A cyclist in Durham is travelling along a straight, flat road at a constant velocity of 12 m/s when she begins to brake uniformly, coming to rest 8.0 s later. Use the Physics Equations Sheet where needed. Take g = 9.81 m/s2.
(a)Calculate the deceleration of the cyclist while braking.(2)
(b)Calculate the distance the cyclist travels while braking.(2)
(c)Show that the same braking distance is obtained using the equation v2 = u2 + 2as.(2)
(d)The cyclist's reaction time before she starts to brake is 0.30 s, during which she continues at 12 m/s. Calculate the total distance travelled from the moment she first sees a hazard to when she stops.(3)
(Total for Question 1 is 9 marks)
2
A crate of mass 25 kg is pulled at constant velocity up a rough ramp inclined at 20 degrees to the horizontal, using a rope parallel to the slope. The coefficient of friction between the crate and the ramp is 0.15. Take g = 9.81 m/s2.
(a)Calculate the component of the crate's weight acting parallel to the slope.(2)
(b)Calculate the normal reaction force acting on the crate.(2)
(c)Calculate the frictional force acting on the crate as it moves up the slope.(2)
(d)The crate moves up the slope at constant velocity. Calculate the tension in the rope.(3)
(Total for Question 2 is 9 marks)
3
A person of mass 70 kg stands on bathroom scales inside a lift in a shopping centre. Take g = 9.81 m/s2. Use the Physics Equations Sheet where needed.
(a)The lift accelerates upwards from rest at 2.0 m/s2. Calculate the reading on the scales (the normal reaction force on the person) during this acceleration.(3)
(b)The lift then travels at a constant velocity. Calculate the reading on the scales during this stage.(2)
(c)As the lift approaches the top floor it decelerates at 1.5 m/s2 while still moving upwards. Calculate the reading on the scales during this deceleration.(3)
(d)Explain, in terms of Newton's laws, why the person feels lighter while the lift decelerates in part (c).(2)
(Total for Question 3 is 10 marks)
4
A stone is thrown horizontally with a speed of 15 m/s from the top of a cliff at Flamborough Head, 20 m above the sea. Air resistance can be ignored. Take g = 9.81 m/s2. Use the Physics Equations Sheet where needed.
(a)Show that the time taken for the stone to fall to sea level is about 2.0 s.(2)
(b)Calculate the horizontal distance travelled by the stone before it lands.(2)
(c)Calculate the vertical component of the stone's velocity as it lands.(2)
(d)Calculate the magnitude and direction of the stone's resultant velocity as it lands.(4)
(Total for Question 4 is 10 marks)
5
Trolley A, of mass 0.80 kg, moves at 3.0 m/s and collides with stationary trolley B, of mass 1.20 kg. The trolleys stick together and move off with a common velocity.
(a)Calculate the momentum of trolley A immediately before the collision.(2)
(b)Use conservation of momentum to calculate the common velocity of the trolleys immediately after the collision.(3)
(c)Calculate the total kinetic energy before the collision, the total kinetic energy after the collision, and hence the kinetic energy lost in the collision.(4)
(d)State and explain what has happened to the kinetic energy lost during the collision.(2)
(Total for Question 5 is 11 marks)
6
An electric motor is used to lift a load of mass 150 kg at a constant speed through a vertical height of 12 m in 20 s. The motor is 75% efficient. Take g = 9.81 m/s2.
(a)Calculate the gain in gravitational potential energy of the load.(2)
(b)Calculate the useful power output of the motor.(2)
(c)Given that efficiency = (useful power output / input power) x 100%, calculate the input power supplied to the motor.(3)
(d)Calculate the power wasted by the motor.(2)
(Total for Question 6 is 9 marks)
7
A statue made from a bronze alloy has a mass of 45 kg and a volume of 5.2 x 10-3 m3. Density of water = 1000 kg/m3. Take g = 9.81 m/s2. Upthrust = density of fluid x volume displaced x g.
(a)Calculate the density of the bronze alloy.(2)
(b)The statue is fully submerged in a tank of water for cleaning. Calculate the upthrust acting on it.(3)
(c)Calculate the apparent weight of the statue while it is fully submerged.(3)
(d)Explain, using the concept of upthrust, why the reading on a support balance holding the statue decreases when it is lowered into the water.(2)
(Total for Question 7 is 10 marks)
8
This question is based on the required practical: investigation of the force-extension characteristics of a spring (Hooke's law). A student hangs a spring vertically from a clamp stand and adds masses one at a time, measuring the total extension with a metre ruler fixed alongside the spring. Her results are: Force / N: 1.0, 2.0, 3.0, 4.0, 5.0 Extension / mm: 12, 25, 37, 50, 61
(a)Describe one step the student could take to improve the accuracy of her extension measurements, and explain how it does this.(2)
(b)Use the student's data to calculate the spring constant, k, of the spring in N/m.(3)
(c)Given that the elastic potential energy stored in a spring is E = 1/2 k x2, calculate the elastic potential energy stored in the spring when it is extended by 61 mm.(3)
(d)Evaluate this method for determining the spring constant of a spring, suggesting improvements that would reduce random and/or systematic error.(6)
(Total for Question 8 is 14 marks)
9
This question is based on the required practical: determination of the Young modulus of a metal wire. A student sets up a copper test wire of original length 2.00 m and diameter 0.28 mm, clamped horizontally and passed over a pulley with a load attached, alongside a fixed reference wire and vernier scale to measure extension. A load of 20 N produces an extension of 5.6 mm, which is within the elastic limit of the wire. Assume the wire has a uniform circular cross-section.
(a)Calculate the cross-sectional area of the wire.(3)
(b)Calculate the strain in the wire.(2)
(c)Calculate the stress in the wire when it supports the 20 N load.(2)
(d)Calculate the Young modulus of the copper wire from this data.(2)
(e)State one precaution taken in this experiment that improves the reliability of the result.(1)
(f)Given that the elastic strain energy stored in a stretched wire is E_stored = 1/2 F x, calculate the elastic strain energy stored in the wire when the 20 N load produces the 5.6 mm extension.(3)
(Total for Question 9 is 13 marks)
10
A uniform ladder of length 5.0 m and weight 120 N rests with its foot on rough ground, 3.0 m horizontally from a smooth vertical wall against which the top of the ladder leans. A window cleaner of weight 700 N stands three-quarters of the way up the ladder from the foot. Take g = 9.81 m/s2.
(a)Calculate the angle between the ladder and the ground.(2)
(b)By taking moments about the foot of the ladder, calculate the normal reaction force exerted by the (smooth) wall on the ladder.(4)
(c)Calculate the frictional force acting at the foot of the ladder.(2)
(d)Calculate the normal reaction force from the ground acting on the ladder.(2)
(e)Calculate the minimum coefficient of friction between the ladder and the ground needed to prevent the ladder slipping.(2)
(Total for Question 10 is 12 marks)
11
This question is based on the required practical: determination of the acceleration due to free fall, g. A steel ball bearing is released from rest by an electromagnet and falls through a measured height, h, between two light gates connected to an electronic timer, which records the total time of fall, t. In one trial, h = 0.800 m and t = 0.404 s.
(a)Show that the average speed of the ball bearing over the fall is about 1.98 m/s.(2)
(b)Given that, starting from rest, g = 2h/t2, calculate the experimental value of g from this trial.(3)
(c)The accepted value of g is 9.81 m/s2. Calculate the percentage difference between the experimental value from part (b) and the accepted value.(2)
(d)Identify one source of random error in this experiment and state how its effect on the result could be reduced.(2)
(e)Explain why using light gates, rather than a hand-operated stopwatch, improves the accuracy of the value obtained for g.(2)
(Total for Question 11 is 11 marks)
12
Figure 2 shows the stress-strain graph for a metal wire, stretched from zero up to fracture. Along the curve, point P marks the limit of proportionality, point E marks the elastic limit, point Y marks the yield point, and the wire fractures at point F, well beyond Y. Within the linear region (up to P), a stress of 2.0 x 108 Pa produces a strain of 1.6 x 10-3.
(a)State what is meant by the limit of proportionality.(1)
(b)State what happens to the wire at the yield point, Y.(1)
(c)Use the gradient of the linear region of the graph to calculate the Young modulus of the wire.(3)
(d)Compare the mechanical behaviour of this ductile metal wire with that of a brittle glass fibre when both are stressed to fracture, explaining the difference in terms of the structure of each material.(6)
(Total for Question 12 is 11 marks)
13
A car of mass 900 kg, travelling at 18 m/s, collides with a barrier and is brought to rest by the energy-absorbing crumple zone in the front of the car.
(a)Calculate the change in momentum of the car during the collision.(2)
(b)The collision brings the car to rest in 0.12 s. Given that force = change in momentum/time, calculate the average force exerted on the car during the collision.(3)
(c)The car is redesigned with a longer crumple zone, so that in an identical collision it now takes 0.20 s to come to rest. Calculate the new average force on the car.(2)
(d)Explain, in terms of impulse, why increasing the duration of the collision reduces the risk of injury to the occupants of the car.(3)
(Total for Question 13 is 10 marks)
Mark scheme · AP4 Mechanics and Materials
Question 1
(a) M1 correct substitution into a = (v - u)/t, oe
(a) A1 a = -1.5 m/s2 (deceleration of 1.5 m/s2), cao
(a) Answer: a = -1.5 m/s2 (deceleration of 1.5 m/s2)
(b) M1 correct substitution into s = (u + v)t/2 or s = ut + 1/2at2, oe
(b) A1 s = 48 m, cao
(b) Answer: 48 m
(c) M1 rearrange v2 = u2 + 2as to make s the subject, oe
(c) A1 cso: 0 = 144 - 3s leading to s = 48 m, consistent with part (b)
(c) Answer: 48 m (shown)
(d) M1 distance during reaction time = 12 x 0.30 = 3.6 m
(d) M1 total distance = reaction distance + braking distance (ft from (b))
(d) A1 51.6 m, cao
(d) Answer: 51.6 m
Question 2
(a) M1 correct substitution into W sin(20), using W = mg = 245 N (or 245.25 N), oe
(a) A1 83.9 N (accept 83.6-84.0 N), cao
(a) Answer: 83.9 N
(b) M1 correct substitution into N = W cos(20), oe
(b) A1 230 N (accept 229-231 N), awrt
(b) Answer: 230.5 N
(c) M1 correct substitution into F = μ N, using N from part (b), ft
(c) A1 34.6 N (ft from (b)), cao
(c) Answer: 34.6 N
(d) M1 recognise equilibrium along the slope: T = weight component (a) + friction (c)
(d) M1 correct substitution using values from (a) and (c), ft
(d) A1 118.5 N (accept 118-119 N, ft), cao
(d) Answer: 118.5 N
Question 3
(a) M1 resultant force equation: R - mg = ma, oe
(a) M1 correct substitution: R = 70(9.81 + 2.0)
(a) A1 827 N (accept 826-828 N), awrt
(a) Answer: 827 N
(b) M1 recognise a = 0, so R = mg
(b) A1 687 N (awrt), cao
(b) Answer: 687 N
(c) M1 resultant force equation for deceleration: mg - R = ma, oe, i.e. R = m(g - a)
(c) M1 correct substitution: R = 70(9.81 - 1.5)
(c) A1 582 N (accept 580-583 N), awrt
(c) Answer: 582 N
(d) B1 resultant force on the person acts downwards, so the normal reaction (support) force from the scales is smaller than the person's weight
(d) B1 by Newton's third law the scales push up on the person with this smaller force, which is sensed as feeling lighter, oe
(d) Answer: The reduced normal reaction force during deceleration is felt as a reduced sensation of weight.
Question 4
(a) M1 rearrange s = 1/2 g t2 to make t the subject: t = √2s/g
(a) A1 cso: t = √2 x 20/9.81 = 2.02 s, which is 2.0 s to 2 s.f.
(a) Answer: t = 2.0 s (2.02 s to 3 s.f.)
(b) M1 x = v x t, using t = 2.02 s (ft from (a))
(b) A1 30.3 m (accept 30-30.4 m), awrt
(b) Answer: 30.3 m
(c) M1 v = g t, using t = 2.02 s (ft from (a))
(c) A1 19.8 m/s (awrt), cao
(c) Answer: 19.8 m/s
(d) M1 use Pythagoras: v_resultant = √vx2 + vy2, using values from (b)/(c), ft
(d) A1 24.8 m/s (accept 24.7-24.9 m/s), awrt
(d) M1 use tan(θ) = vy/vx (or equivalent) to find the angle below the horizontal
(d) Answer: 24.8 m/s at 52.9 degrees below the horizontal
Question 5
(a) M1 p = m v substitution
(a) A1 2.4 kg m/s, cao
(a) Answer: 2.4 kg m/s
(b) M1 apply conservation of momentum: mA uA + mB uB = (mA + mB) v
(b) M1 correct substitution: 0.80(3.0) + 1.20(0) = 2.00 v
(b) A1 v = 1.2 m/s, cao
(b) Answer: 1.2 m/s
(c) M1 KE before = 1/2 (0.80)(3.0)2 = 3.6 J
(c) M1 KE after = 1/2 (2.00)(1.2)2 = 1.44 J, ft from (b)
(c) A1 correct method: energy lost = KE before - KE after
(c) A1 2.16 J, cao/ft
(c) Answer: KE before = 3.6 J; KE after = 1.44 J; energy lost = 2.16 J
(d) B1 momentum is conserved but kinetic energy is not, because the collision is inelastic
(d) B1 the lost KE is transferred to other forms such as heat, sound and/or the permanent deformation of the trolleys, oe
(d) Answer: The collision is inelastic; kinetic energy is transferred to heat, sound and deformation of the trolleys.
Question 6
(a) M1 GPE = mgh substitution
(a) A1 1.77 x 104 J (17700 J, accept 17600-17700 J), cao
(a) Answer: 1.77 x 104 J
(b) M1 P = work done/time, using answer from (a), ft
(b) A1 883 W (accept 880-885 W, ft), awrt
(b) Answer: 883 W
(c) M1 rearrange efficiency equation to make input power the subject
(c) M1 correct substitution: input power = 882.9/0.75, ft from (b)
(c) A1 1180 W (accept 1175-1185 W, ft), awrt
(c) Answer: 1180 W
(d) M1 wasted power = input power - useful power output, ft from (b) and (c)
(d) A1 294 W (accept 290-298 W, ft), awrt
(d) Answer: 294 W
Question 7
(a) M1 density = mass/volume substitution
(a) A1 8650 kg/m3 (accept 8600-8700 kg/m3), awrt
(a) Answer: 8650 kg/m3
(b) M1 recall/apply upthrust = density of water x volume x g
(b) M1 correct substitution: 1000 x 5.2x10-3 x 9.81
(b) A1 51.0 N (accept 50.8-51.2 N), cao
(b) Answer: 51.0 N
(c) M1 true weight = mg = 45 x 9.81 = 441 N
(c) M1 apparent weight = true weight - upthrust, ft from (b)
(c) A1 390 N (accept 388-392 N, ft), cao
(c) Answer: 390 N
(d) B1 the water exerts an upward force (upthrust) on the statue equal to the weight of water displaced
(d) B1 this upthrust supports part of the statue's weight, so the force needed from the balance (and hence the reading) decreases, oe
(d) Answer: Upthrust from the displaced water supports part of the statue's weight, reducing the balance reading.
Question 8
(a) B1 named valid technique, e.g. use a set square (or fiducial marker) held against the ruler at the reading, or view the scale at eye level
(a) B1 linked to reduction of parallax error when reading the extension, oe
(a) Answer: Using a set square against the scale reduces parallax error.
(b) M1 correctly convert extensions to metres (e.g. 12 mm = 0.012 m, 61 mm = 0.061 m)
(b) M1 gradient = change in force/change in extension, using two data points, oe
(b) A1 81.6 N/m (accept 80-83 N/m), awrt
(b) Answer: 81.6 N/m
(c) M1 recall/apply E = 1/2 k x2
(c) M1 correct substitution: k from (b), x = 0.061 m, ft
(c) A1 0.152 J (accept 0.148-0.156 J, ft), awrt
(c) Answer: 0.152 J
(d) L1 (1-2): Basic, isolated statements about sources of error, with little or no reference to the specific method and limited or no suggestions for improvement.
(d) L2 (3-4): Some relevant sources of error and improvements are identified, showing some application to this specific method, but explanations may be incomplete or the link between error and improvement unclear.
(d) L3 (5-6): A range of relevant sources of error is identified with clear, well-linked improvements that show detailed understanding of how each improvement increases accuracy or reduces uncertainty in this specific method.
(d) Answer: See levels of response and indicative content.
Question 9
(a) M1 radius = diameter/2 = 0.14 mm = 1.4 x 10-4 m
(a) M1 correct substitution into A = π r2
(a) A1 6.16 x 10-8 m2 (accept 6.1-6.2 x 10-8 m2), awrt
(c) M1 stress = force/area, using area from (a), ft
(c) A1 3.25 x 108 Pa (accept 3.2-3.3 x 108 Pa, ft), awrt
(c) Answer: 3.25 x 108 Pa
(d) M1 Young modulus = stress/strain, using (b) and (c), ft
(d) A1 1.16 x 1011 Pa (accept 1.1-1.2 x 1011 Pa, ft), awrt
(d) Answer: 1.16 x 1011 Pa
(e) B1 valid precaution, e.g. use of a reference wire alongside the test wire to compensate for the effects of temperature change and to support the vernier scale, or measuring the diameter at several points/orientations along the wire and using a mean value
(e) Answer: Use of a reference wire to compensate for temperature effects (or measuring the diameter at several points and averaging).
(f) M1 recall/apply E_stored = 1/2 F x
(f) M1 correct substitution: F = 20 N, x = 5.6 x 10-3 m
(f) A1 0.056 J (accept 0.055-0.057 J), awrt
(f) Answer: 0.056 J
Question 10
(a) M1 use trigonometry, e.g. cos(θ) = 3.0/5.0 (or equivalent using the 3-4-5 triangle)
(c) M1 recognise horizontal equilibrium: since the wall is smooth, friction at the foot must equal the wall's reaction force, ft from (b)
(c) A1 439 N (ft), cao
(c) Answer: 439 N
(d) M1 recognise vertical equilibrium: N = weight of ladder + weight of person
(d) A1 820 N, cao
(d) Answer: 820 N
(e) M1 μ = friction force/normal reaction force, using (c) and (d), ft
(e) A1 0.535 (accept 0.53-0.54, ft), awrt
(e) Answer: 0.535
Question 11
(a) M1 average speed = h/t substitution
(a) A1 cso: 1.98 m/s
(a) Answer: 1.98 m/s
(b) M1 correct rearrangement/use of g = 2h/t2
(b) M1 correct substitution: g = 2(0.800)/(0.404)2
(b) A1 9.80 m/s2 (accept 9.79-9.81 m/s2), awrt
(b) Answer: 9.80 m/s2
(c) M1 % difference = (|experimental - accepted|/accepted) x 100, ft from (b)
(c) A1 0.10% (accept 0.09-0.11%, ft), awrt
(c) Answer: 0.10%
(d) B1 valid random error, e.g. small variation in the height h measured with a ruler (parallax), or slight variation in the release of the ball bearing each time
(d) B1 valid method to reduce its effect, e.g. repeat the drop several times at the same height and calculate a mean value of t (or g), oe
(d) Answer: Repeat the drop several times and take a mean value of t (or g) to reduce the effect of random error.
(e) B1 light gates start and stop the timing electronically at the instant the ball passes, removing the effect of human reaction time
(e) B1 this removes a (systematic) timing error that would otherwise make the measured time too long or inconsistent, giving a more accurate value of g, oe
(e) Answer: Light gates remove human reaction time error, giving a more accurate value of t and hence g.
Question 12
(a) B1 the point beyond which strain is no longer directly proportional to stress (Hooke's law is no longer obeyed), oe
(a) Answer: The point beyond which stress is no longer directly proportional to strain.
(b) B1 the material begins to deform plastically (extends significantly) with little or no further increase in stress/load, oe
(b) Answer: The wire begins to stretch plastically with little or no increase in stress.
(c) M1 recognise Young modulus = gradient of linear region = stress/strain
(c) M1 correct substitution: 2.0x108/1.6x10-3
(c) A1 1.25 x 1011 Pa (125 GPa), cao
(c) Answer: 1.25 x 1011 Pa
(d) L1 (1-2): Basic comparison with limited reference to structure; may only describe the shapes of the graphs without explanation.
(d) L2 (3-4): Some valid comparisons of behaviour and structure are made, with partial explanation of the link between them.
(d) L3 (5-6): A clear, well-linked comparison of the mechanical behaviour and underlying structure of both materials is given, explaining why each material behaves as it does.
(d) Answer: See levels of response and indicative content.
Question 13
(a) M1 change in momentum = m x change in velocity, substitution
(a) A1 16200 kg m/s (1.62 x 104 kg m/s), cao
(a) Answer: 16200 kg m/s
(b) M1 recall/apply F = change in momentum/time
(b) M1 correct substitution using (a) and t = 0.12 s, ft
(b) A1 1.35 x 105 N, cao/ft
(b) Answer: 1.35 x 105 N
(c) M1 correct substitution using (a) and new t = 0.20 s, ft
(c) A1 8.1 x 104 N, cao/ft
(c) Answer: 8.1 x 104 N
(d) B1 the impulse (change in momentum) needed to stop the car is the same in both cases
(d) B1 since force = change in momentum/time, increasing the collision time for the same change in momentum means the average force is smaller
(d) B1 a smaller average force means a smaller deceleration of the occupants, reducing the risk of injury, oe
(d) Answer: For the same change in momentum, a longer collision time reduces the average force, and hence the deceleration and injury risk to occupants.