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1) objects in motion stay in motion, a body at rest stays at rest, until a force is applied ("law of inertia") 2) change in momentum of a body is equal in magnitude and direction to the force applied to it (force = mass * acceleration) 3) when two bodies interact, they apply forces that are equal to each other, and opposite in direction ("law of action and reaction")
F = ma p = mv Fnet = Δp / Δt (since Δv/Δtime = acceleration) J = F*d = 1 newton for 1 meter = kg * m²/s² F = J/d
kinetic energy = 1/2 * mv²
Change in potential energy is given by U=mgh
dimensional homogeneity - units must be correct, parts added together, left side matches right side, etc.
Displacement is change in position.
s(t) = s0 + t*(v0+vt)/2 s=displacement from origin at time t vt = v0 + a*t if v0 = 0 then s(t) = s0 + t²*(a)/2 so in free-fall, from position 0, you have: s(t) = g * t²/2
conservation of momentum: p1i + p2i = p1f + p2f
for m1 having velocity u1 to the right, m2 initially at rest, ends with velocity v2. x dimension: m1u1 = m1u2cosθ1 + m2v2cosθ2 y dimension: 0 = m1u2sinθ1 - m2v2sinθ2
Glancing blow: If and only if both masses are equal (like billiards), then the angle between the resulting vectors is always 90 degrees.
normal force = force perpendicular to the plane normal force on a block resting on a slope: f = m*g*cos(degrees from horizontal) parallel force = force parallel to the inclined plane it is unbalanced (objects will move down the plane), sometimes called net force f = m*g*sin(θ)
static friction - μS (mu static) = fS/N (fS = force where static friction is overcome N = normal force) must be overcome before the mass moves μS = fs/N = m*g*sin(θ) / m*g*cos(θ) = sin(θ)/cos(θ) = tan(θ) kinetic friction - normal moving friction only one type of friction applies at a time
Hooke's law: F=-kx, k=spring constant, x = displacement
Fulcrum: t = r * f (torque = radius * force) just add the torques for multiple objects on one side of a fulcrum
Vx = Vo*cos(θ) Vy = Vo*sin(θ) - gt x = Vx*t y = Vy*t - (1/2)*g*t^2
projectile follows the shape of a parabola
y = Ax^2 + Bx y = -gx^2/(2(VoCos(θ))^2) + xtan(θ) time of flight: t = 2Vosin(θ)/g max height: H = (Vosin(θ))^2/2g distance: x = sin(2*θ)*Vo^2 / g
Vo = initial velocity Can use 2sin(θ)cos(θ) = sin(2θ) if filling in t with time of flight in the x = Vx*t formula
Vf^2 = Vi^2 + 2ad ?
pressure P = F/A (force/area) hydrostatic gauge pressure: P = pgh, p = density of fluid, g=gravity, h=height (depth) buoyant force Fb = Fup - Fdown
Fb = pgVf, where Vf = volume of displaced fluid, and density * volume = mass, so Fb = mf*g, where mf = mass of displaced fluid
⇒ buoyant force depends on mass of displaced fluid, not the mass of the object
gravitational constant between two bodies F = G * m1 * m2 / r^2
and g = G * m1 / r^2 gE (gravity Earth) = 9.8 m/s^2
no use of forces in the equations
typical equations: d = vo*t + 1/2*a*t^2 d = (vo + vf)/2 * t vf^2 = vo^2 + 2ad vf = v0 + at
coefficient of restitution = ratio of energy conserved after collision
e = (vel. after collision) / (vel. before collision) (for collision with immovable object) e = (Vfa * Vfb) / (Via * Vib) (for collision between objects a and b. f = final, i = initial velocity) e = 1 for perfectly elastic, 0 for perfectly inelastic