High School — Ontario Grade 12

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SPH4U is Ontario's Grade 12 university-preparation physics course: dynamics, energy and momentum, gravitational, electric and magnetic fields, the wave nature of light, and modern physics. Exams combine multi-step problems — forces on inclines, collisions, circular orbits — with conceptual questions on fields, interference and the photoelectric effect. It's a prerequisite for engineering and physical-science programs, and the final evaluation counts for 30% of the course mark.

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Topics typically tested

  • Dynamics & uniform circular motion
  • Energy, work & momentum
  • Gravitational fields & orbits
  • Electric & magnetic fields
  • Wave nature of light
  • Modern physics: quantum & relativity

Sample SPH4U Physics exam questions

Three original questions in the style of a real SPH4U Physics exam, each with a worked solution. Your own practice exam can be built from your course's notes and syllabus.

Question 1Multiple choiceGravitational fields & orbits

A satellite is in a stable circular orbit at an altitude of 6.00×102 km6.00\times10^{2}\ \text{km} above Earth's surface. Take Earth's mass to be 5.98×1024 kg5.98\times10^{24}\ \text{kg}, Earth's radius to be 6.38×106 m6.38\times10^{6}\ \text{m}, and G=6.67×10−11 N⋅m2/kg2G = 6.67\times10^{-11}\ \text{N}\cdot\text{m}^2/\text{kg}^2. What is the orbital speed of the satellite?

  1. A.

    7.91×103 m/s7.91\times10^{3}\ \text{m/s}

  2. B.

    8.27×103 m/s8.27\times10^{3}\ \text{m/s}

  3. C.

    7.56×103 m/s7.56\times10^{3}\ \text{m/s}

  4. D.

    1.07×104 m/s1.07\times10^{4}\ \text{m/s}

Worked solution
Answer:C

For a circular orbit, gravity is the only force acting on the satellite, and it supplies the centripetal force:

GMmr2=mv2r  ⇒  v=GMr\frac{GMm}{r^2} = \frac{mv^2}{r} \;\Rightarrow\; v = \sqrt{\frac{GM}{r}}

The satellite's mass cancels, so the speed depends only on Earth's mass and on rr, the distance from Earth's centre, not from its surface. Add the altitude to Earth's radius:

r=6.38×106+6.00×105=6.98×106 mr = 6.38\times10^{6} + 6.00\times10^{5} = 6.98\times10^{6}\ \text{m} v=(6.67×10−11)(5.98×1024)6.98×106=5.71×107=7.56×103 m/sv = \sqrt{\frac{(6.67\times10^{-11})(5.98\times10^{24})}{6.98\times10^{6}}} = \sqrt{5.71\times10^{7}} = 7.56\times10^{3}\ \text{m/s}

As a check by a second route, the gravitational field strength at that radius is g=GM/r2=8.19 N/kgg = GM/r^2 = 8.19\ \text{N/kg}, and v=gr=(8.19)(6.98×106)v = \sqrt{gr} = \sqrt{(8.19)(6.98\times10^{6})} gives the same 7.56×103 m/s7.56\times10^{3}\ \text{m/s}. The period is then T=2πr/v=5.80×103 sT = 2\pi r/v = 5.80\times10^{3}\ \text{s}, about 97 min, which is typical of a low Earth orbit.

The three distractors are the three standard slips. 7.91×103 m/s7.91\times10^{3}\ \text{m/s} comes from using Earth's radius alone as rr, so the altitude is ignored. 8.27×103 m/s8.27\times10^{3}\ \text{m/s} comes from v=grv = \sqrt{gr} with the surface value g=9.8 m/s2g = 9.8\ \text{m/s}^2; the field is weaker 600 km up, so this overestimates the speed. 1.07×104 m/s1.07\times10^{4}\ \text{m/s} is the escape speed 2GM/r\sqrt{2GM/r}, which is 2\sqrt{2} times the orbital speed; it describes a satellite leaving Earth for good, not one in a stable orbit.

v=7.56×103 m/sv = 7.56\times10^{3}\ \text{m/s} (option C)

Question 2Short answerElectric & magnetic fields

An electron, initially at rest, is accelerated through a potential difference of 2.50×102 V2.50\times10^{2}\ \text{V}. It then enters a region of uniform magnetic field of magnitude 1.50×10−3 T1.50\times10^{-3}\ \text{T} directed perpendicular to its velocity. Take e=1.60×10−19 Ce = 1.60\times10^{-19}\ \text{C} and the mass of the electron to be 9.11×10−31 kg9.11\times10^{-31}\ \text{kg}. (a) Calculate the speed of the electron as it leaves the accelerating region. (b) Calculate the radius of the electron's circular path in the magnetic field.

Worked solution
Answer:

(a) 9.37×106 m/s9.37\times10^{6}\ \text{m/s}; (b) 3.56×10−2 m3.56\times10^{-2}\ \text{m} (about 3.6 cm)

(a) The electric field does work on the electron as it crosses the potential difference, and that work becomes kinetic energy. A charge qq moved through a potential difference ΔV\Delta V gains energy q ΔVq\,\Delta V, so with the electron starting from rest

e ΔV=12mv2  ⇒  v=2e ΔVme\,\Delta V = \tfrac{1}{2}mv^2 \;\Rightarrow\; v = \sqrt{\frac{2e\,\Delta V}{m}} v=2(1.60×10−19)(2.50×102)9.11×10−31=8.78×1013=9.37×106 m/sv = \sqrt{\frac{2(1.60\times10^{-19})(2.50\times10^{2})}{9.11\times10^{-31}}} = \sqrt{8.78\times10^{13}} = 9.37\times10^{6}\ \text{m/s}

The energy gained is e ΔV=4.00×10−17 Je\,\Delta V = 4.00\times10^{-17}\ \text{J}, which is 250 eV: a 250 V accelerating voltage always produces a 250 eV electron. The speed is about 3% of the speed of light, so the classical kinetic energy formula is adequate here.

(b) In the magnetic field the magnetic force F=qvBF = qvB is always perpendicular to the velocity, so it changes the electron's direction but never its speed. The path is a circle, with the magnetic force acting as the centripetal force:

qvB=mv2r  ⇒  r=mvqBqvB = \frac{mv^2}{r} \;\Rightarrow\; r = \frac{mv}{qB} r=(9.11×10−31)(9.37×106)(1.60×10−19)(1.50×10−3)=8.54×10−242.40×10−22=3.56×10−2 mr = \frac{(9.11\times10^{-31})(9.37\times10^{6})}{(1.60\times10^{-19})(1.50\times10^{-3})} = \frac{8.54\times10^{-24}}{2.40\times10^{-22}} = 3.56\times10^{-2}\ \text{m}

A second route skips the speed entirely: substituting vv from part (a) gives r=1B2m ΔVer = \dfrac{1}{B}\sqrt{\dfrac{2m\,\Delta V}{e}}, which evaluates to the same 3.56 cm. The electron's negative charge decides which way it curves (opposite to the right-hand rule for a positive charge); it has no effect on the size of the radius.

Common slips: dropping the factor of 2 in the kinetic energy gives 6.63×106 m/s6.63\times10^{6}\ \text{m/s}, and leaving v2v^2 in the numerator, r=mv2/qBr = mv^2/qB, is dimensionally wrong (the units do not give metres). Notice that the radius grows with ΔV\sqrt{\Delta V} and shrinks as BB increases: a faster electron is harder to bend, and a stronger field bends it more tightly.

(a) v=9.37×106 m/sv = 9.37\times10^{6}\ \text{m/s}; (b) r=3.56×10−2 mr = 3.56\times10^{-2}\ \text{m}, about 3.6 cm

Question 3Worked problemEnergy, work & momentum

Block A, of mass 2.00 kg, is released from rest at the top of a frictionless ramp of height 1.80 m. At the bottom of the ramp it slides along a frictionless horizontal surface and collides with block B, of mass 3.00 kg, which is at rest. The two blocks stick together. The combined blocks then slide into a horizontal spring of force constant 4.50×102 N/m4.50\times10^{2}\ \text{N/m} whose other end is fixed to a wall. Take g=9.8 m/s2g = 9.8\ \text{m/s}^2.

(a) Determine the speed of block A at the bottom of the ramp.

(b) Determine the speed of the two blocks immediately after the collision.

(c) Determine the kinetic energy lost in the collision, and state what has happened to it.

(d) Determine the maximum compression of the spring.

Worked solution
Answer:

(a) 5.94 m/s; (b) 2.38 m/s; (c) 21.2 J lost (60% of the kinetic energy), converted to thermal energy, sound and deformation; (d) 0.250 m

(a) The ramp is frictionless and the normal force does no work, so mechanical energy is conserved from the top of the ramp to the bottom. Gravitational potential energy becomes kinetic energy:

mAgh=12mAvA2  ⇒  vA=2gh=2(9.8)(1.80)=35.28=5.940 m/s≈5.94 m/sm_A g h = \tfrac{1}{2} m_A v_A^2 \;\Rightarrow\; v_A = \sqrt{2gh} = \sqrt{2(9.8)(1.80)} = \sqrt{35.28} = 5.940\ \text{m/s} \approx 5.94\ \text{m/s}

The mass cancels: the speed at the bottom depends only on the drop, not on the shape or length of the ramp.

(b) During the collision the blocks push on each other with large forces, but those forces are internal to the two-block system and the surface is frictionless, so the total momentum is conserved. Kinetic energy is not conserved, because the blocks stick together (a perfectly inelastic collision). Taking the direction of A's motion as positive:

mAvA+mB(0)=(mA+mB)v′  ⇒  v′=(2.00)(5.940)2.00+3.00=2.376 m/s≈2.38 m/sm_A v_A + m_B(0) = (m_A + m_B)v' \;\Rightarrow\; v' = \frac{(2.00)(5.940)}{2.00 + 3.00} = 2.376\ \text{m/s} \approx 2.38\ \text{m/s}

(c) Kinetic energy before and after the collision, keeping the unrounded speeds:

Ek,before=12(2.00)(5.940)2=35.28 J,Ek,after=12(5.00)(2.376)2=14.11 JE_{k,\text{before}} = \tfrac{1}{2}(2.00)(5.940)^2 = 35.28\ \text{J}, \qquad E_{k,\text{after}} = \tfrac{1}{2}(5.00)(2.376)^2 = 14.11\ \text{J} ΔEk=35.28−14.11=21.17 J≈21.2 J\Delta E_k = 35.28 - 14.11 = 21.17\ \text{J} \approx 21.2\ \text{J}

Three-fifths of the kinetic energy is lost, which is the fraction mB/(mA+mB)m_B/(m_A + m_B) that a perfectly inelastic collision with a stationary target always loses. It is not destroyed: it becomes thermal energy in the blocks, sound, and the work done permanently deforming whatever makes them stick.

(d) After the collision the surface is frictionless, so mechanical energy is conserved again. At maximum compression the blocks are momentarily at rest and all of their remaining kinetic energy is stored as elastic potential energy:

12(mA+mB)v′2=12kx2  ⇒  x=2Ek,afterk=2(14.11)450=0.250 m\tfrac{1}{2}(m_A + m_B)v'^2 = \tfrac{1}{2}kx^2 \;\Rightarrow\; x = \sqrt{\frac{2E_{k,\text{after}}}{k}} = \sqrt{\frac{2(14.11)}{450}} = 0.250\ \text{m}

The classic trap is to apply energy conservation across the collision, carrying the full 35.28 J through to the spring: that gives an after-collision speed of 3.76 m/s and a compression of 0.396 m, both wrong. Energy conservation holds on the ramp and at the spring; through the collision only momentum is conserved. A second slip is to use block A's mass alone at the spring, forgetting that both blocks are moving; that gives 0.158 m.

(a) vA=5.94 m/sv_A = 5.94\ \text{m/s}; (b) v′=2.38 m/sv' = 2.38\ \text{m/s}; (c) 21.2 J21.2\ \text{J} lost (60%), to thermal energy, sound and deformation; (d) x=0.250 mx = 0.250\ \text{m}

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