Edexcel A level Maths (9MA0) · Mechanics › Quantities and units
Practise SI units and modelling assumptions. 4 exam-style questions plus unlimited generated ones on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
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State one consequence, for the forces in the model, of modelling the rope as light.
The tension is the same throughout the rope (the weight of the rope is ignored).
Sample questions
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy5 marks
A child pulls a sledge across horizontal snow using a rope. In a model of the motion: • the sledge is modelled as a particle • the rope is modelled as light and inextensible • the snow is modelled as smooth.
(a) Explain what is meant by modelling the sledge as a particle.[1]
(b) State one consequence, for the forces in the model, of modelling the rope as light.[1]
(c) Explain why modelling the snow as smooth is unrealistic, and state how this assumption affects the predicted acceleration of the sledge for a given pulling force.[2]
(d) The sledge has mass 12 kg, and the rope is horizontal with tension 30 N. Using the model, find the acceleration of the sledge.[1]
Show the answer and mark scheme
(a)Answer: The size (dimensions) of the sledge is ignored: its mass is treated as concentrated at a single point.
B1 for the dimensions of the sledge are ignored (its mass acts at a single point), so, e.g., rotation and air resistance can be ignored
Worked solution: Modelling the sledge as a particle means ignoring its size and shape: all its mass is treated as being at one point, so effects such as rotation (and air resistance, which depends on shape) are ignored.
(b)Answer: The tension is the same throughout the rope (the weight of the rope is ignored).
B1 for the tension is the same at every point of the rope (or: the rope's weight/mass is ignored)
Worked solution: A light rope has no mass, so its weight is ignored and the tension is the same all along the rope.
(c)Answer: Real snow exerts some friction on the sledge; ignoring it means the model over-estimates the acceleration.
B1 for in reality there is friction (resistance) between the sledge and the snow
B1 for the model predicts an acceleration that is too large
Worked solution: Real snow exerts a frictional force opposing the motion. A model with no friction leaves out a force that opposes the pull, so it predicts a larger acceleration than the real one.
(d)Answer: 2.5 m s−2
B1 for 2.5 (m s−2) from \(30 = 12a\)
Worked solution: \(F = ma\): \(30 = 12a\), so \(a = 2.5\) m s−2.
Question 2Medium8 marks
A train travels along a straight horizontal track. A student is asked to find the distance it travels in 5 minutes at a constant speed of 72 km h−1. The student writes: distance = speed × time = 72 × 5 = 360 m The train then decelerates uniformly from 72 km h−1 to rest in 40 seconds. The student writes: deceleration = 72 ÷ 40 = 1.8 m s−2
(a) Explain the mistake the student has made in finding the distance, and find the correct distance, in metres.[3]
(b) Find the correct deceleration of the train.[2]
(c) Find the average speed of the train for the whole motion, from the start of the 5 minutes until it comes to rest. Give your answer in m s−1 to 3 significant figures.[3]
Show the answer and mark scheme
(a)Answer: The speed (in km h−1) and time (in minutes) have been used without converting to consistent units. Correct distance: 20 m s−1 × 300 s = 6000 m. m
B1 for the units are not consistent (km h−1 and minutes are mixed; the answer is not in metres)
M1 for 72 km h−1 = 20 m s−1 and 5 minutes = 300 s (or 72 × \(\frac{5}{60}\) = 6 km)
A1 for 6000 m
Worked solution: The student multiplied a speed in km h−1 by a time in minutes, so the units are inconsistent. \(72\text{ km h}^{-1} = \frac{72 \times 1000}{3600} = 20\text{ m s}^{-1}\) and 5 minutes = 300 s, so the distance is \(20 \times 300 = 6000\) m.
(b)Answer: \(0.5\) m s−2
M1 for \(\frac{20}{40}\) (speed in m s−1)
A1 for 0.5 (m s−2)
Worked solution: \(a = \frac{v - u}{t} = \frac{0 - 20}{40} = -0.5\), so the deceleration is 0.5 m s−2.
(c)Answer: \(18.8\) m s−1
M1 for the distance while decelerating: \(\tfrac{1}{2} \times 20 \times 40 = 400\) m (or \(s = \left(\frac{u + v}{2}\right)t\))
M1 for \(\frac{\text{total distance}}{\text{total time}} = \frac{6000 + 400}{300 + 40}\) using their distances
A1 for awrt 18.8 (m s−1)
Worked solution: Distance while decelerating: \(s = \left(\frac{20 + 0}{2}\right) \times 40 = 400\) m. \(\text{average speed} = \frac{6000 + 400}{300 + 40} = \frac{6400}{340} = 18.82\ldots = 18.8\text{ m s}^{-1}\) (3 s.f.)
Question 3Hard12 marks
A skydiver of mass 80 kg falls vertically from rest from a hovering helicopter. In model A, the skydiver is modelled as a particle falling freely under gravity. In model B, the skydiver is modelled as a particle falling under gravity against air resistance of magnitude \(kv^2\) newtons, where \(v\) m s−1 is the speed of the skydiver and \(k\) is a constant. Take \(g = 9.8\) m s−2. Give any answer that uses this value of \(g\) to 2 or 3 significant figures.
(a) Use model A to find the speed of the skydiver 3 seconds after leaving the helicopter.[2]
(b) Use model A to find the distance fallen by the skydiver in these 3 seconds.[2]
(c) Explain why model A becomes less accurate as the skydiver's speed increases.[2]
(d) Find the SI unit of \(k\), giving your answer in terms of kg, m and s.[2]
(e) In model B, the skydiver eventually falls at a constant speed of 50 m s−1. Find the value of \(k\).[3]
(f) State one assumption about \(k\) that is made in model B, and explain why it may not be realistic.[1]
Show the answer and mark scheme
(a)Answer: 29.4 m s−1 (accept 29 m s−1) m s−1
M1 for using \(v = u + at\) with \(u = 0, \ a = 9.8, \ t = 3\)
A1 for 29.4 or 29 (m s−1)
Worked solution: \(v = u + at = 0 + 9.8 \times 3 = 29.4\text{ m s}^{-1}\)
(b)Answer: 44.1 m (accept 44 m) m
M1 for using \(s = ut + \tfrac{1}{2}at^2\) (or \(s = \left(\frac{u + v}{2}\right)t\)) with \(u = 0, \ a = 9.8, \ t = 3\)
A1 for 44.1 or 44 (m)
Worked solution: \(s = ut + \tfrac{1}{2}at^2 = 0 + \tfrac{1}{2} \times 9.8 \times 3^2 = 44.1\text{ m}\)
(c)Answer: Air resistance increases as the speed increases, so the acceleration falls below \(g\); model A (constant acceleration \(g\)) increasingly overestimates the speed.
B1 for air resistance increases with speed
B1 for so the acceleration is less than \(g\) (and model A overestimates the speed / distance)
Worked solution: Model A ignores air resistance. In reality air resistance grows as the skydiver speeds up, so the resultant force, and hence the acceleration, decreases. Model A keeps the acceleration at \(g\), so its predictions become too large.
(d)Answer: kg m−1
M1 for \(k = \frac{\text{force}}{v^2}\) with N = kg m s−2
A1 for kg m−1
Worked solution: kg m s−2 ÷ (m s−1)2 = kg m s−2 ÷ m2 s−2 = kg m−1
(e)Answer: \(k = 0.314\) (accept 0.31)
M1 for the resultant force being zero at constant speed (air resistance balances the weight)
A1 for a correct equation: \(k \times 50^2 = 80 \times 9.8\)
A1 for 0.314 or 0.31
Worked solution: At constant speed the acceleration is zero, so the air resistance balances the weight: \(k \times 50^2 = 80 \times 9.8 \Rightarrow k = \frac{784}{2500} = 0.3136 = 0.314\) (3 s.f.)
(f)Answer: \(k\) is constant; in reality it depends on the skydiver's body position (and changes when the parachute opens).
B1 for \(k\) is constant, with a reason why it may vary, e.g. the skydiver's position or shape changes / a parachute opens
Worked solution: Model B assumes \(k\) is constant, but the air resistance depends on the shape and orientation of the skydiver, which can change during the fall (for example when a parachute opens).
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