MechanicsHigh Yield (1-2 Qs in JEE)

Center of Mass & Momentum Conservation Formulas

Center of mass coordinates, continuous mass distribution, momentum conservation, and explosion of bodies.

7 Formulas•271+ Questions in Bank

Core Formulas & Mathematical Identities

\vec{R}_{cm} = \frac{\sum m_i \vec{r}_i}{\sum m_i} = \frac{\int \vec{r} \, dm}{\int dm}

Mass-weighted average position vector of a multi-particle or continuous system.

Examiner Tip: If net external force is zero, center of mass moves with constant velocity.

y_{cm} = \frac{2R}{\pi} \; (\text{Semicircular ring}), \quad y_{cm} = \frac{4R}{3\pi} \; (\text{Semicircular disc}), \quad y_{cm} = \frac{3R}{8} \; (\text{Solid hemisphere})

Standard heights of center of mass measured from the diametric base.

Examiner Tip: Hollow hemisphere has y_{cm} = R / 2.

\vec{F}_{ext} = 0 \implies \vec{P}_{system} = M \vec{v}_{cm} = \text{constant}

Total linear momentum remains constant in the absence of external forces.

Examiner Tip: Explosions and rifle recoil conserve total momentum.

y_{cm} = \frac{h}{4} \; (\text{Solid Cone}), \quad y_{cm} = \frac{h}{3} \; (\text{Hollow Cone}), \quad y_{cm} = \frac{R}{2} \; (\text{Hemispherical Shell})

Height of center of mass measured from flat circular base along axis of symmetry.

Examiner Tip: Memory mnemonic from base: Shell (R/2) > Solid hemisphere (3R/8); Hollow cone (h/3) > Solid cone (h/4).

\vec{R}_{cm} = \frac{M_{\text{orig}}\vec{r}_1 - M_{\text{removed}}\vec{r}_2}{M_{\text{orig}} - M_{\text{removed}}}

Calculates the shift in center of mass when a symmetrical portion or hole is carved out from a uniform body.

Examiner Tip: For circular disc with circular hole of radius r at distance d: x_{cm} = - (r² d) / (R² - r²).

\mu = \frac{m_1 m_2}{m_1 + m_2}, \quad T = 2\pi \sqrt{\frac{\mu}{k}}

Reduces two interacting bodies connected by a spring into an equivalent single particle problem.

Examiner Tip: When a spring connects m₁ and m₂ on a frictionless surface, natural frequency is ω = √(k / μ).

F_{\text{thrust}} = v_{\text{rel}} \left(-\frac{dm}{dt}\right), \quad v = v_0 + v_{\text{rel}} \ln\left(\frac{m_0}{m}\right) - gt

Tsiolkovsky rocket equation for mass ejection rate dm/dt with exhaust velocity v_rel.

Examiner Tip: For upward acceleration just after liftoff: a = (v_rel / m₀)(|dm/dt|) - g.

Frequently tested misconceptions in Center of Mass & Momentum Conservation

•Center of mass of a system depends on distribution of mass, independent of the coordinate origin.
•Internal forces CANNOT change the velocity of the center of mass.