# How to Calculate Momentum of a System: Various Problems and Facts

The article discusses about how to calculate momentum of a system from the momenta of each object.

The momentum of a system is determined by the motion of the number of objects the system possesses. Some of them move with their velocities within the system, while others rest. Hence, we can calculate the momentum of a system by summing up all momenta of objects within the system.

## Calculate the momentum of a system containing moving objects such as object A of mass 5kg moving at 2m/s, and object B of mass 3kg moving at 5m/s.

Given:

m1 =5kg

m2 = 3kg

u1 = 2m/s

u2 = 5m/s

To Find: Psystem =?

Formula:

Σ Psystem = P1+ P2 + …

Solution:

The momentum of a system is calculated by adding the momenta of two moving objects.

Psystem = P1 + P2

Psystem = m1u1 + m2u2

Substituting all values,

Psystem = 5 x 2 + 3 x 5

Psystem = 25

The momentum of a system containing two moving objects is 25kg.m/s

## What is the Momentum of the System?

The momentum of the system is the motion of the whole mass of the system that includes several objects.

When the system comprises the motion of many objects, the concept of the center of mass is introduced. It is the average position of all the objects within the system, weight as per their mass. So the momentum of the system is the product of the total mass and velocity of the center of mass.

The momentum of system is Pcm = Mvcm

Where vcm is the velocity of the center of mass.

Mvcm = Md/dtrcm ………..(1)

Where rcm is the position of the center of mass.

The formula of center of mass is,

rcm = m1r1/m2r2

Equation (1) becomes,

Mvcm = Md/dt (m1r1/m2r2)

Mvcm = m1d/dtr1 +m2d/dtr2 +…

Mvcm = m1v1 + m2v2

Mvcm = P1+P2

P cm = P1+P2

That is why we say the momentum of the system is the vector sum of all momenta of each object within the system.

## We see three vehicles running such as a car having a mass of 150kg running at 50km/hr, a motorbike having a mass of 80kg running at 80km/hr, and a truck having a mass of 250kg running at 30km/hr. Calculate the momentum of the system where the car and motorbike are moving in the same direction while a truck is in the opposite direction.

Given

m1 = 150kg

m2 = 80kg

m3 = 250kg

v1 = 50km/hr

v2 = 80km/hr

v3 = 30km/hr

To Find: Psystem =?

Formula:

Pcm= P1+P2 + …

Solution:

The momentum of system is calculated as,

Pcm = P1+P2 + …

Psystem = P1 + P2 + P3

Psystem = m1v1+ m2v2 + (-m3v3)

Substituting all values,

Psystem = 150 x 50 + 80 x 80 – 250 x 30

Psystem = 7500 + 6400 – 7500

Psystem = 6400 kg.km/hr

Psystem = 6400 x 1000/3600

Psystem = 1777.7

The momentum of the system that includes running car, motorbike, and truck is 1777.7 kg.m/s

## How to calculate total angular momentum of a system?

The total angular momentum of a system is calculated using its moment of inertia and angular velocity.

While calculating total angular momentum, we need to assume only angularly moving objects within a system. The rotational inertia and angular velocity of an object offer its angular momentum to calculate the total angular momentum by adding all angular momenta of an object within a system.

We discussed only the momentum of an object that moves in a straight path or linearly with velocity v. The system also involves an object moving angularly with velocity ω. An object has linear momentum (P) or Angular Momentum (L) depending on a motion.

For linear motion, an object’s linear momentum is P = mv.

For angular motion, an object’s angular momentum is

L = Iω ………………(*)

Where I = moment of inertia is defined as

the sum of the mass of an object with the system with a square of its distance from the system’s axis of rotation’

I = Σ miri2

I = mr2

The relation between angular velocity ω and linear velocity v is ω = v/r

Substituting values of I and ω into equation (*), we get

L = v/r mr2

=  r *mv or

= r * P ……………… (2)

From the above relation between angular momentum and linear momentum, we can calculate the angular momentum for non-orbital motion objects within the system.

The total angular momentum of a system is the sum of angular momentum of each object within a system.

L = l1 + l2 + lN ………………. (3)

Newton’s second law of motion says

Σ F = d/dt P

Differentiating equation (2) with respect to t,

d/dt L = v * d/dt P

d/dt L = r*Σ F ……….. (4)

Right hand side is the formula of torque which is also called ‘moment of force’ that causes the body to rotate about a fixed axis.

Σ = r * Σ F ……………….. (5)

Using equation (3), (4) and (5),

d/dt Lidli/dt = Σiτi

The above equation expresses the sum of torque on each object within a system produces a net external torque Στ on a system to change its total angular momentum. Hence,

dL/dt= Στ

But if there is no torque exerted on the rotating objects, Στ = 0 , then

dL/dt = 0,

That’s how angular momentum of the system is conserved.

That means the angular momenta of an individual object may change during the collision, but the total angular momentum of a system remains the same.

For the system of two rotating objects,

Lbefore collision = Lafter collision

I1iω1i + I2iω2i = I1fω1f + I2fω2f ………………….. (6)

## Calculate the angular momentum of a system that includes two skaters having a moment of inertia 5kg.m2 and 8kg.m2 sliding angularly at 10 rad/s and 15 rad/s respectively by pulling their hands. Also, if they extend their hands to reduce their motion, the first skater slides at 2rad/s. Then what is the rate at second skater slides?

Given

I1 = 5kg.m2

I2 = 8kg.m2

To Find:

1. Ltotal =?
2. ω2f =?

Formula

1. Ltotal = L1 + L2
2. I1iω1i + I2iω2i = I1fω1f + I2fω2f

Solution:

The angular momentum of a system of both skater is calculated as,

Ltotal = L1 + L2

Ltotal = I1iω1i + I2iω2i

Substituting all values,

Ltotal = 5 x 10 + 8 x 15

Ltotal = 50 + 120

Ltotal = 170

The angular momentum of a system of skaters is 170kg.m2/s.

The final velocity of the second skater is calculated using the law of conservation of angular momentum.

I1iω1i + I2iω2i = I1fω1f + I2fω2f

Substituting all values,

5 x 10 + 8 x 15 = 5 x 2 + 8 x ω2f

170 = 10 + 8ω2f

2f = 160

ω2f = 20

The rate at which the second skater slides after extending hands is 20rad/s.