In the previous section, we saw the method for finding the location of 'C' of any system of particles. In this section we will see the applications of 'C'
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 an 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 't1'
♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t2)}}$
4. If we subtract the 'initial position vector' from the 'final position vector', we will get the displacement (Details here)
• So we can write:
The displacement suffered by 'P' during the time interval of (t2-t1)
= $\mathbf\small{\vec{\Delta r}_P=\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}$
5. If we divide the displacement by the 'time interval during which the displacement took place', we will get the average velocity (Details here)
• Time interval during which the displacement took place = (t2-t1) = Δt
• So we can write:
Average velocity with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{\Delta r}_P}{\Delta t}=\frac{\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{r}_{P(t2)}}{\Delta t}-\frac{\vec{r}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average velocity with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{Q(\Delta t)}=\frac{\vec{r}_{Q(t2)}}{\Delta t}-\frac{\vec{r}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above velocities with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}=\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}$
• This is $\mathbf\small{\sum{m_i \vec{r}_{i(t2)}} }$
• That means:
(i) The position vectors of all the particles at the 'instant when reading of the stop-watch is t2' is taken
(ii) The summation is done using those position vectors
10. But we have Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The summation in the numerator of Eq.7.4 is the same summation that we wrote in (9)
11. Now, Eq,7.4 can be written as: $\mathbf\small{M\,\vec{r}_C=\sum{} \,m_i\,\vec{r}_i}$
• So the summation in (9) can be replaced by $\mathbf\small{M\,\vec{r}_{C(t2)}}$
Where $\mathbf\small{\vec{r}_{C(t2)}}$ is the position vector of 'C' at the 'instant when reading of the stop-watch is t2'
12. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
13. The 'numerator of the second term' on the right side in (12) above, can also be modified in this way. We will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\vec{r}_{C(t1)}}{\Delta t}\right]}$
14. This can be rearranged as:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{M\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}}$
15. But $\mathbf\small{\frac{\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}=\vec{\bar{v}}_{C(\Delta t)}}$
• So (13) becomes: $\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=M\,\,\vec{\bar{v}}_{C(\Delta t)}}$
In general, we can write:
Eq.7.5:
$\mathbf\small{M\,\,\vec{\bar{v}}_{C(\Delta t)}}=m_1 \,\, \vec{\bar{v}}_{1(\Delta t)}+m_2 \,\, \vec{\bar{v}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{\bar{v}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{\bar{v}}_{n(\Delta t)}$
16. In the above discussion, we have considered a time interval of Δt, which is equal to (t2-t1)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average velocities
• If this Δt is very small, the division will give instantaneous velocities
• Then the Eq.7.5 will become:
Eq.7.6:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
• All the velocities in this expression, are instantaneous velocities
• It is the velocity at the instant when the reading in the stop-watch is 't'
17. Eq.7.6 can be written as: $\mathbf\small{M\,\,\vec{v}_{C(t)}=\sum {m_i\;\vec{v}_{i(t)}}}$
• Rearranging this, we get:
Eq.7.7: $\mathbf\small{\vec{v}_{C(t)}=\frac{\sum {m_i\;\vec{v}_{i(t)}}}{M}}$
• This is an effective method to find the velocity with which the 'C' moves
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 an 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 't1'
♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the position vector of P be $\mathbf\small{\vec{r}_{P(t2)}}$
4. If we subtract the 'initial position vector' from the 'final position vector', we will get the displacement (Details here)
• So we can write:
The displacement suffered by 'P' during the time interval of (t2-t1)
= $\mathbf\small{\vec{\Delta r}_P=\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}$
5. If we divide the displacement by the 'time interval during which the displacement took place', we will get the average velocity (Details here)
• Time interval during which the displacement took place = (t2-t1) = Δt
• So we can write:
Average velocity with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{\Delta r}_P}{\Delta t}=\frac{\vec{r}_{P(t2)}-\vec{r}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{v}}_{P(\Delta t)}=\frac{\vec{r}_{P(t2)}}{\Delta t}-\frac{\vec{r}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average velocity with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{v}}_{Q(\Delta t)}=\frac{\vec{r}_{Q(t2)}}{\Delta t}-\frac{\vec{r}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above velocities with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}=\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{r}_{P(t2)}+m_Q\,\,\vec{r}_{Q(t2)}}$
• This is $\mathbf\small{\sum{m_i \vec{r}_{i(t2)}} }$
• That means:
(i) The position vectors of all the particles at the 'instant when reading of the stop-watch is t2' is taken
(ii) The summation is done using those position vectors
10. But we have Eq.7.4: $\mathbf\small{\vec{r}_C=\frac{\sum{} \,m_i\,\vec{r}_i}{M}}$
• The summation in the numerator of Eq.7.4 is the same summation that we wrote in (9)
11. Now, Eq,7.4 can be written as: $\mathbf\small{M\,\vec{r}_C=\sum{} \,m_i\,\vec{r}_i}$
• So the summation in (9) can be replaced by $\mathbf\small{M\,\vec{r}_{C(t2)}}$
Where $\mathbf\small{\vec{r}_{C(t2)}}$ is the position vector of 'C' at the 'instant when reading of the stop-watch is t2'
12. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{r}_{P(t1)}+m_Q\,\,\vec{r}_{Q(t1)}}{\Delta t}\right]}$
13. The 'numerator of the second term' on the right side in (12) above, can also be modified in this way. We will get:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{M\,\vec{r}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\vec{r}_{C(t1)}}{\Delta t}\right]}$
14. This can be rearranged as:
$\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=\frac{M\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}}$
15. But $\mathbf\small{\frac{\,[\vec{r}_{C(t2)}-\vec{r}_{C(t1)}]}{\Delta t}=\vec{\bar{v}}_{C(\Delta t)}}$
• So (13) becomes: $\mathbf\small{m_P \,\, \vec{\bar{v}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{v}}_{Q(\Delta t)}=M\,\,\vec{\bar{v}}_{C(\Delta t)}}$
In general, we can write:
Eq.7.5:
$\mathbf\small{M\,\,\vec{\bar{v}}_{C(\Delta t)}}=m_1 \,\, \vec{\bar{v}}_{1(\Delta t)}+m_2 \,\, \vec{\bar{v}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{\bar{v}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{\bar{v}}_{n(\Delta t)}$
16. In the above discussion, we have considered a time interval of Δt, which is equal to (t2-t1)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average velocities
• If this Δt is very small, the division will give instantaneous velocities
• Then the Eq.7.5 will become:
Eq.7.6:
$\mathbf\small{M\,\,\vec{v}_{C(t)}=m_1 \,\, \vec{v}_{1(t)}+m_2 \,\, \vec{v}_{2(t)}+ \,\,.\,\,.\,\,.+m_i \,\, \vec{v}_{i(t)}+\,\,.\,\,.\,\,.+m_n \,\, \vec{v}_{n(t)}}$
• All the velocities in this expression, are instantaneous velocities
• It is the velocity at the instant when the reading in the stop-watch is 't'
17. Eq.7.6 can be written as: $\mathbf\small{M\,\,\vec{v}_{C(t)}=\sum {m_i\;\vec{v}_{i(t)}}}$
• Rearranging this, we get:
Eq.7.7: $\mathbf\small{\vec{v}_{C(t)}=\frac{\sum {m_i\;\vec{v}_{i(t)}}}{M}}$
• This is an effective method to find the velocity with which the 'C' moves
So we have seen the velocity of the particles in a system. Next we will see acceleration
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 an 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 't1'
♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t2)}}$
4. If we subtract the 'initial velocity vector' from the 'final position vector', we will get the change in velocity
• So we can write:
The 'change in velocity' suffered by 'P' during the time interval of (t2-t1)
= $\mathbf\small{\vec{\Delta v}_P=\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}$
5. If we divide the 'change in velocity' by the 'time interval during which the change took place', we will get the average acceleration (Details here)
• Time interval during which the displacement took place = (t2-t1) = Δt
• So we can write:
Average acceleration with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{\Delta v}_P}{\Delta t}=\frac{\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{v}_{P(t2)}}{\Delta t}-\frac{\vec{v}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average acceleration with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{Q(\Delta t)}=\frac{\vec{v}_{Q(t2)}}{\Delta t}-\frac{\vec{v}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above accelerations with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}=\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}+m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}$
• From Eq.7.6, this is $\mathbf\small{M\,\,\vec{v}_{C(t2)}}$
• Similarly, the numerator of the second term on the right side will become:
$\mathbf\small{M\,\,\vec{v}_{C(t1)}}$
10. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\left[\frac{M\,\,\vec{v}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\,\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\left[\frac{\vec{v}_{C(t2)}-\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
11. From Newton's second law, We have:
Force = mass × acceleration
• In our present case, the acceleration is average acceleration. So we can write:
Average force = mass × average acceleration
• Thus the result in (10) becomes:
$\mathbf\small{\vec{\bar{F}}_{P(\Delta t)}+\vec{\bar{F}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
• That means:
Average force acting on particle P + Average force acting on particle Q
= [Total mass × Average acceleration of the 'C']
In general we can write:
Eq.7.8:
[Total mass × Average acceleration of the 'C']
= $\mathbf\small{\vec{\bar{F}}_{1(\Delta t)}+\vec{\bar{F}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+\vec{\bar{F}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+\vec{\bar{F}}_{n(\Delta t)}}$
11. In the above discussion, we have considered a time interval of Δt, which is equal to (t2-t1)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average acceleration
• If this Δt is very small, the division will give instantaneous acceleration
• Then the Eq.7.8 will become:
Eq.7.9:
[Total mass × Instantaneous acceleration of the 'C']
= $\mathbf\small{\vec{F}_{1(t)}+\vec{F}_{2(t)}+ \,\,.\,\,.\,\,.+\vec{F}_{i(t)}+\,\,.\,\,.\,\,.+\vec{F}_{n(t)}}$
• All the forces in this expression, are instantaneous forces
• It is the force at the instant when the reading in the stop-watch is 't'
So we can write:
The following two quantities are equal:
(i) The product of the 'total mass' and 'acceleration experienced by 'C''
(ii) The vector sum of all the forces acting on the particles in the system
12. Note:
• $\mathbf\small{\vec{F}_{1(t)}}$ is the force acting on 'particle 1' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 1'. We must calculate the resultant of all those forces
Like wise:
• $\mathbf\small{\vec{F}_{2(t)}}$ is the force acting on 'particle 2' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 2'. We must calculate the resultant of all those forces
• Like wise, we can write for all other particles also
13. This leads us to an interesting result. It can be explained with the help of an example:
(i) A system consists of 2 particles P and Q
• Forces acting on P are: $\mathbf\small{\vec{F}_{P1},\,\vec{F}_{P2}\, \text{and}\,\vec{F}_{P3}}$
• In addition to the above forces, the particle Q also exerts a force on P. This force is $\mathbf\small{\vec{F}_{PQ}}$
♦ This force can be considered as an internal force
♦ Because it is exerted between particles inside the system
• So the net force acting on P is: $\mathbf\small{\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ}}$
(ii) Forces acting on Q are: $\mathbf\small{\vec{F}_{Q1},\,\vec{F}_{Q2}\, \text{and}\,\vec{F}_{Q3}}$
• In addition to the above forces, the particle P also exerts a force on Q. This force is $\mathbf\small{\vec{F}_{QP}}$
♦ This force can be considered as an internal force
♦ Because it is exerted between particles inside the system
• So the net force acting on Q is: $\mathbf\small{\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP}}$
[mP+mQ) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP})}$
(iv) But according to Newton's third law, $\mathbf\small{\vec{F}_{PQ}=-\vec{F}_{QP}}$
• So those two internal forces will cancel each other
• Thus we get:
[mP+mQ) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3})}$
■ That means, while applying Eqs.7.8 and 7.9, internal forces in the system have no role to play
14. Consider all the external forces acting on the particles of a system
• Let us denote the vector sum of all the external forces as $\mathbf\small{\vec{F}_{ext}}$
• Also let us denote acceleration of the 'C' as $\mathbf\small{\vec{A}_C}$
• Then Eq.7.9 becomes:
Eq.7.10: $\mathbf\small{\text{M}\;\vec{A}_C=\vec{F}_{ext}}$
• On the left side, we have: Total mass multiplied by the acceleration experienced by 'C'
• On the right side, we have the vector sum of all external forces acting on the system. That is., the net external force
■ So we can write:
• The net external force will produce the following effect:
Movement of the 'C' with an acceleration, as if, all the mass of the system is concentrated at the 'C'
16. It may be noted that, if the net force is not acting at 'C', the system will rotate
• We did not consider any rotational motions of objects in the previous chapters
• There was no reason to consider rotation because, we assumed that the net force acts at the 'C'
1. Consider a projectile shown in fig.7.45(a) below:
• We know that, it's path will be a parabola
♦ This is indicated by the yellow curve
2. At point 'A', the projectile is intact
• But when it reaches point 'B', it explodes into two pieces
• Let us call them 'Part P' and 'Part Q'
3. We know that, the only force acting on a projectile is the 'gravitational force'
• This force acts vertically
• There is no horizontal force (Details here)
4. Let us assume that, at the time of launch, P and Q were glued together
• This is shown in fig.b
• The glued combination will have a definite 'C'
♦ A force of (mP × g) acts on P
♦ A force of (mQ × g) acts on Q
• The glued combination moves as if [(mP × g)+(mQ × g)] is acting at 'C' of the combination
5. The combination explodes when it reaches B
(The explosion may be due to the 'ignition of some explosives' kept at the interface between P and Q)
• Beyond B, the parts P and Q move as a system
• Beyond B:
♦ the force acting on P is (mP × g)
♦ the force acting on Q is (mQ × g)
• So we see that, the forces acting on each part remains the same even after collision
6. P will change course because a force due to explosion is exerted on it
• Q will also change course because a force due to explosion is exerted on it
• But those forces are internal forces and have no contribution on the net force
7. The only external forces are:
• (mP × g) acting on P
• (mQ × g)acting on Q
• These forces remain the same before and after the explosion
■ So force acting at 'C' remains the same before and after the explosion
■ Since there is no change in the external force, the force acting at 'C', does not change
■ So the course of 'C' will not change
■ That means, the location of 'C' will continue as if no explosion have occurred
The following solved example will make this point clear:
Solved example 7.8
A projectile explodes into two pieces P and Q. The explosion occurred at the top most point of it's trajectory. The horizontal distance between the 'launch point' and the 'point of explosion' is x0. The larger piece Q has 3 times the mass of the smaller piece P. The smaller piece P lands back at the launch point.
(a) Where does the 'C' of the system land ?
(b) Where does 'Q' land?
Solution:
1. The 'C' of the projectile always lands at a distance of 'R' from the launch point
This is shown in fig.7.46(a) below:
• Even if an explosion occur, the 'C' of the fragments will land at the same point
This is shown in fig.b
2. Given that, the explosion occurred at a horizontal distance of x0 from the launch point
• Also given that, the explosion occurred at the top of the trajectory
3. The top of a parabolic trajectory is at a horizontal distance of $\mathbf\small{\frac{R}{2}}$ from the launch point
• Thus we get: $\mathbf\small{x_0=\frac{R}{2}}$
$\mathbf\small{\Rightarrow R=2x_0}$
• This is the answer for part (a)
4. Now we have the final position of 'C'
• We are given the final position of one of the pieces:
The smaller piece P lands back at the launch point
(The path followed by P is shown as the green dashed curve in fig.c)
5. So we have the position of 'C'
• And we have the position of one of the pieces
6. The position of 'C' is given by: $\mathbf\small{X=\frac{\sum m_i x_i}{M}}$
• To apply this formula, we want a reference frame
• The reference frame shown in fig.7.46 is that of the projectile
• We will use it for 'C' also
• Thus we have:
♦ X = R = 2x0
♦ m1 = mP
♦ m2 = mQ = 3mP
♦ x1 = xP = 0
♦ x2 = xQ = ? (The path followed by Q is shown as the red dashed curve in fig.c)
♦ M = mP + mQ = 4mP
• Substituting these values, we get:
$\mathbf\small{2x_0=\frac{m_P\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_P\times 0+3m_P\,x_Q}{4m_P}=\frac{3}{4}x_Q}$
$\mathbf\small{\Rightarrow x_Q=\frac{8}{3}x_0}$
• This is the answer for part (b)
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 an 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 't1'
♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t1)}}$
• Consider the instant at which the reading in the stop-watch is 't2'
♦ Let at that instant, the velocity vector of P be $\mathbf\small{\vec{v}_{P(t2)}}$
4. If we subtract the 'initial velocity vector' from the 'final position vector', we will get the change in velocity
• So we can write:
The 'change in velocity' suffered by 'P' during the time interval of (t2-t1)
= $\mathbf\small{\vec{\Delta v}_P=\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}$
5. If we divide the 'change in velocity' by the 'time interval during which the change took place', we will get the average acceleration (Details here)
• Time interval during which the displacement took place = (t2-t1) = Δt
• So we can write:
Average acceleration with which P traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{\Delta v}_P}{\Delta t}=\frac{\vec{v}_{P(t2)}-\vec{v}_{P(t1)}}{\Delta t}}$
$\mathbf\small{\Rightarrow \vec{\bar{a}}_{P(\Delta t)}=\frac{\vec{v}_{P(t2)}}{\Delta t}-\frac{\vec{v}_{P(t1)}}{\Delta t}}$
6. In the same way, we will get:
• Average acceleration with which Q traveled during the time interval Δt =
$\mathbf\small{\vec{\bar{a}}_{Q(\Delta t)}=\frac{\vec{v}_{Q(t2)}}{\Delta t}-\frac{\vec{v}_{Q(t1)}}{\Delta t}}$
7. Let us multiply the above accelerations with the respective masses and then add
• For particle P, we will get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}=\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}}$
• For particle Q, we will get:
$\mathbf\small{m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}}$
8. Adding the two, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}-\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}\right]+\left[\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}-\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
• Rearranging this, we get:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}}{\Delta t}+\frac{m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}}$
$\mathbf\small{=\left[\frac{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}{\Delta t}\right]-\left[\frac{m_P\,\,\vec{v}_{P(t1)}+m_Q\,\,\vec{v}_{Q(t1)}}{\Delta t}\right]}$
9. Consider the 'numerator of the first term' on the right side:
$\mathbf\small{m_P\,\,\vec{v}_{P(t2)}+m_Q\,\,\vec{v}_{Q(t2)}}$
• From Eq.7.6, this is $\mathbf\small{M\,\,\vec{v}_{C(t2)}}$
• Similarly, the numerator of the second term on the right side will become:
$\mathbf\small{M\,\,\vec{v}_{C(t1)}}$
10. So the result in (8) will become:
$\mathbf\small{m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=\left[\frac{M\,\,\vec{v}_{C(t2)}}{\Delta t}\right]-\left[\frac{M\,\,\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\left[\frac{\vec{v}_{C(t2)}-\vec{v}_{C(t1)}}{\Delta t}\right]}$
$\mathbf\small{\Rightarrow m_P \,\, \vec{\bar{a}}_{P(\Delta t)}+m_Q \,\, \vec{\bar{a}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
11. From Newton's second law, We have:
Force = mass × acceleration
• In our present case, the acceleration is average acceleration. So we can write:
Average force = mass × average acceleration
• Thus the result in (10) becomes:
$\mathbf\small{\vec{\bar{F}}_{P(\Delta t)}+\vec{\bar{F}}_{Q(\Delta t)}=M\,\, \vec{\bar{a}}_{C(\Delta t)}}$
• That means:
Average force acting on particle P + Average force acting on particle Q
= [Total mass × Average acceleration of the 'C']
In general we can write:
Eq.7.8:
[Total mass × Average acceleration of the 'C']
= $\mathbf\small{\vec{\bar{F}}_{1(\Delta t)}+\vec{\bar{F}}_{2(\Delta t)}+ \,\,.\,\,.\,\,.+\vec{\bar{F}}_{i(\Delta t)}+\,\,.\,\,.\,\,.+\vec{\bar{F}}_{n(\Delta t)}}$
11. In the above discussion, we have considered a time interval of Δt, which is equal to (t2-t1)
• If the particles are travelling with non-uniform velocities, division by this Δt will give average acceleration
• If this Δt is very small, the division will give instantaneous acceleration
• Then the Eq.7.8 will become:
Eq.7.9:
[Total mass × Instantaneous acceleration of the 'C']
= $\mathbf\small{\vec{F}_{1(t)}+\vec{F}_{2(t)}+ \,\,.\,\,.\,\,.+\vec{F}_{i(t)}+\,\,.\,\,.\,\,.+\vec{F}_{n(t)}}$
• All the forces in this expression, are instantaneous forces
• It is the force at the instant when the reading in the stop-watch is 't'
So we can write:
The following two quantities are equal:
(i) The product of the 'total mass' and 'acceleration experienced by 'C''
(ii) The vector sum of all the forces acting on the particles in the system
12. Note:
• $\mathbf\small{\vec{F}_{1(t)}}$ is the force acting on 'particle 1' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 1'. We must calculate the resultant of all those forces
Like wise:
• $\mathbf\small{\vec{F}_{2(t)}}$ is the force acting on 'particle 2' at the instant 't'
• But this force itself is a vector sum
■ Many forces may be acting on 'particle 2'. We must calculate the resultant of all those forces
• Like wise, we can write for all other particles also
13. This leads us to an interesting result. It can be explained with the help of an example:
(i) A system consists of 2 particles P and Q
• Forces acting on P are: $\mathbf\small{\vec{F}_{P1},\,\vec{F}_{P2}\, \text{and}\,\vec{F}_{P3}}$
• In addition to the above forces, the particle Q also exerts a force on P. This force is $\mathbf\small{\vec{F}_{PQ}}$
♦ This force can be considered as an internal force
♦ Because it is exerted between particles inside the system
• So the net force acting on P is: $\mathbf\small{\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ}}$
(ii) Forces acting on Q are: $\mathbf\small{\vec{F}_{Q1},\,\vec{F}_{Q2}\, \text{and}\,\vec{F}_{Q3}}$
• In addition to the above forces, the particle P also exerts a force on Q. This force is $\mathbf\small{\vec{F}_{QP}}$
♦ This force can be considered as an internal force
♦ Because it is exerted between particles inside the system
• So the net force acting on Q is: $\mathbf\small{\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP}}$
(iii) Once we know all the forces acting on each particles, we can apply Eq.7.9
We get:[mP+mQ) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3}+\vec{F}_{PQ})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3}+\vec{F}_{QP})}$
(iv) But according to Newton's third law, $\mathbf\small{\vec{F}_{PQ}=-\vec{F}_{QP}}$
• So those two internal forces will cancel each other
• Thus we get:
[mP+mQ) × Acceleration of the 'C'] = $\mathbf\small{(\vec{F}_{P1}+\vec{F}_{P2}+\vec{F}_{P3})+(\vec{F}_{Q1}+\vec{F}_{Q2}+\vec{F}_{Q3})}$
■ That means, while applying Eqs.7.8 and 7.9, internal forces in the system have no role to play
14. Consider all the external forces acting on the particles of a system
• Let us denote the vector sum of all the external forces as $\mathbf\small{\vec{F}_{ext}}$
• Also let us denote acceleration of the 'C' as $\mathbf\small{\vec{A}_C}$
• Then Eq.7.9 becomes:
Eq.7.10: $\mathbf\small{\text{M}\;\vec{A}_C=\vec{F}_{ext}}$
• On the left side, we have: Total mass multiplied by the acceleration experienced by 'C'
• On the right side, we have the vector sum of all external forces acting on the system. That is., the net external force
■ So we can write:
• The net external force will produce the following effect:
Movement of the 'C' with an acceleration, as if, all the mass of the system is concentrated at the 'C'
• This is the reason why, in the previous chapters, we were able to do problems by considering, wooden blocks, cars, trucks etc., as 'point masses'
• We did not consider any rotational motions of objects in the previous chapters
• There was no reason to consider rotation because, we assumed that the net force acts at the 'C'
A practical example:
 |
| Fig.7.45 |
♦ This is indicated by the yellow curve
2. At point 'A', the projectile is intact
• But when it reaches point 'B', it explodes into two pieces
• Let us call them 'Part P' and 'Part Q'
3. We know that, the only force acting on a projectile is the 'gravitational force'
• This force acts vertically
• There is no horizontal force (Details here)
4. Let us assume that, at the time of launch, P and Q were glued together
• This is shown in fig.b
• The glued combination will have a definite 'C'
♦ A force of (mP × g) acts on P
♦ A force of (mQ × g) acts on Q
• The glued combination moves as if [(mP × g)+(mQ × g)] is acting at 'C' of the combination
5. The combination explodes when it reaches B
(The explosion may be due to the 'ignition of some explosives' kept at the interface between P and Q)
• Beyond B, the parts P and Q move as a system
• Beyond B:
♦ the force acting on P is (mP × g)
♦ the force acting on Q is (mQ × g)
• So we see that, the forces acting on each part remains the same even after collision
6. P will change course because a force due to explosion is exerted on it
• Q will also change course because a force due to explosion is exerted on it
• But those forces are internal forces and have no contribution on the net force
7. The only external forces are:
• (mP × g) acting on P
• (mQ × g)acting on Q
• These forces remain the same before and after the explosion
■ So force acting at 'C' remains the same before and after the explosion
■ Since there is no change in the external force, the force acting at 'C', does not change
■ So the course of 'C' will not change
■ That means, the location of 'C' will continue as if no explosion have occurred
The following solved example will make this point clear:
Solved example 7.8
A projectile explodes into two pieces P and Q. The explosion occurred at the top most point of it's trajectory. The horizontal distance between the 'launch point' and the 'point of explosion' is x0. The larger piece Q has 3 times the mass of the smaller piece P. The smaller piece P lands back at the launch point.
(a) Where does the 'C' of the system land ?
(b) Where does 'Q' land?
Solution:
1. The 'C' of the projectile always lands at a distance of 'R' from the launch point
This is shown in fig.7.46(a) below:
 |
| Fig.7.46 |
This is shown in fig.b
2. Given that, the explosion occurred at a horizontal distance of x0 from the launch point
• Also given that, the explosion occurred at the top of the trajectory
3. The top of a parabolic trajectory is at a horizontal distance of $\mathbf\small{\frac{R}{2}}$ from the launch point
• Thus we get: $\mathbf\small{x_0=\frac{R}{2}}$
$\mathbf\small{\Rightarrow R=2x_0}$
• This is the answer for part (a)
4. Now we have the final position of 'C'
• We are given the final position of one of the pieces:
The smaller piece P lands back at the launch point
(The path followed by P is shown as the green dashed curve in fig.c)
5. So we have the position of 'C'
• And we have the position of one of the pieces
6. The position of 'C' is given by: $\mathbf\small{X=\frac{\sum m_i x_i}{M}}$
• To apply this formula, we want a reference frame
• The reference frame shown in fig.7.46 is that of the projectile
• We will use it for 'C' also
• Thus we have:
♦ X = R = 2x0
♦ m1 = mP
♦ m2 = mQ = 3mP
♦ x1 = xP = 0
♦ x2 = xQ = ? (The path followed by Q is shown as the red dashed curve in fig.c)
♦ M = mP + mQ = 4mP
• Substituting these values, we get:
$\mathbf\small{2x_0=\frac{m_P\,x_P+m_Q\,x_Q}{m_P+m_Q}=\frac{m_P\times 0+3m_P\,x_Q}{4m_P}=\frac{3}{4}x_Q}$
$\mathbf\small{\Rightarrow x_Q=\frac{8}{3}x_0}$
• This is the answer for part (b)
So we have seen velocity and force on a system of particles. In the next section, we will see momentum
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