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 $\mathbf\small{\vec{v}_{P(t)}}$
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'

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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'

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Conservation of momentum

1. Consider the equation in (5) above. We will write it again:

$\mathbf\small{\vec{\bar{F}}_{System(\Delta t)}=\frac{\vec{p}_{System(t2)}-\vec{p}_{System(t1)}}{\Delta t}}$
• 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

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

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Chapter 5.9 - Conservation of Momentum

In the previous section we saw how to find the apparent weight in a lift. In this section we will see another situation where there is a combined application of the Second and Third laws of motion.

Conservation of momentum

1. A bullet is fired from a gun. See fig.5.24 below:

Fig.5.24

We are given the following information:
• m_G is the mass of the gun
• m_B is the mass of the bullet
• $\mathbf\small{\vec{v_B}}$ is the velocity with which the bullet is fired
• $\mathbf\small{\vec{v_G}}$ is the recoil velocity of the gun
2.Let us assume that, the motion of the bullet is along the x axis
• We can begin the calculations:
• First we do the calculations for the bullet:
x direction:
(i) Initial momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx1}}=m_B \times \vec{v_{Bx1}}=m_B \times 0=0}$
(ii) Final momentum of the bullet in the x direction = $\mathbf\small{\vec{p_{Bx2}}=m_B \times \vec{v_{Bx2}}=m_B \times \vec{v_{B}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Bx}}=(\vec{p_{Bx2}}-\vec{p_{Bx1}})=(m_B  \vec{v_{B}}-0)=m_B  \vec{v_{B}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{By}}}$ = 0

3. Resultant change in momentum $\mathbf\small{\Delta \vec{p_B}}$ is the resultant of  $\mathbf\small{\Delta \vec{p_{Bx}}}$ and  $\mathbf\small{\Delta \vec{p_{By}}}$     
• But  $\mathbf\small{\Delta \vec{p_{By}}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p_B}=\Delta \vec{p_{Bx}}=m_B  \vec{v_{B}}}$ 
4. We can write the following four points:
(i) The bullet suffers a change in momentum of $\mathbf\small{m_B  \vec{v_{B}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_B\, |\vec{v_B}|}$
(iv) It's  direction is towards the positive side of the x axis
5. Now we do the calculations for the gun:
x direction:
(i) Initial momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx1}}=m_G \times \vec{v_{Gx1}}=m_G \times 0=0}$
(ii) Final momentum of the gun in the x direction = $\mathbf\small{\vec{p_{Gx2}}=m_G \times \vec{v_{Gx2}}=-m_G \times \vec{v_{G}}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{Gx}}=(\vec{p_{Gx2}}-\vec{p_{Gx1}})=(-m_G  \vec{v_{G}}-0)=-m_G  \vec{v_{G}}}$
y direction:
• We do not need to worry about the y direction because, there is no motion along that direction
• We can straight away write: $\mathbf\small{\Delta \vec{p_{Gy}}}$ = 0
6. Resultant change in momentum $\mathbf\small{\Delta \vec{p_G}}$ is the resultant of  $\mathbf\small{\Delta \vec{p_{Gx}}}$ and  $\mathbf\small{\Delta \vec{p_{Gy}}}$     

• But  $\mathbf\small{\Delta \vec{p_{Gy}}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p_G}=\Delta \vec{p_{Gx}}=-m_B  \vec{v_{G}}}$ 
7. We can write the following four points:
(i) The gun suffers a change in momentum of $\mathbf\small{-m_G  \vec{v_{G}}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{m_G\, |\vec{v_G}|}$
(iv) It's  direction is towards the negative side of the x axis

8. The bullet suffers a change in momentum because, a force acted on it
Obviously, this force was exerted by the gun 
• Let us denote this force (which is applied on the bullet by the gun) as $\mathbf\small{\vec{F_{BG}}}$

• Let the time duration for which the force acts be Δt

• Consider the product: $\mathbf\small{\vec{F_{BG}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{BG}}\times\Delta t=m_B  \vec{v_{B}}}$
10. The gun suffers a change in momentum because, a force acted on it
Obviously, this force was exerted by the bullet
• Let us denote this force (which is applied on the gun by the bullet) as $\mathbf\small{\vec{F_{GB}}}$

• The time duration for which the force acts will be the same Δt

• Consider the product: $\mathbf\small{\vec{F_{GB}}\times\Delta t}$

11. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{GB}}\times\Delta t=-m_G  \vec{v_{G}}}$
12. So we have the following two information:
(i) The bullet suffers a change in momentum $\mathbf\small{\vec{\Delta p_B}}$:
$\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t=m_B  \vec{v_{B}}}$
(ii) The gun suffers a change in momentum $\mathbf\small{\vec{\Delta p_G}}$:
$\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t=-m_G  \vec{v_{G}}}$
13. Take out the first and second part from 12(i) and 12(ii):
(i) $\mathbf\small{\vec{\Delta p_B}=\vec{F_{BG}}\times\Delta t}$
(ii) $\mathbf\small{\vec{\Delta p_G}=\vec{F_{GB}}\times\Delta t}$
14. Applying Newton's third law, we have:
$\mathbf\small{\vec{F_{BG}}}$ = -$\mathbf\small{\vec{F_{GB}}}$
15. So we can replace $\mathbf\small{\vec{F_{BG}}}$ in 13(i). We get:
$\mathbf\small{\vec{\Delta p_B}=-\vec{F_{GB}}\times\Delta t}$ 
• But from 13(ii), $\mathbf\small{-\vec{F_{GB}}\times\Delta t}$ = $\mathbf\small{-\vec{\Delta p_G}}$   
• So we get: $\mathbf\small{\vec{\Delta p_B}=-\vec{\Delta p_G}}$
■ That is., 'change in momenta' are numerically equal, but have opposite signs
16. Now, the change in momentum for the bullet can be written as:
• Change in momentum of bullet = Final momentum of bullet - Initial momentum of bullet   
• That is., $\mathbf\small{\vec{\Delta p_B}=[\vec{p_{B1}}-\vec{p_{B2}}]}$
17. Similarly, the change in momentum for the gun can be written as:
• Change in momentum of gun = Final momentum of gun - Initial momentum of gun   
• That is., $\mathbf\small{\vec{\Delta p_G}=[\vec{p_{G1}}-\vec{p_{G2}}]}$
18. But according to the result in (15), the two 'change in momenta' are numerically equal, but have opposite signs
• So we can change the sign of (17) and equate it to (16). We get:
$\mathbf\small{[\vec{p_{B1}}-\vec{p_{B2}}]}= {-[\vec{p_{G1}}-\vec{p_{G2}}]}$ 
$\mathbf\small{\Rightarrow \vec{p_{B1}}-\vec{p_{B2}}}= {-\vec{p_{G1}}-\vec{p_{G2}}}$
$\mathbf\small{\Rightarrow \vec{p_{B1}}+\vec{p_{G1}}}= {\vec{p_{B2}}+\vec{p_{G2}}}$
19. On the left side, we have:
• Sum of momenta before the firing of the gun
On the right side, we have:
• Sum of momenta after the firing of the gun
• So we can write: The total momentum of the (bullet + gun) system is conserved
20. The Law of Conservation of momentum states that:
■ The total momentum of an isolated system of interacting particles is conserved

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Note: The 'Recoil of a gun' is essentially a one-dimensional problem. We do not need to use vector notations. But in the above discussion we used them, to get a general understanding of the Law of conservation of momentum

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Solved example 5.10
A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is 80 ms^-1, what is the recoil speed of the gun?
Solution:
• For this problem, there is one-dimensional motion only. So we do not need to use vector notations.
1. Applying the law of conservation of momentum, we have:
p_S1 + p_G1 = p_S2 + p_G2
2. But p_S1 = p_G1 = 0
• So we get: p_S2 = -p_G2
⟹ m_S × v_S2 = m_G × v_G2
3. Substituting the values, we get:
0.020 × 80 = 100 × v_G2
⟹ v_G2 = 0.016 ms^-1.

Solved example 5.11
In a quarry, a stationary rock explodes into 3 parts. A 1 kg part and another 2 kg part moves at right angles to each other. Their velocities are respectively 12 ms^-1 and 8 ms^-1. If the remaining third part moves with a velocity of 4 ms^-1, what is it's mass?
Solution:
• Let the first two parts be named as 'A' and 'B' and the third part be named as 'C'. This is shown in fig.5.25(b) below:

Fig.5.25

• The given data can be written as:
$\mathbf\small{m_A=1\, \text{kg},\: m_B=2\,\,  \text{kg},\: \vec{v_{A2}}=12\, \,ms^{-1},\: \vec{v_{B2}}=8\, \,ms^{-1},\: \vec{v_{C2}}=4\, \,ms^{-1},\: m_C=?}$
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{A1}}+\vec{p_{B1}}+\vec{p_{C1}}}= {\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}}$
2. But the initial momenta are all zero. So we get:
$\mathbf\small{\vec{p_{A2}}+\vec{p_{B2}}+\vec{p_{C2}}=0}$
$\mathbf\small{\Rightarrow \vec{p_{A2}}+\vec{p_{B2}}=-\vec{p_{C2}}}$
• That is., the vector $\mathbf\small{\vec{p_{C2}}}$ must be equal and opposite to the resultant of the two vectors $\mathbf\small{\vec{p_{A2}}}$ and $\mathbf\small{\vec{p_{B2}}}$
3. We will denote the resultant as $\mathbf\small{\vec{R}}$
We can write: $\mathbf\small{\vec{R}=(\vec{p_{A2}}+\vec{p_{B2}})}$
$\mathbf\small{\Rightarrow \vec{R}=-\vec{p_{C2}}}$
4. So let us find $\mathbf\small{\vec{R}}$ first:
• Given that, the two masses A and B move perpendicular to each other
• So their momentum vectors are perpendicular to each other. This is shown in fig.5.25(c) above
5. If two vectors are perpendicular to each other, their resultant can be easily calculated.
• We have:
The magnitude of the resultant = $\mathbf\small{|\vec{R}|=|(\vec{p_{A2}}+\vec{p_{B2}})|=\sqrt{|\vec{p_{A2}}|^2+|\vec{p_{B2}}|^2}}$   
• Where:
$\mathbf\small{|\vec{p_{A2}}|=m_A \times |\vec{v_{A2}}|}$ = 1 × 12 = 12 kg ms^-1.
$\mathbf\small{|\vec{p_{B2}}|=m_B \times |\vec{v_{B2}}|}$ = 2 × 8 = 16 kg ms^-1.
• Thus we get:
$\mathbf\small{|\vec{R}|=\sqrt{12^2+16^2}=20\: kg\:  ms^{-1}}$
6. Now consider the result in (3):
• We have an equality: $\mathbf\small{\vec{R}=-\vec{p_{C2}}}$
• This is shown in fig.5.25(d)
• We can write two points:
(i) Magnitude of ${\mathbf\small\vec{R}}$ is same as the magnitude of $\mathbf{\small\vec{p_{C2}}}$   
(ii) Direction of $\mathbf{\small\vec{R}}$ is opposite to the direction of $\mathbf\small{\vec{p_{C2}}}$
7. The first point is important to us. Based on that, we can write:
$\mathbf\small{|\vec{p_{C2}}|=20\: kg\: ms^{-1}}$
• But $\mathbf\small{|\vec{p_{C2}}|=m_C \times |\vec{v_{C2}}|=m_C \times 4}$
• So we get: (m_c × 4) = 20 ⇒ m_c = 5 kg

Solved example 5.12
Two objects, each of mass 5 kg move in a straight line towards each other with the same speed of 3 ms^-1. After collision, they stick together. What will be the velocity of the combined mass after collision?
Solution:
• For this problem, there is one-dimensional motion only. See fig.5.26 below:

Fig.5.26

• So we do not need to use vector notations.
1. Applying the law of conservation of momentum, we have:
p_A1 + p_B1 = p_(A+B)
⇒ m_A × v_A1 +  m_B × v_B1 = m_(A+B) × v_(A+B)
2. Substituting the values, we get:
⇒ 5 × 3 + 5 × -3 = 10 × v_(A+B)
⇒ 0 = 10 × v_(A+B).
⇒ v_(A+B) = 0

Solved example 5.13
A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.
Solution:
1. Applying the law of conservation of momentum, we have:
$\mathbf\small{\vec{p_{(A+B)}}=\vec{p_{A2}}+\vec{p_{B2}}}$
Where 
• _{$\mathbf\small{\vec{p_{A2}}}$} is the final momentum of the first smaller nucleus A
• _{$\mathbf\small{\vec{p_{B2}}}$} is the final momentum of the second smaller nucleus B
2. But the original nucleus was at rest. So we have: $\mathbf\small{\vec{p_{(A+B)}}}$ = 0
• Substituting this in (1), we get:
$\mathbf\small{\vec{p_{A2}}=-\vec{p_{B2}}}$
3. So we can write:
$\mathbf\small{m_A \times\vec{v_{A2}}=-m_B \times\vec{v_{B2}}}$
4. The result in (3) gives the condition.
• This condition must be satisfied when the given stationary nucleus disintegrates into two smaller nuclei
• From this condition, we have to prove that, A and B move in opposite directions.
5. In (3), we have an equality of two vectors. Then we can write the two points:
(i) Magnitude of the two vectors must be equal    
(ii) Direction of the two vectors must be the same
6. Considering magnitude, we can write:
$\mathbf\small{m_A \times|\vec{v_{A2}|}=-m_B \times|\vec{v_{B2}}|}$
7. Considering direction, we can write:
• Direction of $\mathbf\small{m_A \times\vec{v_{A2}}}$ must be same as the direction of $\mathbf\small{-m_B \times\vec{v_{B2}}}$
• m_A and m_B are scalar quantities. They are masses. So they cannot be negative. 
8. Thus, the directions of the two vectors are decided entirely by the direction of the velocities
• So we can write:
Direction of $\mathbf\small{\vec{v_{A2}}}$ must be same as the diretion of $\mathbf\small{-\vec{v_{B2}}}$
• That is., A must be travelling in the direction which is the exact opposite to the 'direction in which B is travelling'

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In the next section we will see Equilibrium of a particle

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Chapter 5.7 - Newton's Third law

In the previous section we saw impulse. In this section we will see Newton's third law of motion.

We will write it in steps:
1. Consider a tug of war contest. 
• One team pulls to the left while the other team pulls to the right. 
2. 'Pulling' is in fact, 'applying a force'. 
• If one of the team stops applying their force, the other team will not be able to apply their's
• That is., if one of the team stops applying their force, the other team's force will also disappear. 
3. In the tug of war, we see 'pulling'. 
• We see the same effect in 'pushing' also. 
• When we push against a 'massive and stable wall', we will feel that someone is pushing us back with the same force
• This reverse force from the wall will disappear if we stop pushing
4. So we see that forces always occur in pairs. 
• That is., When ever we apply a force on a body, there occurs another force. 
• We can call the applied force as action and the resulting force as reaction
5. What is the magnitude of the reaction?
• We do not need to do any calculations for finding that magnitude
■ The 'magnitude of the reaction' will be exactly same as the 'magnitude of the action'
6. What about the direction?
■ The direction of the reaction is exactly opposite to the direction of the action
7. Newton's third law states that:
■ To every action, there is an equal and opposite reaction

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Let us see some more examples:

Example 1:
1. Apply a compressive force on a spring
• Let the force applied on the spring by the hand be F_SH
• Then F_SH is the action
2. The spring will exert a reactive force on the hand
• This is the force exerted on the hand by the spring. We can call it F_HS
■ According to Newton's third law, we can write F_SH = -F_HS .
Example 2:
1. Consider a stone falling from a height
• The earth is exerting a force on the stone. This force is due to gravity
2. It is an attractive force. That is., the stone is pulled towards the earth
• Let the force applied by the earth on the stone be F_ES
3. According to Newton’s third law, the stone will exert an equal and opposite force on the earth
• That means, the earth is pulled towards the stone
• Let this force be F_SE
4. We can write: F_ES = -F_SE
5. We see the stone moving towards the earth. That means, we can clearly see the effect of F_ES
• But we do not see the earth moving towards the stone. That means, we do not see the effect of F_SE.
• Why is that so? 
Ans: Since the two forces are equal and opposite, we can write:
F_ES = -F_SE
• Applying Newton's second law, we get:
m_E × a_E = -m_S × a_S .
• Where:
    ♦ m_E and m_S are the masses of earth and stone respectively
    ♦ a_E and a_S are the accelerations of earth and stone respectively
• m_E is very large compared to m_E
• So, for the left side to become equal to right side, a_E has to be extremely small
• It will be so small that, we will not be able to see the motion of earth towards the stone

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Now we will see a practical application of the third law. We will write it in steps:
1. Consider a billiard ball striking against a rigid wall. See fig.5.16(a) below:

Fig.5.16

• The red line is a normal to the wall. 
    ♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$. 
• It travels along the red line. So we can write:
(i) The path of the ball is normal to the wall
(ii) The ball will hit the wall normally.
3. After hitting the wall, the ball rebounds
• The velocity after rebound is the same $\mathbf\small{\vec{u}}$. But in the reverse direction.
• The path after rebound is the same normal line shown in red color. This is shown in fig(b)
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=m\, \vec{u}}$
    ♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=-m\, \vec{u}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=(-m\, \vec{u}-m\, \vec{u})=-2m\, \vec{u}}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=m \times 0=0}$

(ii) Final momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y2}}=m \times \vec{v_{y2}}=m \times 0=0}$
■ So change in momentum in the y direction = $\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=(0-0)=0}$.
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of  $\mathbf\small{\Delta \vec{p_x}}$ and  $\mathbf\small{\Delta \vec{p_y}}$     
• But  $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_x}=-2m\, \vec{u}}$ 
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{-2m\, \vec{u}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{2m\, |\vec{u}|}$
(iv) It's  direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
    ♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt 
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=-2m\, \vec{u}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is  along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall
12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$ 
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also

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In the above example, the ball strike the wall along a normal. What if it strikes at an angle?

Let us write the steps:
1. Consider a billiard ball striking against a rigid wall. See fig.5.17(a) below:

Fig.5.17

• The red line is a normal to the wall. 
    ♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$
• The direction of this $\mathbf\small{\vec{u}}$ makes an angle of 30^o with the red line 
• Also note that, the red line is drawn at the point where the ball hits the wall. 
• So we can write:
The ball will hit the wall at an angle of 30^o
3. After hitting the wall, the ball rebounds
• The velocity after rebound is $\mathbf\small{\vec{v}}$
• Let the path after rebound makes the same angle of 30^o with the normal line shown in red color. This is shown in fig(b)
• Also let the magnitudes of the two velocities be the same. That is: $\mathbf\small{|\vec{u}|=|\vec{v}|}$
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=[m|\vec{u}|\cos\,30^o]\hat{i}}$
    ♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=[-m|\vec{v}|\cos\,30^o]\hat{i}}$
■ So change in momentum in the x direction =
 $\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=\left([-m|\vec{v}|\cos\,30^o]\hat{i}-[m|\vec{u}|\cos\,30^o]\hat{i}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-|\vec{v}|-|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=[-m|\vec{u}|\sin\,30^o]\hat{j}}$
(ii) Final momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y2}}=m \times \vec{v_{y2}}=[-m|\vec{v}|\sin\,30^o]\hat{j}}$
• Both the above quantities are negative because, they are directed towards the negative side of the y axis
■ So change in momentum in the y direction =
 $\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=\left([-m|\vec{v}|\sin\,30^o]\hat{j}-[-m|\vec{u}|\sin\,30^o]\hat{j}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{v}|--|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{u}|+|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}=0}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of  $\mathbf\small{\Delta \vec{p_x}}$ and  $\mathbf\small{\Delta \vec{p_y}}$     
• But  $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]}$
(iv) It's  direction:
    ♦ The unit vector '$\mathbf\small{\hat{i}}$' indicates that, it acts parallel to the x axis
    ♦ The '-' sign indicates that it acts towards the negative side of the x axis
• So it's direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
    ♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt 
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is  along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall

---

This is an interesting situation. We can write it as two points:
(i) First we made the ball to strike the wall normally
• Then we made the ball to strike the wall at an angle
(ii) In both cases, the force exerted by the wall on the ball acts along the normal to the wall

---

12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$ 
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also

---

Let us write a summary in three points A, B and C:
A. A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the wall on the ball will be along the normal (at the point of contact) to the wall
• This force is directed 'away from the wall'
B. From (A), we get the direction of the force exerted by the ball on the wall. For that, we apply Newton's third law. We will write the steps again:
A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the ball on the wall will be along the normal (at the point of contact) to the wall
• This force is directed 'towards the wall'
C. We can obtain the 'magnitude of the force' exerted by the ball on the wall, if we have the following three items:
(i) Mass of the ball
(ii) Velocity of the ball
(iii) Time duration for which the ball is in contact with the wall

---

In the above discussion we saw this:
■ What ever be the direction in which the ball strikes the wall, the forces will be along the red line.
• What is so special about this red line?
Let us analyse:
1. When the ball hits the wall, there is a 'surface of contact'
• Since the ball is spherical, this surface of contact will be very small
• It will be like a 'point of contact'
2. If we draw a tangent (at the point of contact) to the sphere, that tangent will lie on the surface of the wall. Details about tangents can be seen here.
• So we can say that, the 'contact surface' is the surface of the wall itself
3. This is true for what ever direction the ball strikes the wall. It is shown in fig.5.18 below:

Fig.5.18

• In fig.5.18(a), the ball hits normally 
    ♦ The 'contact surface' is a point
    ♦ The wall is tangential to that point
    ♦ So the red line is perpendicular to the contact surface
• In fig.5.18(b), the ball hits at an angle
    ♦ Here also, the 'contact surface' is a point
    ♦ The wall is tangential to that point
    ♦ So the red line is perpendicular to the contact surface
4. So we find that: 
■ The forces act along the line perpendicular to the contact surface
5. So what is the speciality of the red line?
Answer can be written in two statements: 
(i) The force always act along the line perpendicular to the contact surface
(ii) When a ball hits a wall at any angle, the red line (in figs.5.16 and 5.17) is the line perpendicular to the contact surface

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• In the above example, the wall is a flat surface. 
• How do we draw the red line when surfaces are not flat? 
• We will see the case when spherical balls collide with each other:
1. In fig.5.19(a), two balls (of same size) A and B are moving with velocities V_A and V_B

Fig.5.19

2. Their paths intersect at P. 
• After some time, let ball A reach P. 
• If ball B also reach there at that same instant, a collision will take place. 
• This is shown in fig.b. 
3. In fig.c, a tangent is drawn at the point of contact
• We can say that, this tangent represents the surface of contact between the balls
• The line joining the 'centers of the balls' will be perpendicular to this surface
4. There will be two forces acting at the time of collision:
(i) F_AB which is the force which A exerts on B
(ii) F_BA which is the force which B exerts on A
5. By Newton's third law, F_AB = - F_BA
• Both the forces act along the line joining the centers of the balls. This is shown in the enlarged view in fig.d
■ So in this case, the 'line joining the ceters of the balls' is the 'red line'

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So we have completed a basic discussion on Newton's third law
• The following points should be always kept in mind:
1. Forces always occur  in pairs
• A 'pair' means two'
• So if we see F_AB, we can be sure that, there will be F_BA
    ♦ If F_BA is not present, there will be no F_AB
    ♦ If F_AB is not present, there will be no F_BA
2. The terms 'action' and 'reaction' may give the wrong impression that:
• Action occurs first, and reaction occurs later, as an 'effect' of the action
■ The fact is that, both occur at the same instant
3. The two forces act on two different bodies
• F_AB acts on B
• F_BA acts on A
■ So even though they are equal and opposite, they cannot cancel each other. 

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We will see solved examples based on the third law, in a later section. In the next section, we will see Apparent weight in a lift

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