Top

# Momentum

What happens when we want to stop a moving car? Do we just take our foot off the accelerator? It is enough to do so? No it is not. We need to apply breaks in order to stop the moving car. This is because the car has gained  something called Momentum. Let us study what is Momentum? What happens to the body after attaining some motion?

## What is Momentum?

These are certain points which describes the basics of momentum:
• It is a vector quantity and has got same direction as that of Velocity.
• It is the product of object's mass and its velocity (moving object).
• The greater the mass or weight the greater is the momentum.
• It is an indication of how tough or difficult it would be to stop a moving object.
Let us quickly recall the Newton's Law of Motion:
All moving bodies will continue to be in the state of rest or motion unless interfered by some external force. The same law applies to momentum as well. That is, if the mass and velocity of an object remain the same then the momentum of the object remains constant, i.e., mv = Constant.

Figure: Depicting the relation of momentum with mass and velocity

## Momentum Definition

Momentum is a property of the body possessed by virtue of its mass and velocity. It is the product of mass of the body and its velocity.
Momentum = mass $\times$ velocity
p = m $\times$ v

S.I Unit of momentum = kgms-1

It is important to note that for a body in motion, the momentum is zero as the velocity is zero.
When the object moves then it gains momentum as the velocity increases. Hence greater the velocity greater is the momentum.

For Example:
It is evident from the fact that it is difficult to stop a truck than a bicycle because of the momentum difference. It can also be visualized from the fact that a truck involved in an accident will do more harm than any bicycle.

The quantity of motion of the body or an object is measured by momentum. It can also be called as “The impulse of the moving object”. It should be noted that momentum is a conserved quantity.

## Momentum Equation

According to newton's second law, the rate of change of momentum always acts in the direction of the resultant force acting on the particle and is directly proportional to this force.
Hence sigma F = $\frac{d p}{d t}$
= $\left \{ \frac{(m)dv}{dt} \right \}+\left \{\frac{(v)dm}{dt} \right \}$
[ As the mass is constant and does not change hence $\frac{(v)dm}{dt}$ = 0 ]
Hence F = m a
Here F is the resultant or net force

## Momentum Formula

The basic formula for momentum of the moving body is:
P = mv ....................(1)
Where,
p = momentum of the body
m = mass
v = velocity of the moving body

Units:
The unit of mass in S.I. system is Kg and
The unit of velocity in S.I. system is m/s
so, the unit of momentum is kgm/s
Also, if velocity is expressed as change of velocity then the formula becomes:
P = m(v1 - v2)...................(2)
Where,
v1 is the final velocity and
v2 is the initial velocity

For a system of particles the formula changes to:
P = $\sigma$ m v
= m1 v1 + m2 v2 + m3 v3 + m4 v4 + . . . . . . . . . . . . .+ mi vi ................(3)

## Units of Momentum

According to the formula of momentum P = m v, the SI unit is Kgm/s. Also expressed in N-s or Newton second.

## Linear Momentum

Let us suppose the file cabinet is in the middle of the room, a room with a smooth floor, we give it a push in order to move it close to the wall and before we realize, it slams into the wall. It is difficult to stop because it has linear momentum.       The measure of an object’s or body’s translation motion is called Linear Momentum.
• Linear momentum is a vector quantity.
• The direction of linear momentum is in the direction of velocity of the object.
• The formula for linear momentum remains the same, p = m v.
But here we also consider it in different reference frames or different axes, viz. Px, Py,Pz etc.
Here,
Px = m Vx
Py = m Vy
Pz = m Vz
Where Vx, Vy, Vz are the velocities in x, y and z directions respectively.

Linear momentum is dependent on the frame of reference:
It is important to note that the an object can have momentum for one frame of reference but the same object if kept in another reference frame can have zero momentum.
Lets us take an example:
An airplane has a velocity of 20 km/s and mass of 100 kg, then the momentum taking the earth as frame of reference is 20 $\times$ 100 = 2000 Kg Km/s. But to the pilot in the cockpit it has a velocity and momentum of zero.

Linear momentum of a system of particles:
for a system of particles with a mass of m1, m2, m3 and so on and a velocity v1, v2, v3 and so on the linear momentum can be expressed as:
P = $\sigma$ m v
= m1 v1 + m2 v2 + m3 v3 + m4 v4 + . . . . . . . . . . . . .+  mi vi  .................(a)

## Change in Momentum

We know that,
P = m v
If we consider the velocity to be the initial and final velocity represented by u and v then momentum change or
The change in momentum would be expressed as:
P = m (v – u)

P = mv – mu
$\Delta P$ = m $\Delta$ vHence, the change in momentum is equal to mass times the change in velocity.
Here mv is the Final Momentum and
mu
is the Initial Momentum
The change in momentum is also called as Impulse.

## Relativistic Momentum

The momentum defined and used in relativistic mechanics is called Relativistic Momentum. In relativistic mechanics the momentum is defined as follows:
P = y mo v . . . . . . . . . . . . . . . . . . . (1)

Here y is the Lorentz factor.

It is equal to,
y = $\frac{1}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$
Here,
v is the object's speed
c is the light's speed
mo is the invariant mass
Also the inverse relation can be expressed as:
v = $\frac{c^{2}P}{\sqrt{(pc)^{2} + (m_{0}c^{2})^{2}}}$
Where P = $\sqrt{(P^{2}x + P^{2}y + P^{2}z)}$

The total energy E of any body is also related to relativistic momentum as:

E2 = (P c)2 + ( m0 c2)2 .......................(1)
Where, P = magnitude of momentum
For mass less particles m0 = 0
Hence E = p c

Figure: Depicting the concept of relativistic momentum

## Impulse and Momentum

The change of momentum is also termed as impulse. It is denoted by “I “.
Hence I = $\Delta$ P

$\frac{dp}{dt}$ = $\frac{d(mv)}{dt}$.
$\frac{dp}{dt}$ = m$\frac{dv}{dt}$ = ma

$\frac{dp}{dt}$
= F
Hence the above equations can be written as: dp = Fdt

So,

$\Delta$ P = F $\Delta$ T = P(final) – P(initial) ............................(1)and
I = $\Delta$ P = change in momentum ...........................(2)The S.I. unit of Impulse is Ns.

## Conservation of Momentum

According to the law of conservation of momentum the momentum of the closed system of objects remains conserved given the fact no external force acts on it.
or
if no external force acts on a system in a particular direction then the total momentum of the system in the direction remains unchanged.

## Momentum Problems

There can be many types of momentum problems. They can be based on the calculations of momentum, impulse, force etc.

The steps to be followed to solve the basic problems of momentum:
1. If simply momentum has to be calculated than calculate the initial and final velocities using the equations of motion.
2. Calculate the momentum by multiplying mass with velocities.
3. In order to calculate the Impulse calculate the force integrating it over time t1 and t2.

Below are given some numericals which helps us to understand more of momentum:

### Solved Examples

Question 1: A ball is thrown whose mass is 1 kg is traveling at 10 meters per second. Calculate its momentum?
Solution:

Given: mass m = 1kg,
Velocity v = 10 m/s.
Momentum p is given by: p = mv
= (1 kg) $\times$ (10m/s)
= 10 kg m/s.

Question 2: A body having 3 kgwt is traveling at 2m/s is subjected to Velocity of 10m/s.
Find: (a) Initial momentum
(b) Final momentum
(c) Change in momentum
(d) Impulse
(e) Calculate its magnitude if force acts for 1 s?
Solution:

Given mass m = 3kg,
Initial Velocity Vi = 2m/s,
Final Velocity Vf = 10m/s
(a) Initial momentum, pi = mvi = 3 $\times$ 2 = 6 kg m/s.
(b) Final momentum, pf = mvf = 3 $\times$ 10 = 30 kg m/s.
(c) Change in momentum, $\Delta$ p = pf - pi = 30 - 6 = 24kgm/s.
(d) Impulse, J = Ft = $\Delta$ p
= 24Ns.
(e)Magnitude of force is given by: F = $\frac{J}{t}$
= $\frac{24 Ns}{1 s}$
= 24N.

## Impulse

When large force acts for a very short time,a force comes in to play. This is called Impulse. It is a vector quantity denoted by $\bar{J}$.
$\bar{dJ}$ = $\bar{F}$ dt
$\therefore$ $\bar{J}$ = $\int_{t_{1}}^{t_{2}}$ $\bar{F}$ dt
$\bar{J}$ = $\bar{F}$ $\Delta{t}$