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size and direction.","correct":false,"explanation":"Direction alone does not define a scalar, since scalars have magnitude only. The correct distinction is that vectors have both size and direction."},{"text":"A scalar has size only; a vector has direction only.","correct":false,"explanation":"A vector cannot be described by direction alone; its magnitude is also essential. The correct answer is that a vector has both size and direction."},{"text":"A scalar has size and direction; a vector has size only.","correct":false,"explanation":"The properties have been reversed. Scalars have magnitude only, while vectors have magnitude together with a specified direction."}]},{"id":"54213917-91af-42b8-be83-fbca7e32d3b6","cardType":"basic","points":1,"front":"How is a vector opposite to the chosen positive direction treated?","back":"It is given a negative sign.","mcqOptions":[{"text":"It is given a negative sign.","correct":true,"explanation":"Choosing one direction as positive means any vector pointing oppositely is represented with a negative sign, indicating its direction relative to that convention."},{"text":"It is given a positive sign.","correct":false,"explanation":"A positive sign represents the chosen positive direction. A vector pointing opposite to that direction must be assigned a negative sign."},{"text":"It is treated as zero.","correct":false,"explanation":"An opposite vector still has a real magnitude and affects the motion. Its direction is represented by a negative sign, not by setting it to zero."},{"text":"Its direction is ignored.","correct":false,"explanation":"Ignoring direction would incorrectly treat the vector like a scalar. The opposite direction is included by assigning the vector a negative sign."}]},{"id":"087434f9-8b7a-4162-ae49-745ef79f04e7","cardType":"cloze","points":null,"front":"Newton’s third-law forces are equal and opposite, and act on [ ].","back":"Newton’s third-law forces are equal and opposite, and act on different objects.","mcqOptions":[{"text":"different objects","correct":true,"explanation":"Newton’s third-law forces arise from an interaction but act on the two separate objects involved, which is why they cannot cancel on one free-body diagram."},{"text":"the same object","correct":false,"explanation":"Forces acting on the same object may contribute to its resultant force, but third-law partners act on different objects. This is a common pairing error."},{"text":"adjacent objects","correct":false,"explanation":"The objects involved need not simply be adjacent; the defining feature is that the two equal and opposite forces act on different objects."},{"text":"objects with equal masses","correct":false,"explanation":"Newton’s third law does not require equal masses. The forces act on different objects, regardless of their masses or acceleration responses."}]},{"id":"41f00b78-51a7-476a-9577-fddb9215de70","cardType":"basic","points":2,"front":"Why do Newton’s third-law force pairs not cancel?","back":"They act on different objects.","mcqOptions":[{"text":"They act on different objects.","correct":true,"explanation":"Third-law forces are equal and opposite but act on separate objects, so they cannot cancel when calculating the resultant force on either single object."},{"text":"They act on the same object.","correct":false,"explanation":"Forces on the same object can cancel if they are equal and opposite. Third-law pairs do not cancel because each force acts on a different object."},{"text":"They act in the same direction.","correct":false,"explanation":"Third-law partners act in opposite directions, not the same direction. Their failure to cancel is because they act on different objects."},{"text":"They act at different times.","correct":false,"explanation":"Third-law forces act simultaneously as part of one interaction. They do not cancel because they act on different objects, not because of timing."}]},{"id":"ebf9bdbf-3a2c-4053-ab75-e575f753551b","cardType":"basic","points":2,"front":"A table pushes a book upward with 12 N. What is the third-law partner force?","back":"The book pushes the table downward with 12 N.","mcqOptions":[{"text":"The book pushes the table downward with 12 N.","correct":true,"explanation":"The contact forces form a third-law pair: the table pushes the book upward and the book pushes the table downward with equal magnitude, 12 N."},{"text":"The table pushes the book upward with 12 N.","correct":false,"explanation":"This repeats the original force rather than identifying its partner. The third-law force is exerted by the book on the table, downward with 12 N."},{"text":"The book pushes the table upward with 12 N.","correct":false,"explanation":"The partner force must act in the opposite direction to the table’s push. Therefore, the book pushes the table downward with 12 N."},{"text":"The book pushes the table downward with 24 N.","correct":false,"explanation":"Third-law forces have equal magnitudes, so doubling the given force is an arithmetic misconception. The partner force is downward with 12 N."}]},{"id":"f3da2351-9c29-4486-9b65-450ae86dc603","cardType":"basic","points":2,"front":"Why are a book’s weight and its normal contact force not a third-law pair?","back":"They both act on the same object: the book.","mcqOptions":[{"text":"They both act on the same object: the book.","correct":true,"explanation":"The book’s weight and the normal contact force both act on the book. Third-law partners must act on two different objects involved in the interaction."},{"text":"They act on different objects: the book and Earth.","correct":false,"explanation":"The book’s weight acts on the book, while Earth experiences the gravitational partner force. The normal force is also on the book, so these two are not partners."},{"text":"They act in the same direction: downward on the book.","correct":false,"explanation":"Weight acts downward, but the normal contact force acts away from the surface, usually upward. They are not a third-law pair because both act on the book."},{"text":"They act at different times: before and after contact.","correct":false,"explanation":"Weight and normal contact force can act simultaneously while the book rests on the table. They are not partners because both forces act on the same object."}]},{"id":"b1b2525e-d88e-4f35-959e-79ef2e6fd303","cardType":"cloze","points":null,"front":"A free-body diagram shows [ ] acting on [ ].","back":"A free-body diagram shows external forces acting on one chosen object.","mcqOptions":[{"text":"A free-body diagram shows external forces acting on one chosen object.","correct":true,"explanation":"A free-body diagram isolates one object and represents all external forces acting on it, allowing the resultant force and motion to be analysed."},{"text":"A free-body diagram shows external forces acting on all interacting objects.","correct":false,"explanation":"Including every interacting object describes a wider system diagram. A free-body diagram isolates one chosen object and shows the external forces acting on it."},{"text":"A free-body diagram shows internal forces acting on one chosen object.","correct":false,"explanation":"Forces within a system are internal and are not the focus of an isolated free-body diagram. It shows external forces acting on one chosen object."},{"text":"A free-body diagram shows external forces acting on the surrounding system.","correct":false,"explanation":"A free-body diagram does not represent the surrounding system as a whole. It isolates one chosen object and shows the external forces acting on that object."}]},{"id":"5cb3e4cf-4ca3-4271-a39f-24c2de23a61c","cardType":"cloze","points":null,"front":"In a free-body diagram, weight acts [ ]; the normal contact force acts [ ].","back":"In a free-body diagram, weight acts vertically downward; the normal contact force acts perpendicular to the surface.","mcqOptions":[{"text":"In a free-body diagram, weight acts vertically downward; the normal contact force acts perpendicular to the surface.","correct":true,"explanation":"Weight is caused by gravity and acts vertically downward. A normal contact force acts at right angles to the contacting surface."},{"text":"In a free-body diagram, weight acts vertically upward; the normal contact force acts perpendicular to the surface.","correct":false,"explanation":"Gravity always pulls objects towards Earth, so weight acts vertically downward. The normal force is correctly identified as perpendicular to the surface."},{"text":"In a free-body diagram, weight acts vertically downward; the normal contact force acts parallel to the surface.","correct":false,"explanation":"The normal force prevents interpenetration and therefore acts at right angles to the surface, not along it. Weight does act vertically downward."},{"text":"In a free-body diagram, weight acts perpendicular to the surface; the normal contact force acts vertically upward.","correct":false,"explanation":"Weight remains vertically downward regardless of the surface orientation. The normal force is perpendicular to the surface, so both directions have been misassigned here."}]},{"id":"18945d2a-5de0-4bf1-beb5-4c82feba3eb2","cardType":"basic","points":1,"front":"Should acceleration be drawn as a force on a free-body diagram?","back":"No - acceleration is not a force.","mcqOptions":[{"text":"No - acceleration is not a force.","correct":true,"explanation":"Acceleration describes how velocity changes, whereas a free-body diagram shows the forces acting on an object. Acceleration is caused by the resultant force but is not drawn as a separate force."},{"text":"Yes - acceleration is a vector, so it should be shown as a force.","correct":false,"explanation":"Being a vector does not make acceleration a force. Free-body diagrams include only forces such as weight, tension and friction; acceleration is the motion resulting from their resultant."},{"text":"Yes - acceleration must balance the resultant force.","correct":false,"explanation":"Acceleration does not balance the resultant force; Newton’s second law links them through $sum F = ma$. The diagram should show the forces producing the acceleration, not acceleration itself."},{"text":"No - acceleration can only be drawn on a velocity-time graph.","correct":false,"explanation":"Acceleration can be represented in several ways, not only on a velocity-time graph. It is excluded from a free-body diagram specifically because that diagram contains forces, not kinematic quantities."}]},{"id":"17928b22-9ef0-4918-a5c2-9d0d9a28e672","cardType":"basic","points":1,"front":"In one dimension, how is the resultant force found?","back":"Add the forces with their signs.","mcqOptions":[{"text":"Add the forces with their signs.","correct":true,"explanation":"Choose a positive direction, give forces in that direction positive signs and opposite forces negative signs, then add them algebraically to obtain the resultant force."},{"text":"Add the magnitudes of all the forces.","correct":false,"explanation":"Adding magnitudes ignores direction, so opposing forces would incorrectly increase the resultant. The correct method is to add the forces algebraically using their signs."},{"text":"Subtract the smaller force from the larger force.","correct":false,"explanation":"This works only for two forces acting in opposite directions and does not handle several forces reliably. In one dimension, all forces should be added with their signs."},{"text":"Multiply the forces by their directions.","correct":false,"explanation":"Direction is represented by a sign, not by multiplying force values by a separate direction. The resultant is found by adding signed force values algebraically."}]},{"id":"0d8408b9-68be-4d55-82b5-b0db21341a9b","cardType":"cloze","points":null,"front":"For constant mass, Newton’s second law is [ ].","back":"For constant mass, Newton’s second law is $\\sum F = ma$.","mcqOptions":[{"text":"$$\\sum F = ma$","correct":true,"explanation":"For constant mass, Newton’s second law states that the resultant force equals mass multiplied by acceleration, written as $\\sum F = ma$."},{"text":"$$\\sum F = \\frac{m}{a}$","correct":false,"explanation":"Dividing mass by acceleration reverses the relationship. Newton’s second law requires mass and acceleration to be multiplied, giving $\\sum F = ma$."},{"text":"$$\\sum F = \\frac{a}{m}$","correct":false,"explanation":"This expression incorrectly places mass in the denominator. Acceleration is resultant force divided by mass, so rearranging gives the correct law $\\sum F = ma$."},{"text":"$$\\sum F = m + a$","correct":false,"explanation":"Mass and acceleration are not added in Newton’s second law. Their product gives the resultant force, so the correct expression is $\\sum F = ma$."}]},{"id":"da5c4209-8f5e-4a9a-9046-36568338efd3","cardType":"basic","points":2,"front":"What does $\\sum F = 0$ imply about the object’s motion?","back":"Acceleration is zero; it may still move at constant velocity.","mcqOptions":[{"text":"Acceleration is zero; it may still move at constant velocity.","correct":true,"explanation":"A zero resultant force gives zero acceleration. The object may be stationary, or it may continue moving in a straight line at constant velocity."},{"text":"The object must be stationary; its velocity is zero.","correct":false,"explanation":"Zero resultant force means zero acceleration, not necessarily zero velocity. An object already moving can continue at constant velocity, so the correct implication is that acceleration is zero."},{"text":"The object is accelerating; its velocity is increasing uniformly.","correct":false,"explanation":"A non-zero acceleration requires a non-zero resultant force. When $\\sum F = 0$, acceleration is zero, although the object may still have a constant non-zero velocity."},{"text":"The object must be slowing down; its velocity is decreasing uniformly.","correct":false,"explanation":"Slowing down requires acceleration opposite to the velocity, so it cannot occur when the resultant force and acceleration are zero. The object may instead move at constant velocity."}]},{"id":"01d79d54-2d36-4e64-8552-89553721d6e8","cardType":"basic","points":2,"front":"A 2.0 kg box has 8.0 N tension right and 2.0 N friction left. Find its acceleration.","back":"3.0 m s^-2 to the right.","mcqOptions":[{"text":"3.0 m s^-2 to the right.","correct":true,"explanation":"The resultant force is $8.0 - 2.0 = 6.0$ N to the right. Using $a = F/m$ gives $6.0/2.0 = 3.0$ m s^-2 to the right."},{"text":"5.0 m s^-2 to the right.","correct":false,"explanation":"This treats the 2.0 N friction as though it were acceleration or removes the wrong amount from the calculation. The resultant is 6.0 N, giving 3.0 m s^-2 right."},{"text":"4.0 m s^-2 to the right.","correct":false,"explanation":"Dividing the 8.0 N tension by the 2.0 kg mass ignores friction. Subtracting 2.0 N first gives a 6.0 N resultant and an acceleration of 3.0 m s^-2 right."},{"text":"3.0 m s^-2 to the left.","correct":false,"explanation":"The tension exceeds the friction, so the resultant force and acceleration point right, not left. The magnitude is 6.0/2.0 = 3.0 m s^-2."}]},{"id":"04954952-8cf4-47d6-9292-c23da95cf19b","cardType":"basic","points":2,"front":"How can the driving force be increased without changing total mass in the trolley practical?","back":"Transfer slotted masses from the trolley to the hanger.","mcqOptions":[{"text":"Transfer slotted masses from the trolley to the hanger.","correct":true,"explanation":"Moving slotted masses from the trolley to the hanger increases the hanging weight and therefore the driving force, while keeping the combined trolley-hanger mass unchanged."},{"text":"Transfer slotted masses from the hanger to the trolley.","correct":false,"explanation":"This reduces the hanging weight, so it decreases the driving force rather than increasing it. To increase the driving force at constant total mass, masses must move to the hanger."},{"text":"Add slotted masses to the trolley only.","correct":false,"explanation":"Adding mass only to the trolley changes the total mass and does not directly increase the hanging weight. The practical transfers existing masses to the hanger to increase driving force."},{"text":"Remove slotted masses from the hanger.","correct":false,"explanation":"Removing masses from the hanger reduces its weight and therefore reduces the driving force. Transferring masses from the trolley to the hanger produces the required increase without changing total mass."}]},{"id":"f6bd5f72-04bf-4a07-968c-6faccc4e98ed","cardType":"basic","points":1,"front":"If friction is negligible, what is the resultant force on the trolley-hanger system?","back":"Approximately the weight of the hanging mass.","mcqOptions":[{"text":"Approximately the weight of the hanging mass.","correct":true,"explanation":"With negligible friction, the trolley’s horizontal resistance is negligible, so the main external driving force on the system is the weight of the hanging mass."},{"text":"Approximately the weight of the trolley.","correct":false,"explanation":"The trolley’s weight acts vertically and is balanced by the normal contact force, so it does not drive the horizontal motion. The resultant is approximately the hanging mass’s weight."},{"text":"Approximately the total weight of the system.","correct":false,"explanation":"The trolley’s vertical weight is balanced by the normal reaction and does not contribute to the horizontal resultant. The unbalanced driving force is approximately the hanging mass’s weight."},{"text":"Approximately zero.","correct":false,"explanation":"Negligible friction does not mean the driving force is zero; the hanging mass still exerts an unbalanced weight. The resultant force is therefore approximately the weight of the hanging mass."}]},{"id":"225437d1-b22b-48ee-a908-8865aad88800","cardType":"cloze","points":null,"front":"On a graph of resultant force against acceleration, the gradient is [ ].","back":"On a graph of resultant force against acceleration, the gradient is the total mass.","mcqOptions":[{"text":"the total mass","correct":true,"explanation":"From Newton’s second law, $F = ma$. On a graph of resultant force against acceleration, the gradient is $\\Delta F/\\Delta a = m$, the total mass of the system."},{"text":"the hanging mass","correct":false,"explanation":"The hanging mass contributes to the driving force but is not generally the mass represented by the graph’s gradient. Since $F = ma$, the gradient is the total mass of the system."},{"text":"the total force","correct":false,"explanation":"The force is plotted on the vertical axis, so it is not the gradient. Dividing the change in force by the change in acceleration gives mass, specifically the total mass."},{"text":"the acceleration","correct":false,"explanation":"Acceleration is the horizontal-axis variable, not the gradient’s physical quantity. Newton’s second law shows that force divided by acceleration gives the total mass."}]},{"id":"1db76a84-de5a-4964-a3fe-731513a4fbf1","cardType":"basic","points":1,"front":"Which equation gives linear momentum?","back":"$p = mv$.","mcqOptions":[{"text":"$p = mv$.","correct":true,"explanation":"Linear momentum is defined as mass multiplied by velocity, so the equation is $p = mv$. Momentum is a vector, meaning its direction is also determined by the velocity."},{"text":"$p = ma$.","correct":false,"explanation":"This confuses momentum with force: $ma$ gives the resultant force, not momentum. The correct relationship is $p = mv$, using velocity rather than acceleration."},{"text":"$p = \\frac{m}{v}$.","correct":false,"explanation":"Dividing mass by velocity gives neither momentum nor a standard dynamics quantity. Momentum increases with both mass and velocity, so the correct equation is $p = mv$."},{"text":"$p = \\frac{v}{m}$.","correct":false,"explanation":"Momentum is proportional to mass, rather than inversely proportional to it. The correct equation multiplies mass by velocity: $p = mv$."}]},{"id":"a71baf58-8e69-4268-8d55-5ea18ec90664","cardType":"basic","points":2,"front":"A 0.150 kg ball moves left at 12 m s^-1. Taking right as positive, find its momentum.","back":"$-1.8\\,\\text{kg m s}^{-1}$.","mcqOptions":[{"text":"$-1.8\\,\\text{kg m s}^{-1}$.","correct":true,"explanation":"Momentum is $p = mv$. The velocity is negative because the ball moves left, so $p = 0.150 \\times (-12) = -1.8\\,\\text{kg m s}^{-1}$."},{"text":"$+1.8\\,\\text{kg m s}^{-1}$.","correct":false,"explanation":"The magnitude is calculated correctly, but the sign is wrong. Taking right as positive makes the ball’s leftward velocity negative, so the momentum is $-1.8\\,\\text{kg m s}^{-1}$."},{"text":"$-0.0125\\,\\text{kg m s}^{-1}$.","correct":false,"explanation":"This comes from dividing mass by speed instead of multiplying them. Momentum is $p = mv$, and the leftward direction gives $0.150 \\times (-12) = -1.8\\,\\text{kg m s}^{-1}$."},{"text":"$-80\\,\\text{kg m s}^{-1}$.","correct":false,"explanation":"This results from dividing speed by mass, reversing the momentum formula. Multiplying $0.150\\,\\text{kg}$ by the signed velocity $-12\\,\\text{m s}^{-1}$ gives $-1.8\\,\\text{kg m s}^{-1}$."}]},{"id":"7df3be9c-85ef-44f3-bbf9-896cda14ada3","cardType":"cloze","points":null,"front":"Average force is the rate of change of momentum: [ ].","back":"Average force is the rate of change of momentum: $F = \\frac{\\Delta p}{\\Delta t}$.","mcqOptions":[{"text":"$$F = \\frac{\\Delta p}{\\Delta t}$","correct":true,"explanation":"Average force equals the change in momentum divided by the time taken for that change. This is the momentum form of Newton’s second law."},{"text":"$$F = \\frac{\\Delta t}{\\Delta p}$","correct":false,"explanation":"The change in time and change in momentum have been inverted. A larger momentum change produces a larger average force, so the correct relationship is $F = \\frac{\\Delta p}{\\Delta t}$."},{"text":"$$F = \\Delta p\\Delta t$","correct":false,"explanation":"Multiplying the changes gives the wrong dependence and incorrect units. Average force is the momentum change per unit time: $F = \\frac{\\Delta p}{\\Delta t}$."},{"text":"$$F = \\frac{p}{t^2}$","correct":false,"explanation":"This uses the momentum itself and squares the time, neither of which represents the average rate of change. The correct equation uses $\\Delta p$ divided by $\\Delta t$."}]},{"id":"8caeba07-3956-490c-a147-761870cfd07a","cardType":"basic","points":2,"front":"How do airbags or crumple zones reduce average impact force?","back":"They increase collision time, so $F = \\frac{\\Delta p}{\\Delta t}$ is smaller.","mcqOptions":[{"text":"They increase collision time, so $F = \\frac{\\Delta p}{\\Delta t}$ is smaller.","correct":true,"explanation":"For the same change in momentum, increasing the collision time increases the denominator in $F = \\frac{\\Delta p}{\\Delta t}$. The average force is therefore reduced."},{"text":"They decrease collision time, so $F = \\frac{\\Delta p}{\\Delta t}$ is smaller.","correct":false,"explanation":"Decreasing the collision time would make the denominator smaller and increase the average force. Airbags and crumple zones instead increase collision time, reducing the force."},{"text":"They increase collision time, so $F = \\frac{\\Delta p}{\\Delta t}$ is larger.","correct":false,"explanation":"Increasing the time does not increase the force when the momentum change is fixed. Since time is in the denominator, a longer collision makes the average force smaller."},{"text":"They decrease momentum change, so $F = \\frac{\\Delta p}{\\Delta t}$ is smaller.","correct":false,"explanation":"The object still undergoes approximately the same momentum change when it is brought to rest. The reduction comes from increasing collision time, which lowers $\\frac{\\Delta p}{\\Delta t}$."}]},{"id":"e9a1f8f0-fab0-4554-b0ea-7f0b7f41ced7","cardType":"basic","points":1,"front":"Under what condition is total momentum conserved?","back":"A closed system with no external resultant force.","mcqOptions":[{"text":"A closed system with no external resultant force.","correct":true,"explanation":"Momentum is conserved when the system is closed and the resultant external force is zero. Internal forces act in equal and opposite pairs and do not change total momentum."},{"text":"An open system with no external resultant force.","correct":false,"explanation":"An open system allows matter to cross its boundary, so momentum can be carried in or out. The required condition is a closed system with no external resultant force."},{"text":"A closed system with a constant external resultant force.","correct":false,"explanation":"A non-zero external resultant force changes the system’s total momentum. Conservation requires a closed system with no external resultant force."},{"text":"A closed system with no internal forces.","correct":false,"explanation":"Internal forces may be present and still cancel in pairs when considering the whole system. It is the absence of an external resultant force, not internal forces, that matters."}]},{"id":"964198cb-f44e-4bed-8be0-b4faf7e9d2cd","cardType":"basic","points":1,"front":"Which equation represents momentum conservation in a one-dimensional collision?","back":"$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.","mcqOptions":[{"text":"$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$.","correct":true,"explanation":"The total momentum before the collision, $m_1u_1 + m_2u_2$, equals the total momentum afterwards, $m_1v_1 + m_2v_2$, when no external resultant force acts."},{"text":"$m_1v_1 + m_2v_2 = m_1u_1 - m_2u_2$.","correct":false,"explanation":"The second object’s initial momentum has been incorrectly subtracted. Momentum conservation adds the signed initial momenta and sets their total equal to the final total."},{"text":"$m_1u_1 + m_2v_2 = m_1v_1 + m_2u_2$.","correct":false,"explanation":"Initial and final velocities have been mixed between the two sides. Both initial momenta belong on one side and both final momenta on the other."},{"text":"$m_1u_1 + m_2u_2 = m_1v_1 - m_2v_2$.","correct":false,"explanation":"The final momentum of the second object has been given the wrong sign in the equation. Velocities already carry direction, so the final momenta are added algebraically."}]},{"id":"f43b29c1-7e32-40a3-92ed-c9e8450dfd8e","cardType":"basic","points":3,"front":"Compare elastic and inelastic collisions in terms of kinetic energy.","back":"

Elastic: no kinetic energy lost.
Inelastic: kinetic energy is lost.
Momentum: conserved if no external resultant force.

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

Elastic: no kinetic energy lost.
Inelastic: kinetic energy is lost.
Momentum: conserved if no external resultant force.

","correct":true,"explanation":"Elastic collisions conserve total kinetic energy, whereas inelastic collisions lose some kinetic energy to deformation, heating or sound. Total momentum remains conserved without an external resultant force."},{"text":"

Elastic: kinetic energy is lost.
Inelastic: no kinetic energy lost.
Momentum: conserved if no external resultant force.

","correct":false,"explanation":"The kinetic-energy descriptions have been reversed. Elastic collisions conserve kinetic energy, while inelastic collisions lose some; momentum is still conserved without an external resultant force."},{"text":"

Elastic: no kinetic energy lost.
Inelastic: kinetic energy is lost.
Momentum: conserved only if kinetic energy is conserved.

","correct":false,"explanation":"Momentum conservation does not depend on kinetic energy being conserved. It depends on there being no external resultant force, including in inelastic collisions."},{"text":"

Elastic: no momentum conserved.
Inelastic: kinetic energy is lost.
Momentum: conserved if no external resultant force.

","correct":false,"explanation":"Elastic collisions conserve momentum as well as kinetic energy when the system has no external resultant force. The collision type determines kinetic-energy loss, not whether momentum is conserved."}]},{"id":"5fb9781b-8f70-4a69-b27c-5a88f87c4498","cardType":"basic","points":2,"front":"A 0.50 kg trolley at 3.0 m s^-1 sticks to a stationary 0.30 kg trolley. Find their common velocity.","back":"$1.9\\,\\text{m s}^{-1}$.","mcqOptions":[{"text":"$1.9\\,\\text{m s}^{-1}$.","correct":true,"explanation":"Momentum before collision is $0.50 \\times 3.0 = 1.5\\,\\text{kg m s}^{-1}$. Dividing by the combined mass, $0.80\\,\\text{kg}$, gives $1.875\\,\\text{m s}^{-1}$, which rounds to $1.9\\,\\text{m s}^{-1}$."},{"text":"$1.5\\,\\text{m s}^{-1}$.","correct":false,"explanation":"This uses the initial momentum value as the final speed and forgets to divide by the combined mass. Momentum conservation gives $1.5 \\div 0.80 = 1.875\\,\\text{m s}^{-1}$, or $1.9\\,\\text{m s}^{-1}$."},{"text":"$2.5\\,\\text{m s}^{-1}$.","correct":false,"explanation":"This divides the initial momentum by only the moving trolley’s mass, ignoring the stationary trolley that joins it. The combined mass is $0.80\\,\\text{kg}$, giving $1.9\\,\\text{m s}^{-1}$."},{"text":"$3.8\\,\\text{m s}^{-1}$.","correct":false,"explanation":"This effectively uses an incorrect mass difference or adds speed rather than conserving momentum. After sticking, the total mass is larger, so the common speed must be below $3.0\\,\\text{m s}^{-1}$: $1.9\\,\\text{m s}^{-1}$."}]}]}],"examCountdown":{"examDate":"2027-05-21","paperLabel":"Component 1"},"heading":"Dynamics","title":"Dynamics Flashcards","topbar":{"backHref":"/a-level/physics/eduqas/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"41132b8f-40d2-5fa4-be35-94765ca8eb95","walled":"$L38"}],"$L39","$L3a"]