Centre of Mass And Collision Notes for NEET by PW

Centre of Mass and Collision can become difficult when a question combines signed coordinates, moving bodies, momentum conservation or changes in kinetic energy. A small mistake in selecting a reference point, assigning a velocity direction or identifying the line of impact can change the entire answer. To solve these questions accurately, you need to understand when to use mass-weighted averages, when momentum is conserved and how the coefficient of restitution affects the final velocities.

For NEET, PW supports Class 11, Class 12 and dropper aspirants through live and recorded classes, handwritten notes, Daily Practice Problems (DPPs), mock tests, previous years' questions (PYQs) and regular assessments. These resources help you revise concepts, practise numerical questions and identify areas that need more work.

Meaning and Definition of Centre of Mass

The centre of mass (COM) is the point at which the entire mass of a system can be considered concentrated when analysing the translational motion of the system.

This representation is useful for studying the total momentum and motion of a system. However, replacing an extended body with a point mass at its centre of mass does not generally preserve its rotational kinetic energy or moment of inertia. These quantities depend on how mass is distributed around an axis.

Therefore, check which physical quantity a question asks you to calculate before applying the point-mass model.

Centre Of Mass Of Discrete Particles

For particles with masses m1, m2, and so on:

xCM = (m1x1 + m2x2 + …) / (m1 + m2 + …)

yCM = (m1y1 + m2y2 + …) / (m1 + m2 + …)

The denominator is always the total mass.

Coordinates must be substituted with their algebraic signs. A particle at negative x contributes a negative term. This is different from calculations where only distance magnitudes are used.

 The vector form is: RCM = (m1r1 + m2r2 + …) / (m1 + m2 + …)

Calculate the x-, y-, and z-components separately to reduce sign errors.

Important Properties For Two Masses

For two point masses, the centre of mass:

• Lies on the straight line joining the two masses.

• Lies closer to the heavier mass.

• Lies at the midpoint when the two masses are equal.

For example, if 1 kg is at x = 0 and 10 kg is at x = 10 m:

xCM = [1(0) + 10(10)] / 11 = 100/11 ≈ 9.1 m

 The result is close to the 10 kg mass, as expected.

 For two equal masses at points (4, 0) and (0, 4), the centre of mass is their midpoint: RCM = (2, 2)

Its distance from the origin is: r = √(2² + 2²)

Symmetric And Extended Bodies

For a uniform symmetric body, the centre of mass lies at its geometrical centre or centroid.

These results apply only to uniform bodies.

The standard method is to replace an extended body with a point mass at its own centre of mass. For example, a uniform rod of length 10 m is represented by a point mass at its midpoint, 5 m from either end.

If two identical masses are symmetrically placed, they may be combined at their midpoint. Use this shortcut only when the symmetry is clear.

Continuous Mass Distribution

When mass is spread continuously over a body, the centre of mass is calculated using integration rather than treating every small part as a separate particle.

For a rod, linear mass density is: λ = dm/dx

Thus: dm = λ dx

For a uniform rod:  λ = M/L

And: M = λL

For a surface distribution: σ = dm/dA

For a volume distribution: ρ = dm/dV

 The continuous centre-of-mass formula is: xCM = ∫x dm / ∫dm

The form of dm depends on the body:

• Rod: dm = λ dx

• Surface: dm = σ dA

• Volume: dm = ρ dV

For a uniform rod extending from 0 to L, integration gives: xCM = L/2

If the density increases with x, the centre of mass shifts towards the positive-x side. For example, if λ is proportional to x, the centre of mass is 2L/3.

Centre Of Mass Motion

The mass-weighted average also applies to the velocities, accelerations and displacements of particles in a system.

vCM = (m1v1 + m2v2 + …) / total mass

aCM = (m1a1 + m2a2 + …) / total mass

dCM = (m1d1 + m2d2 + …) / total mass

The system relations are: Ptotal = MvCM

Fexternal = MaCM

Internal forces cancel in pairs when considering the complete system. Therefore, the acceleration of the centre of mass depends on the net external force.

If the net external force is zero, the centre of mass moves with constant velocity. It can remain at rest even when individual particles move, provided the system's total momentum is zero.

Centre Of Mass And Work–Energy

The centre of mass is useful when calculating the gravitational potential energy of an extended body in a uniform gravitational field.

The gravitational potential energy of an extended body is: U = MghCM

where hCM is the height of its centre of mass.

For a body moving on a movable support, first use horizontal momentum conservation. If a mass m moves with speed v on a support of mass M, the support moves with speed u in the opposite direction:

mu = Mv

The correct signs must be used. The heavier body moves more slowly.

Collision: Basic Principles

A collision occurs when two bodies interact over a short time, producing changes in their velocities. For an isolated system with negligible external impulse, total linear momentum is conserved.

For a one-dimensional collision:

m1u1 + m2u2 = m1v1 + m2v2

The coefficient of restitution is: e = velocity of separation / velocity of approach

For a head-on collision: e = (v2 − v1) / (u1 − u2)

The relative velocities are measured along the line of impact.

For a perfectly inelastic collision: v1 = v2

For equal masses in a head-on elastic collision, the velocities are exchanged. If one equal mass is initially at rest, the moving body stops and the stationary body takes its velocity.

For a mass m moving with speed v colliding elastically with a stationary mass 2m:

• The m mass rebounds with speed v/3.

• The 2m mass moves forward with speed 2v/3.

Kinetic-Energy Loss In Collision

Kinetic-energy loss can be calculated directly using the reduced mass and the coefficient of restitution. 

The loss of kinetic energy is: Loss = 1/2 × μ × urelative² × (1 − e²)

where reduced mass: μ = m1m2 / (m1 + m2)

This formula is important for direct examination applications.

At maximum deformation during an inelastic collision:

• Both bodies have the same velocity.

• Deformation is maximum.

• Kinetic energy is minimum.

For a spring connected between two bodies, maximum compression or extension also occurs when their velocities become equal. The relative kinetic energy changes into spring potential energy: 1/2 × μ × urelative² = 1/2 × kx²

Impulse And Rebound

Impulse is equal to the change in momentum of a body during an interaction: J = Δp = pf − pi

When a ball strikes the ground and reverses direction, the incoming and outgoing momentum magnitudes add. If the ball approaches with speed v and rebounds with speed ev:

J = m(ev + v)

Average force is: Faverage = J / contact time

For a stationary wall, the rebound speed is ev in the opposite direction. In a perfectly elastic collision, e = 1, so the speed remains unchanged while the direction reverses.

Oblique Collision And Line Of Impact

The line of impact is the direction of the common normal force during collision.

For an oblique collision:

1. Resolve velocity into normal and tangential components.

2. Apply restitution along the line of impact.

3. Keep the tangential component unchanged if friction is absent.

For a frictionless surface:

• Normal velocity changes according to e.

• Tangential velocity remains constant.

If the incoming angle with the normal is θ1 and the outgoing angle is θ2: tan θ2 = tan θ1 / e

For e = 1, the angle of incidence equals the angle of reflection.

For NEET, focus on understanding the centre-of-mass formula, standard results for uniform bodies, the effect of external forces and the conservation of momentum during collisions. Practise assigning velocity signs correctly and distinguishing momentum conservation from kinetic-energy conservation. PW supports Class 11, Class 12 and dropper aspirants through live and recorded classes, handwritten notes, DPPs, mock tests, PYQs and regular assessments. Combine these resources with NCERT revision and regular numerical practice to strengthen your problem-solving skills.

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