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 $\mathbf\small{\vec{v}_{P(t)}}$
4. If we multiply this velocity by 'mass of P', we will get the momentum of P at that instant
So we can write: $\mathbf\small{\vec{p}_{P(t)}=m_P\,\vec{v}_{P(t)}}$
5. Similarly, if the velocity of Q at that instant is $\mathbf\small{\vec{v}_{Q(t)}}$, we can write it's momentum:
$\mathbf\small{\vec{p}_{Q(t)}=m_Q\,\vec{v}_{Q(t)}}$
6. Sum of the two momenta will give the 'total momentum of the system'. So we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_P\,\vec{v}_{P(t)}+m_Q\,\vec{v}_{Q(t)}}$
7. In general, if there are n particles, we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
8. Consider the right side of the above equation. We have seen it before
• In Eq.7.6 of the previous section, we wrote:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
9. Comparing (7) and (8), we can write:
Eq.7.11:
$\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
■ 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'
• 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 $\mathbf\small{\vec{p}_{System(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the total momentum of the system be $\mathbf\small{\vec{p}_{System(t2)}}$
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)
= $\mathbf\small{\vec{p}_{System(t2)}-\;\vec{p}_{System(t1)}}$
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 =
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{\Delta p}_{System(\Delta t)}}{\Delta t}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
6. Now we expand the numerator on the right side
Using Eq.7.11, we get:
$\mathbf\small{\vec{p}_{System(t2)}=M\,\,\vec{v}_{C(t2)}}$
$\mathbf\small{\vec{p}_{System(t1)}=M\,\,\vec{v}_{C(t1)}}$
• So the result in (5) becomes:
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{M\,\,\vec{v}_{C(t2)}-M\,\,\vec{v}_{C(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=\frac{M[\vec{v}_{C(t2)}-\,\,\vec{v}_{C(t1)}]}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=M[\vec{a}_{C(\Delta t)}]}$
7. If we consider an interval of time Δt, we will be getting the average force $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}}$
• If we consider an instant 't', we will get the instantaneous force $\mathbf\small{\vec{F}_{System(t)}}$
• Thus we can write:
Eq.7.12: $\mathbf\small{\Rightarrow \vec{F}_{System(t)}=M[\vec{a}_{C(t)}]}$
♦ Where $\mathbf\small{\vec{a}_{C(t)}}$ 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'
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
• Suppose that, the system experiences no external force. Then $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=0}$
• Then the numerator will become zero
• We get: $\mathbf\small{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}=0}$
$\mathbf\small{\Rightarrow \vec{p}_{System(t2)}=\vec{p}_{System(t1)}}$
• 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: $\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
• If the left side is constant, the right side must also be constant
• In the right side, we have 'M' and '$\mathbf\small{\vec{v}_{C(t)}}$'
• '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
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 $\mathbf\small{\vec{v}_{P(t)}}$
4. If we multiply this velocity by 'mass of P', we will get the momentum of P at that instant
So we can write: $\mathbf\small{\vec{p}_{P(t)}=m_P\,\vec{v}_{P(t)}}$
5. Similarly, if the velocity of Q at that instant is $\mathbf\small{\vec{v}_{Q(t)}}$, we can write it's momentum:
$\mathbf\small{\vec{p}_{Q(t)}=m_Q\,\vec{v}_{Q(t)}}$
6. Sum of the two momenta will give the 'total momentum of the system'. So we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_P\,\vec{v}_{P(t)}+m_Q\,\vec{v}_{Q(t)}}$
7. In general, if there are n particles, we can write:
$\mathbf\small{\vec{p}_{System(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
8. Consider the right side of the above equation. We have seen it before
• In Eq.7.6 of the previous section, we wrote:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
9. Comparing (7) and (8), we can write:
Eq.7.11:
$\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
■ 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 $\mathbf\small{\vec{p}_{System(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the total momentum of the system be $\mathbf\small{\vec{p}_{System(t2)}}$
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)
= $\mathbf\small{\vec{p}_{System(t2)}-\;\vec{p}_{System(t1)}}$
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 =
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{\Delta p}_{System(\Delta t)}}{\Delta t}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
6. Now we expand the numerator on the right side
Using Eq.7.11, we get:
$\mathbf\small{\vec{p}_{System(t2)}=M\,\,\vec{v}_{C(t2)}}$
$\mathbf\small{\vec{p}_{System(t1)}=M\,\,\vec{v}_{C(t1)}}$
• So the result in (5) becomes:
$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{M\,\,\vec{v}_{C(t2)}-M\,\,\vec{v}_{C(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=\frac{M[\vec{v}_{C(t2)}-\,\,\vec{v}_{C(t1)}]}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{F}}_{System(\Delta t)}=M[\vec{a}_{C(\Delta t)}]}$
7. If we consider an interval of time Δt, we will be getting the average force $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}}$
• If we consider an instant 't', we will get the instantaneous force $\mathbf\small{\vec{F}_{System(t)}}$
• Thus we can write:
Eq.7.12: $\mathbf\small{\Rightarrow \vec{F}_{System(t)}=M[\vec{a}_{C(t)}]}$
♦ Where $\mathbf\small{\vec{a}_{C(t)}}$ 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:• Suppose that, the system experiences no external force. Then $\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=0}$
• Then the numerator will become zero
• We get: $\mathbf\small{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}=0}$
$\mathbf\small{\Rightarrow \vec{p}_{System(t2)}=\vec{p}_{System(t1)}}$
• 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: $\mathbf\small{\vec{p}_{System(t)}=M\,\,\vec{v}_{C(t)}}$
• If the left side is constant, the right side must also be constant
• In the right side, we have 'M' and '$\mathbf\small{\vec{v}_{C(t)}}$'
• '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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