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GCSE · Physics

P5b Forces, Newton's Laws and Momentum

AQA 8464 · Calculator allowed · about 105 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Forces on the road: from Newton's laws to crash safety

Original text written for Revision Library.

Every time a car speeds up, slows down or turns a corner, its motion is being controlled by resultant forces acting on it, exactly as described by Newton's three laws of motion. Engineers use these laws, together with the equation force = mass x acceleration, to predict how a vehicle will behave and to design safety features that protect the people inside it. Stopping distance calculations show why speed, road conditions and a driver's reaction time all matter so much for road safety. In a crash, a car's momentum has to change very quickly, and it is this rapid change of momentum that produces the large forces which cause injury. Features such as seat belts, airbags and crumple zones are designed to stretch out the time over which this change happens, reducing the forces on passengers. The same ideas about forces and stretching apply on a much smaller scale to a spring: engineers rely on Hooke's law to predict how far a spring will stretch for a given force, right up until it reaches its limit of proportionality.

Figure 1 shows a crate on a rough floor. A rope pulls the crate to the right with a force of 40 N. Friction acts on the crate, opposing its motion, with a force of 15 N to the left.

(a)Calculate the resultant (net) force acting on the crate, including its direction.(2)

(b)The crate starts at rest. Describe and explain what happens to the crate's motion once the rope starts pulling with this resultant force acting on it.(2)

(Total for Question 1 is 4 marks)

A hockey puck is sliding across smooth ice at a constant velocity.

(a)State Newton's First Law.(1)

(b)State the resultant force acting on the puck while it slides at constant velocity.(1)

(c)The puck eventually slows down and stops. Explain, in terms of forces, why this happens.(2)

(Total for Question 2 is 4 marks)

Amara, a cyclist, and her bicycle have a combined mass of 78 kg. The resultant forward force acting on Amara and her bicycle is 156 N. Use the equation force = mass x acceleration (F = m x a).

(a)Calculate Amara's acceleration.(2)

(b)State what would happen to Amara's acceleration if the same resultant force acted on a greater combined mass (for example, if she carried a heavy backpack).(1)

(Total for Question 3 is 3 marks)

Ryan drives a go-kart. The go-kart and Ryan have a combined mass of 150 kg. The engine provides a driving force, and resistive forces (friction and air resistance) act against the motion. Use the equation force = mass x acceleration (F = m x a).

(a)The driving force is 450 N and the total resistive force is 150 N. Calculate the resultant force on the go-kart.(1)

(b)Calculate the acceleration of the go-kart.(2)

(c)Ryan then carries a passenger, increasing the total mass to 200 kg. Assuming the resultant force stays the same as in part (a), calculate the new acceleration.(2)

(Total for Question 4 is 5 marks)

The gravitational field strength on Earth is 9.8 N/kg. Use the equation weight = mass x gravitational field strength (W = m x g).

(a)A bag of cement has a mass of 25 kg. Calculate its weight on Earth.(2)

(b)On the Moon, the gravitational field strength is 1.6 N/kg. Calculate the weight of the same bag of cement on the Moon.(2)

(c)State what happens to the mass of the bag of cement when it is taken to the Moon, and explain your answer.(2)

(Total for Question 5 is 6 marks)

Leah, a swimmer, pushes backwards against the water with her hands and feet in order to swim forwards.

(a)State Newton's Third Law.(1)

(b)Identify the pair of forces described by Newton's Third Law in this situation, stating how their sizes and directions compare.(2)

(c)A common misconception is that Newton's Third Law pairs of forces cancel out. Explain why they do not cancel out in this situation.(1)

(Total for Question 6 is 4 marks)

Figure 2 shows a car of mass 1200 kg travelling along a level road. Four forces act on the car: the driving force from the engine (F(drive)), the combined resistive force from air resistance and friction (F(resist)), the car's weight (W), and the normal contact force from the road (N).

(a)State the name of the pair of forces that act vertically on the car, and state how their sizes compare while the car travels along the flat, level road.(2)

(b)At one instant, F(drive) = 3000 N and F(resist) = 3000 N. Calculate the resultant horizontal force on the car at this instant, and state what this means for the car's motion.(2)

(c)The driver presses the accelerator so that F(drive) increases to 3900 N, while F(resist) stays at 3000 N. Calculate the new resultant force, and use F = m x a to calculate the car's acceleration at this instant.(3)

(Total for Question 7 is 7 marks)

Stopping distance = thinking distance + braking distance.

(a)Define 'thinking distance'.(1)

(b)Define 'braking distance'.(1)

(c)Give two factors that increase a driver's reaction time, and so increase thinking distance.(2)

(d)Give two factors that increase braking distance for a given speed.(2)

(Total for Question 8 is 6 marks)

Priya is driving at 24 m/s and has a reaction time of 0.60 s. Use thinking distance = speed x reaction time.

(a)Calculate Priya's thinking distance.(2)

(b)Priya then reads a text message while driving, which increases her reaction time to 1.5 s. Calculate her new thinking distance at the same speed, and state how much further she travels while thinking, compared with part (a).(3)

(Total for Question 9 is 5 marks)

A car of mass 1000 kg is travelling when the driver brakes. Use the equation force = mass x acceleration (F = m x a).

(a)The brakes provide a braking force of 4000 N. Calculate the deceleration of the car.(2)

(b)The car's tyres are worn, reducing the maximum braking force that can be safely applied to 2500 N. Calculate the new deceleration.(2)

(c)State and explain the effect of the worn tyres on the car's braking distance for the same initial speed.(2)

(Total for Question 10 is 6 marks)

Higher tier only. Kwame, a rugby player of mass 95 kg, runs with a velocity of 6.0 m/s. Use the equation momentum = mass x velocity (p = m x v).

(a)Calculate Kwame's momentum.(2)

(b)State the direction of Kwame's momentum.(1)

(Total for Question 11 is 3 marks)

Higher tier only. Figure 3 shows trolley A (mass 2.0 kg) moving at 3.0 m/s towards stationary trolley B (mass 1.0 kg). They collide and stick together, moving off with a common velocity, v.

(a)State the law of conservation of momentum.(1)

(b)Calculate the total momentum of the system before the collision.(2)

(c)Use conservation of momentum to calculate the common velocity, v, of the two trolleys after the collision.(3)

(Total for Question 12 is 6 marks)

Higher tier only. A car of mass 900 kg is travelling at 12 m/s when it crashes into a wall and comes to rest in 0.15 s. Use the equation force = change in momentum / time taken, i.e. F = (m x change in v) / change in t.

(a)Calculate the magnitude of the change in momentum of the car during the crash.(2)

(b)Calculate the average force on the car during the crash.(2)

(c)The car is fitted with a crumple zone, which increases the collision time to 0.45 s for the same change in momentum. Calculate the new average force on the car.(2)

(d)Explain why increasing the collision time reduces the force on the occupants of the car.(2)

(Total for Question 13 is 8 marks)

Modern cars include safety features such as seat belts, airbags and crumple zones, designed to reduce the risk of injury to passengers during a collision. In a collision, a passenger's speed must be reduced to zero, which requires a resultant force to act on them. Use ideas about force, deceleration and time to explain how these safety features work to reduce the risk of injury.

(Total for Question 14 is 6 marks)

Oliver, a student, carries out the required practical to investigate the extension of a spring. He clamps a spring vertically, measures its natural (unstretched) length of 12.0 cm, then hangs different weights from the bottom, measuring the new total length with a ruler and a set square each time. Figure 4 shows the equipment. His results are given in the table: Force (N): 0, 1.0, 2.0, 3.0, 4.0, 5.0; Length (cm): 12.0, 14.0, 16.0, 18.0, 20.0, 23.5.

(a)State the independent variable and the dependent variable in this investigation.(2)

(b)State one variable that should be controlled during this investigation, and explain why.(2)

(c)State why a set square is used against the ruler when measuring the length of the spring, rather than reading the ruler directly.(1)

(d)Calculate the extension of the spring for each force, using the natural (unstretched) length of 12.0 cm, and complete the extension column of the table.(2)

(e)Using the extension values for forces from 0 to 4.0 N, show that the spring obeys Hooke's law over this range.(2)

(f)Describe what happens at 5.0 N, and name the point on a force-extension graph beyond which Hooke's law no longer applies.(2)

(g)Convert the extension at 4.0 N into metres, then use Hooke's law, F = k x, to calculate the spring constant, k, while the spring is obeying Hooke's law.(3)

(h)Suggest one improvement to the method that would give a more accurate value for the spring constant, and explain why it would improve the result.(2)

(Total for Question 15 is 16 marks)

A spring is stretched within its limit of proportionality. A different spring is stretched beyond its limit of proportionality.

(a)State what is meant by 'elastic deformation'.(1)

(b)State whether the spring stretched beyond its limit of proportionality will return to its original length when the force is removed, and explain your answer.(2)

(Total for Question 16 is 3 marks)

A parcel of mass 5.0 kg is released from rest and falls through the air. Air resistance acts on the parcel as it falls. Take the gravitational field strength as 9.8 N/kg.

(a)Calculate the weight of the parcel, using W = m x g.(2)

(b)Just after release, air resistance is very small, so the resultant force on the parcel can be approximated as equal to its weight. Use F = m x a to calculate the parcel's initial acceleration.(2)

(c)As the parcel speeds up, air resistance increases until it becomes equal in size to the parcel's weight. State the resultant force on the parcel at this point, describe its motion from then on, and name this constant speed.(3)

(d)Explain, in terms of the forces acting, why the parcel's acceleration gradually decreases as its speed increases, before it reaches terminal velocity.(2)

(Total for Question 17 is 9 marks)

Mark scheme · P5b Forces, Newton's Laws and Momentum

Question 1

(a) M1 40 - 15 (subtracts the smaller force from the larger) oe
(a) A1 25 N to the right cao (direction required)
(a) Answer: 25 N to the right
(b) B1 the crate accelerates (starts moving and speeds up) in the direction of the resultant force, i.e. to the right, oe
(b) B1 because there is a non-zero (unbalanced) resultant force acting on it, oe
(b) Answer: The crate accelerates to the right, because a non-zero resultant force is acting on it.

Question 2

(a) B1 an object will remain at rest, or continue moving at a constant velocity in a straight line, unless it is acted on by a resultant (unbalanced) force, oe
(a) Answer: An object stays at rest, or keeps moving at a constant velocity in a straight line, unless a resultant force acts on it.
(b) B1 zero (the forces on the puck are balanced)
(b) Answer: Zero N.
(c) B1 a friction force from the ice acts on the puck, opposing its motion, oe
(c) B1 this creates a resultant (unbalanced) force in the opposite direction to the puck's motion, causing it to decelerate, oe
(c) Answer: Friction from the ice acts against the puck's motion, creating a resultant force that decelerates it until it stops.

Question 3

(a) M1 rearranges and substitutes: a = 156 / 78
(a) A1 2.0 m/s² cao
(a) Answer: 2.0 m/s²
(b) B1 the acceleration would be smaller (decrease), because acceleration is inversely proportional to mass for a given force, oe
(b) Answer: The acceleration would be smaller.

Question 4

(a) B1 300 N (450 - 150)
(a) Answer: 300 N
(b) M1 ft substitutes: a = their (a) / 150
(b) A1 2.0 m/s² cao ft
(b) Answer: 2.0 m/s²
(c) M1 ft substitutes: a = their (a) / 200
(c) A1 1.5 m/s² cao ft
(c) Answer: 1.5 m/s²

Question 5

(a) M1 substitutes: W = 25 x 9.8
(a) A1 245 N cao
(a) Answer: 245 N
(b) M1 substitutes: W = 25 x 1.6
(b) A1 40 N cao
(b) Answer: 40 N
(c) B1 the mass stays the same (unchanged)
(c) B1 because mass is the amount of matter in an object, which does not depend on gravitational field strength, oe
(c) Answer: The mass stays the same, because mass does not depend on gravitational field strength.

Question 6

(a) B1 when 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 Leah pushes the water backwards, and the water pushes Leah forwards, oe
(b) B1 the two forces are equal in size and opposite in direction, acting on different objects
(b) Answer: Leah pushes the water backwards; the water pushes Leah forwards. The forces are equal in size and opposite in direction.
(c) B1 because the two forces act on different objects (one on the water, one on Leah), so they cannot cancel; only forces acting on the same object can combine to give a resultant, oe
(c) Answer: The two forces act on different objects, so they cannot cancel out; only forces on the same object combine to a resultant.

Question 7

(a) B1 weight and the normal contact force (reaction force)
(a) B1 they are equal in size (and opposite in direction), because the car does not accelerate vertically, oe
(a) Answer: Weight and the normal contact force; they are equal in size because the car does not accelerate vertically.
(b) B1 resultant force = 0 N (3000 - 3000)
(b) B1 the car travels at a constant velocity (Newton's First Law), oe
(b) Answer: 0 N; the car travels at a constant velocity.
(c) M1 resultant = 3900 - 3000 = 900 N
(c) M1 substitutes: a = 900 / 1200
(c) A1 0.75 m/s² cao
(c) Answer: 0.75 m/s²

Question 8

(a) B1 the distance a vehicle travels during the driver's reaction time, before the brakes are applied, oe
(a) Answer: The distance travelled during the driver's reaction time, before the brakes are applied.
(b) B1 the distance a vehicle travels while decelerating, from when the brakes are applied until it stops, oe
(b) Answer: The distance travelled while decelerating, from applying the brakes until the vehicle stops.
(c) B1 any one of: tiredness/fatigue, alcohol or drugs, distractions (e.g. mobile phone use), illness
(c) B1 a second, different factor from the list above
(c) Answer: Any two of: tiredness, alcohol or drugs, distractions such as mobile phone use, illness.
(d) B1 any one of: wet, icy or loose road surface, worn or faulty brakes, worn tyres (poor tread), a large mass/heavy load in the vehicle
(d) B1 a second, different factor from the list above
(d) Answer: Any two of: a wet or icy road surface, worn brakes, worn tyres, a heavy load in the vehicle.

Question 9

(a) M1 substitutes: distance = 24 x 0.60
(a) A1 14.4 m cao
(a) Answer: 14.4 m
(b) M1 substitutes: distance = 24 x 1.5
(b) A1 36 m cao
(b) B1 ft difference = 36 - their (a) = 21.6 m further
(b) Answer: 36 m; 21.6 m further than in part (a).

Question 10

(a) M1 substitutes: a = 4000 / 1000
(a) A1 4.0 m/s² cao
(a) Answer: 4.0 m/s²
(b) M1 substitutes: a = 2500 / 1000
(b) A1 2.5 m/s² cao
(b) Answer: 2.5 m/s²
(c) B1 the braking distance increases
(c) B1 because with a smaller deceleration, the car takes longer (and travels further) to reduce its speed to zero, oe
(c) Answer: The braking distance increases, because the smaller deceleration means the car travels further before stopping.

Question 11

(a) M1 substitutes: p = 95 x 6.0
(a) A1 570 kg m/s cao
(a) Answer: 570 kg m/s
(b) B1 the same direction as his velocity (the direction he is running)
(b) Answer: The same direction he is running in.

Question 12

(a) B1 in a closed system, the total momentum before an event (such as a collision) equals the total momentum after the event, provided no external forces act, oe
(a) Answer: Total momentum before a collision equals total momentum after, provided no external forces act.
(b) M1 momentum of A = 2.0 x 3.0 = 6.0 kg m/s; momentum of B = 1.0 x 0 = 0
(b) A1 6.0 kg m/s cao (total)
(b) Answer: 6.0 kg m/s
(c) M1 (m(A) + m(B)) x v = their (b), i.e. 3.0 x v = their (b)
(c) M1 ft rearranges: v = their (b) / 3.0
(c) A1 2.0 m/s cao ft
(c) Answer: 2.0 m/s

Question 13

(a) M1 initial momentum = 900 x 12 = 10800 kg m/s; final momentum = 0
(a) A1 10800 kg m/s cao (magnitude of the decrease)
(a) Answer: 10800 kg m/s (a decrease)
(b) M1 ft substitutes: F = their (a) / 0.15
(b) A1 72000 N cao ft
(b) Answer: 72000 N
(c) M1 ft substitutes: F = their (a) / 0.45
(c) A1 24000 N cao ft
(c) Answer: 24000 N
(d) B1 for the same change in momentum, increasing the time taken decreases the rate of change of momentum, oe
(d) B1 because force = change in momentum / time, a smaller rate of change of momentum means a smaller force, oe
(d) Answer: A longer collision time gives a smaller rate of change of momentum, and since force = change in momentum / time, this means a smaller force.

Question 14

Level 1 (1-2): One or two isolated, correct statements are made, for example that airbags stop a passenger hitting the dashboard or that seat belts hold a passenger in their seat, with little or no reference to force, deceleration or time.
Level 2 (3-4): Identifies that these features increase the time taken for the passenger to come to rest during a collision, and links this to a smaller deceleration or force, but the explanation covers at most one feature in detail or the logical chain from time to force is incomplete.
Level 3 (5-6): A clear, logically sequenced explanation covering at least two safety features, showing that each increases the time taken for the passenger's speed to reach zero during a collision; this produces a smaller deceleration, and since force = mass x acceleration, a smaller deceleration means a smaller force acts on the passenger's body, reducing the risk of injury. Correct use of the terms force, deceleration and time throughout.
Indicative content:
- seat belts stretch slightly, increasing the time taken for the wearer to decelerate to a stop
- airbags inflate to cushion the passenger, increasing the time taken for the head and chest to decelerate
- crumple zones crumple/deform in a collision, increasing the time taken for the car (and its occupants) to decelerate to a stop
- for the same change in speed, increasing the time taken decreases the deceleration
- using force = mass x acceleration, a smaller deceleration means a smaller force acts on the passenger
- a smaller force on the passenger's body reduces the risk of injury
- without these features, the same change in speed would happen in a much shorter time, producing a much larger deceleration and force

Question 15

(a) B1 independent variable: the force (weight) applied to the spring
(a) B1 dependent variable: the extension (or length) of the spring
(a) Answer: Independent: force applied. Dependent: extension of the spring.
(b) B1 the spring used (the same spring must be used throughout, not swapped partway through)
(b) B1 reason: different springs have different spring constants (stiffness), so changing spring would make the extension values not comparable, oe
(b) Answer: The spring used, because different springs have different spring constants, which would make results incomparable.
(c) B1 to line up the ruler at a right angle to the spring, so the exact position of the bottom of the spring can be read accurately, avoiding a parallax error, oe
(c) Answer: To line up the reading at 90 degrees to the spring and avoid a parallax error.
(d) M1 correct method (length - 12.0) applied correctly to at least one row
(d) A1 all six extensions correct: 0, 2.0, 4.0, 6.0, 8.0, 11.5 cm
(d) Answer: 0, 2.0, 4.0, 6.0, 8.0, 11.5 cm
(e) M1 calculates extension / force for at least two forces in this range and shows it is constant (2.0 cm per N each time)
(e) A1 conclusion: extension is directly proportional to force over this range, so Hooke's law is obeyed, oe
(e) Answer: Extension / force = 2.0 cm/N for 1.0, 2.0, 3.0 and 4.0 N, so extension is directly proportional to force: Hooke's law is obeyed.
(f) B1 the extension increases disproportionately: 11.5 cm is more than the 10.0 cm expected if proportionality had continued, oe
(f) B1 this point is called the limit of proportionality
(f) Answer: The extension (11.5 cm) is more than the expected 10.0 cm; this point is called the limit of proportionality.
(g) M1 converts: 8.0 cm = 0.080 m
(g) M1 rearranges and substitutes: k = 4.0 / 0.080
(g) A1 50 N/m cao
(g) Answer: 50 N/m
(h) B1 e.g. repeat each length measurement and calculate a mean, or use smaller force increments close to the expected limit of proportionality, or use a fixed marker/fiducial and view the ruler at eye level
(h) B1 linked reason: reduces random error or pinpoints the limit of proportionality more precisely, giving a more reliable value of k, oe
(h) Answer: E.g. repeat each length measurement and take a mean, reducing random error and giving a more reliable spring constant.

Question 16

(a) B1 the object returns to its original shape and length when the deforming force is removed, oe
(a) Answer: The object returns to its original shape and length when the force is removed.
(b) B1 no, it will not return to its original length
(b) B1 because it has been deformed inelastically (permanently stretched) beyond its limit of proportionality, oe
(b) Answer: No, because it has been deformed inelastically beyond its limit of proportionality.

Question 17

(a) M1 substitutes: W = 5.0 x 9.8
(a) A1 49 N cao
(a) Answer: 49 N
(b) M1 ft substitutes: a = their (a) / 5.0
(b) A1 9.8 m/s² cao ft
(b) Answer: 9.8 m/s²
(c) B1 resultant force = 0 N
(c) B1 the parcel no longer accelerates; it moves at a constant velocity
(c) B1 this constant velocity is called terminal velocity
(c) Answer: 0 N; the parcel moves at a constant velocity, called terminal velocity.
(d) B1 as the parcel's speed increases, air resistance increases, oe
(d) B1 this reduces the resultant force on the parcel, and since acceleration = resultant force / mass, a smaller resultant force gives a smaller acceleration, oe
(d) Answer: As speed increases, air resistance increases, reducing the resultant force; since acceleration = force / mass, the acceleration decreases.

Forces, Newton's Laws and Momentum - Worksheets, Questions and Revision