Equations.
$\vec{v}=\dfrac{d\vec{r}}{dt}$
$\vec{a}=\dfrac{d\vec{v}}{dt}$
$\vec{r}=\vec{r}_0+\displaystyle\int\vec{v}\,dt$
$\vec{v}=\vec{v}_0+\displaystyle\int\vec{a}\,dt$
$x=x_0+v_{x0}t+\tfrac{1}{2}a_xt^2$
$v_x^2=v_{x0}^2+2a_x(x-x_0)$
$\vec{r}_{cm}=\dfrac{\sum m_i\vec{r}_i}{\sum m_i}=\dfrac{1}{M}\displaystyle\int\vec{r}\,dm$
$\vec{a}=\dfrac{\sum\vec{F}}{m}$
$\vec{F}=\dfrac{d\vec{p}}{dt}$
$\vec{F}_{A\text{ on }B}=-\vec{F}_{B\text{ on }A}$
$|\vec{F}_{f,k}|=\mu_k|\vec{F}_n|$
$|\vec{F}_{f,s}|\le\mu_s|\vec{F}_n|$
$\vec{F}_s=-k\Delta\vec{x}$
$F_g=G\dfrac{m_1m_2}{r^2}$
$a_c=\dfrac{v^2}{r}$
$K=\tfrac{1}{2}mv^2$
$W=\displaystyle\int\vec{F}\cdot d\vec{r}$
$\Delta U=-\displaystyle\int\vec{F}\cdot d\vec{r}$
$F_x=-\dfrac{dU}{dx}$
$\Delta U_g=mg\Delta y$
$U_s=\tfrac{1}{2}k(\Delta x)^2$
$U_G=-G\dfrac{m_1m_2}{r}$
$P=\dfrac{dE}{dt}=\vec{F}\cdot\vec{v}$
$\vec{p}=m\vec{v}$
$\vec{J}=\displaystyle\int\vec{F}\,dt=\Delta\vec{p}$
Given.
$\vec{a}_g=-9.80665\,\hat{\jmath}$ m/s$^2$, which may be taken as $10$ N/kg
$G=6.67\times10^{-11}$ N$\cdot$m$^2$/kg$^2$
Air resistance is negligible unless a question says otherwise.
This is the same reference information printed at the top of the paper version of this test. Nothing else is provided.