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kg m s⁻¹
N s

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

kg m s⁻¹
N s

","correct":true,"explanation":"Momentum has units $\\text{kg m s}^{-1}$. Impulse is force multiplied by time, so $\\text{N s}=\\text{kg m s}^{-1}$ and the units are equivalent."},{"text":"

kg m s⁻²
N

","correct":false,"explanation":"These are units of force, not momentum or impulse. Momentum has units kg m s⁻¹, equivalent to N s because impulse is force multiplied by time."},{"text":"

kg m² s⁻²
J

","correct":false,"explanation":"These are units of energy or work, arising from multiplying force by distance. Momentum and impulse instead use kg m s⁻¹ and the equivalent unit N s."},{"text":"

kg m² s⁻³
W

","correct":false,"explanation":"These are units of power, which measure energy transferred per unit time. Momentum and impulse have units kg m s⁻¹, equivalently written as N s."}]},{"id":"0df47d93-1c2e-4d99-8bca-92878988ce81","cardType":"cloze","points":null,"front":"The momentum $p$ of an object is the product of its [ ].","back":"The momentum $p$ of an object is the product of its mass and velocity.","mcqOptions":[{"text":"mass and velocity","correct":true,"explanation":"Momentum is defined by $p=mv$, so it is the product of an object’s mass and its velocity, including the velocity’s direction."},{"text":"mass and acceleration","correct":false,"explanation":"Mass multiplied by acceleration contributes to force, since $F=ma$, rather than momentum. Momentum is the product of mass and velocity, given by $p=mv$."},{"text":"force and time","correct":false,"explanation":"Force multiplied by time gives impulse, which equals the change in momentum rather than momentum itself. Momentum is directly calculated as mass multiplied by velocity."},{"text":"mass and speed","correct":false,"explanation":"Using speed ignores direction and therefore gives only the magnitude of momentum. Momentum is a vector, so its defining product uses mass and velocity."}]},{"id":"8b5b26e5-84d5-40c4-89ab-ccc73d13684b","cardType":"basic","points":1,"front":"Why is momentum classified as a vector quantity?","back":"It depends on velocity, which has both magnitude and direction.","mcqOptions":[{"text":"It depends on velocity, which has both magnitude and direction.","correct":true,"explanation":"Momentum is given by $p=mv$. Mass is scalar, but velocity is a vector with magnitude and direction, so momentum also has both properties."},{"text":"It depends on speed, which has both magnitude and direction.","correct":false,"explanation":"Speed has magnitude only and is therefore a scalar, so it cannot explain momentum’s direction. Momentum is a vector because it depends on velocity, which includes direction."},{"text":"It depends on mass, which has both magnitude and direction.","correct":false,"explanation":"Mass is a scalar quantity and has no direction. Momentum is classified as a vector because it depends on velocity, which possesses both magnitude and direction."},{"text":"It depends on kinetic energy, which has both magnitude and direction.","correct":false,"explanation":"Kinetic energy is a scalar, even though its calculation involves speed. Momentum is a vector because it depends on velocity, whose direction determines the momentum’s direction."}]},{"id":"cafacd2f-f5b8-4697-a538-1624799be700","cardType":"basic","points":2,"front":"How are the horizontal and vertical components of a momentum vector $p$ calculated at an angle $\\theta$ to the horizontal?","back":"

$p_x = p\\cos\\theta$
$p_y = p\\sin\\theta$

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

$p_x = p\\cos\\theta$
$p_y = p\\sin\\theta$

","correct":true,"explanation":"Resolving a vector measured from the horizontal forms a right-angled triangle: the horizontal component is adjacent to $\\theta$, while the vertical component is opposite."},{"text":"

$p_x = p\\sin\\theta$
$p_y = p\\cos\\theta$

","correct":false,"explanation":"This swaps the adjacent and opposite components. Because $\\theta$ is measured from the horizontal, the correct components are $p_x=p\\cos\\theta$ and $p_y=p\\sin\\theta$."},{"text":"

$p_x = \\frac{p}{\\cos\\theta}$
$p_y = \\frac{p}{\\sin\\theta}$

","correct":false,"explanation":"Dividing by the trigonometric ratios would make components exceed the original momentum for many angles. Resolution requires $p_x=p\\cos\\theta$ and $p_y=p\\sin\\theta$."},{"text":"

$p_x = p\\cos\\theta$
$p_y = p\\cos\\theta$

","correct":false,"explanation":"Both components cannot generally use the same trigonometric ratio. With the angle measured from the horizontal, the adjacent horizontal component uses cosine and the opposite vertical component uses sine."}]},{"id":"a140f0c6-26ec-4d2b-8c5b-1eecdee403ae","cardType":"basic","points":2,"front":"Why is the change in momentum $\\Delta p$ usually larger when an object rebounds compared to when it stops?","back":"Velocity changes direction, so one value is negative; subtracting a negative results in a larger vector difference.","mcqOptions":[{"text":"Velocity changes direction, so one value is negative; subtracting a negative results in a larger vector difference.","correct":true,"explanation":"On rebounding, the final velocity has the opposite sign to the initial velocity. Thus $\\Delta p=p_f-p_i$ involves subtracting oppositely directed momenta, producing a larger magnitude."},{"text":"Velocity increases in magnitude, so both values are positive; adding two positives results in a larger scalar difference.","correct":false,"explanation":"A rebound need not increase the object’s speed; its reversed direction is the key factor. Opposite signs in $\\Delta p=p_f-p_i$ produce the larger vector difference."},{"text":"Mass changes direction, so one value is negative; subtracting a negative results in a larger scalar difference.","correct":false,"explanation":"Mass normally remains constant during the collision and does not reverse direction. The velocity reverses, giving initial and final momenta opposite signs and therefore a larger vector difference."},{"text":"Velocity changes direction, so one value is negative; subtracting a negative results in a smaller vector difference.","correct":false,"explanation":"Subtracting a negative increases rather than decreases the resulting magnitude. When velocity reverses, the oppositely signed initial and final momenta give a larger vector difference."}]},{"id":"d4ba6bcd-5bea-4051-a2f5-9496b0097cb4","cardType":"cloze","points":null,"front":"Impulse is equal to the change in [ ] ($J = \\Delta p$).","back":"Impulse is equal to the change in momentum ($J = \\Delta p$).","mcqOptions":[{"text":"momentum","correct":true,"explanation":"The impulse-momentum theorem states that impulse equals the change in momentum: $J=\\Delta p$. Equivalently, impulse is the resultant force integrated over the interaction time."},{"text":"velocity","correct":false,"explanation":"Impulse can cause velocity to change, but $J$ is not generally equal to $\\Delta v$ because mass must be included. It equals the change in momentum, $\\Delta p$."},{"text":"force","correct":false,"explanation":"Force is related to the rate of change of momentum, not directly equated to impulse. Impulse is force multiplied by time and equals the change in momentum."},{"text":"kinetic energy","correct":false,"explanation":"Kinetic energy is scalar and its change is determined by work done, not impulse alone. Impulse is a vector quantity equal to the change in momentum."}]},{"id":"0deee2e2-3d62-4cfc-b8bf-f600a67e9dba","cardType":"basic","points":1,"front":"State the relationship between resultant force, change in momentum, and time.","back":"$$F_{\\text{resultant}} = \\frac{\\Delta p}{\\Delta t}$","mcqOptions":[{"text":"$$F_{\\text{resultant}} = \\frac{\\Delta p}{\\Delta t}$","correct":true,"explanation":"Newton’s second law states that resultant force is the rate of change of momentum. Over a time interval, the average resultant force is therefore $\\Delta p/\\Delta t$."},{"text":"$$F_{\\text{resultant}} = \\Delta p\\Delta t$","correct":false,"explanation":"Multiplying momentum change by time reverses the required rate relationship and gives incorrect units. Resultant force is momentum change divided by the elapsed time."},{"text":"$$F_{\\text{resultant}} = \\frac{\\Delta t}{\\Delta p}$","correct":false,"explanation":"This inverts Newton’s second-law relationship and does not give force units. Resultant force is the rate of momentum change, so $F_{\\text{resultant}}=\\Delta p/\\Delta t$."},{"text":"$$F_{\\text{resultant}} = \\frac{\\Delta p}{(\\Delta t)^2}$","correct":false,"explanation":"Dividing momentum change by squared time introduces an extra time factor and corresponds to neither force nor impulse. Force is obtained by dividing $\\Delta p$ by $\\Delta t$ once."}]},{"id":"5ddb20e3-58f2-478f-b45b-f603dca02c3a","cardType":"basic","points":2,"front":"Explain how safety features like mats or helmets reduce the risk of injury.","back":"They increase the contact time $\\Delta t$ for a given $\\Delta p$, which reduces the average force.","mcqOptions":[{"text":"They increase the contact time $\\Delta t$ for a given $\\Delta p$, which reduces the average force.","correct":true,"explanation":"Impulse satisfies $\\Delta p=F_{\\text{average}}\\Delta t$. For the same momentum change, extending the collision time makes the average force smaller and reduces injury risk."},{"text":"They decrease the contact time $\\Delta t$ for a given $\\Delta p$, which reduces the average force.","correct":false,"explanation":"Reducing contact time would increase, not reduce, the average force for a fixed momentum change. Safety features extend the contact time so the same impulse is delivered less forcefully."},{"text":"They increase the contact time $\\Delta t$ for a given $\\Delta p$, which increases the average force.","correct":false,"explanation":"For fixed $\\Delta p$, average force is inversely proportional to contact time. Increasing the contact time therefore decreases the force rather than increasing it."},{"text":"They decrease the contact time $\\Delta t$ for a given $\\Delta p$, which increases the average force.","correct":false,"explanation":"Safety features do not protect mainly by shortening an impact while increasing force. They lengthen the contact time, reducing the average force needed for the same momentum change."}]},{"id":"6132d8f5-df08-406a-8aca-331322ae25ae","cardType":"basic","points":1,"front":"What physical quantity is represented by the area under a force-time graph?","back":"Impulse (or change in momentum).","mcqOptions":[{"text":"Impulse (or change in momentum).","correct":true,"explanation":"The area under a force-time graph equals the integral of force over time, which is impulse. By the impulse-momentum theorem, impulse equals the change in momentum."},{"text":"Work done (or change in energy).","correct":false,"explanation":"Work is found from the area under a force-distance graph, not a force-time graph. The force-time area gives impulse, which equals the change in momentum."},{"text":"Power (or rate of energy transfer).","correct":false,"explanation":"Power describes the rate of energy transfer and is not obtained from the area under a force-time graph. That area represents impulse, or change in momentum."},{"text":"Force (or rate of change of momentum).","correct":false,"explanation":"Force is represented by the height of the graph at a particular time, not its area. The area accumulates force over time to give impulse."}]},{"id":"de7a803a-d6bc-4f18-b046-e04b2128316f","cardType":"cloze","points":null,"front":"During a typical collision, the force is not constant; it [ ] and then falls back to zero.","back":"During a typical collision, the force is not constant; it rises to a peak and then falls back to zero.","mcqOptions":[{"text":"rises to a peak","correct":true,"explanation":"During a typical collision, the objects deform and the contact force increases, reaching a maximum before the collision ends and the force returns to zero."},{"text":"falls to a minimum","correct":false,"explanation":"A collision force usually increases initially as the objects compress, rather than falling immediately. The correct description is that it rises to a peak before returning to zero."},{"text":"remains at a constant value","correct":false,"explanation":"The force during a real collision changes as compression and rebound occur, so it is not constant. It typically rises to a peak and then falls back to zero."},{"text":"reverses its direction repeatedly","correct":false,"explanation":"The contact force may change in magnitude but does not normally reverse direction repeatedly during one collision. Its typical profile rises to a peak and then falls back to zero."}]},{"id":"e2086038-975a-4ab8-a1c6-45004e6868d1","cardType":"cloze","points":null,"front":"During a typical collision, the force is not constant; it rises to a peak and then [ ].","back":"During a typical collision, the force is not constant; it rises to a peak and then falls back to zero.","mcqOptions":[{"text":"falls back to zero","correct":true,"explanation":"Once the objects separate or stop exerting contact force on each other, the collision force decreases back to zero after reaching its maximum value."},{"text":"continues rising indefinitely","correct":false,"explanation":"The force cannot continue rising indefinitely in a finite collision; the objects eventually stop compressing and separate. The force rises to a peak and then falls back to zero."},{"text":"remains at its peak value","correct":false,"explanation":"A peak is the maximum force, not a constant level maintained after the collision. The force subsequently decreases and falls back to zero as contact ends."},{"text":"becomes equal to the initial force","correct":false,"explanation":"The force does not generally return to its initial value after the peak; when contact ends, it returns to zero. The complete pattern is a rise followed by a fall to zero."}]},{"id":"8876bcac-32b4-46fe-a295-ad0358717672","cardType":"basic","points":1,"front":"State the principle of conservation of momentum.","back":"Total momentum remains constant provided no resultant external force acts on the system.","mcqOptions":[{"text":"Total momentum remains constant provided no resultant external force acts on the system.","correct":true,"explanation":"Momentum is conserved for a system when the resultant external force is zero. Internal forces can transfer momentum between objects but cannot change the system total."},{"text":"Total momentum remains constant provided no internal force acts within the system.","correct":false,"explanation":"Internal forces are allowed and commonly occur during collisions; they cancel in equal and opposite pairs. Conservation requires no resultant external force on the system."},{"text":"Total momentum remains constant provided a resultant external force acts on the system.","correct":false,"explanation":"A non-zero resultant external force changes the system’s total momentum according to force equals rate of change of momentum. Conservation requires no resultant external force."},{"text":"Total momentum remains constant provided the kinetic energy is conserved in the system.","correct":false,"explanation":"Momentum conservation depends on the resultant external force, not on whether kinetic energy is conserved. Inelastic collisions can conserve momentum while losing kinetic energy."}]},{"id":"22a3033f-452c-4421-8e37-51d323b15d3f","cardType":"basic","points":2,"front":"Why do internal forces between colliding objects not change the total momentum of a system?","back":"They produce equal and opposite impulses that cancel out within the system.","mcqOptions":[{"text":"They produce equal and opposite impulses that cancel out within the system.","correct":true,"explanation":"Newton’s third law gives equal and opposite interaction forces, acting for the same time. Therefore the impulses are equal and opposite, so their net effect on total momentum is zero."},{"text":"They produce equal and opposite forces that cancel out within each object.","correct":false,"explanation":"The forces act on different objects, so they do not cancel within each individual object. Their equal and opposite impulses cancel only when considering the whole system."},{"text":"They produce unequal and opposite impulses that cancel out within the system.","correct":false,"explanation":"Newton’s third law requires the interaction forces, and hence their impulses over the same time, to be equal and opposite. Unequal impulses would change the system momentum."},{"text":"They produce equal and opposite impulses that increase the total momentum of the system.","correct":false,"explanation":"Equal and opposite impulses have a zero vector sum, so they transfer momentum internally without increasing the system total. Only a resultant external impulse changes total momentum."}]},{"id":"a6b40dfc-b8b3-4391-97b7-b48b5df9501c","cardType":"basic","points":2,"front":"Distinguish between elastic and inelastic collisions regarding kinetic energy.","back":"

Elastic: kinetic energy is conserved.
Inelastic: kinetic energy is transferred to thermal/sound/deformation.

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

Elastic: kinetic energy is conserved.
Inelastic: kinetic energy is transferred to thermal/sound/deformation.

","correct":true,"explanation":"In an elastic collision, total kinetic energy remains unchanged. In an inelastic collision, some kinetic energy is transferred into internal energy, sound, or deformation."},{"text":"

Elastic: kinetic energy is transferred to thermal/sound/deformation.
Inelastic: kinetic energy is conserved.

","correct":false,"explanation":"This reverses the definitions: elastic collisions conserve kinetic energy, whereas inelastic collisions transfer some kinetic energy into thermal energy, sound, or deformation."},{"text":"

Elastic: momentum is not conserved.
Inelastic: momentum is conserved only if objects stick together.

","correct":false,"explanation":"Both elastic and inelastic collisions conserve total momentum when no resultant external force acts. The distinction concerns kinetic energy, not whether the objects stick together."},{"text":"

Elastic: kinetic energy increases.
Inelastic: kinetic energy remains unchanged.

","correct":false,"explanation":"An isolated collision cannot create extra total kinetic energy in an elastic collision, and an inelastic collision does not leave it unchanged. Elastic conserves it; inelastic transfers some away."}]},{"id":"a3f93776-342e-4e7f-802f-be818f4197fe","cardType":"cloze","points":null,"front":"In a two-dimensional collision, momentum components must be conserved separately in the [ ] and the y-direction.","back":"In a two-dimensional collision, momentum components must be conserved separately in the x-direction and the y-direction.","mcqOptions":[{"text":"x-direction","correct":true,"explanation":"Momentum is a vector, so in two dimensions its perpendicular components are treated independently. Conservation must therefore be applied separately in the x-direction and y-direction."},{"text":"z-direction","correct":false,"explanation":"For the stated two-dimensional setup, the relevant perpendicular components are x and y. The missing term is x-direction, not a third z-direction component."},{"text":"radial direction","correct":false,"explanation":"Radial and tangential components may be useful in circular motion, but they are not the standard rectangular components for this collision. The correct component is the x-direction."},{"text":"resultant direction","correct":false,"explanation":"Momentum conservation is applied to separate perpendicular components rather than only along the single resultant direction. The sentence requires the x-direction before the y-direction."}]},{"id":"af7fafdb-c53c-4d02-8e45-48a97a54b4a3","cardType":"cloze","points":null,"front":"In a two-dimensional collision, momentum components must be conserved separately in the x-direction and the [ ].","back":"In a two-dimensional collision, momentum components must be conserved separately in the x-direction and the y-direction.","mcqOptions":[{"text":"y-direction","correct":true,"explanation":"In two-dimensional motion, the momentum vector is resolved into perpendicular x and y components. Conservation is checked independently in each, so the second component is the y-direction."},{"text":"z-direction","correct":false,"explanation":"A two-dimensional collision is resolved using x and y components, with no third z component needed. The correct missing direction is the y-direction."},{"text":"horizontal direction","correct":false,"explanation":"Horizontal direction may correspond to the x-axis, but the sentence already specifies x-direction. The second independent component is the y-direction."},{"text":"resultant direction","correct":false,"explanation":"The resultant direction combines the components rather than providing the second independent component. Momentum must be conserved separately in the x-direction and y-direction."}]},{"id":"39e0ebd8-44c3-42a7-b429-82fefd368d5d","cardType":"basic","points":2,"front":"If a stationary object is hit by a projectile moving purely in the x-direction, what must be true about the total y-momentum after the collision?","back":"The total y-momentum must be zero (the vertical components of the fragments cancel out).","mcqOptions":[{"text":"The total y-momentum must be zero (the vertical components of the fragments cancel out).","correct":true,"explanation":"The initial projectile has no y-component of momentum, so conservation of momentum requires the fragments’ vertical momentum components to add to zero after the collision."},{"text":"The total y-momentum must equal the projectile’s initial x-momentum (the fragments transfer momentum between directions).","correct":false,"explanation":"Momentum is conserved separately in perpendicular directions, so x-momentum cannot become y-momentum. The initial y-momentum is zero, making the final total y-momentum zero."},{"text":"The total y-momentum must be equal to the object’s weight (the collision creates vertical momentum).","correct":false,"explanation":"Weight is a force, not a momentum, and the short collision does not create a net y-momentum in an isolated system. The final total y-momentum must be zero."},{"text":"The total y-momentum must be opposite to the projectile’s x-momentum (the fragments recoil vertically).","correct":false,"explanation":"Perpendicular momentum components are not opposite versions of one another. Since the initial y-component is zero, the fragments’ vertical components must cancel and give zero total y-momentum."}]},{"id":"9f44b5d8-3ff0-4cd5-b264-c3fe4817823a","cardType":"cloze","points":null,"front":"Conservation of momentum for an isolated system in the x-axis is expressed as $$\\text{[...]}$$.","back":"Conservation of momentum for an isolated system in the x-axis is expressed as $$\\boxed{\\sum p_{x,\\text{before}} = \\sum p_{x,\\text{after}}}$$.","mcqOptions":[{"text":"$$$\\sum p_{x,\\text{before}} = \\sum p_{x,\\text{after}}$$","correct":true,"explanation":"For an isolated system, no external impulse changes the x-component of momentum, so the total x-momentum before the event equals the total x-momentum afterwards."},{"text":"$$$\\sum p_{x,\\text{before}} = 0$$","correct":false,"explanation":"This incorrectly assumes the system starts with no x-momentum. Conservation requires the before and after totals to be equal, even when the initial x-momentum is non-zero."},{"text":"$$$\\sum p_{x,\\text{before}} + \\sum p_{x,\\text{after}} = 0$$","correct":false,"explanation":"Adding the two totals would incorrectly treat the final momentum as opposite to the initial momentum. Conservation states that the two totals are equal, not that their sum is zero."},{"text":"$$$\\sum p_{x,\\text{before}} = \\sum p_{y,\\text{after}}$$","correct":false,"explanation":"This compares momentum components in different directions, which are conserved independently. The x-momentum before must equal the x-momentum after, not the y-momentum."}]},{"id":"ad4e9e48-b0ff-4c64-81bd-947145e9431d","cardType":"basic","points":2,"front":"Why can external forces like gravity often be ignored during high-speed impact calculations?","back":"Impact forces are significantly larger than weight over the very short duration of the collision.","mcqOptions":[{"text":"Impact forces are significantly larger than weight over the very short duration of the collision.","correct":true,"explanation":"During a high-speed impact, the momentum changes over a very short time, producing a large impact force. Gravity acts throughout but contributes only a small impulse during that interval."},{"text":"Impact forces are significantly smaller than weight over the very short duration of the collision.","correct":false,"explanation":"A rapid change in momentum produces a large force, so impact forces are usually much greater than weight. Treating them as smaller reverses the relevant force comparison."},{"text":"Impact forces are approximately equal to weight over the very short duration of the collision.","correct":false,"explanation":"The collision time is too short for gravity to produce a significant impulse compared with the impact. Therefore the impact force is generally far larger than the object’s weight."},{"text":"Impact forces are unrelated to weight over the very short duration of the collision.","correct":false,"explanation":"Although impact and weight arise from different causes, comparing their magnitudes explains why gravity can be neglected. Impact forces are typically significantly larger than weight during the collision."}]},{"id":"8b210964-aa41-48ca-9649-6ec75d9b8780","cardType":"basic","points":1,"front":"What is the common mistake when calculating impulse for an object that changes direction?","back":"Subtracting speeds instead of using velocities with correct positive/negative signs.","mcqOptions":[{"text":"Subtracting speeds instead of using velocities with correct positive/negative signs.","correct":true,"explanation":"Impulse equals the change in momentum, $m(v-u)$, so velocity signs must represent direction. Using speeds alone can make a reversal appear to produce too small a change."},{"text":"Adding masses instead of using velocities with correct positive/negative signs.","correct":false,"explanation":"Mass is not the main issue when an object reverses direction. The common error is subtracting speeds; velocities must carry positive and negative signs so the change in momentum is correct."},{"text":"Multiplying speeds instead of using velocities with correct positive/negative signs.","correct":false,"explanation":"Impulse is based on the difference between final and initial momentum, not a product of speeds. Direction must be included through signed velocities, especially when the object reverses."},{"text":"Ignoring mass instead of using velocities with correct positive/negative signs.","correct":false,"explanation":"Mass is required because impulse is the change in momentum, $m\\Delta v$. However, the specific direction mistake is treating opposite velocities as speeds and losing their signs."}]},{"id":"1f2e8ca6-a08b-45e7-ad88-713569c28b83","cardType":"basic","points":1,"front":"A 0.1 kg ball at 10 m s⁻¹ hits a wall and stops. Calculate the impulse.","back":"$$1.0\\ \\text{N s}$ (or kg m s⁻¹).","mcqOptions":[{"text":"$$1.0\\ \\text{N s}$ (or kg m s⁻¹).","correct":true,"explanation":"The impulse magnitude is the change in momentum: $I=m|v-u|=0.1\\times|0-10|=1.0\\ \\text{N s}$. Its direction is opposite to the ball’s initial motion."},{"text":"$$0.1\\ \\text{N s}$ (or kg m s⁻¹).","correct":false,"explanation":"This uses the mass alone and omits the velocity change. Multiplying the mass by the change from $10$ to $0\\ \\text{m s⁻¹}$ gives the correct impulse magnitude of $1.0\\ \\text{N s}$."},{"text":"$$10\\ \\text{N s}$ (or kg m s⁻¹).","correct":false,"explanation":"This effectively divides by the mass or ignores the factor of $0.1\\ \\text{kg}$. The correct magnitude is $0.1\\times10=1.0\\ \\text{N s}$."},{"text":"$$2.0\\ \\text{N s}$ (or kg m s⁻¹).","correct":false,"explanation":"Doubling the speed change would incorrectly treat stopping as a reversal. The ball changes from $10$ to $0\\ \\text{m s⁻¹}$, so the impulse magnitude is $1.0\\ \\text{N s}$."}]},{"id":"f8e4fa44-ece2-4237-b252-feb9e6005ad3","cardType":"cloze","points":null,"front":"To find the velocity of an object after a 2D collision, you divide its final [ ] by its mass.","back":"To find the velocity of an object after a 2D collision, you divide its final momentum components by its mass.","mcqOptions":[{"text":"momentum components","correct":true,"explanation":"Momentum is a vector, so in two dimensions the final momentum is resolved into perpendicular components. Dividing each component by the mass gives the corresponding velocity component."},{"text":"momentum magnitude","correct":false,"explanation":"Using only the momentum magnitude loses the direction information needed in a two-dimensional collision. The velocity components are found by dividing the corresponding momentum components by mass."},{"text":"force components","correct":false,"explanation":"Force components describe the rate of change of momentum, not the momentum itself. After finding final momentum, divide its x- and y-components by mass to obtain velocity components."},{"text":"impulse components","correct":false,"explanation":"Impulse gives the change in momentum, not necessarily the object’s final momentum unless the initial momentum is also included. Final velocity comes from final momentum components divided by mass."}]},{"id":"7d7b7ae0-737c-4d81-bb8a-7fa912a9b8c6","cardType":"cloze","points":null,"front":"To find the velocity of an object after a 2D collision, you divide its final momentum components by its [ ].","back":"To find the velocity of an object after a 2D collision, you divide its final momentum components by its mass.","mcqOptions":[{"text":"mass","correct":true,"explanation":"The momentum relation is $\\mathbf{p}=m\\mathbf{v}$, so rearranging gives $\\mathbf{v}=\\mathbf{p}/m$. This applies separately to the x- and y-components in a two-dimensional collision."},{"text":"force","correct":false,"explanation":"Dividing momentum by force does not give velocity; it produces a time quantity in this context. Momentum components must be divided by mass because $p=mv$."},{"text":"impulse","correct":false,"explanation":"Impulse is the change in momentum, not the proportionality factor connecting momentum and velocity. The required divisor is mass, as shown by $p=mv$."},{"text":"acceleration","correct":false,"explanation":"Acceleration describes how velocity changes with time and is not the quantity relating momentum directly to velocity. Dividing momentum components by mass gives the velocity components."}]}]}],"examCountdown":{"examDate":"2027-05-12","paperLabel":"Paper 1"},"heading":"Impulse and two-dimensional momentum","title":"Impulse and two-dimensional momentum Flashcards","topbar":{"backHref":"/as-level/physics/edexcel/flashcards"},"topicSidebar":"$L38"}]}],"topicId":"ac9b93e4-ef33-50e7-98b8-e8157e6d39ba","walled":"$L39"}],"$L3a","$L3b"]