Linear Momentum of System of Particles
- The total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of its COM.
- P = p1 + p2 + …. + pn
= m1v1 + m2v2 + …. + mnvn
- P = MV
- pi = momentum of ith particle ,
- P = momentum of system of particles ,
- V = velocity of COM
- Newton’s Second Law extended to system of particles: dP/dt = Fext .
- When the total external force acting on a system of particles is zero (Fext = 0), the total linear momentum of the system is constant (dP/dt = 0 => P = constant). Also the velocity of the centre of mass remains constant (Since P = mv = Constant ).
- If the total external force on a body is zero, then internal forces can cause complex trajectories of individual particles but the COM moves with a constant velocity.
- Example: Decay of Ra atom into He atom & Rn Atom
- Case I – If Ra atom was initially at rest, He atom and Rn atom will have opposite direction of velocity, but the COM will remain at rest.
Example – A bullet of mass m is fired at a velocity of v1, and embeds itself in a block of mass M, initially at rest and on a frictionless surface. What is the final velocity of the block?
Solution: Now if we take bullet and block as a system, then no external force is acting on it. So we can conserve momentum.
(m1v1 + Mv2)before = (m1+ M)vafter
vafter = m1v1/( m1+ M) , as the initial velocity of block M is zero.
- Vector product (cross product) of two vectors a and b is a × b = ab sinΘ = c , where Θ is angle between a & b
- Vector product c is perpendicular to the plane containing a and b.
- If you keep your palm in direction of vector a and curl your fingers to the direction a to b, your thumb will give you the direction of vector product c
Angular velocity & its relation with linear velocity
Every particle of a rotating body moves in a circle. Angular displacement of a given particle about its centre in unit time is defined as angular velocity.