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United StatesPhysics C: MechanicsSyllabus dot point

What is linear momentum as a vector, and how does it relate to a system's mass and velocity and to Newton's second law?

Topic 4.1 Linear Momentum: define linear momentum as the product of mass and velocity, treat it as a vector, and relate the net force to its rate of change.

A focused answer to AP Physics C: Mechanics Topic 4.1, covering linear momentum as a vector equal to mass times velocity, the momentum of a system as the sum of its parts, the relation between momentum and the center-of-mass velocity, and Newton's second law as the rate of change of momentum, with worked examples.

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  1. What this topic is asking
  2. Defining momentum
  3. Momentum of a system
  4. Newton's second law as the rate of change of momentum
  5. Try this

What this topic is asking

The College Board (Topic 4.1) wants you to define linear momentum as the product of mass and velocity, to treat it as a vector, and to relate the net force to its rate of change. Momentum is the quantity that conservation laws and collision analysis are built on, and recognizing that it is a vector, with direction as well as magnitude, is essential to everything in Unit 4.

Defining momentum

Momentum measures the "quantity of motion": a heavy, fast object has large momentum and is hard to stop. Because it is a vector, two objects moving in opposite directions at equal speeds have momenta that cancel when summed. This vector character is what distinguishes momentum problems from energy problems: kinetic energy is a scalar and adds as numbers, while momentum must be added by components, respecting direction. Note also that momentum is linear in speed, so doubling the speed doubles the momentum (but quadruples the kinetic energy).

Momentum of a system

For a system of several objects, the total momentum is the vector sum of the individual momenta:

pβƒ—sys=βˆ‘imivβƒ—i.\vec{p}_{sys} = \sum_i m_i \vec{v}_i.

A beautiful result connects this to the center of mass: since vβƒ—cm=βˆ‘mivβƒ—iM\vec{v}_{cm} = \dfrac{\sum m_i\vec{v}_i}{M}, the total momentum equals the total mass times the center-of-mass velocity,

p⃗sys=Mv⃗cm.\vec{p}_{sys} = M\vec{v}_{cm}.

So the whole system moves, momentum-wise, as if its entire mass were concentrated at the center of mass. This is why momentum and the center-of-mass concept from Unit 2 are two views of the same physics, and it is the foundation of momentum conservation.

Newton's second law as the rate of change of momentum

The deepest statement of Newton's second law is in terms of momentum:

F⃗net=dp⃗dt.\vec{F}_{net} = \frac{d\vec{p}}{dt}.

The net external force equals the time rate of change of the momentum. When the mass is constant this is F⃗net=mdv⃗dt=ma⃗\vec{F}_{net} = m\dfrac{d\vec{v}}{dt} = m\vec{a}, the familiar form. But the momentum form is more general: it correctly handles systems whose mass changes (a rocket ejecting fuel, a cart gathering sand) where ma⃗m\vec{a} alone would be wrong. It also leads directly to the impulse-momentum theorem in the next topic, since integrating the force over time gives the change in momentum.

Try this

Q1. A 0.150.15 kg baseball moves at 4040 m/s. Calculate its momentum. [2 points]

  • Cue. p=mv=(0.15)(40)=6.0p = mv = (0.15)(40) = 6.0 kg m/s in the direction of motion.

Q2. State the general form of Newton's second law in terms of momentum and explain when it differs from F⃗=ma⃗\vec{F} = m\vec{a}. [2 points]

  • Cue. Fβƒ—net=dpβƒ—/dt\vec{F}_{net} = d\vec{p}/dt; it differs from maβƒ—m\vec{a} when the mass changes, such as a rocket losing fuel.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (short FRQ). A 2.02.0 kg cart moves east at 3.03.0 m/s and a 1.01.0 kg cart moves north at 4.04.0 m/s. (a) Determine the momentum of each cart as a vector. (b) Determine the total momentum of the two-cart system (magnitude and direction). (c) Determine the velocity of the center of mass of the system.
Show worked answer β†’

A 4-point FRQ on momentum as a vector.

(a) Individual momenta (1 point): pβƒ—1=m1vβƒ—1=(2.0)(3.0) ı^=6.0 ı^\vec{p}_1 = m_1\vec{v}_1 = (2.0)(3.0)\,\hat{\imath} = 6.0\,\hat{\imath} kg m/s; pβƒ—2=(1.0)(4.0) ȷ^=4.0 ȷ^\vec{p}_2 = (1.0)(4.0)\,\hat{\jmath} = 4.0\,\hat{\jmath} kg m/s.
(b) Total momentum (2 points): pβƒ—=6.0 ı^+4.0 ȷ^\vec{p} = 6.0\,\hat{\imath} + 4.0\,\hat{\jmath}; magnitude 6.02+4.02=7.2\sqrt{6.0^2 + 4.0^2} = 7.2 kg m/s, at tanβ‘βˆ’1(4.0/6.0)=34∘\tan^{-1}(4.0/6.0) = 34^\circ north of east.
(c) Center-of-mass velocity (1 point): vβƒ—cm=pβƒ—M=6.0 ı^+4.0 ȷ^3.0=2.0 ı^+1.33 ȷ^\vec{v}_{cm} = \dfrac{\vec{p}}{M} = \dfrac{6.0\,\hat{\imath} + 4.0\,\hat{\jmath}}{3.0} = 2.0\,\hat{\imath} + 1.33\,\hat{\jmath} m/s, magnitude 2.42.4 m/s.

Markers reward adding momenta as vectors (by components) and dividing by the total mass for the center-of-mass velocity.

AP 2021 (style)1 marksSection I (multiple choice). Object A has momentum of magnitude pp. Object B has twice the mass and half the speed of A. The magnitude of B's momentum is... (A) p/2p/2 (B) pp (C) 2p2p (D) 4p4p. Justify your reasoning.
Show worked answer β†’

A 1-point conceptual MCQ. The answer is (B).

Momentum is p=mvp = mv. For B, pB=(2m)(v/2)=mv=pp_B = (2m)(v/2) = mv = p, equal to A's. Doubling the mass and halving the speed exactly cancel for momentum (which is linear in both), unlike kinetic energy (which goes as v2v^2 and would differ). The trap is to confuse momentum's linear speed dependence with the squared dependence of energy.

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