AP Physics 1 50 Flashcards Intermediate 100% Free

AP Physics 1:: Momentum

Created by Chat Robotics Community  ·  Updated 2026-08-31

Curriculum Overview

Comprehensive, high-yield AP Physics 1 study deck focusing on Momentum. Features 50 rigorous, curriculum-aligned flashcards designed for intermediate-level mastery. Core concepts covered include Momentum, key problem-solving heuristics, foundational formulas, and exam-tested application scenarios. Ideal for active recall review, spaced repetition study, and scoring in the top percentile.

Topics & Key Concepts

MASS This ZERO Since TOTAL LINEAR VECTOR Physics MOMENTUM Momentum

Sample Flashcard Questions & Answers

Showing 8 of 50 cards
Question #1 Active Recall

What is LINEAR MOMENTUM, and what is the mathematical FORMULA relating an object's momentum to its MASS and VELOCITY?

- **A)** Momentum is a SCALAR quantity with no associated direction, unlike velocity which is a vector
- **B)** Momentum is calculated as p = m/v (mass divided by velocity), rather than mass multiplied by velocity
- **C)** Momentum is a VECTOR quantity representing an object's 'quantity of motion'; p = m*v (mass multiplied by velocity) -- momentum points in the SAME direction as the object's velocity
- **D)** This concept has no actual mathematical relationship between an object's momentum, its mass, and its velocity

Answer & Explanation:
**Answer: C)**

p = mv is the foundational momentum equation for this entire unit -- as a VECTOR quantity, momentum inherits its direction directly from velocity's direction, which will matter greatly for collision problems.
Question #2 Active Recall

What is the SI (metric) UNIT for LINEAR MOMENTUM, and how does this unit relate to the base units of mass and velocity?

- **A)** Kilogram-meters-per-second (kg*m/s) -- directly reflecting momentum's definition as mass (kg) multiplied by velocity (m/s)
- **B)** Momentum is measured in NEWTONS, identical to the unit for force, with no distinct unit of its own
- **C)** The SI unit for momentum is the JOULE, rather than kg*m/s
- **D)** This concept has no actual relationship between the SI unit for momentum and the base units of mass and velocity

Answer & Explanation:
**Answer: A)**

The kg*m/s unit directly reflects the p=mv relationship, distinguishing momentum's units from both force (newtons) and energy (joules).
Question #3 Active Recall

What is IMPULSE, and what is the mathematical FORMULA relating impulse to a FORCE and the TIME INTERVAL over which that force acts?

- **A)** This concept has no actual mathematical relationship between impulse, an applied force, and the time interval over which it acts
- **B)** Impulse is the PRODUCT of a force and the time interval over which it acts; J = F*delta_t -- like momentum, impulse is a VECTOR quantity, pointing in the same direction as the applied (net) force
- **C)** Impulse is calculated as J = F/delta_t (force divided by time interval), rather than force multiplied by time interval
- **D)** Impulse depends only on the TIME INTERVAL, with the applied force's magnitude playing no actual role in calculating impulse

Answer & Explanation:
**Answer: B)**

J = F*delta_t is the foundational impulse formula, directly setting up the crucial impulse-momentum theorem that connects force/time to momentum change.
Question #4 Active Recall

What is the IMPULSE-MOMENTUM THEOREM, and what does it state about the relationship between the IMPULSE delivered to an object and the resulting CHANGE in that object's momentum?

- **A)** This theorem has no actual mathematical relationship between impulse delivered to an object and its resulting change in momentum
- **B)** Impulse delivered to an object is always ZERO regardless of how much its momentum actually changes, contradicting the impulse-momentum theorem
- **C)** The impulse-momentum theorem states that impulse always equals an object's TOTAL kinetic energy, rather than its change in momentum
- **D)** The impulse-momentum theorem states that the impulse delivered to an object EQUALS the CHANGE in that object's momentum: J = delta_p = m*delta_v -- this follows directly from combining J=F*delta_t with Newton's second law, F=ma

Answer & Explanation:
**Answer: D)**

The impulse-momentum theorem (J = delta_p) is a powerful problem-solving tool, especially useful for collisions and impacts where force varies rapidly over a very short time interval.
Question #5 Active Recall

How can the IMPULSE-MOMENTUM THEOREM explain WHY extending the TIME of impact (e.g., bending your knees when landing a jump, or a car's crumple zone during a collision) REDUCES the AVERAGE FORCE experienced during that impact?

- **A)** The impulse-momentum theorem states that force and time interval are completely INDEPENDENT of each other for a given momentum change
- **B)** This concept has no actual relationship between the duration of an impact and the average force experienced during that impact
- **C)** Extending the time of impact INCREASES the average force experienced, the reverse of the actual relationship described by the impulse-momentum theorem
- **D)** Since impulse (J = F*delta_t) must equal a FIXED momentum change (delta_p) for a given collision, INCREASING the time interval (delta_t) over which that impulse is delivered REQUIRES a correspondingly SMALLER average force (F) to produce the same impulse

Answer & Explanation:
**Answer: D)**

This inverse force-time relationship (for a fixed impulse) is one of the most important practical applications of this unit, explaining safety features like airbags, crumple zones, and padded landings.
Question #6 Active Recall

What is the LAW OF CONSERVATION OF MOMENTUM, and under what specific CONDITION does the TOTAL momentum of a system of objects remain CONSTANT?

- **A)** Conservation of momentum requires that EACH INDIVIDUAL object's momentum remains SEPARATELY constant, rather than the TOTAL momentum of the system remaining constant
- **B)** Momentum is conserved under ALL circumstances, even when significant NET EXTERNAL forces act on a system
- **C)** The law of conservation of momentum states that the TOTAL momentum of a system remains CONSTANT, provided NO NET EXTERNAL FORCE acts on that system (i.e., the system is 'closed' or 'isolated') -- momentum can be TRANSFERRED between objects WITHIN the system, but the TOTAL remains unchanged
- **D)** This law has no actual relationship to a specific condition (absence of net external force) required for a system's total momentum to be conserved

Answer & Explanation:
**Answer: C)**

The no-net-external-force condition for momentum conservation is THE essential requirement underlying this unit's most powerful problem-solving technique, directly analogous to the mechanical energy conservation principle from the previous unit.
Question #7 Active Recall

Why does momentum conservation apply to a system of TWO objects COLLIDING with each other, even though EACH INDIVIDUAL object experiences a large internal collision force during the impact?

- **A)** The collision forces the two objects exert on EACH OTHER are INTERNAL to the two-object system (equal and opposite, per Newton's third law) and therefore CANCEL OUT when considering the system's TOTAL momentum -- as long as no additional EXTERNAL force (like friction from outside the system) acts, total momentum is conserved
- **B)** This concept has no actual relationship between Newton's third law (internal, equal-and-opposite forces) and why total momentum is conserved during a collision
- **C)** Momentum conservation does NOT actually apply during collisions, since the internal collision forces are too large to ignore
- **D)** Momentum conservation during a collision requires that BOTH objects have IDENTICAL mass, with no other condition being relevant

Answer & Explanation:
**Answer: A)**

This directly connects momentum conservation to Newton's third law (from the earlier Dynamics unit) -- internal action-reaction pairs always cancel when summing the total momentum of a closed system, which is exactly why momentum conservation works for collisions.
Question #8 Active Recall

What is a PERFECTLY ELASTIC COLLISION, and what TWO quantities are BOTH conserved during such a collision?

- **A)** A perfectly elastic collision conserves kinetic energy but NOT momentum, the reverse of the correct definition
- **B)** A perfectly elastic collision conserves momentum but NOT kinetic energy, with kinetic energy always decreasing significantly
- **C)** A collision in which BOTH total momentum AND total kinetic energy are conserved (unchanged before and after the collision) -- objects bounce off each other without any permanent deformation or energy loss to heat/sound
- **D)** This concept has no actual relationship between a collision being 'elastic' and which quantities remain conserved during that collision

Answer & Explanation:
**Answer: C)**

The perfectly-elastic-collision definition (both momentum AND kinetic energy conserved) is a crucial special case, contrasted with the more common perfectly inelastic collision discussed next.

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