Important Formulas

Week 1

Both of these are pages of equations are for constant acceleration

Follow the steps

1. List the knowns

2. List the unknowns

3. Draw a sketch if it helps

4. Pick an equation

   

Practice questions

Use equation 1

Week 2

When air resistance is negligible

s is defined to be the total displacement and x and y are its component along the horizontal and vertical axes, respectively.

How to do a projectile motion?

Analysing projectile motion

Step 1: Break the motion into horizontal and vertical components along the x and y axes. These axes are perpendicular so we can use x=rcos$\theta$ and y=rsin$\theta$

Step 2: Treat the motion as two independent one-dimensional motions, one horizontal and the other vertical. The kinematic equations for horizontal and vertical motion take the forms:

Step 3: Solve for the unknowns in the two separate motions - one horizontal and one vertical.

Step 4: Use time to link x and y

Step 5: Combine the two motions to find the total displacement s and velocity v. Since x and y motions are perpendicular, we determine these vectors by using the techniques outlined in 3.3

For total displacement and velocity

The range R of a projectile on level ground for which air resistance is negligible is given by

where v₀ is the initial speed and $\theta0$ is the initial angle relative to the horizontal.

Practice questions

Week 3

𝐚∝𝐅_net (acceleration is directly proportional to and in the same direction as the net(total) external force acting on a system)

𝐚∝$\frac{1}{m}$ (acceleration is inversely proportional to the mass of the system)

If air resistance is negligible, net force on a falling object is the gravitational force known as weight.

If two surfaces are in contact and moving relative to one another then the friction between them is called kinetic friction.

When objects are stationary, static friction can act between them, the static friction is usually greater than the kinetic friction between the surfaces.

Kinetic friction f_k is given by f_k=𝜇_k⁢𝑁 where 𝜇_k⁢ is the coefficient of kinetic friction.

s is arc length, the distance travelled along a circular path and r is the radius of the curvature of the circular path.

w is the angular velocity where angular rotation is Δ𝜃 in a time Δt.

This has units rad/s

The direction of centripetal acceleration is toward the center of curvature and magnitude 𝑎_c=$\frac{v^2}{r}$=𝑟𝜔²

F_c=m$\frac{v^2}{r}$ =m𝑟𝜔²

where F_c is the force of centripetal acceleration.

Ideal banked curve

v = speed, r = radius, g=gravity

this gives the ideal angle for a banked curve

“where the angle goes for determining N using weight and mass”

$T=\frac{w}{2\sin\left(\theta\right)}$

above is used in a symmetric situation like a mass hanging from two ropes at angles

T=mxg

simple case with no angles

$\frac{rev}{\min}\cdot\frac{2\pi rad}{1rev}\cdot\frac{1\min}{60\sec}=\frac{rad}{\sec}$

$a=g\left(\sin\theta-\mu\cos\theta\right)$

The acceleration on an incline

Week 4

4.4, 7.1, 7.2, 7.3

W = $\left\vert F\right\vert\left(\cos\theta\right)\left\vert d\right\vert=Fd\cos\theta$

where W is work, d is displacement of the system, and $\theta$ is the angle between the force vector F and the displacement vector d.

For multi-way work break it up and deal with it then add.

Work is in Joules

W_net=F_netd

F_net=ma

W_net=mad

𝑊_net =$\frac12mv^2-\frac12m$v₀²

$KE=\frac12mv^2$ is the translational kinetic energy

W_app=F_appdcos($\theta$)=F_appd

Work done against gravity (e.g. lifting an object)

W = Fd = mgh

Work done by gravity

W = Fd = -mgh

This is gravitational potential energy (PE_g)

If gravitational force is released kinetic energy will increase by the same amount.

PRACTICE 7.3 QUESTIONS

7.4, 7.5, 7.6, 8.1

Potential energy of a spring

PE_s = $\frac12kx^2$

where k is the spring’s force constant and x is the displacement from its undeformed position.

𝑊_net=$\frac12mv^2-\frac12m\left(v0\right)^2$ =ΔKE

If only conservative forces act, then

𝑊_net=W_c

where W_c is the total work done by all conservative forces. Thus,

W_c = ΔKE

KE + PE = constant

or

KE_i + PE_i = KE_f + PE_f

Friction needs to be negligible

KE + PE is mechanical energy

spring and gravitational force

KEi⁢+PEi=KEf+PEf
$\frac12mvi^2+mghi+\frac12kxi^2=\frac12mvf^2+mghf+\frac12kxf^2$

Kinetic Energy: KE=$\frac12mv^2$

Gravitational Potential Energy: U=mgh

Elastic Potential Energy: PE=$\frac12kx^2$

Conservation of Mechanical Energy: Kinetic + Potential

KE + PE = constant or KE_i + PE_i = KE_f + PE_f

NON CONSERVATIVE FORCES

W_net = W_nc + W_c =ΔKE

KE_i + PE_i + W_nc = KE_f + PE_f'

Efficiency (Eff) = (useful energy or work output)/(total energy input) = $\frac{Wout}{Ein}$

impulse is F_net△t

Conservation of Momentum Principle

𝐩tot=constant

𝐩tot=𝐩′tot(isolatedsystem)

where isolated means Fnet is 0

Elastic collision

m₀₁v₀₁ + m₀₂v₀₂ = m_f1v_f1 + m_f2v_f2

Inelastic collision

m₀₁v₀₁ + m₀₂v₀₂ = (m_f1 + m_f2) x v_f3

Example 1:The Force to Stop Falling

A 60.0-kg person jumps onto the floor from a height of 3.00 m. If he lands stiffly (with his knee joints compressing by 0.500 cm), calculate the force on the knee joints.

The person’s energy is brought to zero in this situation. The initial PE_g is transformed in to KE as he falls. The work done by the floor reduces this kinetic energy to zero.

W = Fdcos$\theta$ = -Fd

The kinetic energy the person has upon reaching the floor is the amount of potential energy lost by falling through height h:

KE = -ΔPE_g=-mgh

The knee joint compression is so small that we can ignore it.

The work W done by the floor on the person stops the person and brings the person’s kinetic energy to zero:

W = -KE = mgh

-Fd = mgh

F = -(mgh)/d

   = -(60 × 9.8 × -3)/0.005

   = 3.53 × 10⁵ N

v² = u² +2as

where u is final velocity and v is initial velocity and s is compression

SI units

Converting esp between mili

power is w/t

Finding the Speed of a Roller Coaster from its Height

(a) What is the final speed of the roller coaster shown in Figure 7.8 if it starts from rest at the top of the 20.0 m hill and work done by frictional forces is negligible? (b) What is its final speed (again assuming negligible friction) if its initial speed is 5.00 m/s?

The roller coaster loses potential energy as it goes downhill. We neglect friction, so that the remaining force exerted by the track is the normal force, which is perpendicular to the direction of motion and does no work.

The net work on the roller coaster is then done by gravity alone. The loss of gravitational potential energy from moving downward through a distance h equals the gain in kinetic energy.

This written in equation form as -ΔPE_g=ΔKE

a) Initial kinetic energy is 0 so ΔKE= $\frac12mv^2$.

The equation for change in potential energy states that ΔPE_g=mgh. Since his negative in this case ΔPE_g=mg$\left\vert h\right\vert$

mg$\left\vert h\right\vert$ = $\frac12mv^2$

masses cancel

9.8 × 5 = 0.5 x v²

v = 19.8 m/s

b) In this case there is initial kinetic energy so ΔKE=⁢$\frac12$𝑚⁢𝑣²−⁢$\frac12$𝑚⁢𝑣₀²
g$\left\vert h\right\vert$ = $\frac12v^2$ −⁢$\frac12$⁢𝑣₀²

2 x 9.8 × 20 = v² - 5 × 5

v = 20.4m/s

Week 5

𝜏=rF⁢sin⁡𝜃

𝑟⊥=𝑟⁢sin⁡𝜃 so that 𝜏=𝑟⊥⁢𝐹

1. The first step is to determine whether or not the system is in static equilibrium. This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur.

2. It is particularly important to draw a free body diagram for the system of interest. Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known.

3. Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net𝐅=0 and net𝛕=0, depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution. Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then 𝑟=0), or along a line through the pivot point (then 𝜃=0)). Always choose a convenient coordinate system for projecting forces.

4. Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience.

Week 6

∣𝑞_𝑒∣=1.60×10⁻¹⁹C

Using Coulomb’s law, $F=\frac{k\left\vert q1q2\right\vert}{r^2}$ , its magnitude is given by the equation $F=\frac{k\left\vert qQ\right\vert}{r^2}$ , for a point charge (a particle having a charge Q) acting on a test charge q at a distance r. Both magnitude and direction of the Coulomb force field depending on Q and the test charge q.

Specifically, the electric field E is defined to be the ratio of the Coulomb force to the test charge:

$E=\frac{k\left\vert Q\right\vert}{r^2}$

The electric field is thus seen to depend only on the charge Q and the distance r it is completely independent of the test charge q.

1. Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.

2. The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.

3. The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines.

4. The direction of the electric field is tangent to the field line at any point in space.

5. Field lines can never cross.

Week 7

P = $\frac{W}{t}=\frac{E}{t}$

E=Pt

V = $\frac{PE}{q}$

1eV = 1.60 × 10^-19 J

𝑊=𝑞𝑉_AB

𝑞𝐸𝑑=𝑞𝑉_AB

The charge cancels, and so the voltage between points A and B is seen to be

𝑉_AB=𝐸𝑑

𝐸=𝑉_AB/𝑑

are uniform 𝐸- field only

𝑉=$\frac{kQ}{r}$ ⁢(PointCharge)

where k is a constant equal to 9.0×10⁹N·m²/C²

$E=\frac{F}{q}=\frac{kQ}{r^2}$

We define their capacitance 𝐶 to be such that the charge 𝑄 stored in a capacitor is proportional to 𝐶. The charge stored in a capacitor is given by

𝑄=CV.

Unit is farad

𝐶=𝜀₀⁢$\frac{A}{d}$

𝜀₀=8.85⁢×10^–12F/m

𝐶=𝜅𝜀0$\frac{A}{d}$ (parallel plate capacitor with dielectric).

The energy stored in a capacitor, 𝐸cap, is
𝐸_cap=⁢𝑄⁢𝑉/2,

$Ecap=\frac{QV}{2}=\frac{CV^2}{2}=\frac{Q^2}{2C}$

where 𝑄 is the charge and 𝑉 the voltage on a capacitor 𝐶. The energy is in joules for a charge in coulombs, voltage in volts, and capacitance in farads.