Work Done W = Fd\cos\theta \quad [M^1L^2$T^{-2}$] \quad \text{(joule, J)}

Kinetic Energy KE = \frac{1}{2}mv^2 = \frac{p^2}{2m} \quad [M^1L^2$T^{-2}$] \quad \text{(joule, J)}

Work-Energy Theorem $W_{\text{net}} = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

Gravitational Potential Energy PE_g = mgh \quad [M^1L^2$T^{-2}$] \quad \text{(joule, J)}

Spring Potential Energy PE_s = \frac{1}{2}kx^2 \quad [M^1L^2$T^{-2}$] \quad \text{(joule, J)}

Power P = \frac{W}{t} = Fv\cos\theta \quad [M^1L^2$T^{-3}$] \quad \text{(watt, W)}

Vertical Circular Motion — String $v_{\text{top, min}} = \sqrt{gR}, \quad v_{\text{bottom, min}} = \sqrt{5gR}, \quad v_{\text{side, min}} = \sqrt{3gR}$

Vertical Circular Motion — Rod $v_{\text{top, min}} = 0, \quad v_{\text{bottom, min}} = \sqrt{4gR} = 2\sqrt{gR}$

Elastic Collision (1D) $v_1 = \frac{(m_1-m_2)u_1 + 2m_2u_2}{m_1+m_2}, \quad v_2 = \frac{(m_2-m_1)u_2 + 2m_1u_1}{m_1+m_2}$

Perfectly Inelastic Collision $v_{\text{common}} = \frac{m_1u_1 + m_2u_2}{m_1+m_2}$

Maximum KE Loss (Perfectly Inelastic) $\Delta KE_{\text{max}} = \frac{1}{2}\cdot\frac{m_1 m_2}{m_1+m_2}\cdot(u_1-u_2)^2$

Coefficient of Restitution $e = \frac{v_2 - v_1}{u_1 - u_2} \quad (e=1: \text{elastic},\ 0<e<1: \text{partial},\ e=0: \text{perfectly inelastic})$

Unit Conversions

• 1 hp = 746 W
• 1 kWh = $3.6 \times 10^{6}$ J
• Spring constant k: [M^{1}$$L^{0}$$T^{-2}], unit N/m