(a) The linear impulse of a force `vecF` during a time interval `trianglet = t_2 - t_1` can be given as

`vecj = int_(t_1)^(t_2) vecF dt = int_(t_1)^(t_2) (dvecp)/(dt) = int_(vecp_1)^(vecp_2) vec(dp) = vecp_2 - vecp_1`

`vecJ = trianglevecp`

The linear impulse of a force `vecF`acting for a time interval is numerically equal to change in linear momentum during that time interval.

(b) Graphical significance of impulse-momentum equation

The impulse of a force `vecF` during a time interval `trianglet = t_2 - t_1` is given by

Note 1 : Area above t-axis must be taken as (+)ve and that below t-axis, must be taken as (-)ve

Here `A_1, A_2. A_3` ,are area under respective F-t graph, shown in Fig.(7.b)

NOTE 2 :

1. Area under F-t graph gives the impulse of the force F which is equal to the change in linear momentum "due to that force". Hence, area under any force-time graph can be equated with the total change in linear momentum of a particle due to that force.

2. The net linear impulse = net area of `F_(n et)` curve = (magnitude of area of F-t curve above t-axis) - ( magnitude of area of F-t curve below t-axis)

When a force, of relatively higher magnitude, acts for a shorter rime, it is referred as an impulsive force. An impulsive force can change the momentum of a body by a finite magnitude in a very short time interval.

Impulsive force is a relative term.
Note: (a) Gravitational force and spring force are always non-impulsive.
(b) Normal reaction, Tension and Friction are case dependent.
(c) An impulsive force can only be balanced by another impulsive force.

(a) Impulsive Normal Reaction

In case of collision, normal forces at the surface of collision are always impulsive
`N_1` : impulsive
`N_1, N_2` : Non-impulsive (shown in Fig.(9.a))

Consider a ball dropped on a large ball. Both normal forces `N_1` & `N_2` are impulsive, shown in Fig.(9.b).

Consider a large ball colliding with small ball, shown in Fig.(9.c).

`N_1, N_3` : Impulsive

`N_2` : Non - impulsive

(b) Impulsive Friction

If normal force between the two objects is impulsive, then the friction between the two will also be impulsive, shown in Fig.(9.d)

Both normal force `N_1` and `N_2` are impulsive. Friction at both surfaces is impulsive if it exists, shown in Fig.(9.e)

Collision between a large ball and small ball. Friction due to `N_1` & `N_2` are impulsive and due to `N_2` Non- impulsive.

(c) Impulsive Tension

When a string jerks, equal and opposite tensions act suddenly at each end. Consequently equal and opposite impulses act on the object to which the two ends of the string are attached. There are two cases to be considered.

Case 1 : One end of the string is fixed

The impulse which acts at the fixed end of the string cannot change the momentum of the fixed object. The object attached to the free end however will undergo a change in momentum in the direction of the string. The momentum remains unchanged in a direction perpendicular to the string where no impulsive force act. When one end of a string is fixed and other end is connected to a small particle. Fig.(9.f)

Let `vecj` be an external impulse imparted to the particle in the direction shown in the figure. Resolving the impulse `vecj` , along and normal to the string , we have , component along the string = `J costheta` and component normal to the string = `J sintheta` , then `J' = J costheta` ( since the particle cannot move along the string) This impulse, is transmitted through the string , towards the massive support.
However , owing to the large mass of the support the impulse , cannot affect , its movement. The particle flies off, normal to the string , with a momentum `J sintheta`.

Case 2 : Both ends of the string attached to movable objects

In this case equal and opposite impulses act on the two objects, producing equal and opposite changes in momentum. The total momentum of the system therefore remains constant, although the momentum of each individual object is changed in the direction of the string. Perpendicular to the string no impulse acts and hence momentum of each object in this direction remains unchanged. shown in Fig.(9.g)

Let there be two particles A and B of masses `m_1` and `m_2` are respectively attached to the two ends of a light string. Let `vecj` be an external impulse imparted to one of the particles (A say), as shown in figure(9.g). The component of impulse along and normal to the string will be `J costheta` and `J sintheta` respectively. If v be the velocity of the two particles along the string and j' be the impulse communicated along the string then, for particle A,

`J costheta - j' = m_1v` ..... (i) and for particle B J' = `m_2 v` ............(ii)


From equations (i) and (ii) `v = (j costheta)/(m_1 + m_2)`

Further, for particle A, normal to the string `J sintheta =m_1v_1 => jsintheta//m_1`
`thereore` The final velocities , after the jerk , for the two particles A and B will be `sqrt(v_1^2 - v^2)` and v respectively.

NOTE: When two patiicles are attached to the two ends of a string , passing over a smooth peg , nail or a pulley. The above concept [corresponding to case (2)] can be extended, to such a case. The only thing to be remembered, is that, the movement of string will take place through the peg (or nail or pulley).

Let there be a system comprising of mass M (moving with speed v ) and mass dM ( moving with speed u towards M) as shown in the figure (10.a) at time t.

Total momentum at t : `Mvecv + vecu dm`

Mass dM combined with mass M in time dt , so that final mass ( M + dM) moves with velocity v + dv

Total momentum at time t + dt : (M + dM) (`vecv + dvecv`)

So the change in momentum `dvecP` is :

`dvecP = (M + dM) (vecv + dvecv) - (Mvecv + vecu dM )`

= ` M dvecv + vecv dM + dM dvecv - vecu dM`

From Newton's second law , we have

`sum vecF_(ext) = (dvecP)/(dt) = (M dvecv + vecv dM -vecu dM)/(dt)`

We have neglected the terms `(dM dvecv//dt)` in the limit of infinitesimals it is zero. Thus we get

`sum vecF_(exr) M (dvecv)/(dt) - (vecu - vecv) (dM)/(dt)`.....................(i)

Note that the quantity `(vecu-vecv)` is the relative velocity , `vecv_(rel)` of dM w.r.t. M. So we can rearrange eqn. (i)

`M (dvecv)/(dt) = sum vecF_(ext) + v_(rel) (dM)/(dt)`.........................(i i)

Where `sum F_(ext)` denotes the external force on the system ( for a rocket it would include the force of gravity and air resistance) . The force exerted by dM on M is `vecv_(rel) (dM)/(dt)` which represents the rate at which momentum is being transferred to ( or from ) the mass M (for a rocket this term is called the thrust) . The equation (i i) has application in rocket propulsion. It propels itself forward by the ejection of burnt gases. The mass M of the rocket decreased during the process, so `dM//dt < 0`. Another application is the dropping of material (gravel ) onto a conveyer belt. In this case the mass of the loaded conveyor belt increases, so `(dM)/(dt) > 0`