Physics Formula Sheet

Projectile Motion Formulas

x f=x0+v0 t+12a t 2

v f=v0+atv f2=v0

2+2a Δ x

v0x=v0cos (αo)v0 y=v0 sin (α o)

x=(v0 cos (αo )) ty=(v0 sin (α o )) t−12 g t

2

vx=v0cos (αo)v y=v0 sin (α o )−¿Dot ProductA∗B=¿ A∨¿B∨cosθUniform Circular Motion

arad=v2

Rarad=

4 π2RT2

v=2πRT

Gravitya=−g

g=9.8m /s2 Relative Motion vP /A= vP /B+ v B/ A

Conservation of EnergyK i+U i+W NC=K f+U f Total Mechanical EnergyE=∑ U+∑ K Energy of a Spring

W spring=U elas=12k x2

F ( x )=−dUdx

⟹F=−kx

Work and PowerW=Fd {cos (θ ) }

P=Wt

=dWdt

=Fv=τω

Gravitational Potential EnergyU grav=mghDistance (Time and Two Velocities Known)

d=v f−v i

2t

Center of Mass

xcm=m1 x1+m2 x2+…m1+m2+…

ycm=m1 x1+m2 y2+…m1+m2+…

vcm=m1 v1+m2 v2+…m1+m2+…

Power: Watts= Joulessecond

Energy in Rotational Motion

K=12I ω2

Torque

∑ τ=Iα I=∑mr2

τ=r × F=rmvsin(θ) Keep in mind: |a bc d|=ad−bc

a× b=| i j ka1 a2 a3b1 b2 b3

|=|a2 a3b2 b3|i−|a1 a3

b1 b3| j+|a1 a2b1 b2|k

Rotation Motion Equations

θ f=θ i+ωi t+12α 2

ωf=ωi+αtωf

2=ωi2+2α (Δθ)

α is the rotational acceleration

ω= vr

a trans=rα

ac=v2

r or ac=ω2 r

Arc Length: s=rθ

Momentump=m vImpulseMomentum change during time interval

J=Δ p= p2− p1=∫ F ( x )dx

F(x) is a function of force in terms of x.Conservation of MomentumIf the vector sum of external forces is zero, the total momentum of the system is constant.

Angular MomentumL=r × p=r ×m v=rmvsin (θ ) L=IωConservation of Angular MomentumAngular momentum is conserved when there is no net torque. L0=L f

Elastic CollisionKinetic energy after the collision is the same as kinetic energy before. (Conserved)New velocities of objects after an elastic collision:

v f 1=v i1 (m1−m2 )+2m2 v i2

m1+m2

v f 2=v i2 (m2−m1)

+2m1 v i1

m2+m1

Inelastic CollisionKinetic energy after the collision is less than kinetic energy before the collision.

tan (θ )=pb

pa

=mB vB

mA v A

Note: In a collision between two objects of unequal masses, the magnitude of the impulse on the lighter object by the heavier one is equal to the impulse on the heavier object by the lighter one.

Note: For problems involving torque, make sure that for sum of the torque everything is multiplied by the radius. For example, for a problem with two tensions on either side of a pulley:

∑ τ=¿R (T 1−T 2)=Iα ¿

Parallel Axis Theorem

I p=I cm+M d2GravitationForce of gravity between two bodies Force of gravity on object from distance R

F1=F2=G(m1m2

r2) Weight=Fg=

GmE∗m

R2

Acceleration by gravity on planet’s Force of gravity with a known angle on a line

surface: g=GmE

r2 F x=G(m1m2

r2 )cos (θ)

Total Mechanical Energy (Orbital)

E=K+U=−GmEm

2r

Heat and Thermodynamicsp iV i

T i

=p f V f

T f

Change in Internal Energy:

ΔU=Q−W

Monatomic Gas: c=32R Diatomic Gas: c=5

2R

Heat: Q=mc(ΔT ) Efficiency:

e= WQH

=1+QC

QH

=1−|QC

QH|

Work: W=∫V 1

V 2

p∗dV=nRTln (V 2

V 1)

pV=nRT KE=32nRT

Harmonic Motion

Pendulum: T=2π √ Lg

Position:

s (t )=Acos (ωt ) Spring: T=2π √ mk

ω=2πT

f= 1T

Isochoric: No volume change, no work done ΔU=Q Adiabatic: No heat transfer. ΔU=−W

Isobaric: Constant pressure. W=p (V 2−V 1) Efficiency: eCarnot=1−TC

T H

K Eper−molecule=32k BT k B=

RN A

N A=6.02∗1023 RMS Speed: vrms=√(3 kTm )=√ 3 RTM

Math Quick Reference