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One newton is the force required to accelerate a mass of 1 kg at 1 m/s²."},{"text":"joules (J)","correct":false,"explanation":"Joules measure energy or work, which can be calculated by multiplying force by distance. Force itself is measured in newtons (N)."},{"text":"pascals (Pa)","correct":false,"explanation":"Pascals measure pressure, defined as force per unit area, rather than force itself. The correct unit of force is newtons (N)."},{"text":"watts (W)","correct":false,"explanation":"Watts measure power, which is the rate of energy transfer. They do not measure force; the correct unit is newtons (N)."}]},{"id":"2abd1726-0e95-421e-a939-38e0e590580b","cardType":"cloze","points":null,"front":"A scalar has [ ]; a vector has [ ].","back":"A scalar has size only; a vector has size and direction.","mcqOptions":[{"text":"A scalar has size only; a vector has size and direction.","correct":true,"explanation":"A scalar is fully described by its size, while a vector requires both a size and a direction to be completely specified."},{"text":"A scalar has direction only; a vector has size and direction.","correct":false,"explanation":"A scalar does not have a direction; it has size only. A vector differs because it has both size and direction."},{"text":"A scalar has size and direction; a vector has size only.","correct":false,"explanation":"This reverses the definitions. A scalar has size only, whereas a vector must have both size and direction."},{"text":"A scalar has size only; a vector has direction only.","correct":false,"explanation":"Although a vector has a direction, it must also have a size. The correct distinction is size only for scalars, but size and direction for vectors."}]},{"id":"6bc64562-9eb5-4219-9a45-1a6f71f9c7ff","cardType":"basic","points":1,"front":"What direction does friction act relative to motion?","back":"Opposite to motion, or the tendency to move.","mcqOptions":[{"text":"Opposite to motion, or the tendency to move.","correct":true,"explanation":"Friction opposes the relative motion, or attempted relative motion, between surfaces. It therefore acts opposite to the direction of motion or the tendency to move."},{"text":"In the same direction as motion, or the tendency to move.","correct":false,"explanation":"Friction resists motion rather than assisting it. It acts opposite to the motion, or opposite to the direction in which the object tends to move."},{"text":"Perpendicular to motion, or the tendency to move.","correct":false,"explanation":"The normal contact force acts perpendicular to a surface, which may cause this confusion. Friction acts opposite to motion or the tendency to move."},{"text":"Vertically downwards, or the tendency to move.","correct":false,"explanation":"Weight acts vertically downwards, but friction is not necessarily vertical. It acts opposite to motion or the tendency to move along the contact surface."}]},{"id":"8c836f57-d1c0-435b-985c-160c86d887f9","cardType":"basic","points":1,"front":"How are forces along one line combined if they act in the same direction?","back":"They are added.","mcqOptions":[{"text":"They are added.","correct":true,"explanation":"Forces acting along the same line and in the same direction reinforce one another, so their magnitudes are added to find the resultant force."},{"text":"They are subtracted.","correct":false,"explanation":"Subtraction is used when forces act in opposite directions. Forces acting along one line in the same direction are added together."},{"text":"They are averaged.","correct":false,"explanation":"Averaging forces does not give their combined effect. When forces act along one line in the same direction, their magnitudes are added."},{"text":"They are cancelled.","correct":false,"explanation":"Forces cancel only when they are equal and act in opposite directions. Forces acting in the same direction are added together."}]},{"id":"cbac10dd-4320-4eee-ba25-acbefbc29e50","cardType":"basic","points":2,"front":"How are forces along one line combined if they act in opposite directions?","back":"Subtract them; the resultant points towards the larger force.","mcqOptions":[{"text":"Subtract them; the resultant points towards the larger force.","correct":true,"explanation":"Opposing forces reduce one another, so their magnitudes are subtracted. The larger force determines the direction of the resultant force."},{"text":"Add them; the resultant points towards the larger force.","correct":false,"explanation":"Adding magnitudes is appropriate for forces acting in the same direction. Opposing forces must be subtracted, with the resultant directed towards the larger force."},{"text":"Subtract them; the resultant points towards the smaller force.","correct":false,"explanation":"The magnitudes are correctly subtracted, but the direction is reversed. The resultant points towards the larger force because its effect remains after subtraction."},{"text":"Add them; the resultant points towards the smaller force.","correct":false,"explanation":"This uses both the wrong operation and the wrong direction. Opposing forces are subtracted, and the resultant points towards the larger force."}]},{"id":"609156ba-c3e1-4f4a-9b6f-0ab7b69e4c79","cardType":"cloze","points":null,"front":"If the resultant force is [ ], forces are balanced; if it is not zero, forces are [ ].","back":"If the resultant force is zero, forces are balanced; if it is not zero, forces are unbalanced.","mcqOptions":[{"text":"If the resultant force is zero, forces are balanced; if it is not zero, forces are unbalanced.","correct":true,"explanation":"Balanced forces combine to give a zero resultant force. Any non-zero resultant means the forces do not cancel completely and are therefore unbalanced."},{"text":"If the resultant force is non-zero, forces are balanced; if it is not zero, forces are unbalanced.","correct":false,"explanation":"Balanced forces must cancel to produce a zero resultant, not a non-zero one. A non-zero resultant means the forces are unbalanced."},{"text":"If the resultant force is zero, forces are balanced; if it is not zero, forces are balanced.","correct":false,"explanation":"A zero resultant does indicate balanced forces, but a non-zero resultant does not. When the resultant is not zero, the forces are unbalanced."},{"text":"If the resultant force is non-zero, forces are balanced; if it is not zero, forces are balanced.","correct":false,"explanation":"This treats non-zero resultant forces as balanced. Forces are balanced only when their resultant is zero; otherwise, they are unbalanced."}]},{"id":"31bafb2a-2045-41b3-bcad-45e3757adc99","cardType":"basic","points":2,"front":"A box has 40 N right and 15 N left. What is the resultant force?","back":"25 N to the right.","mcqOptions":[{"text":"25 N to the right.","correct":true,"explanation":"The forces oppose each other, so their magnitudes are subtracted: 40 N - 15 N = 25 N. The larger force acts to the right."},{"text":"55 N to the right.","correct":false,"explanation":"This comes from adding 40 N and 15 N despite their opposite directions. Subtracting gives 25 N, directed to the right towards the larger force."},{"text":"25 N to the left.","correct":false,"explanation":"The magnitude is correctly found by subtracting 15 N from 40 N, but the direction is reversed. The 40 N rightward force is larger, so the resultant is rightward."},{"text":"40 N to the right.","correct":false,"explanation":"This ignores the 15 N opposing force. Subtracting it from the 40 N rightward force gives a resultant of 25 N to the right."}]},{"id":"f52a7440-87aa-45fd-80e7-bb8b271b7ac2","cardType":"cloze","points":null,"front":"The equation linking force, mass and acceleration is [ ].","back":"The equation linking force, mass and acceleration is $F = m \\times a$.","mcqOptions":[{"text":"$$F = m \\times a$","correct":true,"explanation":"Newton’s second law states that resultant force equals mass multiplied by acceleration. In SI units, kilograms multiplied by m/s² gives newtons."},{"text":"$$F = m \\div a$","correct":false,"explanation":"Dividing mass by acceleration does not produce force and gives the wrong units. Newton’s second law requires mass to be multiplied by acceleration: $F = m \\times a$."},{"text":"$$F = a \\div m$","correct":false,"explanation":"This incorrectly divides acceleration by mass. Force increases with both mass and acceleration, so the correct relationship is $F = m \\times a$."},{"text":"$$F = m + a$","correct":false,"explanation":"Mass and acceleration cannot be added because they are different physical quantities with different units. They must be multiplied to calculate force: $F = m \\times a$."}]},{"id":"a7fb2021-c162-4342-8029-028c70f1d897","cardType":"basic","points":2,"front":"A 1200 kg car has a 1800 N resultant force forwards. What is its acceleration?","back":"1.5 m/s² forwards.","mcqOptions":[{"text":"1.5 m/s² forwards.","correct":true,"explanation":"Newton’s second law gives $a = F/m = 1800/1200 = 1.5\\text{ m/s}^2$. The acceleration is forwards because the resultant force acts forwards."},{"text":"0.67 m/s² forwards.","correct":false,"explanation":"This comes from reversing the calculation and dividing mass by force. Acceleration is $F/m$, so $1800/1200 = 1.5\\text{ m/s}^2$ forwards."},{"text":"15 m/s² forwards.","correct":false,"explanation":"This is a decimal-place error when dividing 1800 by 1200. Using $a = F/m$ gives $1.5\\text{ m/s}^2$, directed forwards with the resultant force."},{"text":"1.5 m/s² backwards.","correct":false,"explanation":"The magnitude is correct, but acceleration acts in the direction of the resultant force. Since the resultant force is forwards, the acceleration is $1.5\\text{ m/s}^2$ forwards."}]},{"id":"3b4b6d0f-3aaf-477c-a312-2280186975cf","cardType":"cloze","points":null,"front":"Weight is calculated using [ ]; on Earth, $g$ is about [ ].","back":"Weight is calculated using $W = m \\times g$; on Earth, $g$ is about 9.8 N/kg.","mcqOptions":[{"text":"Weight is calculated using $W = m \\times g$; on Earth, $g$ is about 9.8 N/kg.","correct":true,"explanation":"Weight is the gravitational force on a mass, so mass is multiplied by gravitational field strength. Earth’s gravitational field strength is approximately $9.8\\text{ N/kg}$."},{"text":"Weight is calculated using $W = \\frac{m}{g}$; on Earth, $g$ is about 9.8 N/kg.","correct":false,"explanation":"Dividing mass by gravitational field strength does not produce a force. Weight is calculated by multiplication, using $W = m \\times g$, with $g$ about $9.8\\text{ N/kg}$ on Earth."},{"text":"Weight is calculated using $W = m \\times g$; on Earth, $g$ is about 98 N/kg.","correct":false,"explanation":"The formula is correct, but the value contains a factor-of-ten error. Near Earth’s surface, gravitational field strength is approximately $9.8\\text{ N/kg}$, not $98\\text{ N/kg}$."},{"text":"Weight is calculated using $W = \\frac{g}{m}$; on Earth, $g$ is about 0.98 N/kg.","correct":false,"explanation":"Both entries are incorrect: gravitational field strength must be multiplied by mass, and the decimal has been misplaced. The correct formula is $W = m \\times g$, where $g$ is about $9.8\\text{ N/kg}$."}]},{"id":"6acb7ff4-ba5b-4420-91b1-155281da8c1c","cardType":"basic","points":2,"front":"Why can weight change with location while mass does not?","back":"Weight depends on gravitational field strength; mass does not.","mcqOptions":[{"text":"Weight depends on gravitational field strength; mass does not.","correct":true,"explanation":"Weight is a force given by $W = mg$, so it changes when gravitational field strength changes. Mass measures the amount of matter and remains constant between locations."},{"text":"Mass depends on gravitational field strength; weight does not.","correct":false,"explanation":"This reverses the roles of mass and weight. Weight depends on gravitational field strength through $W = mg$, whereas an object’s mass remains unchanged when its location changes."},{"text":"Weight depends on the object’s speed; mass does not.","correct":false,"explanation":"Speed does not determine an object’s weight. Weight changes with gravitational field strength, while mass is an intrinsic property that remains constant regardless of location."},{"text":"Weight remains constant with location; mass changes.","correct":false,"explanation":"This incorrectly treats mass as location-dependent. Mass stays constant, but weight changes because different locations can have different gravitational field strengths acting on that mass."}]},{"id":"d6a3dc3c-9a86-42be-8116-ff475eab1d0a","cardType":"cloze","points":null,"front":"Stopping distance = [ ] + [ ].","back":"Stopping distance = thinking distance + braking distance.","mcqOptions":[{"text":"Stopping distance = thinking distance + braking distance.","correct":true,"explanation":"A vehicle travels during the driver’s reaction time and then continues travelling while the brakes slow it. These distances are thinking distance and braking distance respectively."},{"text":"Stopping distance = reaction time + braking distance.","correct":false,"explanation":"Reaction time is measured in seconds, so it cannot be added directly to a distance. The distance travelled during that time is thinking distance, which is added to braking distance."},{"text":"Stopping distance = thinking distance + braking force.","correct":false,"explanation":"Braking force is measured in newtons and cannot be added to thinking distance. The second component must be braking distance, the distance travelled after braking begins."},{"text":"Stopping distance = reaction distance + stopping time.","correct":false,"explanation":"This mixes a distance with a time and does not use the defined stopping-distance components. Stopping distance is the sum of thinking distance and braking distance."}]},{"id":"d26bc6e5-cc3f-4e3c-98e1-bd2175d01c93","cardType":"basic","points":2,"front":"Which stopping-distance components increase at higher speed?","back":"Thinking distance and braking distance.","mcqOptions":[{"text":"Thinking distance and braking distance.","correct":true,"explanation":"At higher speed, the vehicle travels farther during the same reaction time and also requires more distance for the brakes to remove its greater kinetic energy."},{"text":"Thinking distance but not braking distance.","correct":false,"explanation":"Higher speed increases thinking distance, but it also increases braking distance because the vehicle has more kinetic energy to dissipate before it can stop."},{"text":"Braking distance but not thinking distance.","correct":false,"explanation":"Higher speed increases braking distance, but the vehicle also travels farther during the driver’s reaction time. Therefore, thinking distance increases as well."},{"text":"Neither thinking distance nor braking distance.","correct":false,"explanation":"Both components depend on speed. A faster vehicle travels farther before the brakes are applied and requires a greater distance to slow to rest."}]},{"id":"2fadd73a-5ec5-4d55-9895-f554ba95cea3","cardType":"basic","points":1,"front":"Which part of stopping distance is increased by a longer reaction time?","back":"thinking distance","mcqOptions":[{"text":"thinking distance","correct":true,"explanation":"Thinking distance is the distance travelled between noticing a hazard and applying the brakes. A longer reaction time lets the vehicle travel farther before braking begins."},{"text":"braking distance","correct":false,"explanation":"Braking distance begins only after the brakes are applied, so reaction time does not directly determine it. A longer reaction time increases thinking distance instead."},{"text":"total distance travelled","correct":false,"explanation":"Although the overall stopping distance consequently becomes larger, the specific stopping-distance component directly increased by a longer reaction time is thinking distance."},{"text":"braking force","correct":false,"explanation":"Reaction time affects how long the vehicle travels before braking, not the force produced by the brakes. It therefore increases thinking distance, not braking force."}]},{"id":"f0bbbd44-0518-455a-967d-337c5200149e","cardType":"basic","points":2,"front":"Why do wet or icy roads increase braking distance?","back":"Friction is reduced, so the braking force is smaller.","mcqOptions":[{"text":"Friction is reduced, so the braking force is smaller.","correct":true,"explanation":"Water or ice reduces friction between the tyres and road. This limits the braking force, producing less deceleration and requiring a greater distance to stop."},{"text":"Friction is increased, so the braking force is larger.","correct":false,"explanation":"Greater friction and a larger braking force would normally shorten braking distance. Wet or icy surfaces reduce tyre-road friction, making the braking force smaller and stopping slower."},{"text":"Air resistance is reduced, so the braking force is smaller.","correct":false,"explanation":"The important change is at the contact between the tyres and road, not in air resistance. Wet or icy roads reduce friction and therefore reduce the available braking force."},{"text":"Reaction time is increased, so the braking force is smaller.","correct":false,"explanation":"Reaction time affects thinking distance and does not determine the braking force. Wet or icy roads increase braking distance because reduced tyre-road friction limits the braking force."}]},{"id":"cff5f9a9-70d2-4297-8192-4825ab7b3fc5","cardType":"basic","points":2,"front":"Why does a falling object accelerate at first?","back":"Weight is greater than air resistance, giving a downward resultant force.","mcqOptions":[{"text":"Weight is greater than air resistance, giving a downward resultant force.","correct":true,"explanation":"At first, the object is moving slowly, so air resistance is small. Weight exceeds air resistance, creating a downward resultant force and therefore downward acceleration."},{"text":"Air resistance is greater than weight, giving a downward resultant force.","correct":false,"explanation":"If air resistance were greater than weight, the resultant force would be upward, not downward. Initially weight is greater than air resistance, so the object accelerates downward."},{"text":"Weight and air resistance are equal, giving a downward resultant force.","correct":false,"explanation":"Equal upward air resistance and downward weight would produce zero resultant force and no acceleration. At first, weight is greater, creating a downward resultant force."},{"text":"Weight increases as the object falls, giving a downward resultant force.","correct":false,"explanation":"Near Earth’s surface, the object’s weight remains approximately constant as it falls. It accelerates initially because air resistance is smaller than weight, leaving a downward resultant force."}]},{"id":"960df683-a0d1-47d3-be5c-f4e55bd3b0a7","cardType":"cloze","points":null,"front":"At terminal velocity, air resistance equals [ ], so resultant force is zero and velocity is [ ].","back":"At terminal velocity, air resistance equals weight, so resultant force is zero and velocity is constant.","mcqOptions":[{"text":"At terminal velocity, air resistance equals weight, so resultant force is zero and velocity is constant.","correct":true,"explanation":"At terminal velocity, upward air resistance balances downward weight. The resultant force is therefore zero, so there is no acceleration and the velocity remains constant."},{"text":"At terminal velocity, air resistance equals weight, so resultant force is zero and velocity is zero.","correct":false,"explanation":"Zero resultant force means zero acceleration, not zero velocity. Air resistance balances weight, allowing the object to continue moving at a constant terminal velocity."},{"text":"At terminal velocity, air resistance equals upthrust, so resultant force is zero and velocity is constant.","correct":false,"explanation":"Upthrust is a separate force exerted by a fluid and does not define terminal velocity. At terminal velocity, air resistance equals weight and velocity is constant."},{"text":"At terminal velocity, air resistance equals mass, so resultant force is zero and velocity is increasing.","correct":false,"explanation":"Mass is measured in kilograms and cannot balance a force measured in newtons. Air resistance equals weight at terminal velocity, making the velocity constant rather than increasing."}]},{"id":"04cee6a5-1957-4dee-b3e2-651de47c9db1","cardType":"basic","points":1,"front":"How is extension calculated from the two lengths?","back":"extension = extended length - original length","mcqOptions":[{"text":"extension = extended length - original length","correct":true,"explanation":"Extension is the increase in an object’s length, so it is found by subtracting its original length from its length after the force is applied."},{"text":"extension = original length - extended length","correct":false,"explanation":"Reversing the subtraction produces a negative value when the object becomes longer. Extension is calculated as extended length minus original length."},{"text":"extension = extended length + original length","correct":false,"explanation":"Adding the two lengths gives their combined length, not the change in length. Extension equals extended length minus original length."},{"text":"extension = extended length ÷ original length","correct":false,"explanation":"Dividing the lengths gives a dimensionless ratio rather than an extension measured in units of length. Subtract the original length from the extended length."}]},{"id":"f4ddeecf-5b6f-46b1-a689-d184367f9e3f","cardType":"cloze","points":null,"front":"In the initial straight-line region, extension is [ ] to applied force; this is [ ].","back":"In the initial straight-line region, extension is proportional to applied force; this is Hooke’s law.","mcqOptions":[{"text":"In the initial straight-line region, extension is proportional to applied force; this is Hooke’s law.","correct":true,"explanation":"In the straight-line region, doubling the applied force doubles the extension, so the quantities are proportional. This relationship is known as Hooke’s law."},{"text":"In the initial straight-line region, extension is inversely proportional to applied force; this is Hooke’s law.","correct":false,"explanation":"An inverse relationship would make extension decrease as force increases. In the initial straight-line region, extension is proportional to force, as stated by Hooke’s law."},{"text":"In the initial straight-line region, extension is proportional to applied force; this is Newton’s second law.","correct":false,"explanation":"The proportional relationship between force and extension is Hooke’s law, not Newton’s second law. Newton’s second law instead relates resultant force, mass and acceleration."},{"text":"In the initial straight-line region, extension is constant to applied force; this is Coulomb’s law.","correct":false,"explanation":"Extension does not remain constant when force increases in the straight-line region. It is proportional to applied force, and this relationship is Hooke’s law."}]},{"id":"48a60230-5d5f-4629-b935-381a3a50bd44","cardType":"basic","points":1,"front":"In the extension practical, what variable is deliberately changed?","back":"applied force","mcqOptions":[{"text":"applied force","correct":true,"explanation":"The applied force is the independent variable because the experimenter deliberately changes it, usually by adding masses, and then measures the resulting extension."},{"text":"extension","correct":false,"explanation":"Extension is the dependent variable measured in response to changing the load. The variable deliberately changed by the experimenter is the applied force."},{"text":"original length","correct":false,"explanation":"Original length should be measured initially and then kept fixed for a fair investigation. The applied force is deliberately changed to observe its effect on extension."},{"text":"spring constant","correct":false,"explanation":"The spring constant is a property determined from force and extension data, not the variable deliberately changed. The independent variable is the applied force."}]},{"id":"7fb330f7-ac56-48b8-b166-c7fdd12f7f7a","cardType":"cloze","points":null,"front":"Momentum is calculated by [ ] and is measured in [ ].","back":"Momentum is calculated by $p = m \\times v$ and is measured in kg m/s.","mcqOptions":[{"text":"Momentum is calculated by $p = m \\times v$ and is measured in kg m/s.","correct":true,"explanation":"Momentum equals mass multiplied by velocity, so its SI unit is kilograms multiplied by metres per second, written kg m/s."},{"text":"Momentum is calculated by $p = \\frac{m}{v}$ and is measured in kg m/s.","correct":false,"explanation":"Dividing mass by velocity does not calculate momentum. Momentum is mass multiplied by velocity, $p = m \\times v$, and its unit is kg m/s."},{"text":"Momentum is calculated by $p = m \\times v$ and is measured in kg m/s².","correct":false,"explanation":"The formula is correct, but kg m/s² is the unit of force. Momentum is measured in kg m/s because it is mass multiplied by velocity."},{"text":"Momentum is calculated by $p = m \\times a$ and is measured in N.","correct":false,"explanation":"Multiplying mass by acceleration gives force, not momentum, and produces newtons. Momentum is $p = m \\times v$ and is measured in kg m/s."}]},{"id":"674508a4-9c0e-44a7-8595-3a11937aa1c1","cardType":"basic","points":2,"front":"When is total momentum conserved?","back":"In a closed system with no external resultant force.","mcqOptions":[{"text":"In a closed system with no external resultant force.","correct":true,"explanation":"Total momentum remains constant when the system is closed and no external resultant force acts. Internal forces may transfer momentum between objects but cannot change the total."},{"text":"In a closed system with a non-zero external resultant force.","correct":false,"explanation":"A non-zero external resultant force changes the total momentum of a system by providing an external impulse. Conservation requires a closed system with no external resultant force."},{"text":"In a closed system with no internal forces.","correct":false,"explanation":"The absence of internal forces is not the condition for momentum conservation, and internal interaction forces commonly occur in collisions. The system must be closed with no external resultant force."},{"text":"In an open system with no external resultant force.","correct":false,"explanation":"An open system may exchange matter and momentum with its surroundings, so zero external resultant force alone is insufficient. Conservation requires a closed system with no external resultant force."}]},{"id":"9c9b02e5-f4e1-405f-bb25-2d4760191ad3","cardType":"basic","points":2,"front":"How do airbags, seat belts, helmets or crumple zones reduce injury?","back":"They increase the time for momentum to change, reducing force.","mcqOptions":[{"text":"They increase the time for momentum to change, reducing force.","correct":true,"explanation":"For a fixed momentum change, increasing the time reduces the average force because force equals change in momentum divided by time. This lowers the risk of injury."},{"text":"They decrease the time for momentum to change, reducing force.","correct":false,"explanation":"Decreasing the stopping time increases force because the same momentum change occurs more rapidly. Safety devices instead increase the time for momentum to change."},{"text":"They increase the momentum that must change, reducing force.","correct":false,"explanation":"A larger momentum change in the same time would produce a larger force, not a smaller one. Safety devices reduce force by increasing the stopping time."},{"text":"They decrease the distance over which momentum changes, reducing force.","correct":false,"explanation":"A shorter stopping distance generally means a shorter stopping time and therefore a larger force. Crumple zones and similar devices increase stopping time to reduce force."}]},{"id":"8d0130af-7caa-481e-8300-0e3f8fdc4e50","cardType":"basic","points":1,"front":"What equation gives force from a change in momentum?","back":"$F = \\frac{mv - mu}{t}$","mcqOptions":[{"text":"$F = \\frac{mv - mu}{t}$","correct":true,"explanation":"Force equals the rate of change of momentum. For constant mass, final momentum is $mv$ and initial momentum is $mu$, giving $F = \\frac{mv-mu}{t}$."},{"text":"$F = \\frac{mv + mu}{t}$","correct":false,"explanation":"Adding the initial and final momenta does not give the change in momentum. The change is final minus initial, so $F = \\frac{mv-mu}{t}$."},{"text":"$F = (mv - mu)t$","correct":false,"explanation":"Multiplying momentum change by time reverses the rate relationship and gives incorrect units. Force is momentum change divided by time: $F = \\frac{mv-mu}{t}$."},{"text":"$F = \\frac{m(v - u)}{t^2}$","correct":false,"explanation":"Dividing by time twice confuses force with a further rate of change. Force requires only momentum change divided by time, giving $F = \\frac{mv-mu}{t}$."}]},{"id":"8219cd18-4151-46e3-bdfa-500aa4737e6f","cardType":"basic","points":2,"front":"Why do Newton’s third-law forces not cancel each other out?","back":"They act on different objects.","mcqOptions":[{"text":"They act on different objects.","correct":true,"explanation":"Newton’s third-law force pairs act on two different interacting objects. They therefore cannot cancel when calculating the resultant force on either individual object."},{"text":"They act on the same object.","correct":false,"explanation":"Third-law forces do not act on the same object. They act on different interacting objects, so they are not combined when finding one object’s resultant force."},{"text":"They act at different times.","correct":false,"explanation":"Newton’s third-law forces occur simultaneously, not at different times. They do not cancel because each force acts on a different object in the interaction."},{"text":"They have different magnitudes.","correct":false,"explanation":"Third-law forces have equal magnitudes, so unequal size is not the reason. They do not cancel because the forces act on different objects."}]},{"id":"5d605b90-54e8-4d78-a98b-5acec6aeb9a5","cardType":"cloze","points":null,"front":"A moment is measured in N m and calculated using [ ], where $d$ is the [ ].","back":"A moment is measured in N m and calculated using $M = F \\times d$, where $d$ is the perpendicular distance from the pivot.","mcqOptions":[{"text":"A moment is measured in N m and calculated using $M = F \\times d$, where $d$ is the perpendicular distance from the pivot.","correct":true,"explanation":"A moment equals the force multiplied by the perpendicular distance from the pivot to the force’s line of action, giving the unit N m."},{"text":"A moment is measured in N m and calculated using $M = F \\div d$, where $d$ is the perpendicular distance from the pivot.","correct":false,"explanation":"Dividing force by distance does not calculate a turning effect. The correct relationship is $M = F \\times d$, using the perpendicular distance from the pivot."},{"text":"A moment is measured in N m and calculated using $M = F \\times d$, where $d$ is the distance travelled by the force.","correct":false,"explanation":"The distance travelled is irrelevant to the turning effect. In $M = F \\times d$, $d$ is the perpendicular distance from the pivot to the force’s line of action."},{"text":"A moment is measured in N m and calculated using $M = F + d$, where $d$ is the parallel distance from the pivot.","correct":false,"explanation":"Adding force and distance is dimensionally invalid, and a parallel distance does not determine turning effect. A moment is force multiplied by perpendicular distance from the pivot."}]},{"id":"fcc011ff-4d27-432d-ac93-627568967bb0","cardType":"basic","points":1,"front":"For an object in equilibrium, what does the principle of moments state?","back":"total clockwise moments = total anticlockwise moments","mcqOptions":[{"text":"total clockwise moments = total anticlockwise moments","correct":true,"explanation":"Rotational equilibrium requires zero resultant moment about any pivot. Therefore, the sum of all clockwise moments equals the sum of all anticlockwise moments."},{"text":"total clockwise forces = total anticlockwise forces","correct":false,"explanation":"This confuses force balance with moment balance. The principle of moments states that total clockwise moments equal total anticlockwise moments when an object is in equilibrium."},{"text":"total clockwise moments > total anticlockwise moments","correct":false,"explanation":"A greater clockwise moment would produce a resultant clockwise turning effect, so the object would not be in equilibrium. The clockwise and anticlockwise totals must be equal."},{"text":"total clockwise moments < total anticlockwise moments","correct":false,"explanation":"A greater anticlockwise moment would cause anticlockwise rotation rather than equilibrium. For equilibrium, total clockwise moments must equal total anticlockwise moments."}]},{"id":"b8fa85c7-8832-43a7-879c-0aff9a0cf346","cardType":"basic","points":1,"front":"For a light beam supported at both ends, which support gives the larger upward force?","back":"The support nearer the heavy object.","mcqOptions":[{"text":"The support nearer the heavy object.","correct":true,"explanation":"Taking moments shows that the support closer to the heavy object must provide the larger upward force to balance the object’s turning effect."},{"text":"The support farther from the heavy object.","correct":false,"explanation":"The farther support has a longer lever arm, so it needs less upward force to provide its balancing moment. The nearer support provides the larger force."},{"text":"Both supports give equal upward forces.","correct":false,"explanation":"Equal support forces occur only for a symmetrically placed load on a uniform beam. A heavy object placed off-centre makes the nearer support force larger."},{"text":"The support with the smaller contact area.","correct":false,"explanation":"Contact area affects pressure, not how the beam’s load is shared between supports. The support nearer the heavy object supplies the larger upward force."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"Forces, movement, shape and momentum","title":"Forces, movement, shape and momentum Flashcards","topbar":{"backHref":"/igcse/physics/edexcel/flashcards"},"topicSidebar":"$L36"}]}],"topicId":"f724fa95-ea34-5818-8ccb-a0159705c8a7","walled":"$L37"}],"$L38","$L39"]