These are original Marlbridge practice questions, not past exam questions. The mark points are a suggested Marlbridge rubric to help you check your work; they are not an official scoring guideline. This set is for the calculus-based Physics C: Mechanics course. In every formula, x is in m and t in s, so each numerical coefficient carries whatever unit makes the term correct. Springs are ideal and strings light unless stated, and pendulums swing through small angles. Use g = 9.8 m/s² on Earth. Tarn is a fictional moon. Round final answers to 2 significant figures unless told otherwise.
Question 1 (multiple choice · foundation)
Take +x to the right, origin at equilibrium. Newton’s second law for a cart gives d²x/dt² = −49x. What is the period of its motion?
- (A) 0.13 s
- (B) 0.14 s
- (C) 0.90 s
- (D) 44 s
Answer and explanation
Answer: (C). The constant is ω² = 49 s⁻², so ω = 7.0 rad/s and T = 2π/ω = 2π ÷ 7.0 = 0.90 s.
- (A) uses 2π/ω² = 2π ÷ 49. The constant in the equation is ω², not ω.
- (B) is 1/ω = 1 ÷ 7.0. That treats ω as a frequency in hertz and forgets the 2π.
- (D) is 2πω. The period is 2π divided by ω, not multiplied.
Question 2 (multiple choice · core)
A block of mass m oscillates on one spring of constant k with period T. It is replaced by a block of mass 2m, hung from two of the same springs joined end to end. What is the new period?
- (A) T/√2
- (B) T
- (C) √2 T
- (D) 2T
Answer and explanation
Answer: (D). Two equal springs in series have k_eff = k/2. The new period is 2π√(2m ÷ (k/2)) = 2π√(4m/k) = 2T.
- (A) ignores the doubled mass and treats joining springs end to end as making them stiffer (k_eff = 2k): 2π√(m/2k) = T/√2.
- (B) assumes joining springs end to end makes them stiffer (k_eff = 2k), which would cancel the doubled mass. Series springs are softer.
- (C) includes the doubled mass but ignores the change in spring constant.
Question 3 (multiple choice · core)
A pendulum clock and a clock driven by a mass on a spring both keep correct time at sea level on Earth. Both are taken to a fictional planet where g is 1.5 times as large. Which statement is correct?
- (A) Both clocks run fast.
- (B) The pendulum clock runs fast; the spring clock keeps correct time.
- (C) The pendulum clock runs slow; the spring clock keeps correct time.
- (D) Both clocks keep correct time, because neither period depends on mass.
Answer and explanation
Answer: (B). T_pendulum = 2π√(ℓ/g): with g × 1.5, T falls by a factor 1/√1.5 ≈ 0.82. The pendulum completes more cycles per real second, so the clock runs fast. T_spring = 2π√(m/k) has no g, so the spring clock is unaffected.
- (A) assumes the spring’s period depends on g. Gravity only shifts the spring’s equilibrium.
- (C) gets the direction wrong: a stronger field gives a shorter pendulum period, so the clock gains time.
- (D) mixes up mass and g. Mass cancels for the pendulum, but g does not.
Question 4 (calculation · core)
Take +y upward, origin at equilibrium. A 1500 kg test platform rests on four identical vertical springs, side by side, each with k = 2.0 × 10⁴ N/m. Treat the platform as a block that moves only up and down.
(a) Find the period and frequency of its vertical oscillation. (b) A 300 kg load is fixed to the platform. Find the new period and frequency. (c) By what factor did the period change? Check this using a factor-of-change argument.
Worked solution
- (a) Parallel springs: k_eff = 4 × 2.0 × 10⁴ = 8.0 × 10⁴ N/m. T = 2π√(1500 ÷ 8.0 × 10⁴) = 0.86 s. f = 1/T = 1.2 Hz.
- (b) T = 2π√(1800 ÷ 8.0 × 10⁴) = 0.94 s. f = 1.1 Hz.
- (c) 0.942 ÷ 0.860 = 1.10. Factor of change: mass × 1800/1500 = 1.2, so T × √1.2 = 1.10. They agree.
Suggested mark points (3): 1 for k_eff = 8.0 × 10⁴ N/m; 1 for both periods and frequencies; 1 for the factor √1.2 ≈ 1.10.
Common error: using one spring’s k. That gives a period twice as long, because the four springs share the load.
Question 5 (derivation · core)
A simple pendulum has a bob of mass m on a light string of length ℓ. Let s be the bob’s displacement measured along its arc from the lowest point, positive to the right, and θ = s/ℓ the angle from the vertical.
(a) Find the component of the net force on the bob along the arc, in terms of m, g and θ. (b) Use Newton’s second law along the arc to write a differential equation for s. (c) Show that for small angles the motion is SHM, and find its period. (d) Explain physically why the period does not depend on m.
Worked solution and suggested Marlbridge rubric
(a) The tension is perpendicular to the arc, so it has no component along it. The weight’s component along the arc is −mg sin θ: it points back toward the lowest point.
(b) m d²s/dt² = −mg sin θ = −mg sin(s/ℓ).
(c) For small θ in radians, sin(s/ℓ) ≈ s/ℓ, so d²s/dt² = −(g/ℓ)s. This has the SHM form with ω² = g/ℓ. So ω = √(g/ℓ) and T = 2π√(ℓ/g).
(d) The restoring force is a part of the weight, which is proportional to m. The inertia that resists the change in motion is also m. A heavier bob is pulled back harder in exact proportion to how much harder it is to accelerate, so its motion is the same.
| Point | What earns it |
|---|---|
| 1 | (a) Tangential component −mg sin θ, with tension correctly excluded |
| 1 | (b) Newton’s second law along the arc with s = ℓθ |
| 1 | (c) Small-angle approximation in radians giving d²s/dt² = −(g/ℓ)s |
| 1 | (c) Identifies ω² = g/ℓ and reaches T = 2π√(ℓ/g) |
| 1 | (d) Weight and inertia both proportional to m, so m cancels |
Alternative method. Using torque about the pivot, τ = −mgℓ sin θ with I = mℓ², earns the (a)–(c) points if each step is shown.
Question 6 (constructed response · stretch)
A student hangs a 0.500 kg block from n identical springs side by side (n = 1 to 5) and times 20 cycles for each:
| n | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| T (s) | 0.993 | 0.702 | 0.574 | 0.497 | 0.444 |
(a) Show that the model predicts T² = (4π²m/k) × (1/n), where k is the constant of one spring. (b) State what to plot to get a straight line through the origin. (c) Use the data to find k. (d) Describe a graph that tests the power of n in T ∝ nᵖ without assuming p = −½, and state the value of p you expect.
Worked solution and suggested Marlbridge rubric
(a) n springs in parallel have k_eff = nk. So T = 2π√(m/(nk)), and squaring: T² = (4π²m/k)(1/n).
(b) Plot T² (s²) against 1/n. The model gives a straight line through the origin with slope 4π²m/k.
(c) T² values: 0.986, 0.493, 0.329, 0.247, 0.197 s² at 1/n = 1, 0.50, 0.33, 0.25, 0.20. The best-fit slope is 0.986 s². So k = 4π² × 0.500 ÷ 0.986 = 20 N/m.
(d) Plot ln T against ln n. If T = Cnᵖ, then ln T = ln C + p ln n, a straight line with slope p. The model predicts p = −0.50; these data give a slope of −0.50.
| Point | What earns it |
|---|---|
| 1 | (a) k_eff = nk and correct squaring of the period formula |
| 1 | (b) T² against 1/n, with slope identified as 4π²m/k |
| 1 | (c) Slope found from the data (not a single point) |
| 1 | (c) k = 20 N/m with unit (carry forward an incorrect slope) |
| 1 | (d) Log–log graph with slope = p, and p = −½ predicted |
Alternative method for (c). Fitting T against 1/√n gives slope 2π√(m/k) = 0.993 s, and k = 20 N/m. This earns both (c) points.
Question 7 (constructed response · stretch)
An astronaut on the fictional moon Tarn has two devices. A simple pendulum of length 0.500 m has a period of 2.22 s there. A block on a vertical spring had a period of 0.80 s on Earth.
(a) Find g on Tarn. (b) Find the period of the spring device on Tarn. (c) Find the static stretch of the spring on Earth and on Tarn. (d) A second astronaut says: “On Tarn the spring stretches less, so it acts stiffer and should oscillate faster.” Explain the error. (e) Show how the spring device alone could still be used to measure g on Tarn.
Model answer and suggested Marlbridge rubric
(a) g = 4π²ℓ/T² = 4π² × 0.500 ÷ 2.22² = 4.0 m/s².
(b) T = 2π√(m/k) contains no g, so T is still 0.80 s.
(c) From T = 2π√(d/g), d = gT²/(4π²). On Earth: d = 9.8 × 0.64 ÷ 39.5 = 0.16 m. On Tarn: d = 4.0 × 0.64 ÷ 39.5 = 0.065 m.
(d) k is a property of the spring and does not change. The spring stretches less only because the weight is smaller. The period depends on m/k, which is the same on both worlds. In T = 2π√(d/g), both d and g fall by the same factor, so their ratio, and T, do not change.
(e) Measure the static stretch d on Tarn and the period T. Then g = 4π²d/T² = 4π² × 0.065 ÷ 0.80² ≈ 4.0 m/s², agreeing with (a).
| Point | What earns it |
|---|---|
| 1 | (a) g = 4.0 m/s² from the pendulum |
| 1 | (b) Spring period unchanged, with the reason (no g in 2π√(m/k)) |
| 1 | (c) Both static stretches |
| 1 | (d) k unchanged; d and g scale together so d/g is constant |
| 1 | (e) g = 4π²d/T² from the spring’s sag and period |
How did you do?
- Q1 wrong: re-read “From the equation of motion to the period” in the study guide.
- Q2 or Q4 wrong: revisit effective spring constants and Worked example 1.
- Q3 or Q7 wrong: compare which quantities appear in each period formula; see “Factors of change” and Worked example 3.
- Q5 incomplete: follow the pendulum derivation and Figure 1.
- Q6 incomplete: compare with Worked example 2 and Figure 2.
Then tick off the topic checklist.