Class 11 Physics Rotational Motion | Linear Momentum of System of Particles |

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 = p**_{1}+ p_{2}+ …. + p_{n}

= m_{1}**v _{1} **+ m

**P**= M**V**

Where

- pi = momentum of i
^{th}particle , **P**= momentum of system of particles ,**V**= velocity of COM

- Newton’s Second Law extended to system of particles: d
**P**/dt =**F**._{ext } - When the total external force acting on a system of particles is zero (
**F**), 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 = m_{ext }= 0*v*= 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.

Case II – If Ra atom is having an uniform velocity before , then He and Rn can have complex trajectories but COM will have the same VELOCITY as of Ra atom.

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.

(m_{1}**v _{1} **+ M

.