# Impulse, Collision, and centre of Mass

As the title suggests, the topic can be broadly divided into three sections. In the first section, we define the impulse of a force and state the impulse-momentum theorem. In the second section, we develop the principle of conservation of linear momentum and apply that to study collision between two bodies. In the third and final section, we define centre of mass of a system of particles and determine its position for various cases.

The impulse of a constant force is given by the product of the force and its duration. If the force is F and its duration is ∆ t , the impulse is given by:

I = F ∆ t

Impulse is a vector; its direction is the same as that of of the force. The dimensions of impulse are [MLT-1], and its SI unit is kilogram metre per second (kg-m/s). Suppose the momentum of a particle changes from p1 at time t1 to p2 at time t2 under a force F applied over the interval ∆ t = t2 - t1. It can be shown from Newton's second law of motion that:

I = F ∆ t = p2 - p1 = ∆ p

In words, the impulse of a force acting on a particle is equal to the change in linear momentum of the particle. This is the statement of the impulse-momentum theorem. If the force varies in magnitude and/or direction with time, the deduction becomes a little complicated. However, the impulse-momentum theorem holds. This last result shows that, for a given change in momentum of a particle, the force is inversely proportional to its duration. We should mention impulsive force in this context. Blows, collisions and explosions involve this type of force whose magnitude is very large and duration is extremely small.

Impulse-momentum theorem can be successfully applied to study the motion of a body of variable mass such as a rocket. The fundamental equation of motion of a body of variable mass is:

F = (v - u) dm/dt + m dv/dt

If the motion happens along a straight line, the scalar equivalent of the above equation is:

F = (v - u) dm/dt + m dv/dt

While other symbols have their usual meanings, dm/dt stands for rate of change in mass with time and dv/dt stands for rate of change in velocity with time. Some examples worth studying are a leaking sand-filled wagon on a frictionless track, sand pouring onto a conveyor belt and, of course, motion of a rocket.

Before we state the principle of conservation of linear momentum, a few terms need to be defined. A system is a well-defined collection of particles under investigation. All other particles in the universe except the system compose the surroundings.

Consider two interacting particles 1 and 2. The force exerted by 1 on 2 is F21 and that exerted by 2 on 1 is F12. If the system consists of only 1, 2 forms part of the surroundings and F12 is an external force on the system. Now, let us redefine the system such that it consists of both 1 and 2. In that case, both F21 and F12 have their sources within the system and are called internal forces. A system on which no net external force acts is called an isolated system.

The principle of conservation of linear momentum states that the total linear momentum of an isolated system remains constant. For a two-particle system, the mathematical statement of the principle would be:

m1 u1 + m2 u2 = m1 v1 + m2 v2

The symbols have their usual meanings.

Collision is defined as an event in which two (or more) particles come together for a short time and exert relatively large forces on each other to produce an abrupt change in motion of at least one of them. It usually implies a physical contact between colliding particles such as two billiard balls. However, this is not an essential condition. An alpha particle (He++) directed towards a gold nucleus deviates sharply from its original path even before any physical contact. Yet they are said to have "collided" because of the strong, brief electrostatic force of repulsion between them.

The coefficient of restitution, e, of a collision is a measure of the restitution (tendency to return to original shape) of the involved particles. It is given by the ratio of the velocity of separation after the collision to the velocity of approach before the collision. For a perfectly elastic collision, e = 1. For a perfectly inelastic collision, e = 0. For a partly elastic collision, 0 < e < 1. Most real collisions belong to this last category.

The centre of mass of a system of particles is defined as a point where the whole mass of the system may be assumed to be concentrated such that the point would move in the same way as a single particle subjected to the same external force would move. Let us consider a system of n particles. The masses of the particles are m1, m2, …, mn. Therefore, the total mass of the system of particles is M = m1 + m2 + … + mn. At a given instant, the position coordinates of the particles in space are (x1, y1, z1), (x2, y2, z2), .... (xn, yn, zn). At the same instant, the position coordinates of the centre of mass of the system will be:

xcm = (m1x1 + m2x2 + … + mnxn) / M
ycm = (m1y1 + m2y2 + … + mnyn) / M
Zcm = (m1z1 + m2z2 + … + mnzn) / M

For uniform rigid bodies of simple shape, the centre of mass can be located by inspection alone. For example, the centre of mass of a uniform rod coincides with its mid-point. The centre of mass of a uniform disc coincides with its geometric centre. The centre of mass of a uniform triangular plate coincides with its centroid, which is the intersection of the three medians.

For nonuniform rigid bodies or for uniform rigid bodies of not-so-simple shape, the centre of mass can be located by the method of integration. Examples of interest are nonuniform rod, uniform semicircular wire, semicircular plate, solid hemisphere etc, as we shall see in due course.

#### Basic level videos

Impulse And Impluse-momentum Theroem 47:10 Basic
300 5
Motion Of A Body Of Variable Mass 49:46 Basic
300 5
Motion Of A Rocket 31:59 Basic
200 3
Problems On Motion Of A Rocket 48:01 Basic
300 5
Principle Of Conservation Of Linear Momentum, And Some Applications 51:20 Basic
300 5
Problems On Principle Of Linear Momentum Conservation 45:01 Basic
250 4
Collision 34:47 Basic
200 3
Mathematical Study Of One Dimensional Collision I 1:00:42 Basic
300 5
Mathematical Study Of One Dimensional Collision II 1:05:02 Basic
350 7.5
Mathematical Study Of Two-Dimensional Collision 1:07:16 Basic
350 7.5
More Problems On Collision 1:07:53 Basic
350 7.5
Centre Of Mass 55:41 Basic
300 5
Finding Centres Of Mass Of Uniform Rigid Bodies By Inspection 43:07 Basic
250 4
Finding Centres Of Mass Of Rigid Bodies By Integration 48:59 Basic
300 5
Finding Centres Of Mass Of Rigid Bodies By Integration II 1:16:27 Basic
300 7.5
Motion Of A System Of Particles 1:13:15 Basic
350 7.5

#### Advanced level Videos Note: (CE) Stands for Problems from Competitive Examination Papers

Advanced-Level Problems On Impulse, Collision, And Centre Of Mass I 1:06:39
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass II 55:49
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass III 1:04:40
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Advanced-Level Problems On Impulse, Collosion, And Centre Of Mass IV 1:03:04
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Advanced-Level Problems On Impulse, Collosion, And Centre Of Mass V 1:03:21
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass VI 59:53
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass VII 1:21:44
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass VIII 1:20:36
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Advanced-Level Problems On Impulse, Collosion, And Centre Of Mass IX 53:03
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Advanced-Level Problems On Impulse, Collosion, And Centre Of Mass X 1:27:01
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Advanced-Level Problems On Impulse, Collision, And Centre Of Mass XI 1:26:16
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Problems On Impulse, Collision, And Linear Momentum Conservation I (CE) 1:01:21
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Problems On Impulse, Collision, And Linear Momentum Conservation II (CE) 52:07
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Problems On Impulse, Collision, And Linear Momentum Conservation III (CE) 1:01:52
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