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resultant force and decreases as mass increases, so acceleration is directly proportional to force and inversely proportional to mass."},{"text":"Acceleration is inversely proportional to resultant force and proportional to mass.","correct":false,"explanation":"This reverses both relationships. Acceleration increases with resultant force and decreases with mass, as given by $a = \\dfrac{F}{m}$."},{"text":"Acceleration is proportional to resultant force and proportional to mass.","correct":false,"explanation":"The force relationship is correct, but the mass relationship is reversed. For a fixed force, increasing mass reduces acceleration because $a = \\dfrac{F}{m}$."},{"text":"Acceleration is inversely proportional to resultant force and inversely proportional to mass.","correct":false,"explanation":"Acceleration is not inversely proportional to force. Increasing resultant force increases acceleration, while increasing mass decreases it, according to $a = \\dfrac{F}{m}$."}]},{"id":"dff87b49-2f7b-4c34-ba52-eec36ebb6ffb","cardType":"basic","points":2,"front":"What is the resultant force on an object?","back":"The overall force after all forces are combined; acceleration is in its direction.","mcqOptions":[{"text":"The overall force after all forces are combined; acceleration is in its direction.","correct":true,"explanation":"The resultant force is the vector total of all forces acting on an object. Newton’s second law states that any acceleration is in the direction of this resultant force."},{"text":"The largest force acting on an object; acceleration is in its direction.","correct":false,"explanation":"The largest individual force does not necessarily determine the motion. All forces must be combined, including opposing forces, to find the resultant and its direction."},{"text":"The average force acting on an object; acceleration is opposite to its direction.","correct":false,"explanation":"Resultant force means the combined vector total, not an average, and acceleration is in the same direction as the resultant force rather than opposite to it."},{"text":"The difference in mass after all forces are combined; acceleration is in its direction.","correct":false,"explanation":"Combining forces produces a resultant force, not a change or difference in mass. The resultant’s direction determines the direction of acceleration."}]},{"id":"8e819d9f-eba2-46d1-8317-b917f91410e4","cardType":"basic","points":1,"front":"What equation links resultant force, mass and acceleration?","back":"$$F = ma$","mcqOptions":[{"text":"$$F = ma$","correct":true,"explanation":"Newton’s second law states that resultant force equals mass multiplied by acceleration, represented by $F = ma$."},{"text":"$$F = \\dfrac{m}{a}$","correct":false,"explanation":"This swaps the roles of mass and acceleration. Rearranging Newton’s second law gives $F = ma$, while $\\dfrac{m}{a}$ does not represent force."},{"text":"$$F = \\dfrac{a}{m}$","correct":false,"explanation":"Dividing acceleration by mass is not the force relationship. Acceleration is found using $a = \\dfrac{F}{m}$, which rearranges to $F = ma$."},{"text":"$$F = m + a$","correct":false,"explanation":"Force is not calculated by adding mass and acceleration. Newton’s second law requires multiplication: $F = ma$."}]},{"id":"9652f77b-6827-4c4a-87cb-0c2f391a4ef0","cardType":"basic","points":1,"front":"What is the unit of resultant force in $F = ma$?","back":"Newtons ($\\text{N}$)","mcqOptions":[{"text":"Newtons ($\\text{N}$)","correct":true,"explanation":"The SI unit of resultant force is the newton, symbol $\\text{N}$. One newton is equivalent to one kilogram metre per second squared."},{"text":"Joules ($\\text{J}$)","correct":false,"explanation":"Joules measure energy or work, not force. Resultant force in $F = ma$ is measured in newtons, written as $\\text{N}$."},{"text":"Watts ($\\text{W}$)","correct":false,"explanation":"Watts measure power, the rate of energy transfer. The force calculated using $F = ma$ is measured in newtons, $\\text{N}$."},{"text":"Pascal ($\\text{Pa}$)","correct":false,"explanation":"Pascals measure pressure, which is force per unit area. The resultant force itself is measured in newtons, symbol $\\text{N}$."}]},{"id":"5f2ef8b7-05c1-41ac-b8fb-8c3da89c8771","cardType":"basic","points":1,"front":"What is the unit of mass in $F = ma$?","back":"Kilograms ($\\text{kg}$)","mcqOptions":[{"text":"Kilograms ($\\text{kg}$)","correct":true,"explanation":"Mass must be measured in kilograms when using $F = ma$ in SI units. The kilogram has the symbol $\\text{kg}$."},{"text":"Grams ($\\text{g}$)","correct":false,"explanation":"Grams measure mass, but they are not the SI unit required directly in $F = ma$. Values in grams must be converted to kilograms first."},{"text":"Newtons ($\\text{N}$)","correct":false,"explanation":"Newtons measure force, not mass. Mass in the equation $F = ma$ is measured in kilograms, symbol $\\text{kg}$."},{"text":"Kilometres ($\\text{km}$)","correct":false,"explanation":"Kilometres measure distance, not mass. The mass term in $F = ma$ must be expressed in kilograms, $\\text{kg}$."}]},{"id":"9586b5c1-e56c-4149-a22e-0124549a9236","cardType":"basic","points":1,"front":"What happens to acceleration if resultant force doubles at constant mass?","back":"The acceleration doubles.","mcqOptions":[{"text":"The acceleration doubles.","correct":true,"explanation":"At constant mass, acceleration is directly proportional to resultant force. Therefore, doubling the force doubles the acceleration."},{"text":"The acceleration halves.","correct":false,"explanation":"Halving acceleration would result from doubling mass at constant force, not doubling force at constant mass. Here, direct proportionality makes acceleration double."},{"text":"The acceleration stays the same.","correct":false,"explanation":"Acceleration only stays constant if the ratio $\\dfrac{F}{m}$ stays constant. With constant mass and doubled force, acceleration doubles."},{"text":"The acceleration becomes zero.","correct":false,"explanation":"A larger non-zero resultant force cannot make acceleration zero when mass remains constant. From $a = \\dfrac{F}{m}$, doubling force increases acceleration."}]},{"id":"01a4fc76-5e6d-4cb1-a6da-55b8cd63c03e","cardType":"basic","points":1,"front":"What happens to acceleration if mass doubles at constant resultant force?","back":"The acceleration halves.","mcqOptions":[{"text":"The acceleration halves.","correct":true,"explanation":"At constant resultant force, acceleration is inversely proportional to mass. Doubling the mass therefore halves the acceleration."},{"text":"The acceleration doubles.","correct":false,"explanation":"This confuses the effect of force with the effect of mass. At constant force, doubling mass makes $a = \\dfrac{F}{m}$ half as large."},{"text":"The acceleration stays the same.","correct":false,"explanation":"Acceleration changes when mass changes unless the force changes in the same proportion. With constant force and doubled mass, acceleration halves."},{"text":"The acceleration becomes zero.","correct":false,"explanation":"Doubling mass does not remove the resultant force. It reduces acceleration to half its original value, provided the force remains non-zero."}]},{"id":"37a23261-d022-4e9c-ac04-3567172b72c4","cardType":"basic","points":2,"front":"Why does an object moving at constant velocity have zero resultant force?","back":"Its velocity is not changing, so its acceleration is zero.","mcqOptions":[{"text":"Its velocity is not changing, so its acceleration is zero.","correct":true,"explanation":"Acceleration is the rate of change of velocity. Constant velocity means no change in velocity, so acceleration is zero and Newton’s second law gives zero resultant force."},{"text":"Its speed is increasing, so its acceleration is zero.","correct":false,"explanation":"Increasing speed means velocity is changing, so acceleration is not zero. Constant velocity requires both constant speed and constant direction."},{"text":"Its mass is not changing, so its acceleration is zero.","correct":false,"explanation":"Constant mass does not imply zero acceleration. An object can accelerate while its mass remains constant if a resultant force acts."},{"text":"Its velocity is changing, so its acceleration is zero.","correct":false,"explanation":"A changing velocity is precisely what acceleration measures. Zero acceleration requires velocity to remain constant, producing zero resultant force."}]},{"id":"0fb3846f-bc69-4ad4-b235-65030e1f1444","cardType":"basic","points":2,"front":"A $1200\\ \\text{kg}$ car accelerates at $2.5\\ \\text{m/s}^2$. What resultant force acts on it?","back":"$$F = ma = 1200 \\times 2.5 = \\mathbf{3000\\ \\text{N}}$","mcqOptions":[{"text":"$$F = ma = 1200 \\times 2.5 = \\mathbf{3000\\ \\text{N}}$","correct":true,"explanation":"Using Newton’s second law, multiply the mass by the acceleration: $F = 1200 \\times 2.5 = 3000\\ \\text{N}$."},{"text":"$$F = ma = 1200 \\times 2.5 = \\mathbf{480\\ \\text{N}}$","correct":false,"explanation":"This commonly results from dividing the mass by the acceleration instead of multiplying. The correct calculation is $1200 \\times 2.5 = 3000\\ \\text{N}$."},{"text":"$$F = ma = 1200 \\times 2.5 = \\mathbf{1202.5\\ \\text{N}}$","correct":false,"explanation":"Adding the mass and acceleration gives this value, but Newton’s second law requires multiplication. Therefore, $F = 1200 \\times 2.5 = 3000\\ \\text{N}$."},{"text":"$$F = ma = 1200 \\times 2.5 = \\mathbf{7500\\ \\text{N}}$","correct":false,"explanation":"This value can result from incorrectly multiplying by $2.5$ twice. The equation requires one multiplication: $1200 \\times 2.5 = 3000\\ \\text{N}$."}]},{"id":"7e3755c8-ccf1-4e5e-ad09-c89165fea5e7","cardType":"basic","points":2,"front":"How can the acceleration of a rapidly accelerating vehicle be estimated?","back":"Acceleration: $a = \\dfrac{\\text{change in velocity}}{\\text{time}}$
Then force: $F = ma$","mcqOptions":[{"text":"Acceleration: $a = \\dfrac{\\text{change in velocity}}{\\text{time}}$
Then force: $F = ma$","correct":true,"explanation":"Acceleration can be estimated from the change in velocity divided by the time taken. Multiplying this acceleration by mass then gives the resultant force using $F = ma$."},{"text":"Acceleration: $a = \\dfrac{\\text{time}}{\\text{change in velocity}}$
Then force: $F = ma$","correct":false,"explanation":"The fraction is inverted. Acceleration is the change in velocity divided by time, then the resulting acceleration is multiplied by mass to calculate force."},{"text":"Acceleration: $a = \\dfrac{\\text{change in velocity}}{\\text{mass}}$
Then force: $F = ma$","correct":false,"explanation":"Mass is not used to calculate acceleration from motion data. Acceleration equals change in velocity divided by time, followed by $F = ma$."},{"text":"Acceleration: $a = \\text{change in velocity} \\times \\text{time}$
Then force: $F = \\dfrac{m}{a}$","correct":false,"explanation":"Both relationships are incorrect: acceleration divides velocity change by time, and force equals mass multiplied by acceleration. The correct sequence is $a = \\dfrac{\\text{change in velocity}}{\\text{time}}$, then $F = ma$."}]},{"id":"5043b7c8-a2fa-4e4b-9cf8-50fdea7eaa52","cardType":"basic","points":2,"front":"In the trolley investigation, why must total mass stay constant when changing the pulling force?","back":"To ensure only the force changes, allowing its effect on acceleration to be tested fairly.","mcqOptions":[{"text":"To ensure only the force changes, allowing its effect on acceleration to be tested fairly.","correct":true,"explanation":"Keeping total mass constant controls a key variable. Any change in acceleration can then be attributed to the changed pulling force, making the investigation a fair test."},{"text":"To ensure only the mass changes, allowing its effect on force to be tested fairly.","correct":false,"explanation":"The investigation is changing pulling force, not mass. Keeping total mass constant ensures acceleration changes can be linked fairly to the force."},{"text":"To ensure the acceleration changes, allowing its effect on force to be tested fairly.","correct":false,"explanation":"Acceleration is the measured response, not the control variable. Total mass stays constant so the effect of changing force on acceleration can be isolated."},{"text":"To ensure only the time changes, allowing its effect on mass to be tested fairly.","correct":false,"explanation":"Time is not the variable being isolated in this control. The pulling force is changed while total mass remains constant, allowing its effect on acceleration to be measured."}]},{"id":"4607f368-b5b1-4a7b-aad6-686b1a189cf6","cardType":"basic","points":1,"front":"What graph is produced when acceleration is plotted against force at constant mass?","back":"A straight line through the origin.","mcqOptions":[{"text":"A straight line through the origin.","correct":true,"explanation":"At constant mass, acceleration is directly proportional to resultant force. Direct proportionality produces a straight-line graph passing through the origin."},{"text":"A horizontal line above the origin.","correct":false,"explanation":"A horizontal line would show acceleration staying constant as force changes. Since acceleration increases proportionally with force, the graph should slope upwards through the origin."},{"text":"A straight line crossing the force axis away from the origin.","correct":false,"explanation":"An offset intercept suggests a constant force or systematic error. For direct proportionality between force and acceleration, the ideal line passes through the origin."},{"text":"A curve through the origin.","correct":false,"explanation":"A curve would indicate a non-linear relationship. With constant mass, Newton’s second law predicts direct proportionality and therefore a straight line through the origin."}]},{"id":"5fbbceac-5c6c-42b1-a964-fc68b4b34be6","cardType":"basic","points":1,"front":"Why are repeat readings and mean accelerations used in the trolley investigation?","back":"They help reduce random error.","mcqOptions":[{"text":"They help reduce random error.","correct":true,"explanation":"Repeating readings reveals random variation, and calculating a mean reduces the effect of unusually high or low readings on the final result."},{"text":"They help remove systematic error.","correct":false,"explanation":"Repeats and means reduce random error but do not remove a consistent systematic error, such as a miscalibrated timer or a constant friction effect."},{"text":"They help increase random error.","correct":false,"explanation":"Repeated measurements generally make random variation easier to identify and reduce its influence through averaging; they do not increase random error."},{"text":"They help change the controlled variables.","correct":false,"explanation":"Repeating readings does not change controlled variables. It improves reliability by reducing the influence of random error on the mean acceleration."}]},{"id":"95b44098-bfe4-42b2-b6a9-569945eb6d93","cardType":"basic","points":3,"front":"What does inertial mass measure, and what happens with the same resultant force?","back":"

Measures how difficult it is to change velocity.
A larger inertial mass produces a smaller acceleration.

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

Measures how difficult it is to change velocity.
A larger inertial mass produces a smaller acceleration.

","correct":true,"explanation":"Inertial mass measures an object’s resistance to changes in velocity. For the same resultant force, $a = \\dfrac{F}{m}$ shows that greater mass produces less acceleration."},{"text":"

Measures how quickly an object changes velocity.
A larger inertial mass produces a larger acceleration.

","correct":false,"explanation":"Inertial mass measures resistance to changing velocity, not the speed of that change. With the same force, greater mass gives smaller acceleration, not larger acceleration."},{"text":"

Measures the force needed to stop an object.
A larger inertial mass produces the same acceleration.

","correct":false,"explanation":"Inertial mass concerns resistance to any change in velocity, not only stopping. At constant force, increasing mass reduces acceleration according to $a = \\dfrac{F}{m}$."},{"text":"

Measures the energy stored by an object.
A larger inertial mass produces a zero acceleration.

","correct":false,"explanation":"Inertial mass does not measure stored energy, and a larger mass does not guarantee zero acceleration. The same non-zero force produces a smaller, but generally non-zero, acceleration."}]},{"id":"c49dea23-c35b-4816-952e-c7f2a54e1c0f","cardType":"cloze","points":null,"front":"The equation for acceleration from resultant force and mass is $\\text{[...]}$.","back":"The equation for acceleration from resultant force and mass is $\\boxed{a = \\dfrac{F}{m}}$.","mcqOptions":[{"text":"$$a = \\dfrac{F}{m}$","correct":true,"explanation":"Rearranging Newton’s second law, $F = ma$, by dividing both sides by mass gives acceleration as $a = \\dfrac{F}{m}$."},{"text":"$$a = \\dfrac{m}{F}$","correct":false,"explanation":"This reverses the force-to-mass ratio. Dividing $F = ma$ by mass gives the correct relationship $a = \\dfrac{F}{m}$."},{"text":"$$a = Fm$","correct":false,"explanation":"Acceleration is not found by multiplying force and mass. Rearranging $F = ma$ requires dividing force by mass, giving $a = \\dfrac{F}{m}$."},{"text":"$$a = F + m$","correct":false,"explanation":"Adding force and mass does not rearrange Newton’s second law and also combines quantities with different units. The correct equation is $a = \\dfrac{F}{m}$."}]},{"id":"27ae0b22-27c0-4057-aaa0-0e1d1aca1ebd","cardType":"cloze","points":null,"front":"The equation defining inertial mass is $\\text{[...]}$.","back":"The equation defining inertial mass is $\\boxed{m = \\dfrac{F}{a}}$.","mcqOptions":[{"text":"$$m = \\dfrac{F}{a}$","correct":true,"explanation":"Rearranging $F = ma$ to make mass the subject requires dividing both sides by acceleration, giving $m = \\dfrac{F}{a}$."},{"text":"$$m = \\dfrac{a}{F}$","correct":false,"explanation":"This reverses the required ratio. From $F = ma$, dividing by acceleration gives the correct expression $m = \\dfrac{F}{a}$."},{"text":"$$m = Fa$","correct":false,"explanation":"Mass is not found by multiplying force and acceleration. Rearranging $F = ma$ requires division by acceleration, so $m = \\dfrac{F}{a}$."},{"text":"$$m = F + a$","correct":false,"explanation":"Adding force and acceleration does not isolate mass and combines quantities with incompatible units. The correct rearrangement is $m = \\dfrac{F}{a}$."}]},{"id":"7e15b2bb-24b3-4317-86b6-55001bcf07c7","cardType":"cloze","points":null,"front":"For a fixed acceleration, a larger inertial mass requires a [ ].","back":"For a fixed acceleration, a larger inertial mass requires a greater resultant force.","mcqOptions":[{"text":"greater resultant force","correct":true,"explanation":"Newton’s second law gives $F = ma$. With acceleration fixed, increasing inertial mass increases the resultant force needed in direct proportion."},{"text":"smaller resultant force","correct":false,"explanation":"A smaller resultant force would produce less acceleration for the same mass. With acceleration fixed, Newton’s second law requires a greater resultant force as mass increases."},{"text":"greater acceleration","correct":false,"explanation":"Greater acceleration is not required when the acceleration is fixed; it would contradict the condition in the question. Increasing mass instead requires a greater resultant force."},{"text":"constant resultant force","correct":false,"explanation":"A constant resultant force would give a smaller acceleration when mass increases. For the acceleration to remain fixed, the resultant force must increase with mass."}]},{"id":"d08f5795-4cb6-48a4-bb0a-b6781fde4a6f","cardType":"cloze","points":null,"front":"In the force investigation, acceleration is proportional to force when [ ].","back":"In the force investigation, acceleration is proportional to force when the mass is constant.","mcqOptions":[{"text":"the mass is constant","correct":true,"explanation":"From $F = ma$, acceleration is $a = F/m$. If mass remains constant, increasing force increases acceleration in direct proportion."},{"text":"the acceleration is constant","correct":false,"explanation":"If acceleration is constant, changing force would require mass to change correspondingly. The proportional relationship between acceleration and force is tested when the mass is constant."},{"text":"the force is constant","correct":false,"explanation":"A constant force would not allow acceleration to vary proportionally with force. The investigation requires varying force while keeping the mass constant."},{"text":"the resultant force is zero","correct":false,"explanation":"A zero resultant force produces zero acceleration, so it cannot demonstrate proportional increases in acceleration and force. The mass must remain constant during the investigation."}]},{"id":"bdb4a75a-e55d-46b1-81bd-64b1d9871fe8","cardType":"cloze","points":null,"front":"A $1500\\ \\text{kg}$ car changing velocity from $0$ to approximately $27\\ \\text{m/s}$ in $9.0\\ \\text{s}$ has an estimated acceleration of $\\text{[...]}$.","back":"A $1500\\ \\text{kg}$ car changing velocity from $0$ to approximately $27\\ \\text{m/s}$ in $9.0\\ \\text{s}$ has an estimated acceleration of $\\boxed{\\approx 3.0\\ \\text{m/s}^2}$.","mcqOptions":[{"text":"$$\\approx 3.0\\ \\text{m/s}^2$","correct":true,"explanation":"Acceleration is change in velocity divided by time: $a = \\frac{27\\ \\text{m/s}}{9.0\\ \\text{s}} \\approx 3.0\\ \\text{m/s}^2$. The car’s mass is not needed for this calculation."},{"text":"$$\\approx 0.33\\ \\text{m/s}^2$","correct":false,"explanation":"This reverses the division, using time divided by velocity. Acceleration is change in velocity divided by time, giving $27/9.0 \\approx 3.0\\ \\text{m/s}^2$."},{"text":"$$\\approx 18\\ \\text{m/s}^2$","correct":false,"explanation":"This can result from combining the velocity and time incorrectly rather than dividing velocity change by elapsed time. The correct calculation is $27/9.0 \\approx 3.0\\ \\text{m/s}^2$."},{"text":"$$\\approx 13,500\\ \\text{m/s}^2$","correct":false,"explanation":"Multiplying the car’s mass by its velocity change and dividing by time confuses force-related calculations with acceleration. Acceleration depends here only on $27/9.0 \\approx 3.0\\ \\text{m/s}^2$."}]},{"id":"e11b1d50-851d-4547-8f84-d84d1a8f45ad","cardType":"cloze","points":null,"front":"Inertial mass is measured in [ ], whereas weight is measured in newtons.","back":"Inertial mass is measured in kilograms ($\\text{kg}$), whereas weight is measured in newtons.","mcqOptions":[{"text":"kilograms ($\\text{kg}$)","correct":true,"explanation":"Inertial mass measures resistance to acceleration and uses the SI unit kilogram, written kg. Weight is a force, so it is measured in newtons."},{"text":"grams ($\\text{g}$)","correct":false,"explanation":"Grams are a mass unit, but GCSE physics normally gives inertial mass in kilograms. Weight remains a force measured in newtons, and the required unit is kilograms ($\\text{kg}$)."},{"text":"metres ($\\text{m}$)","correct":false,"explanation":"Metres measure distance, not mass, so they cannot describe an object’s inertia. Inertial mass is measured in kilograms ($\\text{kg}$), while weight is measured in newtons."},{"text":"kilograms per second ($\\text{kg/s}$)","correct":false,"explanation":"Kilograms per second describes a rate involving mass, not mass itself. Inertial mass has the unit kilograms ($\\text{kg}$), whereas weight is measured in newtons."}]}]}],"examCountdown":{"examDate":"2027-05-25","paperLabel":"Paper 1"},"heading":"2.3.2 Newton's second law","title":"5.6.2c Newton's Second Law Flashcards","topbar":{"backHref":"/gcse/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"e2478e08-5483-4aff-ac5f-a334c106560c","walled":"$L39"}],"$L3a","$L3b"]

2.3.2 Newton's second law Flashcards

2.3.2 Newton's second law Flashcards