Centre of Mass and Collisions – Ourphysicsworld.com

Centre of Mass, Momentum, and Impulse – Complete Physics Guide

1. Introduction to Centre of Mass

In classical mechanics, everyday physical objects are not mere point particles. Real objects possess finite size, distinct physical geometry, and complex internal structures consisting of vast collections of atoms. When forces are applied to real macroscopic objects, the resulting motion is rarely a simple translation. Instead, bodies frequently undergo simultaneous translational motion along a trajectory while executing complex rotational motion about an internal axis.

Analyzing the motion of every individual constituent particle within an extended system using Newton’s second law is computationally impractical. Calculating force vectors, accelerations, and particle-to-particle interactions for millions of individual mass points introduces an overwhelming number of equations of motion.

To overcome this complexity, classical mechanics introduces the concept of the Centre of Mass (COM). The Centre of Mass is a uniquely defined geometric point that allows an entire multi-particle system or extended rigid body to be represented as a single ideal point particle.

When a resultant external force acts on a system of total mass $M$, the overall translation of the body can be described entirely by the motion of its Centre of Mass. The Centre of Mass moves through space precisely as if the entire system’s mass were concentrated at that single point and all external forces were applied directly to it.

By employing this central abstraction, we can separate pure translation from rotation, enabling complex mechanical systems to be analyzed with exceptional precision using elementary Newtonian principles.

Figure 1: Representation of an extended body reduced to its Centre of Mass.

2. What is Centre of Mass?

The Centre of Mass (COM) of a system of particles or a rigid body is defined as the unique point in space at which the entire mass of the system is assumed to be concentrated, such that the vector sum of the mass moments of all constituent particles relative to this point equals zero.

Definition Formula: $\mathbf{r}_{cm} = \frac{\sum m_i \mathbf{r}_i}{M}$

S.I. Unit: Meter (m)

Dimensional Formula: $[M^0 L^1 T^0]$

Nature: Vector quantity (specifies position in coordinate space relative to an origin).

Key Characteristics:

It is a mathematical position; physical mass does not need to exist at the exact COM position (e.g., ring, hollow sphere).
Its physical location depends solely on the distribution of mass within the system.
It is independent of the choice of coordinate frame origin.
For symmetrical objects with uniform mass density, the COM lies on the geometric center/axis of symmetry.

To understand the concept deeply, consider a multi-particle system distributed in space. Every particle possesses a mass $m_i$ and a position vector $\mathbf{r}_i$. The product $m_i \mathbf{r}_i$ represents the mass moment of that particle about the origin.

Summing these mass moments across all $n$ particles and dividing by the total mass $M$ yields the weighted average position of the system. This average position is the Centre of Mass.

If an external force is applied, the Centre of Mass follows a trajectory governed by Newton’s second law. Internal forces operating between internal particles equal out in opposite pairs and produce zero net acceleration of the COM.

Thus, regardless of how violently internal components move, explode, or interact, the motion of the Centre of Mass remains governed exclusively by external forces acting on the total mass.

Figure 2: Centre of Mass of a uniform thin ring located in empty space at its center.

3. Centre of Gravity

The Centre of Gravity (COG) of a body is defined as the specific point through which the resultant gravitational force (total weight) of the Earth acts, regardless of the body’s orientation.

Definition Formula: $\mathbf{r}_{cog} = \frac{\sum (m_i g_i \mathbf{r}_i)}{\sum (m_i g_i)}$

S.I. Unit: Meter (m)

Dimensional Formula: $[M^0 L^1 T^0]$

Nature: Vector position quantity.

Key Characteristics:

Depends on both mass distribution and the local gravitational field intensity.
Requires a non-zero gravitational field to be defined.
Coincides with the Centre of Mass in a uniform gravitational field.
Deviates from the Centre of Mass in large-scale or non-uniform gravitational fields.

While Centre of Mass deals strictly with the spatial distribution of mass, Centre of Gravity accounts for gravitational pull acting on each mass element. In a uniform gravitational field where $g$ is constant across all points of a body, the gravitational acceleration terms cancel out mathematically, rendering COG identical to COM.

However, for massive structures such as tall mountains or satellites in space, the gravitational pull decreases with distance from Earth. The lower mass elements experience slightly higher gravitational force than the upper mass elements, causing the COG to lie slightly lower than the COM.

Figure 3: Shift between COM and COG in a tall structure within a non-uniform field.

4. Position of Centre of Mass of Two Particles

Consider a system consisting of two point particles of masses $m_1$ and $m_2$ separated by a total distance $d$ along a straight line.

Position Formulae:

$r_1 = \left( \frac{m_2}{m_1 + m_2} \right) d$

$r_2 = \left( \frac{m_1}{m_1 + m_2} \right) d$

S.I. Unit: Meter (m)

Dimensional Formula: $[M^0 L^1 T^0]$

Characteristics: Inverse mass proportion ratio ($m_1 r_1 = m_2 r_2$).

Vertical Step-by-Step Derivation:

Let mass $m_1$ be placed at position $x_1 = 0$ (Origin).

Let mass $m_2$ be placed at position $x_2 = d$.

The general position formula for COM is:

$x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$

Substitute $x_1 = 0$ and $x_2 = d$:

$x_{cm} = \frac{m_1(0) + m_2(d)}{m_1 + m_2}$

$x_{cm} = \frac{m_2 d}{m_1 + m_2}$

This gives distance $r_1$ from mass $m_1$:

$r_1 = \left( \frac{m_2}{m_1 + m_2} \right) d$

The distance $r_2$ from mass $m_2$ is:

$r_2 = d – r_1$

$r_2 = d – \left( \frac{m_2 d}{m_1 + m_2} \right)$

$r_2 = d \left[ 1 – \frac{m_2}{m_1 + m_2} \right]$

$r_2 = d \left[ \frac{m_1 + m_2 – m_2}{m_1 + m_2} \right]$

$r_2 = \left( \frac{m_1}{m_1 + m_2} \right) d$

Special Cases:

If $m_1 = m_2 = m$, then $r_1 = r_2 = d / 2$ (Midpoint).

If $m_1 \gg m_2$, then $r_1 \approx 0$ and $r_2 \approx d$ (COM lies on $m_1$).

Figure 4: Position of Centre of Mass for a two-particle system.

5. Position of Centre of Mass of Large Number of Particles

For a multi-particle system consisting of $n$ particles with masses $m_1, m_2, …, m_n$ located at 3D coordinates $(x_i, y_i, z_i)$, the total mass is $M = \sum m_i$.

Coordinate Equations:

$x_{cm} = \frac{\sum m_i x_i}{M}$

$y_{cm} = \frac{\sum m_i y_i}{M}$

$z_{cm} = \frac{\sum m_i z_i}{M}$

S.I. Unit: Meter (m)

Dimensional Formula: $[M^0 L^1 T^0]$

Characteristics: Scalar coordinate components of the 3D position vector.

Vertical Step-by-Step Derivation:

Let position vector of $i$-th particle be:

$\mathbf{r}_i = x_i \hat{i} + y_i \hat{j} + z_i \hat{k}$

The vector COM equation is defined as:

$\mathbf{r}_{cm} = \left(\frac{1}{M}\right) \sum m_i \mathbf{r}_i$

Substitute $\mathbf{r}_i$ into the summation:

$\mathbf{r}_{cm} = \left(\frac{1}{M}\right) \sum m_i (x_i \hat{i} + y_i \hat{j} + z_i \hat{k})$

Expand summation across scalar components:

$\mathbf{r}_{cm} = \left( \frac{\sum m_i x_i}{M} \right) \hat{i} + \left( \frac{\sum m_i y_i}{M} \right) \hat{j} + \left( \frac{\sum m_i z_i}{M} \right) \hat{k}$

Equating components along axes yields individual coordinates:

$x_{cm} = \frac{\sum m_i x_i}{M}$

$y_{cm} = \frac{\sum m_i y_i}{M}$

$z_{cm} = \frac{\sum m_i z_i}{M}$

Figure 5: Discrete multi-particle mass distribution in 2D space.

6. Concept of Motion of Centre of Mass with Mathematical Expression

When constituent particles move, the position vector of the Centre of Mass changes with time, establishing velocity and acceleration vectors for the COM.

Velocity Equation: $\mathbf{v}_{cm} = \frac{\sum m_i \mathbf{v}_i}{M}$

Acceleration Equation: $\mathbf{a}_{cm} = \frac{\sum m_i \mathbf{a}_i}{M}$

S.I. Units: m/s (Velocity), m/s² (Acceleration)

Dimensional Formulae: $[M^0 L^1 T^{-1}]$ (Velocity), $[M^0 L^1 T^{-2}]$ (Acceleration)

Vertical Derivation of Motion Equations:

Differentiating position vector $\mathbf{r}_{cm}$ with respect to time $t$:

$\frac{d\mathbf{r}_{cm}}{dt} = \frac{d}{dt} \left[ \left(\frac{1}{M}\right) \sum m_i \mathbf{r}_i \right]$

$\mathbf{v}_{cm} = \left(\frac{1}{M}\right) \sum m_i \left( \frac{d\mathbf{r}_i}{dt} \right)$

$\mathbf{v}_{cm} = \left(\frac{1}{M}\right) \sum m_i \mathbf{v}_i$

$M \mathbf{v}_{cm} = \sum m_i \mathbf{v}_i = \mathbf{P}_{total}$

Differentiating velocity $\mathbf{v}_{cm}$ with respect to time $t$:

$\frac{d\mathbf{v}_{cm}}{dt} = \frac{d}{dt} \left[ \left(\frac{1}{M}\right) \sum m_i \mathbf{v}_i \right]$

$\mathbf{a}_{cm} = \left(\frac{1}{M}\right) \sum m_i \left( \frac{d\mathbf{v}_i}{dt} \right)$

$\mathbf{a}_{cm} = \left(\frac{1}{M}\right) \sum m_i \mathbf{a}_i$

$M \mathbf{a}_{cm} = \sum m_i \mathbf{a}_i$

$M \mathbf{a}_{cm} = \mathbf{F}_1 + \mathbf{F}_2 + … + \mathbf{F}_n$

$M \mathbf{a}_{cm} = \mathbf{F}_{ext}$

Figure 6: Translational velocity vector of Centre of Mass.

7. Law of Conservation of Linear Momentum and Its Derivation

The Law of Conservation of Linear Momentum states that if the net external force acting on a system is zero, the total linear momentum of the system remains conserved (constant) over time.

Mathematical Statement: If $\mathbf{F}_{ext} = 0$, then $\mathbf{P}_{total} = \text{Constant}$

Initial & Final State: $\mathbf{P}_{initial} = \mathbf{P}_{final}$

S.I. Unit: kg·m/s

Dimensional Formula: $[M^1 L^1 T^{-1}]$

Nature: Fundamental vector conservation law.

Vertical Step-by-Step Derivation:

According to Newton’s Second Law for a multi-particle system:

$\mathbf{F}_{ext} = \frac{d\mathbf{P}_{total}}{dt}$

Substitute the condition $\mathbf{F}_{ext} = 0$:

$0 = \frac{d\mathbf{P}_{total}}{dt}$

Integrating both sides with respect to time:

$\int 0 \, dt = \int d\mathbf{P}_{total}$

$\mathbf{P}_{total} = \text{Constant}$

In terms of constituent particles:

$m_1 \mathbf{v}_1 + m_2 \mathbf{v}_2 + … + m_n \mathbf{v}_n = \text{Constant}$

In terms of Centre of Mass velocity:

$M \mathbf{v}_{cm} = \text{Constant}$

$\mathbf{v}_{cm} = \text{Constant}$

Figure 7: Internal explosion where momentum is conserved and v_cm remains unaltered.

8. Impulse and Its Mathematical Expression

Impulse (J) is defined as the overall effect of a force acting on a body over a finite time interval, equal to the change in linear momentum produced by the force.

Definition Formula: $\mathbf{J} = \int \mathbf{F} \, dt = \mathbf{F}_{avg} \cdot \Delta t = \Delta \mathbf{p}$

S.I. Unit: Newton-second (N·s) or kg·m/s

CGS Unit: dyne·s or g·cm/s

Dimensional Formula: $[M^1 L^1 T^{-1}]$

Nature: Vector quantity parallel to the change in momentum.

Characteristics:

Measures the total accumulated effect of force over time.
Equal to the area under the Force versus Time ($F-t$) graph.
Directly dictates the change in an object’s velocity and momentum.

Vertical Derivation (Impulse-Momentum Theorem):

From Newton’s second law, force is the rate of change of momentum:

$\mathbf{F} = \frac{d\mathbf{p}}{dt}$

Rearranging differential terms:

$d\mathbf{p} = \mathbf{F} \, dt$

Integrating from initial time $t_1$ to final time $t_2$:

$\int_{\mathbf{p}_1}^{\mathbf{p}_2} d\mathbf{p} = \int_{t_1}^{t_2} \mathbf{F} \, dt$

$\mathbf{p}_2 – \mathbf{p}_1 = \int_{t_1}^{t_2} \mathbf{F} \, dt$

$\Delta \mathbf{p} = \mathbf{J}$

$\mathbf{J} = \mathbf{F}_{avg} (t_2 – t_1)$

$\mathbf{J} = \mathbf{F}_{avg} \cdot \Delta t$

Figure 8: Force-Time graph showing Area under curve = Impulse.

9. Impulsive Force

An Impulsive Force is defined as a relatively large force that acts on a body for an extremely short time interval, producing a finite and measurable change in linear momentum.

Definition Formula: $\mathbf{F}_{avg} = \frac{\Delta \mathbf{p}}{\Delta t}$

S.I. Unit: Newton (N)

CGS Unit: dyne

Dimensional Formula: $[M^1 L^1 T^{-2}]$

Nature: Vector quantity.

Key Characteristics:

Magnitude is extremely high, while contact duration $\Delta t$ is extremely brief.
Instantaneous values fluctuate rapidly and are difficult to measure directly.
Action is evaluated using total impulse or area under $F-t$ curve.
Increasing contact time $\Delta t$ significantly reduces impact force $F_{avg}$, a principle applied in safety equipment like airbags and helmets.

Real-world instances include batting a cricket ball, kicking a resting football, or catching a fast ball.

Figure 9: Impulsive contact during bat hitting a ball.

10. Conceptual Understanding Questions

Q1: Does the Centre of Mass of a body always lie inside its physical material? Explain with an example.

Ans: No, the Centre of Mass does not necessarily lie inside the physical material of a body. It is a mathematical point determined by mass distribution. For instance, in a uniform circular ring, hollow sphere, or horseshoe magnet, the COM lies in empty space at their geometric center where no physical mass exists.

Q2: Can the Centre of Mass of a body move if no external force acts on it?

Ans: If the Centre of Mass was initially moving with a constant velocity, it will continue moving with that exact same velocity even if no external force acts on it. However, if it was initially at rest, it cannot start moving without an external force.

Q3: Under what condition do the Centre of Mass and Centre of Gravity of a body coincide perfectly?

Ans: They coincide perfectly when the body is located in a uniform gravitational field where the acceleration due to gravity ($g$) is constant across all points of the body.

Q4: A projectile explodes in mid-air into several fragments. What path does the Centre of Mass of the fragments follow?

Ans: The forces of explosion are internal forces. Therefore, the Centre of Mass continues along its original parabolic trajectory undisturbed until the first fragment hits the ground.

Q5: Why does a cricketer pull their hands backward while catching a fast-moving ball?

Ans: Pulling hands backward increases the contact time interval ($\Delta t$) for stopping the ball. From $F_{avg} = \frac{\Delta p}{\Delta t}$, a larger time interval reduces the impact force felt by the fielder’s hands.

Q6: Is linear momentum conserved during an internal explosion?

Ans: Yes, because explosive forces are purely internal ($\mathbf{F}_{ext} = 0$). The initial total linear momentum equals the vector sum of momenta of all fragments after explosion.

Q7: Can a system have zero linear momentum but non-zero kinetic energy?

Ans: Yes. For example, two equal masses moving with equal speeds in opposite directions have zero total linear momentum ($\mathbf{p}_1 + \mathbf{p}_2 = 0$), but their total kinetic energy is positive scalar sum ($K_1 + K_2 \neq 0$).

Q8: What is the area under a Force vs Time graph equal to?

Ans: The area under a Force-Time ($F-t$) graph represents total Impulse, which equals net change in linear momentum ($\Delta p$).

Q9: Why are glass vessels wrapped in straw or bubble wrap during transportation?

Ans: Bubble wrap increases impact contact duration ($\Delta t$) during jerks, reducing average force exerted on glass, preventing breakage.

Q10: If two particles move toward each other under mutual gravitational attraction, how does their Centre of Mass move?

Ans: Gravitational force between them is an internal force. If they were initially at rest, their Centre of Mass remains completely stationary.

11. Solved Numerical Problems

P1: Two masses of 2 kg and 6 kg are separated by a distance of 12 m. Find the distance of the Centre of Mass from the 2 kg mass.

Solution:

$m_1 = 2 \text{ kg},\, m_2 = 6 \text{ kg},\, d = 12 \text{ m}$

$r_1 = \left( \frac{m_2}{m_1 + m_2} \right) d$

$r_1 = \left( \frac{6}{2 + 6} \right) \cdot 12$

$r_1 = \left(\frac{6}{8}\right) \cdot 12 = 9 \text{ m}$

Answer: 9 m from 2 kg mass.

P2: Three particles of masses 1 kg, 2 kg, and 3 kg are placed at coordinates (0,0), (2,0), and (0,3) respectively. Find COM coordinates.

Solution:

Total Mass $M = 1 + 2 + 3 = 6 \text{ kg}$

$x_{cm} = \frac{1(0) + 2(2) + 3(0)}{6} = \frac{4}{6} = 0.67 \text{ m}$

$y_{cm} = \frac{1(0) + 2(0) + 3(3)}{6} = \frac{9}{6} = 1.5 \text{ m}$

Answer: (0.67 m, 1.5 m).

P3: A force of 50 N acts on a body of mass 10 kg for 0.2 seconds. Calculate the impulse imparted to the body.

Solution:

$F = 50 \text{ N},\, \Delta t = 0.2 \text{ s}$

$J = F \cdot \Delta t$

$J = 50 \cdot 0.2 = 10 \text{ N}\cdot\text{s}$

Answer: 10 N·s.

P4: A ball of mass 0.5 kg moving at 20 m/s hits a wall and rebounds at 15 m/s. Calculate the magnitude of impulse.

Solution:

$v_1 = +20 \text{ m/s},\, v_2 = -15 \text{ m/s},\, m = 0.5 \text{ kg}$

$\Delta p = m(v_2 – v_1)$

$\Delta p = 0.5 \cdot (-15 – 20)$

$\Delta p = 0.5 \cdot (-35) = -17.5 \text{ kg}\cdot\text{m/s}$

Answer: 17.5 N·s.

P5: Two bodies of masses 3 kg and 5 kg move with velocities 4 m/s and 2 m/s toward each other. Find velocity of COM.

Solution:

$v_1 = +4 \text{ m/s},\, v_2 = -2 \text{ m/s}$

$v_{cm} = \frac{m_1 v_1 + m_2 v_2}{m_1 + m_2}$

$v_{cm} = \frac{3(4) + 5(-2)}{3 + 5}$

$v_{cm} = \frac{12 – 10}{8} = \frac{2}{8} = 0.25 \text{ m/s}$

Answer: 0.25 m/s in direction of 3 kg body.

P6: A bomb of mass 12 kg at rest explodes into two pieces of masses 4 kg and 8 kg. If velocity of 4 kg piece is 12 m/s, find velocity of 8 kg piece.

Solution:

$P_{initial} = 0$

$m_1 v_1 + m_2 v_2 = 0$

$4(12) + 8(v_2) = 0$

$48 + 8 v_2 = 0 \implies v_2 = -6 \text{ m/s}$

Answer: 6 m/s in opposite direction.

P7: A body of mass 2 kg is acted upon by force $F = (6t) \text{ N}$. Calculate impulse between $t = 0$ and $t = 3 \text{ s}$.

Solution:

$J = \int_{0}^{3} 6t \, dt$

$J = 6 \left[ \frac{t^2}{2} \right]_{0}^{3} = 3 [3^2 – 0^2]$

$J = 3 \cdot 9 = 27 \text{ N}\cdot\text{s}$

Answer: 27 N·s.

P8: A bullet of mass 0.05 kg is fired horizontally with 200 m/s into a stationary wooden block of mass 1.95 kg resting on a smooth surface. Find the final common velocity.

Solution:

$m_1 = 0.05 \text{ kg},\, v_1 = 200 \text{ m/s},\, m_2 = 1.95 \text{ kg},\, v_2 = 0$

$P_{initial} = (0.05 \cdot 200) + (1.95 \cdot 0) = 10 \text{ kg}\cdot\text{m/s}$

$P_{final} = (m_1 + m_2) v = (0.05 + 1.95) v = 2.0 v$

$2.0 v = 10 \implies v = 5 \text{ m/s}$

Answer: 5 m/s.

P9: An impulsive force acts on a body, changing its momentum from $10 \text{ kg}\cdot\text{m/s}$ to $40 \text{ kg}\cdot\text{m/s}$ in $0.02 \text{ s}$. Find the average impulsive force.

Solution:

$\Delta p = 40 – 10 = 30 \text{ kg}\cdot\text{m/s}$

$\Delta t = 0.02 \text{ s}$

$F_{avg} = \frac{\Delta p}{\Delta t} = \frac{30}{0.02} = 1500 \text{ N}$

Answer: 1500 N.

P10: A shell of mass 4 kg is moving with $20 \text{ m/s}$ and explodes into two equal pieces. One piece continues in the same direction with $30 \text{ m/s}$. Find velocity of the second piece.

Solution:

$M = 4 \text{ kg},\, V = 20 \text{ m/s} \implies P_{initial} = 4 \cdot 20 = 80 \text{ kg}\cdot\text{m/s}$

$m_1 = 2 \text{ kg},\, v_1 = 30 \text{ m/s}$

$P_{final} = m_1 v_1 + m_2 v_2 \implies 80 = (2 \cdot 30) + (2 \cdot v_2)$

$80 = 60 + 2 v_2 \implies 2 v_2 = 20 \implies v_2 = 10 \text{ m/s}$

Answer: 10 m/s in original direction.