Systems Of Particles And Rotational Motion | Class 11 Physics Notes
By ConceptScroll Team · Published on 17 July 2026 · 4 min read
Systems Of Particles And Rotational Motion – this guide gives you a concise, exam-ready overview of Systems Of Particles And Rotational Motion from Class 11 Physics, written by ConceptScroll editors and reviewed against the latest NCERT textbook.
Linear Momentum of a System of Particles
Linear momentum is a fundamental quantity in mechanics representing the product of mass and velocity. For a system of particles, the total linear momentum P is the vector sum of the momenta of all individual particles: P = Σ m_i v_i. This total momentum can be expressed as P = M V, where M is the total mass and V is the velocity of the center of mass. Newton's second law applied to a system states that the rate of change of total momentum equals the net external force acting on the system: dP/dt = F_ext. Internal forces between particles cancel out due to Newton's third law, hence do not affect the total momentum. This principle leads to the law of conservation of linear momentum: if no external force acts on the system, the total linear momentum remains constant. This concept is crucial in analyzing collisions, explosions, and other interactions in physics. The chapter also discusses the impulse-momentum theorem for systems, relating the change in momentum to the impulse applied.
📊 Diagram: Diagrams illustrate two particles moving with velocities v₁ and v₂, their momentum vectors, and the resultant total momentum vector. Another figure shows forces acting on a system and the cancellation of internal forces.
🧪 Activity: Activity: Using a collision experiment with carts on a track to verify conservation of linear momentum.
🔗 Connection: This section prepares for the study of motion of the center of mass and extends to rotational motion concepts.
Frequently asked questions
6.1 Give the location of the centre of mass of a (i) sphere, (ii) cylinder, (iii) ring, and (iv) cube, each of uniform mass density. Does the centre of mass of a body necessarily lie inside the body ?
For each of the given bodies with uniform mass density, the centre of mass (CM) is located at the geometric centre due to symmetry:
(i) Sphere: The CM is at the centre of the sphere.
(ii) Cylinder: The CM is at the midpoint of the axis of the cylinder, i.e., at the centre along its length and at the centre of the circular cross-section.
(iii) Ring: The CM is at the centre of the ring (the centre of the circle formed by the ring).
(iv) Cube: The CM is at the geometric centre of the cube.
Reg
6.2 In the HCl molecule, the separation between the nuclei of the two atoms is about 1.27 Å (1 Å = 10-10 m). Find the approximate location of the CM of the molecule, given that a chlorine atom is about 35.5 times as massive as a hydrogen atom and nearly all the mass of an atom is concentrated in its nucleus.
Let the mass of hydrogen atom = m Mass of chlorine atom = 35.5 m Distance between nuclei = d = 1.27 × 10^{-10} m
Taking hydrogen at origin (x=0), chlorine at x=d,
Centre of mass (CM) position from hydrogen nucleus:
x_CM = (m × 0 + 35.5 m × d) / (m + 35.5 m) = (35.5 d) / 36.5
x_CM = (35.5 / 36.5) × 1.27 × 10^{-10} m ≈ 1.236 × 10^{-10} m
So, the CM lies approximately 1.236 Å from the hydrogen nucleus towards chlorine.
6.3 A child sits stationary at one end of a long trolley moving uniformly with a speed V on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner, what is the speed of the CM of the (trolley + child) system ?
The speed of the centre of mass (CM) of the system (trolley + child) remains the same as the initial speed V of the trolley.
Explanation: Since there are no external horizontal forces acting on the system (assuming smooth floor and no friction), the velocity of the CM cannot change due to internal motions of the child on the trolley. The child running inside the trolley is an internal motion and does not affect the CM velocity.
Therefore, the speed of the CM of the system remains V.
6.4 Show that the area of the triangle contained between the vectors a and b is one half of the magnitude of a × b.
The area of the parallelogram formed by vectors a and b is |a × b|.
Since a triangle is half of a parallelogram, the area of the triangle formed by vectors a and b is:
Area = (1/2) |a × b|
Proof: The magnitude of the cross product |a × b| = |a||b| sin θ, where θ is the angle between a and b.
The area of the parallelogram with sides a and b is base × height = |a||b| sin θ = |a × b|.
Therefore, the area of the triangle = (1/2) × area of parallelogram = (1/2) |a × b|.
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