Forces can be split into two groups: contact forces and non-contact forces.
(a)Define what is meant by a contact force.(1)
(b)State whether each of the following is a contact force or a non-contact force: (i) friction (ii) magnetic force (iii) air resistance (iv) gravitational force.(4)
(Total for Question 1 is 5 marks)
2
A cyclist sets off from a lamp post, travels 500 m due north, then turns around and travels 500 m due south back to the same lamp post.
(a)State the difference between a scalar quantity and a vector quantity, giving one example of each.(2)
(b)State the cyclist's total distance travelled and total displacement for the whole journey.(2)
(Total for Question 2 is 4 marks)
3
A swimmer pushes backwards against the water with their arms in order to move forwards through a pool.
(a)State Newton's Third Law.(1)
(b)Identify the reaction force to the swimmer's push on the water, stating its size and direction compared with the swimmer's push, and explain why the swimmer moves forwards.(2)
(Total for Question 3 is 3 marks)
4
A hardback book rests on a level table and is not moving.
(a)State Newton's First Law.(2)
(b)State the two forces acting on the book and explain why the book remains stationary.(2)
(Total for Question 4 is 4 marks)
5
The gravitational field strength near the Earth's surface is g = 9.8 N/kg. Use the Physics Equations Sheet if needed: weight (N) = mass (kg) x gravitational field strength (N/kg), i.e. W = m g.
(a)Calculate the weight of a shopping bag of mass 6.0 kg on Earth.(2)
(b)The same bag is taken to the Moon, where the gravitational field strength is 1.6 N/kg. Calculate its weight on the Moon.(2)
(c)State what happens to the mass of the bag on the Moon compared with on Earth.(1)
(Total for Question 5 is 5 marks)
6
A crate is pulled along a warehouse floor. A horizontal driving force of 340 N acts on the crate in the forward direction. Friction acts on the crate with a force of 210 N in the backward direction.
(a)Calculate the resultant force acting on the crate, including its direction.(2)
(b)Describe the effect of this resultant force on the motion of the crate.(2)
(Total for Question 6 is 4 marks)
7
A delivery drone hovers above a garden. Two vertical forces act on it: an upward thrust force of 18.5 N from its rotors, and its weight of 16.0 N acting downwards.
(a)Calculate the resultant force on the drone, including its direction.(2)
(b)Describe the motion of the drone as a result of this resultant force.(1)
(Total for Question 7 is 3 marks)
8
Use the Physics Equations Sheet: work done (J) = force (N) x distance moved in the direction of the force (m), i.e. W = F s. A gardener pushes a lawnmower with a steady force of 85 N through a distance of 12 m.
(a)Calculate the work done by the gardener.(2)
(b)State the main energy transfer that takes place as the gardener does this work.(1)
(Total for Question 8 is 3 marks)
9
A skydiver jumps from a plane and falls for several seconds before opening their parachute.
(a)State the two forces acting on the skydiver as they fall, before the parachute opens.(2)
(b)Explain, in terms of these two forces, why the skydiver's acceleration decreases as their falling speed increases, until they reach terminal velocity.(3)
(Total for Question 9 is 5 marks)
10
Stopping distance = thinking distance + braking distance. A car is travelling at 24 m/s. The driver's reaction time is 0.60 s. Use the Physics Equations Sheet if needed: distance (m) = speed (m/s) x time (s).
(a)Define 'thinking distance' and 'braking distance'.(2)
(b)Calculate the thinking distance for this car.(2)
(c)State two factors that could increase the driver's reaction time.(2)
(d)State two factors, other than reaction time, that could increase the braking distance of the car.(2)
(Total for Question 10 is 8 marks)
11
A student carried out the required practical investigation into the extension of a spring. The spring was clamped vertically with a ruler fixed alongside it. Weights were added one at a time and the extension was measured each time. Force (N): 0, 1.0, 2.0, 3.0, 4.0, 5.0 Extension (mm): 0, 8, 16, 24, 32, 44
(a)Identify the anomalous result in the table.(1)
(b)Use the equation F = k e (valid only within the limit of proportionality) and the result for a force of 3.0 N to calculate the spring constant k, in N/m.(3)
(c)State one variable that should be controlled to make this a valid test of the effect of force on extension.(1)
(d)State one hazard in this experiment and a way of reducing the risk it presents.(1)
(Total for Question 11 is 6 marks)
12
Newton's Second Law can be written as: resultant force (N) = mass (kg) x acceleration (m/s2), i.e. F = m a.
(a)Define inertial mass, referring to the equation F = m a.(2)
(Total for Question 12 is 2 marks)
13
Use the Physics Equations Sheet: resultant force (N) = mass (kg) x acceleration (m/s2), i.e. F = m a. A cyclist and her bicycle have a combined mass of 78 kg.
(a)Calculate the resultant force needed to accelerate the cyclist and bicycle at 1.5 m/s2.(2)
(b)The cyclist's brakes later provide a braking force of 210 N, which is the only horizontal force acting. Calculate the deceleration produced, giving your answer to 2 significant figures.(3)
(Total for Question 13 is 5 marks)
14
A student carried out the required practical investigation into the effect of force on the acceleration of a trolley. A string over a pulley connected the trolley to hanging masses, which provided the accelerating force. The total mass of the trolley system (trolley plus hanging masses) was kept constant. A light gate and data logger recorded the acceleration for each force used. Force (N): 0.5, 1.0, 1.5, 2.0, 2.5 Acceleration (m/s2): 0.21, 0.40, 0.62, 0.79, 1.02
(a)Describe the relationship between resultant force and acceleration shown by these results.(2)
(b)Explain why it is important to keep the total mass of the trolley system constant throughout this investigation.(2)
(c)Using the result for a force of 2.0 N, calculate the total mass of the trolley system. Use the Physics Equations Sheet: F = m a.(3)
(d)Suggest one improvement the student could make to increase the accuracy of the acceleration values obtained.(2)
(Total for Question 14 is 9 marks)
15
The elastic potential energy stored in a stretched spring, while it is not permanently deformed, can be calculated using: elastic potential energy (J) = 0.5 x spring constant (N/m) x (extension (m))2, i.e. Ee = 0.5 k e2.
(a)A spring has a spring constant of 125 N/m and is extended by 0.024 m, within the limit of proportionality. Calculate the elastic potential energy stored in the spring.(3)
(Total for Question 15 is 3 marks)
16
A car travelling at high speed on a wet road has to perform an emergency stop. Explain, using ideas about forces and energy, why increasing the speed of a car has a much greater effect on its braking distance than the same increase in speed would have on its thinking distance. You may refer to the equations distance = speed x time, kinetic energy = 0.5 x mass x speed2, and work done = force x distance in your answer.
(Total for Question 16 is 6 marks)
17
Use the Physics Equations Sheet: momentum (kg m/s) = mass (kg) x velocity (m/s), i.e. p = m v. A 1200 kg car travels at 18 m/s.
(a)Calculate the momentum of the car.(2)
(b)This car then collides with a stationary 900 kg car, and the two vehicles move off together immediately after the collision. Using conservation of momentum, calculate their common velocity immediately after the collision, giving your answer to 3 significant figures.(3)
(Total for Question 17 is 5 marks)
18
The force needed to change an object's momentum can be calculated using: force (N) = change in momentum (kg m/s) / time taken (s), i.e. F = (m x change in velocity) / time taken.
(a)In a crash test, a 70 kg crash-test dummy travelling at 14 m/s is brought to rest in 0.080 s by a seatbelt. Calculate the average force exerted on the dummy by the seatbelt.(3)
(b)Explain, in terms of momentum, why an airbag reduces the risk of injury to a person in a crash compared with the person hitting the steering wheel directly.(2)
(Total for Question 18 is 5 marks)
19
A trampolinist of mass 55 kg lands on a trampoline and momentarily comes to rest after descending through a height of 0.80 m while stretching the trampoline surface. Assume all of the gravitational potential energy transferred is stored as elastic potential energy in the trampoline (ignore air resistance and other energy losses). Gravitational field strength g = 9.8 N/kg.
(a)Show that the gravitational potential energy transferred as the trampolinist descends is about 430 J. Use the Physics Equations Sheet: gravitational potential energy (J) = mass (kg) x gravitational field strength (N/kg) x height (m), i.e. GPE = m g h.(3)
(b)The trampoline behaves like a very stiff spring with spring constant k = 5400 N/m. Using Ee = 0.5 k e2 and the energy value of 431 J from part (a), calculate the extension e of the trampoline surface, giving your answer to 2 significant figures.(3)
(Total for Question 19 is 6 marks)
Mark scheme · P5 Forces
Question 1
(a) B1 a force that only acts when two objects are physically touching (in contact), oe
(a) Answer: A force that acts only when two objects are physically touching.
(b) B1 friction = contact force
(b) B1 magnetic force = non-contact force
(b) B1 air resistance = contact force
(b) B1 gravitational force = non-contact force
(b) Answer: (i) contact (ii) non-contact (iii) contact (iv) non-contact
Question 2
(a) B1 scalar has magnitude only; vector has magnitude and direction, oe
(a) B1 correct example of each, e.g. speed/distance (scalar) and velocity/displacement/force (vector)
(a) Answer: Scalar = magnitude only (e.g. speed). Vector = magnitude and direction (e.g. velocity).
(b) B1 distance = 1000 m, cao
(b) B1 displacement = 0 m (or zero), cao
(b) Answer: Distance = 1000 m; displacement = 0 m
Question 3
(a) B1 whenever two objects interact, they exert equal and opposite forces on each other, oe
(a) Answer: When two objects interact, they exert equal and opposite forces on each other.
(b) B1 the water pushes back on the swimmer's arms/hands, equal in size and opposite in direction to the swimmer's push, oe
(b) B1 this reaction force acts forwards on the swimmer, so the swimmer accelerates/moves forwards, oe
(b) Answer: The water exerts an equal and opposite (forward) force on the swimmer, which pushes the swimmer forwards.
Question 4
(a) B1 if the resultant force on an object is zero
(a) B1 the object stays at rest, or continues to move at a constant velocity in a straight line, oe
(a) Answer: If the resultant force on an object is zero, the object stays at rest or continues to move at a constant velocity in a straight line.
(b) B1 weight (acting downwards) and the normal contact force from the table (acting upwards), oe
(b) B1 these forces are equal in size and opposite in direction, so the resultant force is zero and the book stays at rest, oe, ft from named forces
(b) Answer: Weight (down) and normal contact force from the table (up); these are balanced (equal and opposite), so the resultant force is zero and the book stays still.
Question 5
(a) M1 correct substitution: W = 6.0 x 9.8
(a) A1 58.8 N, cao (allow 59 N to 2 s.f.)
(a) Answer: 58.8 N
(b) M1 correct substitution: W = 6.0 x 1.6
(b) A1 9.6 N, cao
(b) Answer: 9.6 N
(c) B1 the mass stays the same (6.0 kg); mass does not depend on gravitational field strength, oe
(c) Answer: The mass stays the same (6.0 kg).
Question 6
(a) M1 340 - 210
(a) A1 130 N forward (in the direction of the driving force), cao
(a) Answer: 130 N forward
(b) B1 the forces are unbalanced (resultant force is not zero), ft direction from 6a
(b) B1 the crate accelerates in the forward direction (direction of the resultant force), oe
(b) Answer: The crate accelerates forwards because the forces on it are unbalanced.
Question 7
(a) M1 18.5 - 16.0
(a) A1 2.5 N upwards, cao
(a) Answer: 2.5 N upwards
(b) B1 the drone accelerates upwards because the forces are unbalanced, ft direction from 7a
(b) Answer: The drone accelerates upwards.
Question 8
(a) M1 85 x 12
(a) A1 1020 J, cao
(a) Answer: 1020 J
(b) B1 energy is transferred (by work done) from the gardener's chemical energy store to the thermal energy stores of the lawnmower/ground/surroundings (due to friction), oe
(b) Answer: Energy is transferred from the gardener's chemical energy store to thermal energy stores (due to friction).
Question 9
(a) B1 weight, acting downwards (due to gravity)
(a) B1 air resistance (drag), acting upwards
(a) Answer: Weight (downwards) and air resistance (upwards).
(b) B1 as speed increases, air resistance increases
(b) B1 air resistance gets closer to (and eventually equals) weight, so the resultant force decreases, so acceleration decreases, oe
(b) B1 at terminal velocity, air resistance = weight, resultant force = zero, so the skydiver falls at a constant velocity
(b) Answer: As speed increases, air resistance increases and reduces the resultant force, reducing acceleration, until air resistance equals weight and the skydiver falls at a constant (terminal) velocity.
Question 10
(a) B1 thinking distance = the distance travelled during the driver's reaction time, before the brakes are applied, oe
(a) B1 braking distance = the distance travelled while the vehicle is decelerating under the braking force, oe
(a) Answer: Thinking distance = distance travelled during the driver's reaction time. Braking distance = distance travelled while braking.
(b) M1 24 x 0.60
(b) A1 14.4 m, cao
(b) Answer: 14.4 m
(c) B1 any one valid factor, e.g. tiredness/fatigue, alcohol or drugs, being distracted (e.g. using a mobile phone), illness/age
(c) B1 any second different valid factor from the list above
(c) Answer: Any two of: tiredness, alcohol/drugs, distraction (e.g. phone use), illness.
(d) B1 any one valid factor, e.g. wet/icy/loose road surface, worn or faulty brakes, worn tyres/low tread, greater vehicle mass/heavier load, higher speed
(d) B1 any second different valid factor from the list above
(d) Answer: Any two of: poor road/weather conditions, worn brakes or tyres, greater mass, higher speed.
Question 11
(a) B1 the extension of 44 mm at a force of 5.0 N (expected to be about 40 mm based on the pattern)
(a) Answer: 44 mm at 5.0 N
(b) M1 convert extension to metres: 24 mm = 0.024 m
(b) M1 correct rearrangement and substitution: k = 3.0 / 0.024
(b) A1 125 N/m, cao
(b) Answer: 125 N/m
(c) B1 any valid control variable, e.g. use the same spring throughout, measure extension from the same fixed point each time, oe
(c) Answer: Use the same spring each time (and measure from the same fixed reference point).
(d) B1 hazard: masses/weights could fall and injure feet; reduce risk by placing a soft mat/foam below or standing clear (or: spring/clamp could release suddenly - wear eye protection)
(d) Answer: Falling masses could injure feet; place a soft mat below the weights or stand clear.
Question 12
(a) B1 inertial mass is a measure of how difficult it is to change the velocity of an object (its resistance to acceleration), oe
(a) B1 it is given by the ratio of force to acceleration, m = F / a
(a) Answer: Inertial mass measures how difficult it is to change an object's velocity; it is given by m = F/a.
Question 13
(a) M1 78 x 1.5
(a) A1 117 N, cao
(a) Answer: 117 N
(b) M1 correct rearrangement: a = F / m
(b) M1 correct substitution: a = 210 / 78
(b) A1 2.7 m/s2, awrt 2.7
(b) Answer: 2.7 m/s2 (2 s.f.)
Question 14
(a) B1 acceleration increases as force increases (positive correlation)
(a) B1 acceleration is (directly) proportional to resultant force, for a constant total mass, oe
(a) Answer: Acceleration is directly proportional to the resultant force (for a fixed total mass).
(b) B1 total mass also affects acceleration, since F = m a, oe
(b) B1 if mass changed between trials, it would not be possible to tell whether a change in acceleration was due to the change in force alone (not a valid/fair test), oe
(b) Answer: Mass also affects acceleration (F = ma), so keeping it constant ensures any change in acceleration is due only to the change in force, making it a fair test.
(c) M1 correct rearrangement: m = F / a
(c) M1 correct substitution: m = 2.0 / 0.79
(c) A1 2.5 kg, awrt 2.5
(c) Answer: 2.5 kg (2 s.f.)
(d) B1 valid improvement identified, e.g. use light gates/a data logger instead of manual timing to remove human reaction-time error, oe
(d) B1 explanation of how this improves accuracy/reduces error, e.g. repeat each force value and calculate a mean acceleration to reduce the effect of random error
(d) Answer: Use light gates and a data logger (removes human reaction-time error), and/or repeat each measurement and take a mean to reduce random error.
Question 15
(a) M1 correct substitution: Ee = 0.5 x 125 x 0.0242
(a) M1 correct evaluation of 0.0242 = 0.000576 and 0.5 x 125 = 62.5
(a) A1 0.036 J, cao
(a) Answer: 0.036 J
Question 16
Level 1 (1-2): Simple, largely unlinked statements are made, e.g. that a faster car takes longer to stop. Little or no reference to forces or energy.
Level 2 (3-4): Some correct links are made between speed and thinking distance and/or between speed and braking distance, with partial use of relevant ideas (e.g. distance = speed x time, or kinetic energy), but the explanation of why braking distance is affected more than thinking distance is incomplete.
Level 3 (5-6): A clear, logically structured explanation is given that correctly links thinking distance to distance = speed x time (so it increases in direct/linear proportion to speed), and links braking distance to kinetic energy (0.5 x mass x speed^2) and the work done by the brakes (force x distance), explaining that braking distance increases with the square of speed. The answer explains why this means braking distance is affected far more than thinking distance when speed increases, and may also link the wet road to reduced friction/braking force increasing the braking distance further.
Indicative content:
Thinking distance = speed x reaction time, so thinking distance increases in direct (linear) proportion to speed
Braking distance depends on the kinetic energy of the car, KE = 0.5 x mass x speed^2, so kinetic energy increases with the square of speed
The brakes do work to remove this kinetic energy: work done = braking force x braking distance, so for a roughly constant braking force, braking distance is proportional to kinetic energy, i.e. proportional to speed^2
Doubling the speed doubles the thinking distance but roughly quadruples the braking distance (since it depends on speed squared)
This means an increase in speed has a much greater effect on braking distance than on thinking distance
A wet road reduces the friction between the tyres and the road, reducing the maximum braking force available, which further increases the braking distance and the risk of an accident at high speed
Question 17
(a) M1 1200 x 18
(a) A1 21600 kg m/s, cao
(a) Answer: 21600 kg m/s
(b) M1 total momentum before collision = 21600 + 0 = 21600 kg m/s, ft from 17a
(b) M1 correct rearrangement using combined mass: v = 21600 / (1200 + 900)
(b) A1 10.3 m/s, awrt 10.3
(b) Answer: 10.3 m/s (3 s.f.)
Question 18
(a) M1 change in momentum = 70 x 14 = 980 kg m/s
(a) M1 correct substitution: F = 980 / 0.080
(a) A1 12250 N (or 1.2 x 104 N to 2 s.f.), awrt 12250
(a) Answer: 12250 N
(b) B1 the airbag increases the time taken for the person's momentum to reach zero (the change in momentum happens over a longer time), oe
(b) B1 since force = change in momentum / time, a longer time for the same change in momentum means a smaller force is exerted on the person, reducing the risk of injury, oe
(b) Answer: The airbag increases the time taken to stop the person's momentum, which reduces the force needed to produce that change in momentum, lowering the risk of injury.
Question 19
(a) M1 correct substitution: GPE = 55 x 9.8 x 0.80
(a) M1 correct evaluation: 55 x 9.8 = 539, then 539 x 0.80 = 431.2 J
(a) A1 431 J shown, rounding to about 430 J as given, cso (answer printed in question, full working must be shown)
(a) Answer: 431 J (about 430 J)
(b) M1 correct rearrangement: e2 = 431 / (0.5 x 5400), ft from 19a
(b) M1 correct evaluation: e2 = 431 / 2700 = 0.1596, then e = √0.1596