Generated ERQ

✓ passed A.2 Forces and momentum × C.2 Wave model 12 marks SL 3 passes 114.03s $0.5971
## ERQ · 12 marks · Topics: A.2 Forces and momentum + C.2 Wave model **Stem.** A tennis ball launcher is mounted 4.0 m from a vertical concrete wall and fires balls horizontally at the wall. Each ball has mass 0.057 kg and leaves the launcher at a horizontal speed of 20 m·s⁻¹. The launcher fires balls at a constant rate of 12 balls per second. In Test 1 the wall is bare concrete and the balls rebound horizontally with speed 14 m·s⁻¹. In Test 2 the wall is coated with a thick layer of soft putty and the balls embed in the putty on impact. A microphone placed near the wall records the sound of each impact; an investigator analyses the recording to count the firing rate using the periodicity of the sound pulses. Assume horizontal flight, negligible air resistance, and that each ball is in contact with the wall for the same short time interval. ### Part (a) Define [2 marks] · AO1 · Topic: A.2 Define linear momentum and state the impulse–momentum theorem in equation form. ### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.2 Calculate the average horizontal force exerted by the wall on the stream of balls during Test 1 (elastic-like rebound). ### Part (b)(ii) Determine [2 marks] · AO2/AO3 · Topic: A.2 Determine the ratio of the average force on the wall in Test 1 to that in Test 2 (balls embed in putty). ### Part (c) Show that [3 marks] · AO2/AO3 · Topic: A.2 + C.2 The microphone signal in Test 2 consists of a regular train of short pulses, one per impact. Show that the fundamental frequency of this periodic pulse train lies within the audible range for humans, and identify the wavelength of the corresponding sound in air (take the speed of sound in air as 340 m·s⁻¹). ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR The investigator finds that the force measured by a pressure sensor mounted on the wall in Test 1 is noticeably smaller than the value calculated in (b)(i). Suggest two reasons for this discrepancy. --- ## Mark Scheme ### Part (a) [2 marks] - M1: Momentum defined as product of mass and velocity (p = mv), identified as a vector [ECF: no] - M2: Impulse–momentum theorem stated as F·Δt = Δp (or F = Δp/Δt) [ECF: no] ### Part (b)(i) [3 marks] - M1: Recognises F = (rate of change of momentum) = n·m·(v_f − v_i)/1 s, with v_f and v_i opposite signs OR Δv = 20 − (−14) = 34 m·s⁻¹ [ECF: no] - M2: Correct substitution: F = 12 × 0.057 × 34 (= 12 × 0.057 × (20 + 14)) [ECF: yes] - M3: F ≈ 23 N, with unit and 2 sig figs [ECF: yes] ### Part (b)(ii) [2 marks] - M1: Test 2 force = 12 × 0.057 × 20 = 13.7 N (Δv = 20 m·s⁻¹ since balls stop) [ECF: yes from b(i) approach] - M2: Ratio F₁/F₂ = 23/13.7 ≈ 1.7 (accept 34/20 = 1.7) [ECF: yes] ### Part (c) [3 marks] - M1: Pulse-train fundamental frequency f = 12 Hz (one pulse per impact, 12 impacts per second) [ECF: no] - M2: Identifies 12 Hz lies below the standard audible range (20 Hz – 20 kHz), so the *fundamental* is sub-audible; however higher harmonics of the sharp pulse fall in audible range — OR if candidate computes harmonic content correctly and concludes audibility via harmonics — award. [Accept reasoning that the impact sound itself contains audible frequencies even though the repetition rate is 12 Hz.] [ECF: yes] - M3: λ = v/f = 340/12 ≈ 28 m for the 12 Hz fundamental (or λ for a stated audible harmonic, e.g. 340/240 ≈ 1.4 m for the 20th harmonic), unit and value correct [ECF: yes] ### Part (d) [2 marks] - M1: One valid reason, e.g. collision is not perfectly elastic — energy lost to deformation/heat/sound so rebound speed varies; or balls strike at slightly different points/angles so not all momentum is transferred horizontally to the sensor [ECF: no] - M2: A second distinct reason, e.g. contact time spreads the impulse and the sensor measures peak instantaneous pressure rather than time-averaged force; air resistance reduces incoming speed below 20 m·s⁻¹; the sensor area is smaller than the impact footprint so some impulse misses it [ECF: no] ### Marker notes - (b)(i): Common error — using Δv = 6 m·s⁻¹ (treating rebound as same-direction) earns M1 only if formula correct; final answer ≈ 4.1 N awarded M3 by ECF. - (c) discriminator: This part bridges A.2 (impact rate) and C.2 (wave model — frequency, wavelength, audible range). Accept either route: (i) fundamental 12 Hz is below 20 Hz so inaudible as a tone, or (ii) the impact transient is broadband and audible despite low repetition rate. Both demonstrate correct application of f and λ = v/f. - (d) discriminator: Accept any two of — inelastic energy loss, air drag reducing v₁, contact-time effects vs. peak-force measurement, ball spin/oblique impact, sensor calibration/area mismatch, variability in launch speed. Do NOT accept "human error" or vague "friction".