AP Physics C: Unit 4 and Unit 7: Linear Momentum and Oscillations

0.0(0)
studied byStudied by 1 person
learnLearn
examPractice Test
spaced repetitionSpaced Repetition
heart puzzleMatch
flashcardsFlashcards
Card Sorting

1/29

encourage image

There's no tags or description

Looks like no tags are added yet.

Study Analytics
Name
Mastery
Learn
Test
Matching
Spaced

No study sessions yet.

30 Terms

1
New cards

Momentum

Product of an object’s mass and velocity. Given by p = mv. The standard unit is the kg*m/s. It is a vector that always points in the same direction as the object’s velocity. Direction is very important for this. A force is required to change this, either by changing the object’s velocity’s magnitude or direction. ΣF = Δp/Δt is another way to write Newton’s Second Law. The actual Newton’s Second Law is given by ΣF = dp/dt.

2
New cards

Impulse

What occurs when a large force acts over a short time interval. Designated by the letter J and given by J = ΣFΔt, J = ∫ΣFdt, or J = Δp. The standard unit is the N*s. It is the same unit as for momentum, and it is a vector quantity that always points in the same direction as the net force. It is equal to the area under a force vs. time graph.

3
New cards

Center of Mass

Point representing the average location of the mass distribution of an object. Objects can be treated like their entire mass is concentrated here. For uniform, symmetrical objects, this is located at the geometric center. For non-uniform or non-symmetrical objects, this can be found experimentally by balancing the object. The dot in FBD’s represents this. For objects in projectile motion, it is this that follows a parabolic trajectory. The position, velocity, or acceleration of this for a system of objects is given by rCM = Σmiri / Σm, where r can be replaced with v for velocity or a for acceleration. Its position stays constant if there is no net external force, even if the objects are subject to internal forces. Additionally, if this is moving at a constant velocity, it will remain constant, even if the objects in the system are exposed to internal forces, like a collision or an explosion. Using integrals, this can also be defined as (∫r dm) / (∫dm).

4
New cards

Conservation of Momentum

Total momentum of a system of objects during an explosion or collision remains constant provided that no net external force acts on the system. It is given by pi = pf. If two objects collide but do not stick together, this can be broken down into m1v1i + m2v2i = m1v1f + m2v2f. If the two objects do stick together, this can be broken down into m1v1i + m2v2i = (m1 + m2)vf.

5
New cards

Recoil

The backwards jerk of a firearm after it is discharged. It is due to conservation of momentum.

6
New cards

(Perfectly) Elastic Collision

Collision in which the kinetic energy is conserved, so Ki = Kf. A collision between two magnets is an example.

7
New cards

Inelastic Collision

Collision in which kinetic energy is not conserved, so Ki ≠ Kf. Kinetic energy is either lost as some of it is converted into sound and thermal energy, or gained due to work being done on the system or potential energy being released into the system.

8
New cards

Perfectly inelastic collision

Collision in which the objects stick together and the maximum kinetic energy is lost.

9
New cards

Impulse-Momentum Theorem

The total momentum of a system of objects changes if an impulse acts on the system. It is given by ∑Ft = Δp. If an impulse acts in the same direction of an object’s momentum, the momentum will increase. If an impulse acts in the opposite direction of an object’s momentum, the momentum will either decrease or change direction.

10
New cards

Multi-Dimensional Collisions

Collisions that occur in two or three dimensions. The initial and final momentum of each object must be broken into components. The momentum must be conserved in each direction.

11
New cards

Periodic Motion

Any motion that repeats itself in a fixed time period.

12
New cards

Simple Harmonic Motion

Type of periodic motion where an object oscillates back and forth, exhibiting sinusoidal motion. In order for oscillating motion to be considered this, it must have a restoring force constantly directed towards equilibrium directly proportional to its displacement from equilibrium. For a mass-spring system, the restoring force is given by Fs = -kx. For a simple pendulum, this is given by mgθ. The acceleration of an object in this type of motion is given by d²x/dt² = -ω²x, where ω is the angular frequency given by 2π/T.

13
New cards

Equilibirum Position

Centerpoint of a simple harmonic oscillator.

14
New cards

Amplitude (A)

Maximum displacement from equilibrium position.

15
New cards

Period (T)

Time required for one full oscillation from the initial point back to the same point. Measured in seconds.

16
New cards

Frequency (f)

Number of cycles per second. Measured in inverse seconds, or hertz (Hz).

17
New cards

Angular Frequency (ω)

Number of degrees or radians covered per second. Measured in radians per second. It is given by ω = 2πf or ω = 2π/T.

18
New cards

Position of an object in simple harmonic motion

x = Acos(2πft) or x = Acos(ωt). The cosine function may change to a sine function if the object is at the equilibrium position at time t = 0.

19
New cards

Simple Pendulum

Small bob swinging back and forth on a string of negligible mass. The total mechanical energy E of the system at any point is the sum of the kinetic and gravitational energies, which is given by E = (1/2)mv² + mgh. At the highest points on either end, velocity is equal to zero and displacement is at a maximum, so the mechanical energy formula simplifies to E = mghmax. At the equilibrium point, the displacement is zero and the velocity is maximized, so the mechanical energy formula simplifies to E = (1/2)mv²max.

20
New cards

Period of a simple pendulum oscillating at small angles

Tp = 2π√(L/g), or fp = (1/2π)√(g/L). The mass of the pendulum does not matter for this. If the pendulum is on Earth, these formulas can be simplified. Since √g is about equal to π, the equations simplify to Tp = 2√L and fp = (1/2√L).

21
New cards

Mass-Spring oscillator

If an object oscillates horizontally on a spring, its mechanical energy is given by E = (1/2)mv² + (1/2)kx². At the endpoints, velocity is zero and displacement is maximized, so the mechanical energy formula simplifies to E = (1/2)kA². At the equilibrium point, velocity is maximized and displacement is zero, so the mechanical energy formula simplifies to E = (1/2)mv²max.

22
New cards

Period of a mass-spring oscillator

Ts = 2π√(m/k) or fs = (1/2π)√(k/m). These formulas stay the same even if the spring is oriented vertically. The new equilibrium position can be found by setting the spring force equal to the gravitational force.

23
New cards

Velocity and acceleration of an oscillator

At the endpoints, velocity is zero, but acceleration is maximized. At the equilibrium point, velocity is maximized but acceleration is zero. They can be modeled by the equations v = -Aωsin(ωt + ϕ) and a = -Aω²cos(ωt + ϕ), where A is the amplitude of the oscillator, ω is the angular frequency, and ϕ is the phase angle, or the vertical shift.

24
New cards

Linear Density

Mass per unit length of an object. Given by λ = dm/dx.

25
New cards

Areal Density

Mass per unit area of an object. Given by σ = dm/dA.

26
New cards

Volumetric Density

Mass per unit volume of an object. Given by ρ = dm/dV.

27
New cards

Coefficient of Restitution

Measure of the elasticity of a collision. Given by e = (|v1f - v2f|) / (|v1i - v2i|). The higher this is, the more elastic the collision is, with the maximum value being 1 for an elastic collision.

28
New cards

Physical Pendulum

Rigid object suspended at a point that oscillates at small angles, like a swinging rod. The period is given by T = 2π√((I) / mgd), where I is the rotational inertia of the object, m is the mass, and d is the distance from the axis of rotation to the center of mass.

29
New cards

Torsional Pendulum

Object hanging on a string that gets twisted, causing the object to rotate back and forth. The period is given by T = 2π√((I) / (β)), where I is the rotational inertia of the object and β is the torsion constant.

30
New cards

Parallel Axis Theorem

Gives the relationship between the rotational inertia of any point on a rotating object and the rotational inertia about the center of mass. It is given by I = ICM + md², where d is the distance from the center of mass to the point of rotation.