$$\bar{x} = \frac{\sum m_i x_i}{M}, \qquad \bar{y} = \frac{\sum m_i y_i}{M}, \qquad \bar{z} = \frac{\sum m_i z_i}{M}$$
Motion of the centre of mass:
$$M\mathbf{a}{\text{cm}} = \mathbf{F}{\text{ext}}$$
The centre of mass behaves as if all the external force were applied to a single particle of mass $M$.
Example 1: Three masses: 1 kg at $(0,0,0)$, 2 kg at $(3,0,0)$, 3 kg at $(0,4,0)$.
$M = 6$ kg.
$\bar{x} = (0+6+0)/6 = 1$; $\bar{y} = (0+0+12)/6 = 2$; $\bar{z} = 0$.
Centre of mass: $(1, 2, 0)$.
For two masses $m_1$ and $m_2$ connected by an inextensible string over a smooth pulley:
| Quantity | Formula |
|---|---|
| Acceleration | $a = \dfrac{(m_1-m_2)g}{m_1+m_2}$ (if $m_1 > m_2$) |
| Tension | $T = \dfrac{2m_1 m_2 g}{m_1+m_2}$ |
Derivation: For $m_1$ descending and $m_2$ ascending:
$m_1 g - T = m_1 a$ and $T - m_2 g = m_2 a$.
Add: $(m_1-m_2)g = (m_1+m_2)a$.
Example 2: $m_1 = 4$ kg, $m_2 = 2$ kg:
$a = \dfrac{2g}{6} = g/3 \approx 3.27$ m/s$^2$; $T = \dfrac{8g}{6} = 4g/3 \approx 13.07$ N.
Mass $m_1$ hangs; $m_2$ is on a smooth incline at angle $\theta$:
For $m_2$ sliding up: $m_1 a = m_1 g - T$ and $T - m_2 g\sin\theta = m_2 a$.
$$a = \frac{(m_1 - m_2\sin\theta)g}{m_1+m_2}, \qquad T = \frac{m_1 m_2 g(1+\sin\theta)}{m_1+m_2}$$
Total momentum: $\mathbf{p}{\text{total}} = \sum m_i \mathbf{v}_i = M\mathbf{v}{\text{cm}}$.
The rate of change of total momentum equals the total external force:
$$\frac{d\mathbf{p}{\text{total}}}{dt} = \mathbf{F}{\text{ext}}$$
Internal forces (particle-to-particle interactions) cancel by Newton’s third law and do not affect total momentum.
Example 3: A 70 kg person jumps off a 50 kg trolley. If the person’s velocity is $3\mathbf{i}$ m/s, find the trolley’s velocity (initially at rest).
Conservation of momentum (no external horizontal force):
$\mathbf{0} = 70(3\mathbf{i}) + 50\mathbf{v}{\text{trolley}} \Rightarrow \mathbf{v}{\text{trolley}} = -4.2\mathbf{i}$ m/s.
KEY TAKEAWAY: The centre of mass moves as if the total external force acts on a single point mass. For connected particles, write Newton’s second law for each particle separately, then solve the system.
EXAM TIP: In pulley problems, choose a consistent positive direction for acceleration at the start: e.g., positive = heavier mass moving down. Apply this consistently in both equations.
COMMON MISTAKE: Forgetting that the tension is the same throughout a light inextensible string over a smooth pulley. It is NOT equal to $m_1 g$ or $m_2 g$ unless the system is in equilibrium.
