{ "format": "recallweave-deck/1", "title": "Momentum and collisions: what stays, what changes", "attribution": "Original lesson by HAMON contributors, 2026. Conceptual references: OpenStax, University Physics Volume 1, sections 9.2–9.4 (https://openstax.org/books/university-physics-volume-1/pages/9-2-impulse-and-collisions); MIT OpenCourseWare 8.01SC, Chapter 15: Collision Theory (https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter15.pdf). Numerical examples, wording and transfer prompts are original.", "license": "Original question wording, explanations and transfer prompts: CC0-1.0. Referenced sources retain their own terms; no textbook figures or passages are reproduced.", "concepts": [ "Signed momentum", "System momentum", "Collision models", "Energy accounting", "Model boundaries" ], "items": [ { "id": "momentum-sign", "prompt": "Cart A has mass 2 kg and velocity -3 m/s, where right is positive. What is its signed momentum?", "options": [ "+6 kg m/s", "-6 kg m/s", "+9 kg m/s", "-9 kg m/s" ], "concept": "Signed momentum", "prerequisites": [], "answer": 1, "explanation": "Momentum is mass times signed velocity: p = 2 × (-3) = -6 kg m/s. The minus sign describes leftward momentum under the stated direction convention. Squaring the velocity belongs in kinetic energy, not momentum.", "transfer": "Reverse the chosen positive direction without changing the cart's physical motion. Explain what changes in its momentum and what stays the same in its kinetic energy." }, { "id": "zero-total-moving", "prompt": "A 1 kg cart moves at +4 m/s and a 2 kg cart moves at -2 m/s. What are their total signed momentum p and total translational kinetic energy K?", "options": [ "p = 0 kg m/s; K = 0 J", "p = +8 kg m/s; K = 12 J", "p = 0 kg m/s; K = 12 J", "p = 0 kg m/s; K = 8 J" ], "concept": "Signed momentum", "prerequisites": [], "answer": 2, "explanation": "The momenta are +4 and -4 kg m/s, so their sum is zero. Their kinetic energies are (1/2) × 1 × 4² = 8 J and (1/2) × 2 × (-2)² = 4 J. Kinetic energy is nonnegative for each positive mass, so the total is 12 J even though the total momentum is zero.", "transfer": "Invent a different pair of moving carts with zero total momentum. Check whether their total kinetic energy can be zero." }, { "id": "system-boundary", "prompt": "Treat both carts together as one fixed-mass system. The net external impulse during their interaction is zero. Which statement must hold?", "options": [ "Each cart keeps its individual momentum.", "The sum of the carts' signed momenta is unchanged.", "The sum of their translational kinetic energies is unchanged.", "Both carts must be stationary afterward." ], "concept": "System momentum", "prerequisites": [ "Signed momentum" ], "answer": 1, "explanation": "Internal interaction forces transfer momentum between the carts. With zero net external impulse on the combined fixed-mass system, those internal transfers do not change the total momentum between the initial and final states. Individual momenta can change, and translational kinetic energy need not be conserved.", "transfer": "Choose a system boundary for a cart striking a wall. Which interaction becomes external if the wall is left outside the system?" }, { "id": "stick-weighted-velocity", "prompt": "Cart A (2 kg, +3 m/s) approaches cart B (1 kg, 0 m/s) from the left. They undergo a completely inelastic collision with zero net external impulse. What common velocity do they have afterward?", "options": [ "+1 m/s", "+1.5 m/s", "+3 m/s", "+2 m/s" ], "concept": "Collision models", "prerequisites": [ "System momentum" ], "answer": 3, "explanation": "The total initial momentum is 2 × 3 + 1 × 0 = 6 kg m/s. Sticking requires one shared final velocity, so (2 + 1)v = 6 and v = +2 m/s. The ordinary average +1.5 m/s ignores the different masses.", "transfer": "Predict how the shared velocity changes if the initially stationary cart's mass increases while the incoming cart stays the same." }, { "id": "elastic-unequal", "prompt": "Cart A (2 kg, +3 m/s) approaches cart B (1 kg, 0 m/s) from the left. An ideal elastic collision conserves momentum and translational kinetic energy, and the carts separate afterward. Which pair of final velocities (vA, vB) is consistent with this event?", "options": [ "(+1, +4) m/s", "(+2, +2) m/s", "(-1, +8) m/s", "(+3, 0) m/s" ], "concept": "Collision models", "prerequisites": [ "System momentum" ], "answer": 0, "explanation": "For (+1, +4), the final momentum is 2 × 1 + 1 × 4 = 6 kg m/s and the kinetic energy is (1/2) × 2 × 1² + (1/2) × 1 × 4² = 9 J, matching the initial values. The relative velocity changes from +3 to -3 m/s, so the carts separate. The unchanged (+3, 0) pair also satisfies the two conservation equations, but it is still approaching and does not describe the stated separating collision event.", "transfer": "Why do two conservation equations still need an event/separation condition in this example? Explain the unchanged algebraic branch." }, { "id": "elastic-equal-masses", "prompt": "Two 1 kg carts approach each other: A, initially on the left, has uA = +4 m/s; B has uB = -2 m/s. After an ideal one-dimensional elastic collision, what are (vA, vB)?", "options": [ "(-4, +2) m/s", "(+1, +1) m/s", "(-2, +4) m/s", "(+4, -2) m/s" ], "concept": "Collision models", "prerequisites": [ "System momentum" ], "answer": 2, "explanation": "In an ideal one-dimensional elastic collision of equal masses, the velocities exchange. The final pair (-2, +4) retains momentum +2 kg m/s and kinetic energy 10 J. The relative velocity changes from +6 to -6 m/s. Simply negating both incoming velocities would reverse the total momentum here.", "transfer": "Choose two equal-mass incoming velocities whose sum is not zero. Show why swapping them preserves momentum but negating them does not." }, { "id": "inelastic-energy", "prompt": "Cart A (2 kg, +3 m/s) meets cart B (1 kg, 0 m/s). They stick and move together at +2 m/s with no net external impulse. How much translational kinetic energy is converted to other forms during this idealized event?", "options": [ "0 J", "6 J", "9 J", "3 J" ], "concept": "Energy accounting", "prerequisites": [ "Collision models" ], "answer": 3, "explanation": "The initial translational kinetic energy is (1/2) × 2 × 3² = 9 J. The combined 3 kg mass moving at +2 m/s has (1/2) × 3 × 2² = 6 J. The difference is 3 J. Total momentum remains +6 kg m/s, while that 3 J is converted to energy forms outside this endpoint model's translational accounting.", "transfer": "Repeat the initial and final kinetic-energy calculation for a different pair that sticks. State separately what momentum conservation tells you." }, { "id": "energy-not-destroyed", "prompt": "Two 1 kg carts approach at +3 m/s and -3 m/s. They stick and stop, reducing their total translational kinetic energy from 9 J to 0 J. Which interpretation is justified?", "options": [ "The 9 J of energy has been destroyed.", "Momentum conservation has failed because the carts stopped.", "The 9 J has changed into other energy forms; this endpoint model does not resolve their distribution.", "Exactly 9 J must have become sound, with no other energy change." ], "concept": "Energy accounting", "prerequisites": [ "Collision models" ], "answer": 2, "explanation": "The initial momenta +3 and -3 kg m/s cancel, so stopping together preserves their total momentum. Translational kinetic energy can change into internal motion, deformation and other energy forms without destroying total energy. The endpoint calculation alone cannot say how the 9 J is divided among those forms.", "transfer": "What extra observations would be needed to estimate how kinetic energy was distributed into deformation, internal energy or sound?" }, { "id": "approach-ordering", "prompt": "A begins to the left of B with a positive gap. Both keep their incoming velocities until contact. Which pair (uA, uB) makes the gap shrink so a future encounter is possible?", "options": [ "(0, 0) m/s", "(-3, -1) m/s", "(+1, +3) m/s", "(-1, -3) m/s" ], "concept": "Model boundaries", "prerequisites": [ "Signed momentum" ], "answer": 3, "explanation": "Let the positive gap be xB - xA. Its rate of change before contact is uB - uA. For (-1, -3), this is -3 - (-1) = -2 m/s, so the gap shrinks. B moves left faster than A and can catch it. Motion in the same direction does not by itself rule out a collision.", "transfer": "Choose two positive velocities and then two negative velocities that make the gap shrink. Use uB - uA to justify both." }, { "id": "equal-velocity-gap", "prompt": "A begins to the left of B with a positive gap. Both move at +2 m/s and have no acceleration before contact. What should a collision explorer report?", "options": [ "Their gap stays constant, so these inputs do not produce a future collision.", "They must immediately stick because their velocities are equal.", "They undergo an elastic collision that exchanges their masses.", "A collision occurs after one second for every positive gap." ], "concept": "Model boundaries", "prerequisites": [ "Signed momentum" ], "answer": 0, "explanation": "Equal incoming velocities give a zero gap-change rate. Starting from a positive gap, the carts remain separated under constant velocities. A completely inelastic formula describes what follows an encounter; it cannot create an encounter when these motion assumptions provide none.", "transfer": "Give one additional physical condition that could create a later encounter despite initially equal velocities. Explain which current assumption it would change." }, { "id": "external-impulse", "prompt": "The two-cart system initially has total momentum +6 kg m/s. During an interaction it receives a known net external impulse of +2 kg m/s. What follows?", "options": [ "Its final momentum is +6 kg m/s because internal forces cancel.", "Its final momentum is +4 kg m/s.", "Its final momentum is +8 kg m/s; the zero-external-impulse collision formulas do not apply unchanged.", "No final total momentum can be inferred even though the net impulse is given." ], "concept": "System momentum", "prerequisites": [ "Signed momentum" ], "answer": 2, "explanation": "The change in total momentum equals the net external impulse. Therefore Pfinal = +6 + (+2) = +8 kg m/s. Internal transfers still cancel in the combined accounting, but the external impulse changes the total. The zero-external-impulse formulas are not an unchanged solution to this different problem.", "transfer": "State what the known net impulse determines about the combined system and what additional information you would need for the two individual final velocities." }, { "id": "minimum-kinetic-energy", "prompt": "For two fixed positive masses and a fixed total momentum, why is the shared velocity of a completely inelastic endpoint the state with the least translational kinetic energy?", "options": [ "Any relative motion adds nonnegative kinetic energy above the center-of-mass motion, and this extra term is zero only at a shared velocity.", "A shared velocity always means both carts have zero velocity.", "Momentum conservation permits only one possible pair of final velocities.", "Kinetic energy is signed, so positive and negative contributions cancel at a shared velocity." ], "concept": "Energy accounting", "prerequisites": [ "Collision models" ], "answer": 0, "explanation": "Write M = mA + mB, P for the fixed total momentum, and μ = mA mB / M. The total translational kinetic energy is P²/(2M) + (μ/2)(vA - vB)². The first term is fixed. Because both masses are positive, the second is nonnegative and vanishes exactly when vA = vB = P/M. A shared velocity need not be zero.", "transfer": "Use the energy decomposition to explain why nonzero total momentum prevents the two-cart system from having zero total translational kinetic energy." } ] }