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13+ · Physics

CE.S-P1 Physics: Forces and Energy

ISEB CE ISEB CE 13+ Science (Physics) · Calculator allowed · about 90 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

For each situation below, state whether the main force acting is a contact force or a non-contact force.

(a)A fridge magnet holding a shopping list onto a fridge door.(1)

(b)Ravi pushing a shopping trolley along a supermarket aisle.(1)

(c)A parachutist falling towards the ground before her parachute opens.(1)

(d)A trainer's rubber sole gripping a school running track.(1)

(Total for Question 1 is 4 marks)

Sam is comparing the mass and weight of an object.

(a)State the difference between mass and weight, including the unit used to measure each quantity.(2)

(b)Sam has a mass of 8 kg. Calculate his weight on Earth. Use the equation: weight = mass x gravitational field strength, and take gravitational field strength g = 10 N/kg.(2)

(Total for Question 2 is 4 marks)

In a tug of war, Team A pulls with a force of 850 N and Team B pulls with a force of 720 N in the opposite direction, along the same straight line.

(a)Calculate the resultant force acting on the rope.(2)

(b)State which team will win the tug of war, and in which direction the rope will accelerate.(1)

(Total for Question 3 is 3 marks)

A shopping trolley of mass 25 kg is pushed by Aisha with a resultant force of 40 N. Use the equation: force = mass x acceleration.

(a)Calculate the acceleration of the trolley.(2)

(b)Aisha then loads the trolley with shopping, doubling its total mass to 50 kg, but keeps pushing with the same resultant force of 40 N. Calculate the new acceleration.(2)

(c)State what your answers to (a) and (b) show about the relationship between mass and acceleration when the resultant force stays the same.(1)

(Total for Question 4 is 5 marks)

A skydiver, Jamal, jumps from a plane and falls before opening his parachute.

(a)Which of the following correctly describes the forces acting on Jamal just after he jumps, before he reaches terminal velocity?(1)

A) Weight and air resistance are equal, so he falls at a constant speed
B) Weight is greater than air resistance, so he accelerates downwards
C) Air resistance is greater than weight, so he accelerates upwards
D) There are no forces acting on him at all

(b)Explain, in terms of forces, why Jamal eventually reaches a constant (terminal) velocity as he continues to fall.(3)

(Total for Question 5 is 4 marks)

A uniform beam is pivoted at its centre. A force of 60 N pushes down on the beam at a distance of 0.5 m from the pivot, on the left-hand side.

(a)Calculate the force, F, that must be applied at a distance of 0.3 m from the pivot on the right-hand side for the beam to balance. Use the equation: moment = force x distance from pivot.(3)

(b)State the rule (law of moments) used to answer part (a).(1)

(Total for Question 6 is 4 marks)

Priya and Tom sit on a seesaw that is pivoted at its centre. Priya has a weight of 400 N and sits 1.5 m from the pivot. Tom has a weight of 500 N.

(a)Calculate the distance from the pivot at which Tom must sit for the seesaw to balance.(4)

(b)Tom then moves to sit 1.4 m from the pivot, without changing his weight. State, with a calculation, what will happen to the seesaw.(1)

(Total for Question 7 is 5 marks)

Olivia wants to find the density of a small stone. The stone has a mass of 45 g.

(a)Describe how Olivia could use a measuring cylinder of water to find the volume of the stone.(2)

(b)The initial volume of water was 40 cm^3. After lowering the stone in, the new volume was 55 cm^3. Calculate the volume of the stone.(1)

(c)Calculate the density of the stone, giving your answer in g/cm^3. Use the equation: density = mass / volume.(2)

(Total for Question 8 is 5 marks)

The density of sea water is 1.03 g/cm^3. A piece of cork has a density of 0.24 g/cm³, and a piece of iron ore has a density of 5.2 g/cm^3.

(a)State, with a reason, whether the cork will float or sink in sea water.(2)

(b)State, with a reason, whether the iron ore will float or sink in sea water.(1)

(Total for Question 9 is 3 marks)

A farmer weighing 750 N is standing on the ground. Use the equation: pressure = force / area.

(a)The farmer's wellington boots have a total sole area of 0.025 m^2. Calculate the pressure he exerts on the ground, giving your answer in N/m² (Pa).(2)

(b)The farmer then puts on a pair of snow shoes with a much larger total area of 0.5 m^2. His weight stays the same. Calculate the pressure he now exerts on the ground.(2)

(c)Explain why wearing the snow shoes stops him sinking as easily into soft snow.(1)

(Total for Question 10 is 5 marks)

Aisha pulls a sledge with a constant force of 45 N over a horizontal distance of 12 m. Use the equation: work done = force x distance.

(a)Calculate the work done by Aisha on the sledge.(2)

(b)State the two conditions that must both be true for work to be done on an object by a force.(1)

(Total for Question 11 is 3 marks)

A parcel of mass 3.5 kg is lifted onto a shelf 1.8 m above the ground. Take gravitational field strength g = 10 N/kg, and use the equation: gravitational potential energy (GPE) = mass x g x height.

(a)Calculate the gain in gravitational potential energy of the parcel.(3)

(b)The parcel is instead lifted onto a higher shelf, doubling the height above the ground. State the effect this has on its gain in gravitational potential energy.(1)

(Total for Question 12 is 4 marks)

A go-kart of mass 120 kg travels at a speed of 8 m/s. Use the equation: kinetic energy (KE) = 0.5 x mass x speed^2.

(a)Calculate the kinetic energy of the go-kart.(3)

(b)The go-kart then speeds up to 16 m/s, double its original speed. Calculate its new kinetic energy.(2)

(Total for Question 13 is 5 marks)

A roller coaster car of mass 400 kg starts from rest at the top of a hill 20 m above the ground. Take g = 10 N/kg.

(a)Assuming no energy is lost to friction or air resistance, calculate the maximum possible speed of the car at the bottom of the hill. (Hint: all the gravitational potential energy lost is transferred to kinetic energy.)(4)

(b)In reality, the car reaches a speed of only 18 m/s at the bottom of the hill. Calculate the amount of energy transferred to the surroundings (as heat and sound) because of friction and air resistance.(2)

(Total for Question 14 is 6 marks)

A crane does 24000 J of work lifting a load in 15 seconds. Use the equation: power = work done / time taken.

(a)Calculate the power developed by the crane.(2)

(b)The crane is upgraded so that it can develop a power of 2000 W. Calculate how long it would take the upgraded crane to do the same 24000 J of work.(2)

(Total for Question 15 is 4 marks)

An electric motor is supplied with 500 J of energy. It usefully transfers 350 J of this as kinetic energy, and the rest is wasted as heat and sound.

(a)Calculate the efficiency of the motor, giving your answer as a percentage. Use the equation: efficiency = (useful energy output / total energy input) x 100%.(3)

(b)State one way, other than a design change to the motor, in which energy could be wasted as heat inside a motor.(1)

(Total for Question 16 is 4 marks)

A spring is loaded with a series of forces and the extension is measured each time. The results are shown in the table below.
Force (N): 0, 1, 2, 3, 4, 5, 6
Extension (mm): 0, 8, 16, 24, 32, 44, 60

(a)State the force at which the spring reaches its limit of proportionality.(1)

(b)Use a pair of values from within the region where the spring obeys Hooke's law to calculate the spring constant, k, in N/m. Use the equation: force = spring constant x extension.(3)

(c)Using your value of k from part (b), predict the extension, in mm, that a force of 3 N would produce if the spring continued to obey Hooke's law throughout.(1)

(d)Explain why the actual extension recorded for a force of 6 N is more than double the extension recorded for a force of 3 N.(2)

(Total for Question 17 is 7 marks)

A cyclist's motion is recorded on a speed-time graph as she rides from home to school.
From 0 to 10 s, her speed increases steadily from 0 m/s to 8 m/s.
From 10 s to 40 s, her speed stays constant at 8 m/s.
From 40 s to 50 s, her speed decreases steadily from 8 m/s back to 0 m/s.

(a)Calculate the acceleration of the cyclist during the first 10 seconds.(2)

(b)Calculate the total distance travelled by the cyclist during the whole 50 second journey, using the area under the speed-time graph.(3)

(Total for Question 18 is 5 marks)

In a physics lab, a trolley of mass 2 kg is released from rest at the top of a smooth ramp 1.25 m high. It then travels across a rough, horizontal floor, where a constant friction force of 4 N acts on it until it comes to rest. Take g = 10 N/kg.

(a)Calculate the speed of the trolley at the bottom of the ramp, assuming the ramp itself is frictionless.(4)

(b)Calculate the distance the trolley travels across the rough floor before it comes to rest, using the work done against friction.(3)

(Total for Question 19 is 7 marks)

A student wants to investigate how the size of the force applied to a spring affects its extension. Describe a method the student could use to carry out this investigation, including how to make it a fair and reliable test and how the results could be used to reach a valid conclusion about the spring's behaviour.

(Total for Question 20 is 6 marks)

Mark scheme · CE.S-P1 Physics: Forces and Energy

Question 1

(a) B1 non-contact (force), cao
(a) Answer: Non-contact force
(b) B1 contact (force), cao
(b) Answer: Contact force
(c) B1 non-contact (force), cao
(c) Answer: Non-contact force
(d) B1 contact (force), cao
(d) Answer: Contact force

Question 2

(a) B1 mass is the amount of matter (substance) in an object, measured in kilograms (kg), and does not change with location, oe
(a) B1 weight is the force of gravity acting on an object, measured in newtons (N), and depends on the gravitational field strength, oe
(a) Answer: Mass is the amount of matter in an object, measured in kilograms (kg); weight is the force of gravity acting on it, measured in newtons (N).
(b) M1 correct substitution, 8 x 10
(b) A1 80 N, cao
(b) Answer: 80 N

Question 3

(a) M1 850 - 720
(a) A1 130 N, cao
(a) Answer: 130 N
(b) B1 Team A, and the rope accelerates towards Team A, because their pulling force is larger, oe
(b) Answer: Team A will win; the rope accelerates towards Team A because their force is larger.

Question 4

(a) M1 a = F / m, so 40 / 25
(a) A1 1.6 m/s², cao
(a) Answer: 1.6 m/s²
(b) M1 a = 40 / 50
(b) A1 0.8 m/s², cao
(b) Answer: 0.8 m/s²
(c) B1 for a constant resultant force, acceleration decreases as mass increases (acceleration is inversely proportional to mass), oe
(c) Answer: Acceleration decreases as mass increases; acceleration is inversely proportional to mass for a fixed resultant force.

Question 5

(a) B1 B, cao
(a) Answer: B
(b) B1 as his falling speed increases, the air resistance (drag) acting on him also increases, oe
(b) B1 air resistance keeps increasing until it becomes equal in size to his weight, oe
(b) B1 at this point the resultant force on him is zero, so his acceleration is zero and he falls at a constant (terminal) velocity, oe
(b) Answer: Air resistance increases as his speed increases, until it equals his weight; the resultant force is then zero, so his acceleration is zero and he falls at a constant, terminal velocity.

Question 6

(a) M1 moment of 60 N force = 60 x 0.5 (= 30 Nm)
(a) dM1 set the two moments equal, F x 0.3 = 30, and rearrange for F
(a) A1 100 N, cao
(a) Answer: 100 N
(b) B1 for a beam in equilibrium (balanced), the total clockwise moment about the pivot equals the total anticlockwise moment, oe
(b) Answer: When a beam is balanced, the sum of the clockwise moments about the pivot equals the sum of the anticlockwise moments.

Question 7

(a) M1 Priya's moment = 400 x 1.5 (= 600 Nm)
(a) M1 set up Tom's moment equal to Priya's, 500 x d = 600
(a) dM1 rearrange, d = 600 / 500
(a) A1 1.2 m, cao
(a) Answer: 1.2 m
(b) B1 Tom's side goes down, because his new moment (500 x 1.4 = 700 Nm) is now greater than Priya's moment (600 Nm), ft
(b) Answer: Tom's side goes down, since his new moment, 500 N x 1.4 m = 700 Nm, is now greater than Priya's 600 Nm.

Question 8

(a) B1 partly fill the measuring cylinder with water and record the initial volume reading, oe
(a) B1 carefully lower the stone into the water so it is fully submerged, record the new (higher) volume reading, and subtract the initial reading from the new reading to find the volume of the stone (displacement method), oe
(a) Answer: Record the initial volume of water in the cylinder, lower the stone in until fully submerged, record the new volume, then subtract the initial volume from the new volume to find the volume of the stone.
(b) B1 15 cm³, cao
(b) Answer: 15 cm³
(c) M1 45 / their (b)
(c) A1 3 g/cm³, cao, ft from (b)
(c) Answer: 3 g/cm³

Question 9

(a) B1 float, cao
(a) B1 because the density of cork (0.24 g/cm³) is less than the density of sea water (1.03 g/cm³), oe
(a) Answer: Float, because its density (0.24 g/cm³) is less than the density of sea water (1.03 g/cm³).
(b) B1 sink, because its density (5.2 g/cm³) is greater than the density of sea water (1.03 g/cm³), cso
(b) Answer: Sink, because its density (5.2 g/cm³) is greater than the density of sea water (1.03 g/cm³).

Question 10

(a) M1 750 / 0.025
(a) A1 30000 N/m² (30000 Pa), cao
(a) Answer: 30000 Pa (30 kPa)
(b) M1 750 / 0.5
(b) A1 1500 N/m² (1500 Pa), cao
(b) Answer: 1500 Pa
(c) B1 the same weight (force) is spread over a much larger area, which reduces the pressure exerted on the snow, making him less likely to sink through its surface, oe
(c) Answer: The same force is spread over a larger area, so the pressure on the snow is much lower, making him less likely to break through the surface.

Question 11

(a) M1 45 x 12
(a) A1 540 J, cao
(a) Answer: 540 J
(b) B1 a force must act on the object, and the object must move some distance in the direction of that force, oe
(b) Answer: A force must act on the object, and the object must move a distance in the direction of the force.

Question 12

(a) M1 GPE = mgh
(a) M1 correct substitution, 3.5 x 10 x 1.8
(a) A1 63 J, cao
(a) Answer: 63 J
(b) B1 the gain in GPE also doubles, because GPE is directly proportional to height, oe
(b) Answer: The GPE doubles, since GPE is directly proportional to height.

Question 13

(a) M1 KE = 0.5 x m x v²
(a) M1 correct substitution, 0.5 x 120 x 8²
(a) A1 3840 J, cao
(a) Answer: 3840 J
(b) M1 0.5 x 120 x 16²
(b) A1 15360 J, cao
(b) Answer: 15360 J

Question 14

(a) M1 GPE lost = mgh = 400 x 10 x 20 (= 80000 J)
(a) M1 set KE = GPE, 0.5 x 400 x v² = 80000
(a) dM1 rearrange, v² = 80000 / 200 (= 400)
(a) A1 20 m/s, cao
(a) Answer: 20 m/s
(b) M1 actual KE = 0.5 x 400 x 18² (= 64800 J)
(b) A1 15200 J, cao (80000 - 64800)
(b) Answer: 15200 J

Question 15

(a) M1 24000 / 15
(a) A1 1600 W, cao
(a) Answer: 1600 W
(b) M1 t = work / power, 24000 / 2000
(b) A1 12 s, cao
(b) Answer: 12 s

Question 16

(a) M1 350 / 500
(a) M1 x 100
(a) A1 70%, cao
(a) Answer: 70%
(b) B1 friction between moving parts, e.g. bearings or gears, oe
(b) Answer: Friction between the motor's moving parts, such as its bearings or gears.

Question 17

(a) B1 4 N, cao
(a) Answer: 4 N
(b) M1 select a valid pair from the proportional region, e.g. 4 N and 32 mm, and convert extension to metres, 32 mm = 0.032 m
(b) M1 k = F / x, so 4 / 0.032
(b) A1 125 N/m, cao (accept 0.125 N/mm)
(b) Answer: 125 N/m
(c) B1 24 mm, ft from their k in (b)
(c) Answer: 24 mm
(d) B1 by 6 N, the spring has been stretched beyond its limit of proportionality, oe
(d) B1 so it no longer obeys Hooke's law, and it extends by more than the proportional amount for each extra newton of force applied, oe
(d) Answer: By 6 N, the spring has passed its limit of proportionality and no longer obeys Hooke's law, so it stretches by more than double the extension at 3 N.

Question 18

(a) M1 a = change in speed / time = 8 / 10
(a) A1 0.8 m/s², cao
(a) Answer: 0.8 m/s²
(b) M1 identify the three sections as two triangles and one rectangle, and find each area: 0.5 x 10 x 8 (= 40 m), 30 x 8 (= 240 m), 0.5 x 10 x 8 (= 40 m)
(b) M1 add the three areas together
(b) A1 320 m, cao
(b) Answer: 320 m

Question 19

(a) M1 GPE lost = mgh = 2 x 10 x 1.25 (= 25 J)
(a) M1 set KE = GPE, 0.5 x 2 x v² = 25
(a) dM1 rearrange, v² = 25 / 1 (= 25)
(a) A1 5 m/s, cao
(a) Answer: 5 m/s
(b) M1 work done against friction = KE at bottom of ramp = 25 J, ft from (a)
(b) M1 distance = work done / friction force, 25 / 4
(b) A1 6.25 m, cao, ft
(b) Answer: 6.25 m

Question 20

Level 1 (1-2): A basic outline is given, such as hanging masses on a spring and measuring how far it stretches, with little detail about equipment, variables or how a conclusion would be reached.
Level 2 (3-4): A clearer method is described, naming suitable equipment (e.g. clamp stand, ruler or pointer, masses), identifying the independent and dependent variables, and mentioning repeating readings, though the description of a fair test or of reaching a conclusion may be incomplete.
Level 3 (5-6): A full, logically ordered method is described, correctly identifying the independent variable (force applied), dependent variable (extension) and at least one control variable, describing how to measure extension accurately (e.g. using a fixed ruler/pointer and measuring from the natural length), stating that readings should be repeated and averaged to improve reliability, and explaining how plotting a graph of force against extension (and identifying where it stops being a straight line) would be used to reach a conclusion about the spring's behaviour and its limit of proportionality.
Indicative content:
- Independent variable: the force applied to the spring (e.g. by adding a known number of standard masses, each with weight found using weight = mass x g).
- Dependent variable: the extension of the spring, found by measuring the spring's stretched length and subtracting its original (natural) length.
- Suitable equipment: a clamp stand to hold the spring vertically, a fixed ruler or pointer/set square to measure length accurately at eye level, and a set of masses and a mass hanger.
- Control variables: use the same spring throughout, add masses one at a time in equal steps, and allow the spring to settle before each reading.
- Repeat each force reading (at least twice more) and calculate a mean extension at each force, to improve the reliability of the results and reduce the effect of random errors.
- Take a sensible range of readings, including some values likely to be beyond the point where the spring stops obeying Hooke's law, so the limit of proportionality can be identified.
- Plot a graph of force (y-axis) against extension (x-axis); while the graph is a straight line through the origin, the spring obeys Hooke's law and the gradient gives the spring constant.
- The point where the graph starts to curve away from the straight line is the limit of proportionality; beyond this point extension is no longer directly proportional to force, allowing a valid conclusion to be drawn about the spring's behaviour.
- Include a safety note, such as not overloading the spring so far that it snaps, or protecting feet/the floor from falling masses.

Physics: Forces and Energy - Worksheets, Questions and Revision