Question

Separating motion of a system of particles into motion of the centre of mass and motion about the centre of mass, show p = pi' + miv where pi is the momentum of ith particle (of mass mi) and Pi' = mi vi'. Note Vi' is the velocity of the ith particle relative to the centre of the mass.Also prove using the definition of the centre of mass Σpi' = 0.

Answer 1

(a)Let's take a system in which contains  i moving particles.
Mass of the ith particle = mi
Velocity of the ith particle = vi

so, momentum of the ith particle, pi = mi.vi

Let us consider , velocity of the centre of mass is V

The velocity of the ith particle with respect to the centre of mass of the system, v'i is given as:

v’i = vi – V ….... (1)

Multiplying mi throughout equation (1), we get:

mi v’i = mi vi – mi V

p’i = pi – ­mi V hence proved //

Where, pi’ = mivi’ = Momentum of the  ith particle with respect to the centre of mass of the system

∴pi = p’i ­+ mi V

We have the relation: p’i = mivi’
Taking the summation of momentum of all the particles with respect to the centre of mass of the system, we get:
$\Sigma p'_i=\Sigma m_iv'_i=\Sigma m_i\frac{dr'_i}{dt}$

where, $r'_i$ is the position vector of ith particle with respect to centre of mass.

by definition of centre of mass,
$\Sigma m_ir_i=0$

so, $\Sigma m_i\frac{dr_i}{dt}=0$

so, $\Sigma p_i =0$ proved.

Answer 2

Answer:

(a)Let's take a system in which contains  i moving particles.

Mass of the ith particle = mi

Velocity of the ith particle = vi

so, momentum of the ith particle, pi = mi.vi

Let us consider , velocity of the centre of mass is V

The velocity of the ith particle with respect to the centre of mass of the system, v'i is given as:

v’i = vi – V ….... (1)

Multiplying mi throughout equation (1), we get:

mi v’i = mi vi – mi V

p’i = pi – ­mi V hence proved //

Where, pi’ = mivi’ = Momentum of the  ith particle with respect to the centre of mass of the system

ANSWER

∴pi = p’i ­+ mi V