In a laboratory demonstration, trolley A of mass 0.80 kg moves at a constant velocity of 2.5 m/s on a frictionless track towards stationary trolley B of mass 1.20 kg. The trolleys collide and stick together, moving off as one combined object.
(a)Define momentum, and state the principle of conservation of momentum.(2)
(b)Calculate the momentum of trolley A before the collision.(2)
(c)Calculate the common velocity of the two trolleys immediately after the collision.(3)
(d)Calculate the total kinetic energy of the system before and after the collision, and use your values to determine whether the collision is elastic.(3)
(Total for Question 1 is 10 marks)
2
A car of mass 1200 kg drives at constant speed around a flat circular bend of radius 45 m. Use the Physics Equations Sheet where needed (a = v2/r for centripetal acceleration). Take g = 9.81 m/s2.
(a)State what is meant by angular velocity, and calculate the angular velocity of a wheel on the car that completes one full rotation in 28 ms (2.8 x 10-2 s).(2)
(b)The car travels around the bend at a constant speed of 15 m/s. Using a = v2/r, calculate the centripetal acceleration of the car, and hence the centripetal force acting on it.(4)
(c)The centripetal force is provided entirely by friction between the tyres and the road, with coefficient of friction μ = 0.65 between tyre and road. Calculate the maximum speed at which the car can take the bend without skidding.(3)
(d)Explain, in terms of Newton's laws, why a passenger sitting on the outside of the turn feels pushed outwards against the car door as the car goes round the bend.(2)
(Total for Question 2 is 11 marks)
3
A bucket of water of mass 0.60 kg is swung in a vertical circle of radius 0.90 m at the end of a light rope. Take g = 9.81 m/s2. Use the Physics Equations Sheet where needed.
(a)State the condition, in terms of the tension in the rope, for the water to just remain in the bucket as it passes the top of the circle.(1)
(b)Show that the minimum speed of the bucket at the top of the circle, for the water to just stay in, is about 3.0 m/s.(3)
(c)The bucket passes the bottom of the circle with speed 4.5 m/s. Calculate the tension in the rope at this point.(4)
(d)State and explain the effect on the tension at the bottom of the circle if the bucket were instead swung faster.(2)
(Total for Question 3 is 10 marks)
4
A mass attached to a spring oscillates with simple harmonic motion (SHM) of amplitude 0.12 m and angular frequency ω = 8.0 rad/s. Use the Physics Equations Sheet where needed (vmax = ω x A and amax = ω2 x A).
(a)State two conditions that must both be satisfied for a system to be undergoing simple harmonic motion.(2)
(b)Using vmax = ω x A and amax = ω2 x A, calculate the maximum speed and the maximum acceleration of the mass.(4)
(c)Using T = 2pi/ω and T = 2pi x √m/k, calculate the period of oscillation, and hence the spring constant k of the spring if the mass is 0.25 kg.(3)
(Total for Question 4 is 9 marks)
5
REQUIRED PRACTICAL. A student carries out the required practical investigation into simple harmonic motion, using a mass-spring system to determine the spring constant k from the relationship T2 = (4pi2/k) x m, where T is the period of oscillation and m is the mass attached to the spring.
(a)Describe how the student should measure the period of oscillation of the mass-spring system accurately, including how random error in each timing is reduced.(3)
(b)The student obtains the following processed data for a graph of T2 (s2) against m (kg): at m = 0.10 kg, T2 = 0.247 s2; at m = 0.50 kg, T2 = 1.234 s2. Using these two points, calculate the gradient of the graph of T2 against m, and hence determine the spring constant k.(5)
(c)Explain how the student could reduce the percentage uncertainty in the final value obtained for k.(2)
(Total for Question 5 is 10 marks)
6
A mass of 0.25 kg oscillates with SHM with amplitude 0.12 m and angular frequency ω = 8.0 rad/s (the same system as in question 4). Use the Physics Equations Sheet where needed (total energy E = 1/2 x m x ω2 x A2, and v = ω x √A2 - x2).
(a)Calculate the total energy of the oscillation.(3)
(b)Using v = ω x √A2 - x2, calculate the speed of the mass, and hence its kinetic energy, when its displacement from equilibrium is x = 0.06 m.(4)
(c)Without further calculation of v, state the potential energy of the system at this displacement, and comment on how the total energy is shared between kinetic and potential energy as the mass moves from the equilibrium position to maximum displacement.(3)
(Total for Question 6 is 10 marks)
7
A footbridge over a river in a town centre is found to sway noticeably when large groups of people cross it walking in step. Engineers are considering two possible modifications: (i) fitting mechanical dampers to the structure, or (ii) redesigning the bridge to change its natural frequency of vibration. Discuss, using your knowledge of forced oscillations, resonance and damping, how each approach would help to reduce unwanted large-amplitude oscillations of the bridge, and evaluate which approach you consider to be more effective.
(Total for Question 7 is 6 marks)
8
REQUIRED PRACTICAL. An aluminium block of mass 1.2 kg has an electrical heater and a thermometer embedded in holes drilled into it, and is connected to a 12 V supply drawing a constant current of 4.0 A.
(a)Describe how this apparatus could be used to determine the specific heat capacity of aluminium, including how heat losses to the surroundings are minimised.(3)
(b)In one run, the temperature of the block rises from 18.0 degC to 42.5 degC in 15 minutes. Using Q = VIt and Q = mc x (temperature change), calculate the value obtained for the specific heat capacity of aluminium from this experiment.(4)
(c)The accepted value for the specific heat capacity of aluminium is 900 J/(kg K). Suggest one reason for the difference between this accepted value and the value calculated in part (b), and suggest one improvement to the experiment that would reduce this effect.(2)
(Total for Question 8 is 9 marks)
9
An ice cube of mass 25 g (0.025 kg) at 0 degC is dropped into a drink. The specific latent heat of fusion of ice is 3.34 x 105 J/kg.
(a)State what is meant by the specific latent heat of fusion of a substance.(1)
(b)Calculate the energy needed to completely melt the ice cube, and hence calculate the minimum power of a heater that would melt this mass of ice in 2.0 minutes, assuming all of the heater's output goes into melting the ice.(4)
(c)Explain, in terms of the behaviour of molecules, why the temperature of the ice-water mixture remains constant while the ice is melting, even though energy is continuously being supplied.(2)
(Total for Question 9 is 7 marks)
10
A fixed mass of an ideal gas occupies a volume of 2.40 x 10-3 m3 at a pressure of 1.05 x 105 Pa and a temperature of 290 K. Use the Physics Equations Sheet where needed (pV = nRT, with R = 8.31 J/(mol K); N = n x NA, with NA = 6.02 x 1023 /mol).
(a)State Boyle's law.(1)
(b)The gas is compressed to a volume of 1.60 x 10-3 m3 and heated to a temperature of 340 K. Using pV = nRT, calculate the new pressure of the gas.(4)
(c)Using the initial conditions (p = 1.05 x 105 Pa, V = 2.40 x 10-3 m3, T = 290 K), calculate the number of moles of gas present, and hence the number of gas molecules.(3)
(Total for Question 10 is 8 marks)
11
A container of volume 0.020 m3 holds 5.4 x 1024 molecules of nitrogen gas at a pressure of 2.0 x 105 Pa. Each nitrogen molecule has a mass of 4.65 x 10-26 kg. Use the Physics Equations Sheet where needed (pV = (1/3) x N x m x <c2>; average molecular kinetic energy Ek = (3/2) x k x T, with the Boltzmann constant k = 1.38 x 10-23 J/K).
(a)State two assumptions of the kinetic theory model of an ideal gas.(2)
(b)Using pV = (1/3) x N x m x <c2>, calculate the root-mean-square speed of the nitrogen molecules.(4)
(c)Using Ek = (3/2) x k x T, calculate the average translational kinetic energy of a gas molecule at a temperature of 300 K.(3)
(d)A second container at the same temperature holds hydrogen gas instead of nitrogen. Without further calculation, explain which gas has the higher root-mean-square speed, and why.(2)
(Total for Question 11 is 11 marks)
12
A bullet of mass 8.0 g (0.008 kg) travelling horizontally at 320 m/s embeds itself in a stationary wooden block of mass 2.0 kg that is free to move. Use the Physics Equations Sheet where needed.
(a)Calculate the common velocity of the bullet and block immediately after the bullet embeds itself in the block.(3)
(b)Calculate the kinetic energy of the bullet immediately before the impact, and the kinetic energy of the bullet-block system immediately after the impact. Hence calculate the loss of kinetic energy during the collision.(3)
(c)Assume the lost kinetic energy is converted into internal (thermal) energy shared between the bullet and the region of wood it embeds in, with half of this energy absorbed by the bullet. The specific heat capacity of the bullet material is 130 J/(kg K). Estimate the rise in temperature of the bullet.(3)
(Total for Question 12 is 9 marks)
Mark scheme · AP6 Further Mechanics and Thermal Physics
Question 1
(a) B1 momentum = mass x velocity (a vector quantity), oe
(a) B1 in a closed system (no external resultant force), the total momentum before an interaction equals the total momentum after the interaction, oe
(a) Answer: Momentum = mass x velocity; total momentum of a closed system is conserved (unchanged) provided no external resultant force acts.
(b) M1 correct substitution into p = mv
(b) A1 p = 2.0 kg m/s, cao
(b) Answer: 2.0 kg m/s
(c) M1 conservation of momentum applied: total momentum before = total momentum after, oe
(c) M1 combined mass = 0.80 + 1.20 = 2.00 kg used correctly, ft from (b)
(c) A1 v = 1.0 m/s, in the original direction of motion of A, cao
(c) Answer: 1.0 m/s (in the direction A was originally travelling)
(d) M1 KE before = 1/2 x 0.80 x 2.52 = 2.5 J
(d) M1 KE after = 1/2 x 2.00 x 1.02 = 1.0 J, ft from (c)
(d) A1 conclusion: KE has decreased (from 2.5 J to 1.0 J), so the collision is not elastic (kinetic energy is not conserved), oe
(d) Answer: KE before = 2.5 J; KE after = 1.0 J; collision is inelastic (1.5 J of KE is lost, e.g. to heat and sound)
Question 2
(a) B1 angular velocity is the rate of change of angular displacement (angle turned per unit time), unit rad/s, oe
(a) A1 ω = 2 x π / 0.028 = 224 rad/s (awrt), using ω = 2pi/T
(a) Answer: 224 rad/s (3 s.f.)
(b) M1 correct substitution into a = v2/r
(b) A1 a = 5.0 m/s2, cao
(b) M1 F = ma used with the value of a found, ft
(b) A1 F = 6000 N, cao
(b) Answer: a = 5.0 m/s2, F = 6000 N
(c) M1 friction force provides centripetal force: μ x m x g = m x v2/r, oe
(c) M1 rearranged correctly to v = √&μ; x g x r
(c) A1 v = 16.9 m/s (awrt), cao
(c) Answer: 16.9 m/s (3 s.f.)
(d) B1 the passenger's inertia tends to keep them moving in a straight line (Newton's first law), oe
(d) B1 the door/seat provides the centripetal force needed to keep the passenger moving in a circle, directed towards the centre; by Newton's third law the passenger pushes back on the door, felt as being pushed outward, oe
(d) Answer: The passenger tends to travel in a straight line due to inertia; the car door supplies the centripetal (inward) force to keep them moving in a circle, and the reaction to this is felt as an outward push - there is no real outward (centrifugal) force.
Question 3
(a) B1 the tension (force of the bucket on the water) is zero, so the weight alone provides the centripetal force, oe
(a) Answer: Tension = 0 N (weight alone provides the centripetal force)
(b) A1 cso: v = √9.81 x 0.90 = 2.97 m/s, rounding to about 3.0 m/s as given
(b) Answer: v = 2.97 m/s (approx 3.0 m/s, as given)
(c) M1 at the bottom, T - mg = mv2/r, oe
(c) M1 centripetal term calculated: mv2/r = 0.60 x 4.52/0.90 = 13.5 N
(c) M1 weight term calculated: mg = 0.60 x 9.81 = 5.89 N (awrt), and added to centripetal term
(c) A1 T = 19.4 N (awrt), cao
(c) Answer: 19.4 N (3 s.f.)
(d) B1 the tension would be greater (increase), oe
(d) B1 a higher speed requires a greater centripetal force (F = mv2/r increases with v2), and this extra force must be provided by additional tension (beyond that supporting the weight), oe
(d) Answer: Tension increases, because a greater speed requires a greater centripetal force, which the rope must supply in addition to supporting the weight of the bucket.
Question 4
(a) B1 the acceleration (or restoring force) is proportional to the displacement from equilibrium, oe
(a) B1 the acceleration (or restoring force) is always directed towards the equilibrium position (opposite in direction to the displacement), oe
(a) Answer: Acceleration is proportional to displacement from equilibrium, and is always directed towards the equilibrium position (a = -ω2 x).
(b) M1 correct substitution into vmax = ω x A
(b) A1 vmax = 0.96 m/s, cao
(b) M1 correct substitution into amax = ω2 x A
(b) A1 amax = 7.68 m/s2, cao
(b) Answer: vmax = 0.96 m/s; amax = 7.68 m/s2
(c) M1 T = 2pi/ω = 2pi/8.0 = 0.785 s (awrt)
(c) M1 rearrangement of T = 2pi x √m/k to make k the subject (e.g. k = 4pi2 x m/T2, or equivalently k = m x ω2), with correct substitution
(c) A1 k = 16.0 N/m, cao
(c) Answer: T = 0.785 s; k = 16.0 N/m
Question 5
(a) B1 use a fiducial marker (fixed reference point) at the equilibrium position, viewed at eye level, to judge when the mass passes through the same point each time and avoid parallax error, oe
(a) B1 time a large number of oscillations (e.g. 20) rather than one, then divide the total time by the number of oscillations, to reduce the percentage/random error in the period, oe
(a) B1 repeat each timing (e.g. three times) and calculate a mean value of T, oe
(a) Answer: Use a fiducial marker to define one full oscillation, time 20 oscillations rather than one and divide by 20, and repeat and average the timings.
(b) M1 gradient = (change in T2)/(change in m), using the two given data points
(c) B1 use a wider range of masses (larger spread of m values), which increases the range spanned by the data and reduces the percentage uncertainty in the gradient, oe
(c) B1 take measurements at more values of m across the range and draw a line of best fit, rather than using only two points, to average out random error, oe
(c) Answer: Use a larger range of masses and take more data points, drawing a line of best fit rather than joining just two points, to reduce the percentage uncertainty in the gradient.
Question 6
(a) M1 correct substitution into E = 1/2 x m x ω2 x A2
(a) M1 correct arithmetic combination, e.g. 0.5 x 0.25 x 64 = 8.0
(a) A1 E = 0.115 J (awrt), cao
(a) Answer: 0.115 J (3 s.f.)
(b) M1 correct substitution into v = ω x √A2 - x2
(b) A1 v = 0.831 m/s (awrt), cao
(b) M1 KE = 1/2 x m x v2 substitution, ft from v
(b) A1 KE = 0.0864 J (awrt), cao
(b) Answer: v = 0.831 m/s; KE = 0.0864 J
(c) B1 PE = E_total - KE = 0.115 - 0.0864 = 0.0288 J (awrt), ft from (a) and (b)
(c) B1 KE is maximum at the equilibrium position (x = 0) and decreases to zero at maximum displacement (x = A); PE does the opposite, oe
(c) B1 the total energy (KE + PE) remains constant throughout the oscillation, oe
(c) Answer: PE = 0.0288 J; KE and PE are continuously exchanged as the mass oscillates, with KE maximum (and PE zero) at equilibrium, and PE maximum (and KE zero) at maximum displacement, while the total energy stays constant.
Question 7
Level 3 (5-6): A detailed and coherent discussion that explains resonance occurring when the driving frequency (footsteps) matches (or is close to) the natural frequency of the bridge, causing a large amplitude of forced oscillation; explains that dampers increase the degree of damping, dissipating vibrational energy as heat and reducing (and broadening) the amplitude peak at resonance; explains that changing the bridge's natural frequency (e.g. by altering its stiffness or mass) moves the resonant frequency away from typical walking frequencies so resonance is avoided; a reasoned, well-justified evaluation is given comparing the two approaches, using correct physics terminology throughout.
Level 2 (3-4): A reasonable discussion showing understanding of resonance, with at least one of the two modifications (damping or changing natural frequency) explained correctly and linked to a reduction in amplitude; the evaluation may be brief, one-sided, or only partially justified; generally correct physics with some gaps in detail or terminology.
Level 1 (1-2): Basic, largely unlinked statements about resonance and/or damping are made, with limited explanation and little or no reference to the bridge context; no meaningful evaluation attempted.
Level 0 (0): No relevant content, or the answer does not relate to the question.
Indicative content:
Resonance occurs when the frequency of a periodic driving force (the footsteps of people walking in step) matches, or is very close to, the natural frequency of the bridge, producing a maximum amplitude of oscillation.
With little damping present, the amplitude of oscillation at resonance can become very large, which explains the noticeable swaying.
Fitting dampers increases the degree of damping in the system, dissipating vibrational energy as heat (internal energy); this reduces the amplitude of oscillation at resonance and broadens (flattens) the resonance peak, so the response is less sensitive to the exact driving frequency.
Changing the structure (its mass and/or stiffness) alters the bridge's natural frequency, so that it no longer coincides with the typical range of frequencies produced by pedestrians walking, meaning resonance is not driven under normal use.
Evaluation: changing the natural frequency removes the risk of resonance for expected walking frequencies but may not protect against a wider range of driving frequencies (e.g. running, wind gusts, or unusual crowd behaviour); dampers reduce the amplitude across a whole range of driving frequencies, giving more general protection.
A well-justified conclusion might argue that using dampers (or a combination of both approaches) is the more robust practical solution, since pedestrian step frequency cannot be precisely controlled, whereas changing natural frequency alone only guards against resonance at one specific frequency.
Question 8
(a) B1 lag (insulate) the block with insulating material to minimise heat loss to the surroundings, oe
(a) B1 record the initial temperature, switch on the heater for a measured time t while recording the current I and potential difference V, then record the final (maximum) temperature reached, oe
(a) B1 calculate the electrical energy supplied Q = VIt and use Q = mc x (temperature change) to find c, oe
(a) Answer: Insulate (lag) the block to reduce heat loss; measure initial and final temperature while supplying a known electrical energy Q = VIt over a measured time; calculate c from Q = mc x deltaT.
(b) M1 Q = VIt = 12 x 4.0 x 900 = 43200 J (using t = 15 x 60 = 900 s)
(b) M1 temperature change = 42.5 - 18.0 = 24.5 K (or degC)
(b) M1 correct rearrangement, c = Q/(m x deltaT), with correct substitution, ft
(b) A1 c = 1470 J/(kg K) (awrt), cao, unit given
(b) Answer: 1470 J kg-1 K-1 (3 s.f.)
(c) B1 some of the electrical energy supplied is lost to the surroundings (e.g. by conduction, convection or radiation from the block, heater leads or thermometer) rather than all going into raising the temperature of the block, so the calculated c (based on the full electrical energy supplied) is an overestimate, oe
(c) B1 suitable improvement, e.g. improve the lagging/insulation around the block, or apply a cooling correction (continue timing/temperature readings after switching off the heater and correcting for heat lost), oe
(c) Answer: Heat is lost to the surroundings rather than fully raising the block's temperature, giving an overestimate of c; improve insulation (better lagging) or use a cooling correction to account for the heat lost.
Question 9
(a) B1 the energy required to change the state of 1 kg of a substance from solid to liquid, without any change in temperature, oe
(a) Answer: The energy needed to melt 1 kg of a substance (solid to liquid) at constant temperature.
(b) M1 correct substitution into Q = mL, using mass in kg
(b) A1 Q = 8350 J, cao
(b) M1 P = Q/t, using t = 2.0 x 60 = 120 s, ft
(b) A1 P = 69.6 W (awrt), cao
(b) Answer: Q = 8350 J; P = 69.6 W (3 s.f.)
(c) B1 the energy supplied does work against the intermolecular forces holding molecules in the fixed lattice structure, increasing their potential energy, rather than increasing their kinetic energy, oe
(c) B1 since temperature is a measure of the average kinetic energy of the molecules, and this does not change during melting, the temperature remains constant until all the ice has melted, oe
(c) Answer: Energy supplied breaks intermolecular bonds (increasing potential energy) rather than increasing molecular kinetic energy, so since temperature depends on average kinetic energy, it stays constant until melting is complete.
Question 10
(a) B1 for a fixed mass of gas at constant temperature, the pressure is inversely proportional to the volume (pV = constant), oe
(a) Answer: For a fixed mass of gas at constant temperature, pressure is inversely proportional to volume (pV = constant).
(b) M1 recognises that n and R are constant, so p1V1/T1 = p2V2/T2, oe
(b) M1 correct substitution of all given values
(b) M1 correctly rearranged to make p2 the subject
(b) A1 p2 = 1.85 x 105 Pa (awrt), cao
(b) Answer: 1.85 x 105 Pa (3 s.f.)
(c) M1 correct substitution into n = pV/RT
(c) A1 n = 0.105 mol (awrt)
(c) A1 N = n x NA = 6.30 x 1022 molecules (awrt), ft
(c) Answer: n = 0.105 mol; N = 6.30 x 1022 molecules
Question 11
(a) B1 any one correct assumption, e.g. molecules undergo perfectly elastic collisions with each other and the container walls, oe
(a) B1 any other correct assumption, e.g. the volume of the molecules is negligible compared with the volume of the container; the duration of a collision is negligible compared with the time between collisions; there are negligible intermolecular forces except during collisions; molecules move in continuous random motion, oe (any two distinct assumptions for 2 marks)
(a) Answer: Any two of: collisions are perfectly elastic; molecular volume is negligible compared to the container volume; collision time is negligible compared to time between collisions; negligible forces between molecules except during collisions; molecules move in continuous random motion.
(b) M1 rearranged correctly to <c2> = 3pV/(Nm)
(b) M1 correct substitution of all given values
(b) A1 <c2> = 4.78 x 104 m2/s2 (awrt), ecf allowed to next mark
(b) A1 crms = √<c2> = 219 m/s (awrt), cao, unit given
(b) Answer: crms = 219 m/s (3 s.f.)
(c) M1 correct substitution into Ek = (3/2) x k x T
(c) M1 correct arithmetic combination, e.g. 1.5 x 1.38 x 10-23 = 2.07 x 10-23
(c) A1 Ek = 6.21 x 10-21 J (awrt), cao, unit given
(c) Answer: 6.21 x 10-21 J (3 s.f.)
(d) B1 hydrogen has the higher rms speed, oe
(d) B1 at the same temperature both gases have the same average kinetic energy per molecule (Ek = (3/2)kT depends only on T); since Ek = 1/2 x m x <c2>, the much smaller mass of a hydrogen molecule (compared to nitrogen) means <c2> (and hence rms speed) must be greater to give the same Ek, oe
(d) Answer: Hydrogen has the higher rms speed, because at the same temperature both gases have equal average molecular kinetic energy, and the lower mass of hydrogen molecules means they must move faster to have the same kinetic energy.
Question 12
(a) M1 conservation of momentum applied: m_bullet x u = (m_bullet + M_block) x v, oe
(a) M1 correct substitution: 0.008 x 320 = (0.008 + 2.0) x v
(a) A1 v = 1.27 m/s (awrt), cao
(a) Answer: 1.27 m/s (3 s.f.)
(b) M1 KE before = 1/2 x 0.008 x 3202 = 409.6 J (awrt), cao
(b) M1 KE after = 1/2 x 2.008 x v2 = 1.63 J (awrt), ft from (a)
(b) A1 KE loss = 408 J (awrt), ft
(b) Answer: KE before = 409.6 J; KE after = 1.63 J; KE lost = 408 J (3 s.f.)
(c) M1 energy absorbed by bullet = half of KE loss = 204 J (awrt), ft from (b)
(c) M1 correct substitution into deltaT = Q/(m x c), using the bullet's mass (0.008 kg)
(c) A1 deltaT = 196 K (or degC) (awrt), cao, ft
(c) Answer: Approximately 196 K rise in temperature (3 s.f.)