The definition of linear momentum for a particle is extended to a system of particles by summing the momentum of individual particles. However, this sum is a vector sum of momentums. We need to either employ vector addition or equivalent component summation with appropriate sign convention as discussed earlier. Linear momentum is, thus, defined as :

Definition 1: Momentum of a system of particles

The linear momentum of a system of particles is a vector sum of linear momentum of individual particles.

Figure 1: Particles moving with diferent velocities.

Momentum of a system of particles

p = m 1 v 1 + m 2 v 2 + ................ + m n v n p = m 1 v 1 + m 2 v 2 + ................ + m n v n

⇒ p = ∑ m i v i ⇒ p = ∑ m i v i

From the concept of "center of mass", we know that :

M v COM = m 1 v 1 + m 2 v 2 + ................ + m n v n M v COM = m 1 v 1 + m 2 v 2 + ................ + m n v n

Comparing two equations,

P = M v COM P = M v COM

It means that linear momentum of a system of momentum is equal to the product of total mass and the velocity of the COM of the system of particles.

Just like the case for a single particle, the first differentiation of the total linear momentum gives the external force on the system of particles :

F Ext. = ⅆ P ⅆ t = M a COM F Ext. = ⅆ P ⅆ t = M a COM

This is the same result that we had obtained using the concept of center of mass (COM) of the system of particles. The application of the concept of linear momentum to a system of particles, however, is useful in the expanded form, which reveals the important aspects of "internal" and "external" force :

F Ext. = ⅆ P ⅆ t = m 1 a 1 + m 2 a 2 + ................ + m n a n F Ext. = ⅆ P ⅆ t = m 1 a 1 + m 2 a 2 + ................ + m n a n

The right hand expression represents the vector sum of all forces on individual particles of the system.

F Ext. = ⅆ P ⅆ t = F 1 + F 2 + ................ + F n F Ext. = ⅆ P ⅆ t = F 1 + F 2 + ................ + F n

This relation is slightly ambiguous. Left hand side symbol, " F Ext. F Ext. " represents net external force on the system of particles. But, the individual forces on the right hand side represent all forces i.e. both internal and external forces. This means that :

F Ext. = ⅆ P ⅆ t = F 1 + F 2 + ................ + F n = F Ext. + F Int. F Ext. = ⅆ P ⅆ t = F 1 + F 2 + ................ + F n = F Ext. + F Int.

It is not difficult to resolve this apparent contracdiction. Consider the example of six billiard balls. We strike one of the ball by the stick, imparting velocity to it. The moving ball may collide with another ball. The these balls after collision, in turn, may collide with other balls and so on. The point is that the motion (i.e. velocity and acceleration) of the balls in this illustration are determined by the "internal" contact forces.

Figure 2: Particles moving with diferent velocities.

Momentum of a system of particles

It happens (law of nature) that the motion of the "center of mass" of the system of particles depend only on the external force - even though the motion of the constituent particles depend on both "internal" and "external" forces. This explanation, however, does not resolve the ambiguity of the equation above.

We again look at the process involved in the example of billiard balls. The forces arising from the collision is always a pair of forces. Actually all force exists in pair. This is the fundamental nature of force. Even the external force like force due to gravity on a projectile is one of the pair of forces. When we study projectile motion, we consider force due to gravity as the external force. We do not consider the force that the projectile applies on Earth. We think projectile as a separate system. In nutshell, we consider a single external force with respect to certain object or system and its motion.

However, the motion within a system is all inclusive i.e both pair forces are considered. It means that internal forces always appear in equal and opposite pair. The net internal force, therefore, is always zero within a system.

F Int. = 0 F Int. = 0

This is how the ambiguity in the relation above is resolved.

If no external force is involved, then change in linear momentum of the system is zero :

F Ext. = ⅆ P ⅆ t = 0 ⇒ ⅆ P = 0 F Ext. = ⅆ P ⅆ t = 0 ⇒ ⅆ P = 0

The relation that has resulted is for infinitesimally small change in linear momentum. By extension, a finite change in the linear momentum of the system is also zero :

Δ P = 0 Δ P = 0

This is what is called conservation of linear momentum. We can state this conservation rule in following two different ways.

Definition 2: Conservation of linear momentum

When external force is zero, the change in the linear momentum of a system of particles between any two states (intervals) is zero.

Mathematically,

Δ P = 0 Δ P = 0

We can also state the law as :

Definition 3: Conservation of linear momentum

When external force on a system of particles is zero, the linear momentum of a system of particles can not change.

Mathematically,

P i = P f P i = P f

In expanded form,

m 1 v 1i + m 2 v 2i + ................ + m n v ni = v 1f + m 2 v 2f + ................ + m n v nf m 1 v 1i + m 2 v 2i + ................ + m n v ni = v 1f + m 2 v 2f + ................ + m n v nf

We can interpret this conservation of linear momentum in terms of the discussion of billiards balls as above. The system of billiard balls has certain linear momentum, say, "Pi", as one of the balls is moving. The ball, then, may collide with other balls. This process may repeat as well. The contact forces between the balls may change the velocities of individual balls. But, these changes in velocities take place in magnitude and direction such that the linear momentum of the system of particles remains equal to "Pi".

Figure 3: Particles moving with diferent velocities.

Momentum of a system of particles

This situation continues till one of the billiard balls strikes the side of the table. In that case, the force by the side of the table constitutes external to the system of six billiard balls. Also, we must note that when one of the ball goes into the corner hole, the linear momentum of the system changes. This fact adds to the condition for the conservation of linear momentum. Conservation of linear momentum of a system of particles is, thus, subject to two conditions :

In short, it means that the system of particles is a closed system. The definition of the conservation principle is, thus, modified as :

Definition 4: Conservation of linear momentum

The linear momentum can not change in a closed system of particles.

Example 1

Problem : A bullet of mass "m" , moving with a velocity "u", hits a wooden block of mass "M", placed on a smooth horizontal surface. The bullet passes the wooden block and emerges out with a velocity "v". Find the velocity of bullet with respect to wooden block.

Figure 4: A bullet of mass "m" , moving with a velocity "u", hits a wooden block of mass "M".

Bullet piercing through a wooden block

Solution : To determine relative velocity of the bullet with respect to wooden block, we need to find the velocity of the wooden block. Let the velocity of block be " v 1 v 1 ", then relative velocity of the bullet is :

Figure 5: The bullet passes the wooden block and emerges out with a velocity "v".

Bullet piercing through a wooden block

v rel = v - v 1 v rel = v - v 1

The important aspect of the application of conservation of linear momentum is that we should ensure that the system of particles/ bodies is not subjected to net external force in the direction of motion.

We note here that there is no external force in horizontal direction on the system comprising of bullet and wooden block. We can, therefore, find the velocity of the wooden block, using conservation of linear momentum.

Δ P = m v + m v 1 - m u = 0 Δ P = m v + m v 1 - m u = 0

⇒ v 1 = m ( u - v ) M ⇒ v 1 = m ( u - v ) M

and

⇒ v rel = v - m ( u - v ) M ⇒ v rel = v - m ( u - v ) M

⇒ v rel = M v - m ( u - v ) M ⇒ v rel = M v - m ( u - v ) M

Conservation of linear momentum in component form

Linear momentum is a vector quantity. It, then, follows that we can formulate conservation principle in a given direction. In three mutually perpendicular directions of a rectangular coordinate system.

P xi = P xf P yi = P yf P zi = P zf P xi = P xf P yi = P yf P zi = P zf

These are very important result. In most of the situations, we are required to analyze motion in linear direction i.e. one direction. In that case, we are required to analyze force, momentum etc in that direction only. This is a great simplification of analysis paradigm as we shall find out in solving problems.

Example 2

Problem : A ball of mass "m", which is moving with a speed " v 1 v 1 " in x-direction, strikes another ball of mass "2m", placed at the origin of horizontal planar coordinate system. The lighter ball comes to rest after the collision, whereas the heavier ball breaks in two equal parts. One part moves along y-axis with a speed " v 2 v 2 ". Find the direction of the motion of other part.

Figure 6: A ball collides with another ball at the origin of coordinate system.

Collision of balls

Solution : Answer to this question makes use of component form of conservation law of momentum. Linear momentum before strike is :

P xi = m v 1 P yi = 0 P xi = m v 1 P yi = 0

Let the second part of the heavier ball moves with a speed " v 3 v 3 " at an angle "θ" as shown in the figure. Linear momentum in two directions after collision :

Figure 7: The ball at the origin breaks up in two parts.

Collision of balls

P xf = m v 3 cos θ P yf = m v 2 - m v 3 sin θ P xf = m v 3 cos θ P yf = m v 2 - m v 3 sin θ

Applying conservation of linear momentum :

⇒ m v 3 sin θ = m v 1 ⇒ m v 3 sin θ = m v 1

and

m v 2 - m v 3 sin θ = 0 m v 2 - m v 3 sin θ = 0

⇒ m v 3 sin θ = m v 2 ⇒ m v 3 sin θ = m v 2

Taking ratio, we have :

tan θ = v 2 v 1 ⇒ θ = tan - 1 ( v 2 v 1 ) tan θ = v 2 v 1 ⇒ θ = tan - 1 ( v 2 v 1 )