Conservation of momentum and angular momentum
The total linear and angular momentum of a system remains constant in the absence of external forces or torques, including during collisions and rotational motion.
Why this shows up in the exam
NEET frequently tests your understanding of conservation principles in both translational and rotational contexts.
How NEET tests this
Learn the idea
When no external force acts, the total linear momentum stays unchanged; similarly, when no external torque acts, the total angular momentum stays unchanged. The key insight is that angular momentum about a point depends only on the perpendicular distance of the line of motion from that point, not on the particle’s position along the line.
🧠 Memory hook: Zero push, zero spin – like a frictionless puck sliding straight; its spin about any fixed point never changes because the distance to the track stays the same.
Get this one clearly and it pays off every single time it shows up in the paper. 🎯
Formulas & facts to keep ready
- Linear momentum p = m v
- Conservation of linear momentum: ΣF_ext = 0 ⇒ Σp is constant
- Angular momentum L = r × p = m v r⊥
- Conservation of angular momentum: Στ_ext = 0 ⇒ ΣL is constant
- For a rigid body about a fixed axis I ω = constant
- Total L of a system = Σ r_i × p_i
How to approach it
- 1Check whether the net external force (or torque) on the system is zero
- 2Choose a convenient origin or axis for calculating L
- 3Find the perpendicular distance r⊥ from that point to the line of motion or to the rotation axis
- 4Use L = m v r⊥ (or I ω) and see if r⊥ changes; if not, L is conserved
Worked example — watch it click
A particle of mass m moves in an XY plane with a velocity 'v' along the straight line AB. If the angular momentum of the particle with respect to origin O is Lₐ when it is at A and Lʙ when it is at B then: [Image of a line AB passing through points A and B, and intersecting the y-axis at C, with origin O]
- ✅Lₐ = Lʙ
- B)The relationship between Lₐ and Lʙ depends upon the slope of the line
- C)Lₐ < Lʙ
- D)Lₐ > Lʙ
The concept behind this problem
The worked example asks for the relation between L at two points on the same straight path; since the perpendicular distance from O to the line AB is fixed, the angular momentum about O is the same at A and B.
Step by step
- 1Angular momentum L = r × p = mvr⊥, where r⊥ is perpendicular distance from origin to line of motion.
- 2Since particle moves along straight line AB, the perpendicular distance from O to line AB is constant (equals OC in the diagram).
- 3Therefore L = mv(OC) remains constant throughout motion.
- 4Lₐ = Lʙ.
Watch out
Students often forget that only the perpendicular distance matters and incorrectly think L varies as the particle moves along the line.
Common slip-ups that cost marks
- •Using the full position vector r instead of its perpendicular component
- •Assuming internal forces produce a net external torque
- •Changing the reference point to one where an external torque is present
🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.
Practise it
These are real questions from past NEET papers that test this exact idea.
A particle of mass m moves in an XY plane with a velocity 'v' along the straight line AB. If the angular momentum of the particle with respect to origin O is Lₐ when it is at A and Lʙ when it is at B then: [Image of a line AB passing through points A and B, and intersecting the y-axis at C, with origin O]
Push further
More challenging7 harder questions built from the past papers above — a step up in difficulty, with distractors designed so you can't get there by elimination. Written and checked by our reviewers, not from a real paper.
A turntable of moment of inertia I is rotating with angular velocity ω. A person of mass m jumps onto the turntable at its edge, which is at a distance R from the center. What is the final angular velocity of the turntable with the person on it?
More from Motion of System of Particles and Rigid Body
Moment of inertia and radius of gyration
Moment of inertia quantifies how mass is distributed with respect to an axis of rotation, and the radius of gyration is a measure related to this distribution.
Torque and rotational equilibrium
Torque is the rotational analogue of force, causing angular acceleration, and equilibrium occurs when the net torque on a body is zero.
Center of mass: definition and calculation
The center of mass is the point representing the mean position of the mass in a system, and can be calculated for discrete particles or continuous bodies.
Rotational kinematics and dynamics
Rotational kinematics describes the motion of rotating bodies, while dynamics relates torque, angular acceleration, and rotational kinetic energy.
Centre of Mass of Discrete Particles
For discrete particles, the centre-of-mass position is the vector sum of each mass times its position divided by total mass; this point governs translation even when the particles move relative to one another.
Centre-of-Mass Shift and Internal Rearrangement
When external force is zero, total momentum and centre-of-mass velocity remain constant; therefore mass displacements relative to a fixed frame obey the mass-weighted displacement constraint.