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Wednesday, April 17, 2019

Chapter 7.8 - Force Acting on a System of Particles

In the previous section, we saw the velocity and acceleration of the 'C'. In this section we will see it's momentum and force

1. Let a system consist of two particles P and Q
• Let both P and Q be in motion
2. Consider any instant during that motion
• In earlier sections of this chapter, we have seen that, all particles in the system need not be having the same velocity at any instant (Details here)
• So let us assume that, at that instant, P and Q are having different velocities
3. Consider the instant at which the reading in the stop-watch is 't'
    ♦ Let at that instant, the velocity vector of P be vP(t)
4. If we multiply this velocity by 'mass of P', we will get the momentum of P at that instant
So we can write: pP(t)=mPvP(t)  
5. Similarly, if the velocity of Q at that instant is vQ(t), we can write it's momentum:
pQ(t)=mQvQ(t)
6. Sum of the two momenta will give the 'total momentum of the system'. So we can write:
pSystem(t)=mPvP(t)+mQvQ(t)
7. In general, if there are n particles, we can write:
pSystem(t)=m1v1(t)+m2v2(t)+...+mivi(t)+...+mnvn(t)
8. Consider the right side of the above equation. We have seen it before
• In Eq.7.6 of the previous section, we wrote:
MvC(t)=m1v1(t)+m2v2(t)+...+mivi(t)+...+mnvn(t)
9. Comparing (7) and (8), we can write:
Eq.7.11:
pSystem(t)=MvC(t)
■ Based on this equation, we can write:
The total momentum of a system of particles is equal to the product of two items:
(i) The total mass of the system
(ii) Velocity of the 'C'

Force on a system

1. Let there be n particles in a system
• Let the system be in motion
2. Consider the instant at which the reading in the stop-watch is 't1'
    ♦ Let at that instant, the total momentum of the system be pSystem(t1)
• Consider the instant at which the reading in the stop-watch is 't2'
    ♦ Let at that instant, the total momentum of the system be pSystem(t2)
4. If we subtract the 'initial momentum vector' from the 'final momentum vector', we will get the change in momentum
• So we can write:
The 'change in momentum' suffered by the system during the time interval of (t2-t1
pSystem(t2)pSystem(t1)
5. If we divide the 'change in momentum' by the 'time interval during which the change took place', we will get the average force (Details here)
• Time interval during which the displacement took place = (t2-t1) = Δt
• So we can write:
Average force experienced by the system during the time interval Δt =
ˉFSystem(Δt)=ΔpSystem(Δt)Δt=pSystem(t2)pSystem(t1)Δt
6. Now we expand the numerator on the right side
Using Eq.7.11, we get:
pSystem(t2)=MvC(t2)
pSystem(t1)=MvC(t1)
• So the result in (5) becomes:
ˉFSystem(Δt)=MvC(t2)MvC(t1)Δt
ˉFSystem(Δt)=M[vC(t2)vC(t1)]Δt
ˉFSystem(Δt)=M[aC(Δt)]
7. If we consider an interval of time Δt, we will be getting the average force ˉFSystem(Δt)
• If we consider an instant 't', we will get the instantaneous force FSystem(t)
• Thus we can write: 
Eq.7.12: FSystem(t)=M[aC(t)]
    ♦ Where aC(t) is the acceleration experienced by the 'C' at the instant 't'
• This is in a form familiar to us: Force = mass × acceleration 
8. Based on Eq.7.12, we can write:
• The force experienced by a system of particles is equal to the product of two items:
(i) The total mass of the system
(ii) Acceleration of the 'C'
• Thus we succeeded in extending Newton's second law to a 'system of particles'

Conservation of momentum

1. Consider the equation in (5) above. We will write it again:
ˉFSystem(Δt)=pSystem(t2)pSystem(t1)Δt
• Suppose that, the system experiences no external force. Then ˉFSystem(Δt)=0
• Then the numerator will become zero
• We get: pSystem(t2)pSystem(t1)=0
pSystem(t2)=pSystem(t1)
• That means: Final momentum = Initial momentum  
• That means: There is no change in momentum
• That means: Momentum is a constant
■ Thus we get the law of conservation of momentum for the system of particles. It can be stated as:
If the total external force acting on a system of particles is zero, the total linear momentum of that system is constant
2. But from Eq.7.11, we have: pSystem(t)=MvC(t)
• If the left side is constant, the right side must also be constant
• In the right side, we have 'M' and 'vC(t)
• 'M' is already a constant because, we assume that, the mass of the system do not change
■ So we can write:
If the total external force acting on a system of particles is zero, the velocity of the 'C' of that system will be constant
• 'Constant velocity' implies that, both magnitude and direction of the velocity does not change
• So we can write:
If the total external force acting on a system of particles is zero, the 'C' of that system will travel at a constant speed along a straight line

In the next section, we will some examples where the 'C' moves with constant velocity

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