Physics 1

Question 1

When you see projectile motion, THINK:

Answer 1

1) Horizontalvelocityneverchanges(aslongasyouareignoringairresistance)
2) Horizontalaccelerationalways=0
3) Verticalaccelerationalways=10m/s2downward
4) Verticalbehaviorisexactlysymmetrical(i.e.,ifignoringairresistance,aprojectile’supward trip is identical to its downward trip)
5) Timeintheairdependsontheverticalcomponentofvelocityonly
6) Rangedependsonboththeverticalandhorizontalcomponentsofvelocity
Q22. Why? Explain the dependence of range on both the x- and y components of velocity.
7) Time is always the same for both the x and y components of the motion.

Question 2

Manipulating Linear Equations

Answer 2

X = (1⁄2)at2

2) V = √(2gh) or V = √(2ax (use when asked for final velocity, or given drop height)
3) t air = 2V/g “round trips” or total time in the air

Question 3

Air resistance

Answer 3

**usually ignored but these factors can affect:
The following factors affect the magnitude of air resistance:
1) Cross-sectional Area: greater cross-sectional area = more air resistance
2) Shape: less aerodynamic = more air resistance
3) Velocity: increased velocity = more air resistance.
o Always assume air resistance is being ignored, unless it specifically states otherwise in the question stem or passage.

Question 4

Terminal Velocity

Answer 4

At terminal velocity, mg = Fair. At terminal velocity, the object has stopped accelerating; the forces
of gravity and air resistance are now balanced.

Question 5

Gravity

Answer 5

Definition: Gravity is a field that exists between any two objects with mass.
o THINK OF A FIELD AS: Field = an invisible influence capable of exerting a force on a mass or
charge (The “charge” part will make more sense after we study electric fields)

Question 6

Universal Law of Gravitation

Answer 6

Formula: F = Gm1m2/r2
The Universal Law of Gravitation is true everywhere. Near earth, however, we make an assumption that gravity is a constant 10 m/s2, despite the fact that this law shows that gravity actually varies ever so slightly with height. Based on the near-earth assumption, we can simplify the equation to: F = mg

Question 7

Gravitational Potential Energy

Answer 7

PE=mgh (near earth)
Anything with mass can have gravitational potential energy. For example, fluids have mass, so they can also have potential energy. However, because they don’t always move as a single unit, it is more useful to replace the mass term in the formula with density ρ (mass/volume) to give PE per unit volume of fluid: PE = ρgh

PE= -Gm1m2/r (in space, or near the earth if one is NOT assuming g = 10m/s2)

Question 8

Friction (def and kinetic vs static)

Answer 8

REMEMBER: Friction opposes sliding, NOT motion

If there’s sliding, it’s kinetic friction; if there’s no sliding, it’s static friction.

Question 9

Formulas for Friction

Answer 9

Ff=usFN or Ff=usmgcosθ
Ff=ukFN or Ff=ukmgcosθ
▪ Whereθistheanglebetweentheforce,mg,andalineperpendiculartothesurface.Thealternative formulas shown above for static and kinetic friction are both derived by substituting the formula for normal force for FN in the first formula.

Question 10

Maximum static friction

Answer 10

In cases of static friction, the friction created before an object begins to slide will always remain equal to the net applied force which the friction is opposing. For example, if you push on a boulder with 20 N of force, there will be 20 N of static frictional force opposing you. If you increase the force you apply to 100 N, the static friction will also increase to 100N. This continues up to the “maximum static friction.” Once this value is exceeded, the object will begin to slide and we then have a case of kinetic friction, NOT static.

Question 11

Inclined planes (formulas)

Answer 11

F = mgsinθ ; Force down an inclined plane, parallel to the surface
o FN = mgcosθ ; Normal Force on an inclined plane
o Vf = √(2gh) ; Velocity of a particle at the base of an inclined plane
o a=gsinθ; acceleration down an inclined plane

Question 12

Tension Force

Answer 12

Tension is the force in a rope, string, cable, etc. In most cases, you can ignore tension by replacing it with a force vector on the object to which the rope, string, or cable is attached.

Question 13

Springs

Answer 13

Hooke’s Law: Springs, and many other items such as resilient solids, rubber, and even bonds between
atoms, follow Hooke’s Law.

F = k∆x (where ∆x is the displacement of the spring from its equilibrium point, NOT the overall length of the spring)

Question 14

Calculating the Spring Constant from Hanging Weights

Answer 14

To calculate the spring constant, solve for k using Hooke’s Law. For ∆x, enter the displacement from the equilibrium point for one trial, or the difference in displacement between two trials. For F, use the force applied in one trial, or the difference in the force applied between two trials. CAUTION: It is a common mistake to plug in the mass of the block hanging on a spring for the force. You need to convert that mass into a force using F = mg.

Question 15

Elastic PE

Answer 15

The potential energy stored in a compressed spring (or in any other object that follows Hooke’s Law)

PE = (1/2)k∆x2

Question 16

Pendulums

Answer 16

Definition: A pendulum is any weight (often called a “bob”) attached by a rod, string, wire, etc., to a fixed overhead point, and capable of freely swinging from side to side.

Question 17

Maximum PE and KE in pendulum bob

Answer 17

oPotential Energy (PE) is at a maximum at the maximum height of the bob, and is at a minimum at the bottom of the pendulum’s arc. Kinetic energy (KE) is at a maximum at the bottom of the pendulum’s arc and is at a minimum at the maximum height of the bob.

o Gravitational potential energy is usually assumed to be zero for a pendulum bob at the lowest point of its arc. In other words, at that point we assume that h = 0. (PE=mgh)

Question 18

Simple Harmonic Motion (SHM)

Answer 18

Anything that oscillates back and forth, and can be represented by a sine
wave graphically, constitutes Simple Harmonic Motion.

Examples: car shock absorber, bungee jumping, waves sloshing in a container, anything in circular motion when views from the side

Question 19

Formulas for pendulum/mass on a spring (SHM)

Answer 19

o T = 2pi√(m/k) [mass on a spring]

o T = 2pi√(L/g) [pendulum]

Question 20

Density and Specific Gravity

Answer 20

Density:
o D = m/v
o Density of water: 1000kg/m3 or 1.0g/cm3; recall that 1cm3 =1mL, 1L of water=1kg, and 1 mL of water = 1 g

Specific Gravity: A ratio that describes how dense something is compared to water.
o SG = Dsubstance/DH2O
oFor objects floating in liquids, the fraction of the object submerged = the ratio of the density of the object to the density of the liquid. If the liquid in which it is submerged is water, the fraction submerged is equal to the specific gravity.

Question 21

Archimedes Principle

Answer 21

Any object displaces an amount of fluid exactly equal to its own volume (if fully submerged), or to the volume of whatever fraction of the object is submerged (if floating). The weight of the displaced fluid is exactly equal to the buoyant force pushing up on the object.

need an example here

Question 22

The Buoyant force

Answer 22

Equation:
o Fbuoyant = ρvg ; Where v is the volume of fluid displaced, NOT the total volume of the fluid, and ρ is the
density of the fluid, NOT the object

o Restating an important principle from above, the buoyant force is always exactly equal to the weight of the amount of fluid displaced by the object.

Question 23

Apparent Weight

Answer 23

This is an important point. The apparent weight of a submerged object is the actual weight minus the buoyant force:

▪ Apparent Weight (AW) = Actual Weight (aW) – Buoyant Force (Fbuoyant)

The difference between the actual weight and the apparent weight tells you: 1) the buoyant force, and 2) the weight of that volume of fluid.

Question 24

Fluid Pressure

Answer 24

General Pressure Formula:
o P = F/A

Units– Pascals, mmHG, atm or torr
1x10^5 pascals = 1atm= 760 mmHG= 760 torr

Fluid Pressure Formula:
o P=ρgh

Question 25

Pascals Law

Answer 25

Pressure is transmitted in all directions, undiminished, through a contained, incompressible fluid. Put another way, if pressure increases at any point in a confined, incompressible fluid, it increases by that same amount at every other point within that fluid.

**Pressure is a constant at any vertical depth within the same fluid

Question 26

Types of Fluid Flow

Answer 26

o Ideal, Non-Viscous Flow: This is how ideal, non-viscous fluids flow. There is assumed to be no friction (drag) between the fluid and the walls of the pipe, or between fluid molecules themselves. Fluid near the wall of the pipe flows with the same velocity as fluid at the center of the pipe. This is assumed on the MCAT if they do not specify otherwise.

o Poiseuille Flow: This is how real, viscous fluids flow in pipes. Real fluids exhibit laminar flow (defined below) and have a leading edge that is parabolic in shape.

o Laminar Flow: Fluid flows in pipes in concentric sheets, each with different velocities. The fastest flow is at the exact center of the pipe and the slowest is at the interface with the wall of the pipe.

o Turbulence: At low velocities real fluids exhibit laminar flow. As velocity increases, and especially for non-viscous fluids, flow becomes turbulent—meaning that although the net flow is still in one direction, there are random eddies, changes in direction, changes in velocity, and so forth.

Question 27

Flow Rate

Answer 27

Q= AV
Areainthisequationisalwaystotalcross-sectionarea.Therefore,ifalargepipesplitsinto
to smaller pipes you must add the cross-sections of both new pipes to get the new area.

Question 28

Biology Application of Flow Rate

Answer 28

often used to describe fluid flow in the cardiovascular system. The variable A is the cross-sectional area of the blood or lymph vessel, and V is velocity. Q is a function of cardiac output.

Cardiac output = stroke volume x heart rate

Question 29

Bernoulli’s Equation

Answer 29

K = P + ρgh + (1⁄2)ρv2

The sum of these three forms of energy in an ideal fluid is always equal to a constant (K). Energy is transferred from one form to the other, but the sum of the components will never change.

• This principle is commonly applied to the flow of an ideal fluid through a horizontal pipe, in which case, if fluid velocity increases, pressure decreases.
• Another application: water exiting a spigot
  equation: V = √(2gh)

MCAT applies this equation to both gases and liquids because they are both fluids

Question 30

Intermolecular Forces

Answer 30

Surface Tension: The intensity of intermolecular forces, per unit length, at the surface of a liquid.

Capillary Action: Cohesive vs. Adhesive forces

cohesion: IMF between molecules of a liquid bring molecules to one another
adhesion: IMF between molecules of liquid and molecules of the container (H2O sticking to the side of the glass)