is reached at 8.00 days.
- Check with a second halving. 160 kBq comes at 16.0 days, another 8.00 days later.
Use the following information to answer the next question.
Binding Energy per Nucleon of Selected Nuclei
20.
Both the fusion of very light nuclei and the fission of very heavy nuclei release energy because, in each case, the products
A.have a greater binding energy per nucleon than the reactantsB.have a smaller binding energy per nucleon than the reactantsC.contain more nucleons in total than the reactantsD.have a greater total mass than the reactants
Show the answer to question 20
The answer is A
- Both reactions move toward the iron peak of the graph. Products that are more tightly bound per nucleon have less mass, and the lost mass is the energy released, E = mc².
- B reads the graph backwards. C breaks conservation of nucleons. D is the mass trap, because the products have less total mass.
ScholarAB practice, original itemsPhysics 30
Official practice from Alberta Education
Real released diploma questions and the exam blueprint are free on alberta.ca. Your school can also sign you in to the practice tests on the digital exam platform.