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24:[["$","$L33",null,{"feature":"flashcards"}],"$L34",["$","$L35",null,{"href":"/a-level/physics/ocr/a/newton-s-laws-of-motion/flashcards","label":"Newton's laws of motion","subjectCode":"a_level_physics__ocr_a","surface":"flashcards"}],["$","$L36",null,{"runner":["$","div",null,{"className":"min-h-0 flex-1 overflow-hidden bg-background","children":["$","$L37",null,{"decks":[{"id":"3d422786-9fa5-40f7-aae6-a0ceb403f4e4","title":"Newton's laws of motion Flashcards","topicId":"cb209ff7-c7a8-5da6-bd54-72ef7da264df","topicSlug":"newton-s-laws-of-motion","cards":[{"id":"5cee1916-9a29-466b-aca7-e8272d708ced","cardType":"basic","points":1,"front":"What is the resultant force on an object moving in a straight line at a steady speed?","back":"Zero (forces are balanced).","mcqOptions":[{"text":"Zero (forces are balanced).","correct":true,"explanation":"Steady speed in a straight line means constant velocity, so acceleration is zero. Newton's second law therefore requires the resultant force to be zero, with all forces balanced."},{"text":"Constant (forces are unbalanced).","correct":false,"explanation":"A constant non-zero resultant force would produce acceleration, changing the object's velocity. Straight-line motion at steady speed has zero acceleration, so the resultant force is zero."},{"text":"Forwards (driving force is greater).","correct":false,"explanation":"A forward resultant force would increase the object's speed rather than maintain steady motion. Constant velocity requires balanced forces and hence a zero resultant force."},{"text":"Backwards (resistive force is greater).","correct":false,"explanation":"A backward resultant force would reduce the object's speed rather than keep it steady. Since the velocity is constant, the forces are balanced and the resultant is zero."}]},{"id":"5a8bec9e-3fd7-4cf4-9397-25ed73e8f647","cardType":"cloze","points":null,"front":"Newton's first law states that an object remains at rest or constant velocity unless acted on by a [ ].","back":"Newton's first law states that an object remains at rest or constant velocity unless acted on by a resultant external force.","mcqOptions":[{"text":"resultant external force","correct":true,"explanation":"Only a resultant external force changes an object's velocity. If no resultant external force acts, an object remains at rest or continues moving with constant velocity."},{"text":"single external force","correct":false,"explanation":"An individual external force may be balanced by another force and therefore cause no acceleration. Newton's first law specifically requires a resultant external force to change velocity."},{"text":"resultant internal force","correct":false,"explanation":"Internal forces occur between parts of the chosen system and cannot change the momentum of that system as a whole. A resultant external force is required."},{"text":"balanced external force","correct":false,"explanation":"A balanced external force is offset by other forces, giving zero resultant force and no change in velocity. The required condition is a resultant external force."}]},{"id":"e7b037b0-326e-4e91-b33c-42e421f91011","cardType":"basic","points":1,"front":"According to Newton's second law, what is the resultant force $F$ equal to in terms of momentum?","back":"The rate of change of momentum ($F = \\frac{\\Delta p}{\\Delta t}$).","mcqOptions":[{"text":"The rate of change of momentum ($F = \\frac{\\Delta p}{\\Delta t}$).","correct":true,"explanation":"Newton's second law defines resultant force as the rate at which momentum changes. Over a time interval, this is expressed as $F = \\frac{\\Delta p}{\\Delta t}$."},{"text":"The change of momentum per displacement ($F = \\frac{\\Delta p}{\\Delta x}$).","correct":false,"explanation":"Dividing the momentum change by displacement does not give force and has incompatible units. Resultant force is the rate of change of momentum, $F = \\frac{\\Delta p}{\\Delta t}$."},{"text":"The change of momentum times time ($F = \\Delta p\\Delta t$).","correct":false,"explanation":"Multiplying momentum change by elapsed time is a common rearrangement error. Force is found by dividing momentum change by time: $F = \\frac{\\Delta p}{\\Delta t}$."},{"text":"The momentum divided by elapsed time ($F = \\frac{p}{\\Delta t}$).","correct":false,"explanation":"Momentum divided by time is only valid in a special case where the initial momentum is zero. Generally, force equals the rate of change of momentum, using $\\Delta p$."}]},{"id":"83260d1d-6cc1-4057-b82a-1280d87d0484","cardType":"basic","points":1,"front":"Under what specific condition is $F = ma$ a valid version of Newton's second law?","back":"When the mass is constant.","mcqOptions":[{"text":"When the mass is constant.","correct":true,"explanation":"From $F = \\frac{d(mv)}{dt}$, constant mass can be taken outside the derivative, giving $F = m\\frac{dv}{dt} = ma$."},{"text":"When the velocity is constant.","correct":false,"explanation":"Constant velocity makes acceleration and resultant force zero, but it is not the condition used to derive the general form $F = ma$. The required condition is constant mass."},{"text":"When the acceleration is constant.","correct":false,"explanation":"Constant acceleration is not required for $F = ma$; acceleration may vary with time. The equation follows from Newton's momentum law specifically when the mass is constant."},{"text":"When the momentum is constant.","correct":false,"explanation":"Constant momentum implies zero resultant force, but does not justify replacing the momentum law with $F = ma$ for a variable-mass system. The necessary condition is constant mass."}]},{"id":"a33ac1e8-ec36-45fe-878d-2e5f05485a47","cardType":"basic","points":3,"front":"State three characteristics of a Newton's third law force pair.","back":"

Equal magnitude
Opposite direction
Same type

","mcqOptions":[{"text":"

Equal magnitude
Opposite direction
Same type

","correct":true,"explanation":"Newton's third law pairs are equal in magnitude and opposite in direction, and both forces arise from the same interaction, so they are the same type."},{"text":"

Equal magnitude
Same direction
Same type

","correct":false,"explanation":"The forces do not point in the same direction; Newton's third law requires them to act in opposite directions. They have equal magnitudes and are the same type of force."},{"text":"

Different magnitude
Opposite direction
Same type

","correct":false,"explanation":"A third law pair cannot have different magnitudes, even when the objects accelerate differently. The forces have equal magnitudes, opposite directions and the same interaction type."},{"text":"

Equal magnitude
Opposite direction
Different type

","correct":false,"explanation":"The paired forces come from one interaction and must therefore be the same type, such as two gravitational forces. They have equal magnitudes and opposite directions."}]},{"id":"dc93ed65-5978-4cf8-9ce5-837398e9ac41","cardType":"basic","points":2,"front":"Why do Newton's third law force pairs never cancel each other out when calculating a resultant force?","back":"Because the two forces act on different objects.","mcqOptions":[{"text":"Because the two forces act on different objects.","correct":true,"explanation":"A resultant force is calculated by combining forces acting on one chosen object. Third law partners act on different objects, so they cannot be added together or cancel."},{"text":"Because the two forces have different magnitudes.","correct":false,"explanation":"Third law forces actually have equal magnitudes, but equality alone does not explain why they cannot cancel. They do not cancel because they act on different objects."},{"text":"Because the two forces act at different times.","correct":false,"explanation":"Third law forces act simultaneously, not at different times. They cannot cancel in one object's resultant because the two forces act on different objects."},{"text":"Because the two forces are of different types.","correct":false,"explanation":"Third law forces arise from the same interaction and are the same type. Their failure to cancel in a resultant occurs because they act on different objects."}]},{"id":"7e02ffa1-ef87-48ef-9437-a30bb0ef235e","cardType":"cloze","points":null,"front":"Linear momentum is defined by the equation $\\text{[...]}$ and is measured in units of kg m s^-1.","back":"Linear momentum is defined by the equation $\\boxed{p = mv}$ and is measured in units of kg m s^-1.","mcqOptions":[{"text":"$$p = mv$","correct":true,"explanation":"Linear momentum is the product of an object's mass and velocity, so it is defined by $p = mv$. Momentum is a vector in the velocity's direction."},{"text":"$$p = \\frac{m}{v}$","correct":false,"explanation":"Dividing mass by velocity does not represent momentum and gives the wrong dimensions. Linear momentum is found by multiplying mass by velocity: $p = mv$."},{"text":"$$p = mv^2$","correct":false,"explanation":"The expression $mv^2$ is associated with twice the kinetic energy, not momentum. Linear momentum contains one power of velocity and is defined by $p = mv$."},{"text":"$$p = \\frac{1}{2}mv$","correct":false,"explanation":"The expression $\\frac{1}{2}mv$ incorrectly imports the factor of one-half from the kinetic-energy formula. Linear momentum is the full product $p = mv$."}]},{"id":"0b7fc6ab-3a50-4f8c-ba8a-db444821b21a","cardType":"cloze","points":null,"front":"Linear momentum is defined by the equation $p = mv$ and is measured in units of [ ].","back":"Linear momentum is defined by the equation $p = mv$ and is measured in units of kg m s^-1.","mcqOptions":[{"text":"kg m s^-1","correct":true,"explanation":"From $p = mv$, multiplying mass in kilograms by velocity in metres per second gives the momentum unit kg m s^-1."},{"text":"kg m s^-2","correct":false,"explanation":"The unit kg m s^-2 is the newton, which measures force because it includes acceleration. Momentum is mass multiplied by velocity, giving kg m s^-1."},{"text":"kg m^2 s^-2","correct":false,"explanation":"The unit kg m^2 s^-2 measures energy, not momentum, and results from a squared velocity term. Momentum has units kg m s^-1."},{"text":"kg m^-1 s^-1","correct":false,"explanation":"The unit kg m^-1 s^-1 incorrectly places distance in the denominator. Multiplying mass by velocity gives metres in the numerator, so momentum uses kg m s^-1."}]},{"id":"d989400a-c9e5-412f-896d-90afe8a0ca14","cardType":"basic","points":2,"front":"If two objects have the same speed but opposite velocities, how do their momenta compare?","back":"They have the same magnitude but opposite directions (momenta are different).","mcqOptions":[{"text":"They have the same magnitude but opposite directions (momenta are different).","correct":true,"explanation":"Momentum is a vector, $p = mv$. Equal masses at equal speeds give equal momentum magnitudes, while opposite velocities make the momentum directions opposite."},{"text":"They have the same magnitude and the same direction (momenta are identical).","correct":false,"explanation":"Treating momentum as a scalar ignores the direction of velocity. Their momenta have the same magnitude but opposite directions, so the momentum vectors are different."},{"text":"They have different magnitudes but the same direction (momenta are different).","correct":false,"explanation":"Reversing velocity changes the direction of momentum, not its magnitude. The momenta therefore have the same magnitude but point in opposite directions."},{"text":"They both have zero magnitude and no direction (momenta are identical).","correct":false,"explanation":"Opposite vector directions do not make each individual momentum zero. Each object has non-zero momentum of equal magnitude, but the two momenta point in opposite directions."}]},{"id":"c811c427-f78a-462b-8dd0-51cdf465d263","cardType":"cloze","points":null,"front":"Impulse is calculated as the product of [ ] ($F\\Delta t$).","back":"Impulse is calculated as the product of force and time ($F\\Delta t$).","mcqOptions":[{"text":"force and time","correct":true,"explanation":"Impulse equals force multiplied by the time interval for which it acts: $J = F\\Delta t$. For varying force, this corresponds to the force-time graph’s area."},{"text":"mass and velocity","correct":false,"explanation":"Mass multiplied by velocity gives momentum, not impulse directly. Impulse is the product of force and time and equals the change in momentum."},{"text":"force and distance","correct":false,"explanation":"Force multiplied by distance gives work done when force acts along the displacement. Impulse instead equals force multiplied by the duration of its action."},{"text":"power and time","correct":false,"explanation":"Power multiplied by time gives energy transferred, not impulse. Impulse is calculated from force multiplied by the time interval, $F\\Delta t$."}]},{"id":"d9821068-5177-4174-b9a6-65a234bf8b4e","cardType":"basic","points":1,"front":"What is the physical relationship between impulse and momentum?","back":"Impulse is equal to the change in momentum ($F\\Delta t = \\Delta p$).","mcqOptions":[{"text":"Impulse is equal to the change in momentum ($F\\Delta t = \\Delta p$).","correct":true,"explanation":"Newton’s second law gives $F = \\frac{\\Delta p}{\\Delta t}$. Multiplying by the interaction time shows that impulse $F\\Delta t$ equals the change in momentum $\\Delta p$."},{"text":"Impulse is equal to the final momentum ($F\\Delta t = p$).","correct":false,"explanation":"Impulse is not generally equal to the final momentum because an object may already have momentum. It equals final momentum minus initial momentum, which is the change in momentum."},{"text":"Impulse is equal to the rate of change of momentum ($F\\Delta t = \\Delta p/\\Delta t$).","correct":false,"explanation":"Force is the rate of change of momentum, whereas impulse is force multiplied by time. Consequently, impulse equals the change in momentum rather than its rate of change."},{"text":"Impulse is equal to the change in kinetic energy ($F\\Delta t = \\Delta E_k$).","correct":false,"explanation":"Momentum and kinetic energy are distinct quantities with different units. Impulse changes an object’s momentum according to $F\\Delta t = \\Delta p$, not its kinetic energy directly."}]},{"id":"269fcc7f-5b9e-4da8-b427-dc08ed869c9a","cardType":"basic","points":1,"front":"How is impulse represented on a force-time graph?","back":"The area under the graph.","mcqOptions":[{"text":"The area under the graph.","correct":true,"explanation":"Each small interval contributes impulse $F\\Delta t$, so summing across the interaction gives the integral $\\int F\\,dt$. Geometrically, this is the area under the force-time graph."},{"text":"The gradient of the graph.","correct":false,"explanation":"The gradient of a force-time graph represents the rate of change of force, not impulse. Impulse is obtained from the area under the graph."},{"text":"The maximum height of the graph.","correct":false,"explanation":"Maximum force gives only the peak force and does not account for how long the force acts. Impulse is represented by the complete area under the graph."},{"text":"The intercept on the time axis.","correct":false,"explanation":"The time-axis intercept indicates when the force is zero, not the accumulated impulse. Impulse is represented by the area under the force-time graph."}]},{"id":"c50e951a-bc8c-4fa8-aaec-7d1065323006","cardType":"basic","points":1,"front":"What does an area below the time axis on a force-time graph represent?","back":"A negative impulse (force acting in the opposite direction).","mcqOptions":[{"text":"A negative impulse (force acting in the opposite direction).","correct":true,"explanation":"Area on a force-time graph equals impulse. Below the time axis the force is negative relative to the chosen positive direction, so the resulting impulse is negative."},{"text":"A positive impulse (force acting in the chosen direction).","correct":false,"explanation":"A positive impulse comes from an area above the time axis under the usual sign convention. Area below it represents negative impulse because the force acts oppositely."},{"text":"A zero impulse (no overall change in momentum).","correct":false,"explanation":"Zero impulse requires positive and negative areas to cancel, or no net area at all. An uncancelled area below the axis represents a negative impulse."},{"text":"A negative kinetic energy (energy acting in the opposite direction).","correct":false,"explanation":"Area under a force-time graph represents impulse, not kinetic energy directly. An area below the axis specifically gives negative impulse from force in the opposite direction."}]},{"id":"777236ac-253b-4ae0-abe0-1a4800b004cc","cardType":"basic","points":2,"front":"Using the concept of impulse, explain how crumple zones reduce the risk of injury.","back":"They increase the time interval ($\\Delta t$) for the change in momentum, which decreases the average force ($F$).","mcqOptions":[{"text":"They increase the time interval ($\\Delta t$) for the change in momentum, which decreases the average force ($F$).","correct":true,"explanation":"For a fixed momentum change, $F_{\\text{avg}}\\Delta t = \\Delta p$. Crumple zones lengthen the collision time, so the same impulse is delivered with a smaller average force."},{"text":"They decrease the time interval ($\\Delta t$) for the change in momentum, which increases the average force ($F$).","correct":false,"explanation":"Crumple zones lengthen rather than shorten the collision. Since the momentum change is fixed, increasing the time interval decreases the average force experienced by the occupants."},{"text":"They increase the time interval ($\\Delta t$) for the change in momentum, which increases the average force ($F$).","correct":false,"explanation":"For a given change in momentum, force and collision time are inversely related, not directly related. Increasing the time interval therefore decreases the average force."},{"text":"They decrease the time interval ($\\Delta t$) for the change in momentum, which decreases the average force ($F$).","correct":false,"explanation":"Crumple zones do not protect occupants by making the collision more abrupt. They increase the stopping time, which reduces the average force for the same momentum change."}]},{"id":"3e181171-d75c-4223-9d85-1d43126bc84e","cardType":"basic","points":1,"front":"A 2 kg mass accelerates at 5 m s^-2. What is the resultant force?","back":"10 N ($F = 2 \\times 5$).","mcqOptions":[{"text":"10 N ($F = 2 \\times 5$).","correct":true,"explanation":"Newton’s second law gives $F = ma$. Multiplying the mass by the acceleration gives $F = 2 \\times 5 = 10\\,\\text{N}$."},{"text":"7 N ($F = 2 + 5$).","correct":false,"explanation":"Adding mass and acceleration is dimensionally invalid and produces the numerical slip $2 + 5 = 7$. Newton’s second law requires multiplication, giving $10\\,\\text{N}$."},{"text":"2.5 N ($F = 5 \\div 2$).","correct":false,"explanation":"Dividing acceleration by mass reverses the required operation in $F = ma$. The correct calculation multiplies $2\\,\\text{kg}$ by $5\\,\\text{m s}^{-2}$ to give $10\\,\\text{N}$."},{"text":"0.4 N ($F = 2 \\div 5$).","correct":false,"explanation":"Dividing mass by acceleration does not follow Newton’s second law and gives incompatible units. Multiplying mass and acceleration instead gives a resultant force of $10\\,\\text{N}$."}]},{"id":"1fed0e3f-a5db-4fb2-b74e-f6c932521750","cardType":"cloze","points":null,"front":"Impulse is measured in units of [ ], which is equivalent to kg m s^-1.","back":"Impulse is measured in units of N s, which is equivalent to kg m s^-1.","mcqOptions":[{"text":"N s","correct":true,"explanation":"Impulse is force multiplied by time, so its unit is $\\text{N s}$. Since $1\\,\\text{N} = 1\\,\\text{kg m s}^{-2}$, this equals $\\text{kg m s}^{-1}$."},{"text":"N","correct":false,"explanation":"A newton measures force alone and is equivalent to $\\text{kg m s}^{-2}$. Impulse includes multiplication by time, so its unit is $\\text{N s}$."},{"text":"N s$^{-1}$","correct":false,"explanation":"Newtons per second measure the rate at which force changes, not impulse. Impulse multiplies force by time, giving $\\text{N s}$ or $\\text{kg m s}^{-1}$."},{"text":"kg m s$^{-2}$","correct":false,"explanation":"The unit $\\text{kg m s}^{-2}$ is equivalent to a newton and therefore measures force. Multiplying force by time gives the impulse unit $\\text{N s}$."}]},{"id":"714cbe53-55de-4e7b-89ba-f4a66e6bf973","cardType":"cloze","points":null,"front":"Impulse is measured in units of N s, which is equivalent to [ ].","back":"Impulse is measured in units of N s, which is equivalent to kg m s^-1.","mcqOptions":[{"text":"kg m s^-1","correct":true,"explanation":"The impulse-momentum theorem equates impulse with change in momentum. Momentum is mass multiplied by velocity, so its SI units are kg m s^-1, equivalent to N s."},{"text":"kg m s^-2","correct":false,"explanation":"kg m s^-2 is the unit of force, obtained before multiplying force by time. Multiplication by seconds gives the correct impulse unit, kg m s^-1."},{"text":"kg m^2 s^-2","correct":false,"explanation":"kg m^2 s^-2 is the unit of energy or work, not impulse. Impulse has the same units as momentum, so the correct unit is kg m s^-1."},{"text":"kg s^-1","correct":false,"explanation":"kg s^-1 represents a rate of mass transfer and omits the required length dimension. Impulse is equivalent to change in momentum, measured in kg m s^-1."}]},{"id":"21054070-40fc-4c2f-a83b-ea7dcfa38ee7","cardType":"basic","points":2,"front":"How can the impulse of a varying force be estimated from a set of discrete data points?","back":"By calculating the sum of the areas of trapezia between data points.","mcqOptions":[{"text":"By calculating the sum of the areas of trapezia between data points.","correct":true,"explanation":"Impulse equals the area under a force-time graph. Joining adjacent data points and summing trapezium areas provides a suitable numerical estimate of that area."},{"text":"By calculating the sum of the gradients of chords between data points.","correct":false,"explanation":"Chord gradients estimate how quickly force changes with time, not the area under the graph. Impulse is estimated by summing the areas of trapezia between adjacent points."},{"text":"By calculating the mean of the force values between data points.","correct":false,"explanation":"A mean force alone does not include the time intervals and therefore does not give impulse. The correct method sums the areas of trapezia between data points."},{"text":"By calculating the sum of the force changes between data points.","correct":false,"explanation":"Adding force changes gives the difference between the final and initial forces, not force multiplied by time. Impulse is found by summing trapezium areas under the graph."}]},{"id":"42f697a2-8f19-4676-80bf-fd76743ef9fb","cardType":"cloze","points":null,"front":"Because velocity is a [ ], linear momentum also has both magnitude and direction.","back":"Because velocity is a vector, linear momentum also has both magnitude and direction.","mcqOptions":[{"text":"vector","correct":true,"explanation":"Velocity has both magnitude and direction, making it a vector. Multiplying velocity by the scalar mass therefore produces momentum that is also a vector."},{"text":"scalar","correct":false,"explanation":"A scalar has magnitude but no direction, whereas velocity includes direction as well as magnitude. Velocity is therefore a vector, which makes momentum a vector too."},{"text":"tensor","correct":false,"explanation":"Tensors can require multiple directional components beyond those of an ordinary vector. At this level velocity is classified as a vector, so momentum also has magnitude and direction."},{"text":"derived unit","correct":false,"explanation":"A derived unit is a measurement unit formed from base units, not a type of physical quantity. Velocity is a vector because it has magnitude and direction."}]},{"id":"98838fd2-827c-48cf-a503-8f6d0e06ce57","cardType":"cloze","points":null,"front":"Because velocity is a vector, linear momentum also has both [ ].","back":"Because velocity is a vector, linear momentum also has both magnitude and direction.","mcqOptions":[{"text":"magnitude and direction","correct":true,"explanation":"Momentum is mass multiplied by velocity. Since mass is scalar and velocity is vector, momentum retains the velocity’s magnitude-dependent size and its direction."},{"text":"speed and direction","correct":false,"explanation":"Speed describes how fast an object moves, not a property possessed by momentum itself. As a vector, momentum has magnitude and direction."},{"text":"mass and direction","correct":false,"explanation":"Mass helps determine momentum’s magnitude but is not one of the defining components of a vector. Linear momentum has magnitude and direction."},{"text":"magnitude and speed","correct":false,"explanation":"Magnitude and speed are both scalar measures, so this pairing omits the directional feature of momentum. A vector such as momentum has magnitude and direction."}]},{"id":"b822bd33-be9c-492a-ab3e-2511d7d0e699","cardType":"basic","points":1,"front":"In which direction does the resultant force act relative to the momentum change?","back":"In the same direction as the change in momentum.","mcqOptions":[{"text":"In the same direction as the change in momentum.","correct":true,"explanation":"Newton’s second law states that resultant force equals the rate of change of momentum. A positive time interval means the force points in the same direction as the momentum change."},{"text":"In the opposite direction to the change in momentum.","correct":false,"explanation":"This reverses the vector relationship in Newton’s second law. Since force is the rate of change of momentum, the resultant force acts in the same direction as that change."},{"text":"In a perpendicular direction to the change in momentum.","correct":false,"explanation":"Perpendicular force and momentum can occur in circular motion, but force is still parallel to the momentum change. The resultant force acts in the same direction as that change."},{"text":"In the same direction as the initial momentum.","correct":false,"explanation":"The initial momentum does not determine the force direction; a force can slow or turn an object. The resultant force follows the direction of the change in momentum."}]},{"id":"f93fefdb-0d61-43f3-9af8-55699ec802f9","cardType":"basic","points":1,"front":"An object's momentum changes from -10 kg m s^-1 to +15 kg m s^-1. Calculate the change in momentum.","back":"+25 kg m s^-1 ($15 - (-10)$).","mcqOptions":[{"text":"+25 kg m s^-1 ($15 - (-10)$).","correct":true,"explanation":"Change in momentum is final momentum minus initial momentum. Subtracting the negative initial value gives $15 - (-10) = +25$ kg m s^-1."},{"text":"+5 kg m s^-1 ($15 + (-10)$).","correct":false,"explanation":"This adds the two signed momenta instead of calculating final minus initial. The correct change is $15 - (-10) = +25$ kg m s^-1."},{"text":"-25 kg m s^-1 ($-10 - 15$).","correct":false,"explanation":"This reverses the subtraction by calculating initial minus final momentum. Change is defined as final minus initial, giving $15 - (-10) = +25$ kg m s^-1."},{"text":"+15 kg m s^-1 ($15 - 0$).","correct":false,"explanation":"This treats the object as if it started from zero momentum and ignores its negative initial value. Using final minus initial gives $15 - (-10) = +25$ kg m s^-1."}]},{"id":"e6e869ff-c567-4a20-b6a6-779ae33e09e0","cardType":"cloze","points":null,"front":"A [ ] shows one object by itself with all external forces acting on it.","back":"A free-body diagram shows one object by itself with all external forces acting on it.","mcqOptions":[{"text":"free-body diagram","correct":true,"explanation":"A free-body diagram isolates one chosen object and represents every external force acting on it with labelled arrows, allowing the resultant force to be determined."},{"text":"force-pair diagram","correct":false,"explanation":"A force-pair diagram would focus on Newton’s third-law forces acting on different objects. A free-body diagram instead shows all external forces acting on one isolated object."},{"text":"motion diagram","correct":false,"explanation":"A motion diagram represents an object’s positions or velocities at successive times rather than the forces on it. External forces on one object are shown using a free-body diagram."},{"text":"vector triangle","correct":false,"explanation":"A vector triangle combines vectors geometrically but does not necessarily identify every external force on one object. That specific representation is called a free-body diagram."}]},{"id":"91493343-e16f-43fb-a034-2b834684bd70","cardType":"cloze","points":null,"front":"A free-body diagram shows one object by itself with all [ ] acting on it.","back":"A free-body diagram shows one object by itself with all external forces acting on it.","mcqOptions":[{"text":"external forces","correct":true,"explanation":"A free-body diagram isolates the selected object, so it includes forces exerted on that object by its surroundings. These are the external forces acting on it."},{"text":"internal forces","correct":false,"explanation":"Internal forces act between parts within the chosen system and are not shown when the whole system is isolated. A free-body diagram displays all external forces."},{"text":"resultant forces","correct":false,"explanation":"The resultant is the single vector sum calculated from the individual forces, not the full set initially drawn. A free-body diagram shows all external forces acting on the object."},{"text":"contact forces","correct":false,"explanation":"Contact forces are included, but non-contact forces such as weight must also be shown. A complete free-body diagram contains all external forces acting on the object."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Paper 1"},"heading":"Newton's laws of motion","title":"Newton's laws of motion Flashcards","topbar":{"backHref":"/a-level/physics/ocr/a/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"cb209ff7-c7a8-5da6-bd54-72ef7da264df","walled":"$L39"}],"$L3a","$L3b"]