Generated ERQ

✓ passed B.1 Thermal energy transfers × A.2 Forces and momentum 12 marks SL 3 passes 106.75s $0.6178
## ERQ · 12 marks · Topics: B.1 Thermal energy transfers + A.2 Forces and momentum **Stem.** A physics student investigates frictional heating in a "collision-slide" apparatus. A copper block A of mass 0.250 kg travels along a horizontal copper rail at 4.00 m s⁻¹ and collides head-on with an identical stationary copper block B. The collision is perfectly inelastic — the blocks stick together and then slide as a single object along the same rail until friction brings them to rest. A thermistor embedded in the combined block records a temperature rise of ΔT = 0.18 K after the blocks stop. The coefficient of kinetic friction between the combined block and the rail is μ_k = 0.35. Take the specific heat capacity of copper as c = 385 J kg⁻¹ K⁻¹ and g = 9.81 m s⁻². Assume the rail itself does not heat up appreciably during the short sliding phase. ### Part (a) Outline [2 marks] · AO1 · Topic: B.1 Outline, in terms of particle behaviour, how the kinetic friction between the sliding block and the rail produces a rise in the internal energy of the block. ### Part (b)(i) Calculate [3 marks] · AO2 · Topic: A.2 + B.1 Using conservation of momentum for the collision **and** the work–energy relation for the subsequent slide, calculate the distance over which the combined block slides on the rail before stopping. ### Part (b)(ii) Calculate [3 marks] · AO2 · Topic: A.2 + B.1 The frictional force acts as an external force during the slide, decelerating the block while simultaneously transferring energy to thermal form. Calculate the time taken for the combined block to come to rest after the collision, using the impulse–momentum theorem. ### Part (c) Calculate [2 marks] · AO2 · Topic: B.1 Calculate the temperature rise that would be predicted if **all** of the thermal energy generated during the slide were deposited in the combined block (total mass 0.500 kg). ### Part (d) Suggest [2 marks] · AO3 · ASSUMPTIONS DISCRIMINATOR The student's measured temperature rise (0.18 K) is smaller than the value calculated in (c). Suggest **two physically distinct** reasons for this discrepancy. --- ## Mark Scheme ### Part (a) [2 marks] - M1: The block's surface asperities do work against the rail; this work increases the random kinetic energy of the lattice ions/atoms of the block. [ECF: no] - M2: The increased mean random kinetic energy of the particles is observed macroscopically as a rise in temperature / internal energy. [ECF: no] ### Part (b)(i) [3 marks] - M1: Apply conservation of momentum: m v₀ = (2m) v_f ⇒ v_f = v₀/2 = 2.00 m s⁻¹. [ECF: no] - M2: Equate post-collision KE to frictional work: ½(2m)v_f² = μ_k(2m)g·d ⇒ d = v_f²/(2μ_k g). [ECF: yes] - M3: d = (2.00)² / (2 × 0.35 × 9.81) = 0.58 m (2 s.f.). [ECF: yes] ### Part (b)(ii) [3 marks] - M1: Frictional force on combined block: F = μ_k(2m)g = 0.35 × 0.500 × 9.81 = 1.72 N. [ECF: no] - M2: Impulse–momentum: F·t = (2m)v_f ⇒ t = (2m)v_f / F (or equivalently t = v_f/(μ_k g)). [ECF: yes] - M3: t = (0.500 × 2.00) / 1.72 = 0.58 s (2 s.f.). [ECF: yes] ### Part (c) [2 marks] - M1: Equate thermal energy to frictional work: Q = μ_k(2m)g·d = ½(2m)v_f² = 1.00 J; apply Q = M c ΔT_pred with M = 0.500 kg. [ECF: yes] - M2: ΔT_pred = 1.00 / (0.500 × 385) = 5.2 × 10⁻³ K (2 s.f.). [ECF: yes] ### Part (d) [2 marks] - M1 (1 mark): Identifies **one** valid physical reason with brief justification. Examples: - Some thermal energy is conducted into the rail (assumption in stem is idealised), so less energy heats the block. - Energy is lost as sound/acoustic vibration during the inelastic collision and during the slide. - Some kinetic energy of the collision is stored as permanent plastic deformation of the blocks, not converted to heat. - Thermal energy is radiated/convected from the block surface to the surroundings during the measurement interval before the thermistor equilibrates. - M2 (1 mark): Identifies a **second, physically distinct** reason with brief justification (drawn from a different mechanism than M1). [ECF: no] ### Marker notes - Award only M2 in part (d) if the two reasons invoke genuinely different physical mechanisms. Paraphrases of the same mechanism (e.g. "heat goes to the rail" and "the rail warms up") score M1 only. - Part (b)(i) alternative: Using kinematics v_f² = 2a·d with a = μ_k g also acceptable, provided v_f from momentum is used (M1). - Part (b)(ii) alternative: t = v_f/a with a = μ_k g = 3.43 m s⁻² gives t = 0.58 s — full credit if impulse–momentum reasoning is shown verbally. - (c) accept ΔT in range 5.1 × 10⁻³ K to 5.3 × 10⁻³ K. - (d) discriminator: accept conduction to rail, sound emission, plastic deformation of blocks, radiative/convective loss to surroundings, non-uniform heating (thermistor reads local rather than mean T), thermistor thermal lag. Do **not** accept "friction" or "air resistance" as a reason (friction is the source modelled; air resistance is negligible at these speeds and was not invoked in the model).