I know that without presence of any "External" force momentum is always conserved. But how do we distinguish between "External" force and "Internal" force where all are "Force"?
[Physics] How do we define what is “External” force or “Internal” force in the context of momentum conservation
definitionforcesfree-body-diagramnewtonian-mechanics
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You are studying the very basics of mechanics. In that case, as @dmckee♦ quoted, we will discuss it on the fundamental level.
Momentum is the characteristic property of a moving body. There are many dynamical variables associated with the motion of an object, like the displacement, velocity, acceleration etc. But, these quantities are just only variables of the motion. They cannot tell you the aftermath of the motion. For example, you heard someone telling a man on the pavement is hit by a car. You cannot get into a clear image unless you know the velocity of the car. Now, if you hear that the man is hit by an object at a velocity of $5 m/s$, still you didn't get a clear idea about the situation. You need to know how big that object is. If the object is a small mass like a stone, it will not hurt that much. But if it is like the size of a truck, then the injury will be maximum. So mass and velocity has to combine to give rise to a new quantity that could explain the impact a moving object could make on another. This way, we could visualize, how the energy or force is being transferred from one body to another. This new quantity which is the characteristic of a moving body is what we call the momentum of the body.
Momentum is defined as
$$\vec{p}=m\vec{v}$$
So momentum is of great importance in collision theories and all places where all we need to study about energy transfer. As you can see, the momentum is not a fundamental quantity like the displacement. It's a derived quantity. The velocity is a vector quantity and hence a scalar (mass) times a vector (velocity) gives you a new vector (momentum) in the same direction as the original. So the momentum (we are talking bout the linear momentum) is a vector quantity whose direction is determined by the velocity of the moving body. The peculiarity of a vector quantity is that it requires both direction and magnitude to represent the quantity completely. Unlike the solo quantities like displacement, velocity etc, the momentum (which is mass times velocity) tells you the effect of motion.
You can see that, when the velocity of a massive body is zero (i.e., it is at rest), then the momentum is also zero. It doesn't matter whether it's mass is huge or not; since it is not moving, it's momentum is zero. That's why we say momentum is the characteristic property of motion. Also, since the object is at rest, it will have zero kinetic energy an so the object cannot transfer it's energy to another one. So, the momentum tells you the direction along which energy transfer takes place. Let's see an example.
Suppose you are a football player. You have a ball and you just kick and it will move. So, the ball gains kinetic energy as it's velocity increased. But, from where, the kinetic energy came from? It's from the kinetic energy of your leg. When the leg collides with the ball, at the same instant, the leg imparts a momentum on the ball and by that way, the object transferred it's energy. So energy is transferred by transferring momentum. Your leg is moving with a certain velocity. So it has a well defined momentum. If it hit the ball, then the momentum is transferred to the ball. It's initial velocity was zero. So the ball gains momentum and it has now got a velocity and since it has got a velocity, we say the kinetic energy of the ball increased from zero. So energy is transferred from the leg to the ball. Also, to observe the vector property of momentum, you can see that the momentum of the ball is in the same direction as the velocity of your leg. That is, the ball fly off in the same direction as you kicked.
Now, what if the object having a less mass and less velocity hit on a wall? For example, a person running at $5 m/s$ hit on a wall. The wall will not move, of course. This means he cannot somehow impart his momentum to the wall to make it move. But something that has a large momentum like a $5000 kg$ truck travelling at a velocity of about $70 km/hr$ could impart a momentum to the wall. Since the wall is fixed to the ground (i.e., it is not designed to move) it will break.
Thus we have the fundamentals of momentum. Now we invoke the concept of force. Force is defined as something that could change the state of motion of a body. A body at rest or having constant motion is said to be in the state of inertia. Inertia is the tendency of a body to continue in it's state, whatever be it is, while force is something that tends to change that state. So inertia of a body tells you the amount of force it could oppose. For example, if you push a ball with your hand, it will move. If you push with the same amount of force on a large massive cart, it will not move. So larger objects have a greater tendency to resist force. Otherwise speaking, large force is required to make a massive object move. So mass is a measure of inertia. Hence something that changes the state of inertia (force) should also depend on mass. So force is proportional to mass. Now, when you apply a force the momentum of the object changes (since the velocity of the object changes). Hence force is also proportional to change in velocity of the object. Higher the change in velocity, greater will be the force. Hence we say the force is proportional to change in momentum with respect to time.
So, we write
$$F\propto change\space in\space momentum\space w.r.t \space time=\frac{\vec{p_2}-\vec{p_1}}{t_2-t_1}$$ $$F\propto \frac{(m\vec{v_2}-m\vec{v_1})}{t_2-t_1}$$
where $\vec{v_1}$ is the initial velocity (velocity just before the force is applied) and $\vec{v_2}$ is the final velocity (the velocity just before the force is withdrawn). The time interval ($t_2-t_1$) indicates the time up to which the force is applied. Mass is not changing with time. The only quantity that varies with time on applying a force is the momentum, which means the velocity is alone changing with time. The change in velocity with respect to time is acceleration.
So, $$F\propto \frac{m(\vec{v_2}-\vec{v_1})}{(t_2-t_1)}=m\vec{a}$$
where $\vec{a}$ is the acceleration of the body. (We have taken the constant of proportionality as unity for consistency in the units). It is also a vector quantity. So, the force is mass times acceleration of a body. Since the momentum depends on mass, force also depends on mass, as clearly understood from the above examples. Also, if the velocity of a body decreases on applying a force (i.e., $\vec{v_2}<\vec{v_1}$), the acceleration will be negative. In such a case, the force opposes the motion. Now, if the applied force increases the velocity of the body (i.e., $\vec{v_2}>\vec{v_1}$), then the acceleration is positive and so the force favors the motion of the body. Whatever be the case, the force always points in the direction of change in momentum. That can be understood by analyzing in which direction the object is accelerating.
Now, if the force applied is zero, the acceleration will be zero, as on withdrawing the applied force, the mass will not vanish. Velocity is not changing with time. That means, velocity is a constant of motion and hence the body moves with constant velocity. Since the velocity is constant, the momentum is also constant. So, if no external force acts on a body, then the total linear momentum will be a constant. This statement is known as the law of conservation of momentum.
In that case, if there are no external force on a system of dynamical objects, then the total momentum before an event will be total momentum after an event. i.e., if we have two objects in our isolated system, then the sum of the momenta of the two bodies at any time will be a constant value. The individual value could change, but the total value is a constant.
Consider your example 1 with two billiard balls, mass $0.16 \,\rm kg$, colliding with a red ball falling down with a speed of $5\, \rm m\,s^{-1}$ and colliding with a stationary white ball.
Applying conservation of linear momentum (assuming there are no external force and the collision is elastic) results in the red ball momentarily stopping and the white ball moving downwards at $5\, \rm m\,s^{-1}$.
The impulse (change of momentum) on each ball is $0.16 \times 5 = 0.8 \, \rm N\,s$.
During the collision the gravitational force would have had a effect on the balls but to know how much one must know the collision time.
Peter Bohacek has produced many Direct Measurement Videos and the relevant one for this answer is Billiard Ball Collision three consecutive frames from which are shown below.
This shows that the collision time is less than $0.001 \, \rm s$.
Going back to the falling red and white billiard balls, in a time of $0.001 \, \rm s$ the impulse due to gravity on one of the balls is $\text{~}\, 0.16 \times 10 \times 0.001 = 0.016 \,\rm Ns $ which is very much smaller than the impulse on the balls due to the collision, $ 0.8 \, \rm N\,s$.
So the assumption of a very short collision time resulting in very little effect on the outcome of the collision is a good one.
More time could have been spent analysing the video to get a more accurate upper bound for the collision time which might be the basis of a nice assignment?
Best Answer
You define a system which you are interested in.
If there is no net external force acting on the system then linear momentum is conserved.
You can identify internal forces as they must occur in equal in magnitude but opposite in direction pairs - Newton's third law.
So you find a force in the system $\:\mathbf{f}_{12}\:$ which is the force on part $1$ of the system due to part $2$ of the system which has its equal in magnitude opposite in direction twin, $\:\mathbf{f}_{21}\:$ force on part $2$ of the system due to part $1$ of the system.
There is no such pairing of forces within the system for external forces which are forces on the system due to something outside the system so their Newton's third law pair would be a force on something outside the system due to force produced by system.