GEN PHYS 1 | Formulas and Solving Tricks
Projectile Motion (WW1)
Solving Tips
Write down the needed components for each the x and y components.
Acceleration is always -9.8 m/s² (due to gravity)
For projectiles launched horizontally, the initial vertical velocity always 0 m/s.
For projectiles launched at an angle,
the vertical velocity at the maximum height is 0 m/s and is also considered as the final velocity.
the overall airtime can be solved by multiplying the solved time by 2
Time is always the same for both components.
The x-component needs two available components to be solved.
The y-component needs three available components to be solved.
Check which y-component is neither given nor the RTF to decide which formula should be used.
Equations
Horizontal Motion
\Delta x=v_{ix}t
Vertical Motion
v_{y}=v_{iy}+at | missing ∆y
\Delta y=v_{iy}t+\frac12at^2 | missing vy
v_{y}^2=v_{iy}^2+2a\Delta y | missing t
Overall Magnitude of Velocity
v^2=v_{x}^2+v_{y}^2
\theta=\tan^{-1}\left(\frac{v_{y}}{v_{x}}\right)
Launched at an Angle
Note: If overall magnitude of velocity is given
v_{y}=v\sin\theta
v_{x}=v\cos\theta
t_{total}=2\left(\frac{v_{iy}}{9.8}\right)=2\left(\frac{v_{i}\sin\theta}{9.8}\right)
h_{\max}=v_{iy}t-\frac12gt^2=\frac{v_{iy}^2}{2g}=\frac{\left(v_{i}\sin\theta\right)^2}{2g}
R=\frac{v_{i}^22\sin\theta\cos\theta}{9.8}=\frac{v_{i}^2\left(\sin2\theta\right)}{9.8}
Forces (WW1)
Formulas
F=ma
F_{x}=F\cos\theta
Fy=F\sin\theta
F_{net}=\sqrt{\Sigma F_{x}^2+\Sigma F_{y}^2}
F_{net}=-F_{eq}
f=\mu N | N = normal force
Tips
Draw free body diagrams for both the individual components and the overall system when needed.
When you need to create an x-axis and y-axis, draw it in such a way it aligns with majority of the forces to minimize calculations.
If it provides a force applied, the force and acceleration applies to the whole system.
Work, Energy, and Power (WW2)
Unit Vector
A=A_{x}i+A_{y}j+A_{z}k
Dot Product
C=\left|A\Vert B\right|\cos\theta
C=\left(A_{x}i\right)\left(B_{x}i\right)+\left(A_{y}j\right)\left(B_{y}j\right)+\left(A_{z}k\right)\left(B_{z}k\right)
Properties
Commutative: A\cdot B=B\cdot A
Distributive: A\left(B+C\right)=A\cdot B+A\cdot C
Scalar Product: kA\cdot B=k\left(A\cdot B\right)
Vector Product
C=AB\sin\theta
Work
W=Fd\cos\theta
W=\left|F\right|\cdot\left|d\right|
Kinetic Energy
KE=\frac12mv^2
W=\Delta KE
Gravitational Potential Energy
GPE=mgh
W=-\Delta GPE
Elastic Potential Energy
EPE=\frac12kx^2
Work-Energy Theorem Proof
Kinetic Energy
W=Fd (assume the angle is equal to 0)
=mad
v_{f}^2=v_{i}^2+2ad
a=\frac{v_{f}^2-v_{i}^2}{2d}
W=md\left(\frac{v_{f}^2-v_{i}^2}{2d}\right)
=m\left(\left(\frac12\right)\left(v_{f}^2-v_{i}^2\right)\right)
W=\frac12mv_{f}^2-\frac12mv_{i}^2
KE=\frac12mv^2
W=KE_{f}-KE_{i}=\Delta KE
Potential Energy
W=Fd
=mgd | mg is equal to the weight
=mg\left(y_{i}-y_{f}\right)
=mgy_{i}-mgy_{f}=-mgy_{f}+mgy_{i}=-\left(mgy_{f}-mgy_{i}\right)
W=-\Delta PE
Law of Conservation of Energy
\Delta KE=-\Delta PE
\frac12mv_{f}^2-\frac12mv_{i}^2=mgy_{i}-mgy_{f}
\frac12mv_{f}^2+mgy_{f}=\frac12mv_{i}^2+mgy_{i}
KE_{f}+PE_{f}=KE_{i}+PE_{i}
Power
Average Power
P=\frac{W}{t}
When Force and Velocity are Constant
P=Fv\cos\theta=F\cdot v
Instantaneous Power
P=\overrightarrow{F}\cdot\overrightarrow{v}
Center of Mass, Momentum, Impulse, and Conservation of Energy
Center of Mass
Mathematical Formula
x_{com}=\frac{\Sigma m_{i}x_{i}}{\Sigma m_{i}}
Momentum of a System
\overrightarrow{p}_{total}=\Sigma m_{i}v_{i}
Motion of the Center of Mass
\overrightarrow{v}_{com}=\frac{\overrightarrow{p}_{total}}{M}
Newton’s 2nd Law
\overrightarrow{F}_{net}=M\overrightarrow{a}_{com}
Momentum
p=mv
Momentum Derivation Proof
\Sigma F=ma
a=\frac{\Delta v}{\Delta t}
\Sigma F=m\frac{\Delta v}{\Delta t}
\Sigma F= d/dt(m)
\Sigma F= d/dt(mv)
p=mv
\Sigma Fdt=mdv
Impulse
J=\Sigma F\Delta t
Impulse Derivation Proof
\Sigma F= dv/dt(m)
\Sigma Fdt=mdv
\Sigma F\Delta t=m\Delta v
Momentum-Impulse Theorem Proof
\Sigma F\Delta t=m\Delta v
\Sigma F\Delta t=m\left(v_{f}-v_{i}\right)
\Sigma F\Delta t=mv_{f}-mv_{i}
\Sigma F\Delta t=p_{f}-p_{i}
\Sigma F\Delta t=\Delta p
Conservation of Momentum
m_1u_1+m_2u_2=m_1v_1+m_2v_2
Elastic Collisions
Conservation of Momentum
m_1u_1+m_2u_2=m_1v_1+m_2v_2
Conservation of Kinetic Energy
\frac12m_1u_1^2+\frac12m_2u_{^{}2}^2=\frac12m_1v_1^2+\frac12m_2v_2^2
Inelastic Collisions
Conservation of Momentum
m_1u_1+m_2u_2=\left(m_1+m_2\right)v_{f}
Coefficient of Restitution
e=\frac{u_2-u_1}{v_1-v_2}
Derivative Formula Using Limits (For Time vs Position)
\lim_{\Delta t\to0}f\left(x\right)=\frac{f\left(t+\Delta t\right)-f\left(t\right)}{\Delta t}
Shortcut
Multiply each term to the power it is raised to (e.g. Multiply x² by 2).
Note: Constants are raised to 0.
Deduct 1 from each term’s exponent.
Remove the original function’s constant term.