Chapter 7 - Systems of Particles And Rotational Motion

In the previous section, we completed a discussion on two-dimensional collision. In this chapter, we will see rotational motion.

• In the previous chapters, we analyzed many cases in which ‘objects’ are involved. Let us recall some examples that we have seen:
    ♦ An object acted upon by a net force moves from one position to another
    ♦ A projectile moving in a parabolic path is acted upon by various forces
    ♦ An object at a height possess potential energy
    ♦ An object moving with a certain velocity possess kinetic energy
    ♦ Collision between objects causes changes in their original velocities     
so on . . .
• In all those cases, we considered the objects to be point masses
    ♦ It is true that, we drew them not as points, but as squares, rectangles, spheres, cars, trucks, etc.,
    ♦ But we drew the vectors from their centers
    ♦ We assumed that, all the mass of an object is concentrated at the 'center point' 

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■ ‘Point mass’ can be explained as follows:
• If an object is said to have a mass ‘m’ kg, all that mass is assumed to be concentrated at a point
• That point is a ‘point mass’  
• Even though it is a ‘point’, it's mass is not zero
• Considering objects to be point masses helps to simplify calculations

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• So we can do calculations by two methods:
(i) Considering an object to be a point mass
(ii) Considering the actual distribution of mass in the object
• On many occasions, the results obtained by the two methods will not be appreciably different
■ If we consider the ‘actual distribution of mass’ in an object, then it is called an extended object
• While considering an extended object, it’s size becomes important
    ♦ It cannot be considered as a point
• So basically, an ‘extended object’ is a ‘system of particles’
• In some problems in physics, it is compulsory to consider objects as ‘extended objects’
• In fig.7.1(a) below, a cube is shown. 

Fig.7.1

• It is a uniform cube. That is., the mass is distributed uniformly through out the entire volume of the cube. 
• In such cases, we can consider the cube to be a point mass located at the 'geometric center' of the cube. 
• The geometric center of a cube is easy to locate. Recall that, in many diagrams that we saw in the previous chapters, we drew velocity vectors, force vectors etc., through the geometric center of the objects. 
• Now consider fig.b. Geometrically, it is a cube. But more mass is concentrated towards the left side.
• That is., the 'left side is heavier' and 'right side is lighter'. We cannot consider the mass to be concentrated at the geometric center. If we make such an assumption, the results will not be correct 
• In such cases, we need to define a new ‘point’
■ This point is called the ‘center of mass’
• So we have two points:
    ♦ A point which is the geometric center
    ♦ A point which is the center of mass
• Unlike the ‘geometric center’, the ‘center of mass’ is not easy to find
• However in later sections, we will see methods to find it for 'simple objects'

Significance of Rigid bodies

Following the above discussion, we encounter a new ‘situation’. It can be explained as follows:
Consider a scenario in steps:
1. We have to solve a problem involving an extended object
2. We start off by calculating the center of mass of that object
• An example is shown in fig.7.2(a) below:

Fig.7.2

• The center of mass is denoted by the red '❌' mark
3. But when a force act on the object, it gets deformed
• As a result, the center of mass shifts to a new position. This is shown in fig.7.2(b)
• We will have to recalculate the new position

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Such a difficulty can be overcome by assuming the object to be ‘rigid’

■ A rigid body is a body with a perfectly definite and unchanging shape
• Consider any one 'pair of points' in that body
    ♦ The distance between the points in that pair, will not change

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• However, no real body is truly rigid. Because, all of them deforms under the influence of forces
• But in many cases, those deformations are negligible
• We can confidently do problems by considering them to be rigid
• We will learn about advanced cases (where deformations also have to be considered) in higher classes

Motion of a rigid body

1. Consider a rigid block shown in fig.7.3(a) below:

Fig.7.3

• It moves with uniform velocity $\mathbf\small{\vec{v}}$ on a horizontal surface
2. Consider any two points P and Q in the body
We can write three detailed steps as follows:
(i) Let  $\mathbf\small{\vec{v}_P}$ be the velocity with which the point P moves
(ii) Let  $\mathbf\small{\vec{v}_Q}$ be the velocity with which the point Q moves
(iii) Then $\mathbf\small{\vec{v}_P=\vec{v}_Q=\vec{v}}$
• The result in (iii) is obvious. In fact we do not even need to write the three detailed steps. We know it is true
3. The same block is shown in fig.b
• But this time it moves with a non-uniform velocity
• For example, if it is moving with an acceleration 'a', we can say that, it is moving with a non-uniform velocity
• In such a situation, the velocity of the block will be different at different instances
4. A point 'A' is marked on the horizontal surface on which the block moves
• Let the velocity of the block at the instant when it just passes 'A' be $\mathbf\small{\vec{v}_{(A)}}$
• We can write three detailed steps as follows:
(i) Let $\mathbf\small{\vec{v}_{P(A)}}$ be the 'velocity with which the point P moves' at the instant when the block just passes 'A'
(ii) Let $\mathbf\small{\vec{v}_{Q(A)}}$ be the 'velocity with which the point Q moves' at the instant when the block just passes 'A'
(iii) Then $\mathbf\small{\vec{v}_{P(A)}=\vec{v}_{Q(A)}=\vec{v}_{(A)}}$
• The result in (iii) is obvious. In fact we do not even need to write the three detailed steps. We know it is true

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If at any instant, all particles of a body have the same velocity, then that body is said to be in pure translational motion

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Now we will see another type of motion:
1. Consider the side view of a cylinder in fig.7.4(a) below:

Fig.7.4

• It is rolling towards the right, on a horizontal surface. This 'rolling' is indicated by the white curved arrow

2. A point 'A' is marked on the horizontal surface on which the cylinder rolls
• Consider the instant at which the cylinder passes 'A' 
• A normal is drawn to the horizontal surface at A. It is shown in red color
• Also, this normal passes through the center O of the cylinder
3. Consider any two points P and Q on this normal
We can write three detailed steps as follows:
(i) Let $\mathbf\small{\vec{v}_{P(A)}}$ be the 'velocity with which the point P moves' at the instant when the cylinder just passes 'A'
(ii) Let $\mathbf\small{\vec{v}_{Q(A)}}$ be the 'velocity with which the point Q moves' at the instant when the cylinder just passes 'A'
(iii) Then $\mathbf\small{\vec{v}_{P(A)}\neq \vec{v}_{Q(A)}}$
4. In 3(iii), We have an inequality
• Unlike the previous cases that we saw about the block, this result is not so obvious. We need to give a bit of explanation
• The explanation can be written based on the second fig.b, as follows:
(i) In fig.b, a circle is drawn through P
• The center of this circle is 'O', which is the center of the cylinder
    ♦ $\mathbf\small{\vec{v}_{P(A)}}$ will be tangential to this circle
    ♦ $\mathbf\small{\vec{v}_{P(A)}}$ will be perpendicular to the red line
• Another circle is drawn through Q. The center of this circle also is 'O'
    ♦ $\mathbf\small{\vec{v}_{Q(A)}}$ will be tangential to this circle
    ♦ $\mathbf\small{\vec{v}_{Q(A)}}$ will be perpendicular to the red line
(ii) Since both the velocities are perpendicular to the red line, we can write:
• $\mathbf\small{\vec{v}_{P(A)}}$ and $\mathbf\small{\vec{v}_{Q(A)}}$ have the same direction
(iii) But their magnitudes will be different. Let us see the reason:
• Let $\mathbf\small{\omega_{(A)}}$ be the instantaneous angular velocity when the cylinder just passes 'A'
• This instantaneous angular velocity will be same for both P and Q
• It is the linear speeds $\mathbf\small{|\vec{v}_{P(A)}|}$ and $\mathbf\small{|\vec{v}_{Q(A)}|}$ which differ
• This difference is due to the 'greater radius' of the circle through P
• We saw a detailed explanation in a previous chapter
(v) For two vectors to be equal, both their magnitudes and directions should be equal
• In our present case, magnitudes are not equal
• This explains the result in 3(iii) that we saw above
5. Let us go back to fig.(a). Besides P and Q, one more point R is also marked
• It is on a violet line passing through O
6. In fig.c, a circle is drawn through R
• The center of this circle is O
• The instantaneous velocity $\mathbf\small{\vec{v}_{R(A)}}$ for R is tangential to this circle
• Also this $\mathbf\small{\vec{v}_{R(A)}}$ is perpendicular to the violet line
• Obviously, the direction of $\mathbf\small{\vec{v}_{R(A)}}$ is different from those of $\mathbf\small{\vec{v}_{P(A)}}$ and $\mathbf\small{\vec{v}_{Q(A)}}$
• Thus we can write: $\mathbf\small{\vec{v}_{P(A)}\neq \vec{v}_{Q(A)}\neq \vec{v}_{R(A)}}$
7. The rolling cylinder is at point 'A' that we marked on the horizontal surface
• If we mark another point 'B' to the right of 'A', after a short time, the cylinder will surely reach that 'B' 
• So we can say that the cylinder is in translational motion
• But all the particles in the cylinder are not moving with the same velocity
• So the cylinder is not in pure translational motion 
• It moves with: [Translation + ''something else']

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In the next section, we will see what this 'something else' is.

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Chapter 6.25 - Two-Dimensional Collision - Examples

In the previous section, we discussed two-dimensional collision. We also saw a solved example. In this section, we will see a few more solved examples.

Solved example 6.40
Two objects 'A' and 'B' have masses 5 kg and 2.5 kg respectively. 'A' moves with a velocity of 4.5 ms^-1 towards 'B' which is initially at rest. After the collision, object 'A' moves in a direction which makes 30^o with it's original direction. Object 'B' moves in a direction which makes -30^o with the original direction of 'A'. Find the final velocities of 'A' and 'B'. Is the collision elastic or inelastic?
Solution:
1. Fig.6.80 below shows the collision:

Fig.6.80

2. The table is shown below:

• All the cells in columns 1 to 5 can be filled up (using the given data) except (A,3) and (B,3)
    ♦ (A,3) contains the unknown $\mathbf\small{|\vec{v_{Af}}|}$
    ♦ (B,3) contains the unknown $\mathbf\small{|\vec{v_{Bf}}|}$
• So there are two unknowns. We need to form two equations only
• Also note that, there are two 'question marks' in the fig.6.80 above
3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow [22.5+0]=[5|\vec{v_{Af(x)}}|+2.5|\vec{v_{Bf(x)}}|]}$
$\mathbf\small{\Rightarrow [22.5]=[5|\vec{v_{Af}}| \cos \theta_{Af}+2.5|\vec{v_{Bf}}| \cos \theta_{Bf}]}$
$\mathbf\small{\Rightarrow [22.5]=[5|\vec{v_{Af}}| \cos 30+|\vec{v_{Bf}}| \cos (-30)]}$
Dividing both sides by 2.5, we get:
$\mathbf\small{[9]=[2|\vec{v_{Af}}| \cos 30+|\vec{v_{Bf}}| \cos (-30)]}$
$\mathbf\small{\Rightarrow [9]=[1.732|\vec{v_{Af}}|+0.866|\vec{v_{Bf}}| ]}$
4. Considering momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow [0]=[5|\vec{v_{Af(y)}}|+2.5|\vec{v_{Bf(y)}}|]}$
$\mathbf\small{\Rightarrow [0]=[5|\vec{v_{Af}}| \sin 30+|\vec{v_{Bf}}| \sin (-30)]}$
Deviding both sides by 2.5, we get:
$\mathbf\small{\Rightarrow [0]=[2|\vec{v_{Af}}| \sin 30+|\vec{v_{Bf}}| \sin (-30)]]}$
$\mathbf\small{\Rightarrow [0]=[2|\vec{v_{Af}}| 0.5+|\vec{v_{Bf}}| (-0.5)]}$
$\mathbf\small{\Rightarrow [0]=[|\vec{v_{Af}}|-.5|\vec{v_{Bf}}|]}$
5. So we have two equations:
(i) From (3) we have: $\mathbf\small{1.732|\vec{v_{Af}}|+0.866|\vec{v_{Bf}}|=9}$
(ii) From (4) we have: $\mathbf\small{|\vec{v_{Af}}|-0.5|\vec{v_{Bf}}|=0}$
• To solve the equations, the following steps can be used:
    ♦ Multiply (i) by '0.5'
    ♦ Multiply (ii) by 0.866
    ♦ Add the results 
• We get:
v_Af = 2.598 ms^-1
v_Bf = 5.196 ms^-1
6. Let us check whether it is an elastic collision or not:
• Applying law of conservation of kinetic energy, we have:
$\mathbf\small{\frac{1}{2}m_A|\vec{v}_{Ai}|^2+\frac{1}{2}m_B|\vec{v}_{Bi}|^2=\frac{1}{2}m_A|\vec{v}_{Af}|^2+\frac{1}{2}m_B|\vec{v}_{Bf}|^2}$
• Let us first calculate the LHS:
$\mathbf\small{\frac{1}{2}m_A|\vec{v}_{Ai}|^2+\frac{1}{2}m_B|\vec{v}_{Bi}|^2=\frac{1}{2}\times 5 \times 4.5^2+\frac{1}{2}\times 2.5 \times0^2=50.625\; \text{J}}$
• Next we calculate the RHS:
$\mathbf\small{\frac{1}{2}m_A|\vec{v}_{Af}|^2+\frac{1}{2}m_B|\vec{v}_{Bf}|^2=\frac{1}{2}\times 5 \times 2.598^2+\frac{1}{2}\times 2.5 \times 5.196^2=50.628\; \text{J}}$
■ We get: LHS = RHS
So kinetic energy is conserved and so it is an elastic collision

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• We saw solved example 6.39 in the previous section
• We saw solved example 6.40 above in the present section
• Both of those examples involves only two unknowns
• Such problems can be easily solved using two equations that we obtain from the law of conservation of momentum
■ The calculations become even more easier if the two objects stick together after collision
• Let us see such a problem: 

Solved example 6.41
Car A has a mass of 1800 kg. It moves towards the North with a velocity of 15 ms^-1. Car B has a mass of 1500 kg. It moves in a South-East direction, making an angle of 30^o with the East-West direction. It has a velocity of 10 ms^-1. After collision, they stick together. Find the velocity (magnitude and direction) of the combined mass after the collision
Solution:
1. Fig.6.81 below shows the collision:

Fig.6.81

• East-West direction can be taken as the x-axis
2. The table is shown below:

• The first 5 columns can be filled up using the given data
• However, be careful while filling column 4
    ♦ Velocity of 'A' makes an angle of 90^o with the East-West direction (the x-axis)
    ♦ So we have '90' in the cell (A,4)
    ♦ Velocity of 'B' makes an angle of 30^o with the x-axis
    ♦ So we have '30' in the cell (B,4)
    ♦ sin 30 is not negative
    ♦ But we must put a negative value in the cell (B,7)
    ♦ This is because, that velocity is towards the negative side of the y-axis
• After collision, both cars move together. So they will be having the same velocity
    ♦ Thus we have: $\mathbf\small{|\vec{v}_{Af}|=|\vec{v}_{Bf}|}$
    ♦ We can put: $\mathbf\small{|\vec{v}_{Af}|=|\vec{v}_{Bf}|=|\vec{v}_{f}|}$ 
    ♦ We see this in the cells (A,3) and (B,3)
• After collision, since they move together with the same velocity, they will be making the same angle with the x axis
    ♦ Thus we have: $\mathbf\small{\theta_{Af}=\theta_{Bf}}$
    ♦ We can put: $\mathbf\small{\theta_{Af}=\theta_{Bf}=\theta_{f}}$ 
    ♦ We see this in the cells (A,5) and (B,5)
• After filling the first five columns, we see that there are two unknowns. So we need to form two equations only
• Also note that, there are two 'question marks' in the fig.6.81 above
3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow [0+12990]=[1800|\vec{v_{f(x)}}|+1500|\vec{v_{f(x)}}|]}$
$\mathbf\small{\Rightarrow [12990]=[(1800+1500)|\vec{v_{f(x)}}|]}$
$\mathbf\small{\Rightarrow [12990]=[3300|\vec{v_{f(x)}}|]}$
$\mathbf\small{\Rightarrow |\vec{v}_{f(x)}|=\frac{12990}{3300}=3.936\;\text{ms}^{-1}}$
4. Considering momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow [27000-7500]=[1800|\vec{v_{f(y)}}|+1500|\vec{v_{f(y)}}|]}$
$\mathbf\small{\Rightarrow [19500]=[(1800+1500)|\vec{v_{f(y)}}|]}$
$\mathbf\small{\Rightarrow [19500]=[3300|\vec{v_{f(y)}}|]}$
$\mathbf\small{\Rightarrow |\vec{v}_{f(y)}|=\frac{19500}{3300}=5.91\;\text{ms}^{-1}}$
5. Now we can find the magnitude of the resultant
$\mathbf\small{|\vec{v}_{f}|=\sqrt{|\vec{v}_{fx}|^2+|\vec{v}_{fy}|^2}=\sqrt{3.936^2+5.91^2}=7.1\;\text{ms}^{-1}}$
6. Direction of this resultant is given by: $\mathbf\small{\tan \theta_f=\frac{|\vec{v}_{fy}|}{|\vec{v}_{fx}|}=\frac{5.91}{3.936}=1.501}$
$\mathbf\small{\Rightarrow \theta_f=\tan^{-1}1.501=56.33^\text{o}}$
7. Thus we can write:
■ The final velocity shown in fig.6.81 has a magnitude of 7.1 ms^-1 and a direction which makes 56.33^o with the x-axis

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• Next we will see problems with 3 unknowns.  Such problems involve somewhat lengthy calculations
■ But those calculations will be greatly simplified if the two colliding objects have the same mass
• The Solved examples 6.42 and 6.43 given below will demonstrate this idea

Solved example 6.42
Prove that an elastic collision between equal masses in 2 dimensions always results in the objects bouncing off each other at a 90^o degrees angle (assume one of the objects to be initially stationary). 
Solution:
1. Fig.6.82 below shows a 2-dimensional collision between two spheres 'A' and 'B'

Fig.6.82

• Both the spheres have the same mass. So we can write: m_A = m_B = m
• We have to prove that $\mathbf\small{(\theta_{Af}+\theta_{Bf})=90^\text{o}}$
2. The table is shown below:

3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow \sum |\vec{p}_{i(x)}|=\sum |\vec{p}_{f(x)}|}$
$\mathbf\small{\Rightarrow [m|\vec{v}_{Ai(x)}|+0]=[m|\vec{v}_{Af(x)}|+m|\vec{v}_{Bf(x)}|]}$
$\mathbf\small{\Rightarrow [|\vec{v}_{Ai(x)}|]=[|\vec{v}_{Af(x)}|+|\vec{v}_{Bf(x)}|]}$
$\mathbf\small{\Rightarrow [|\vec{v}_{Ai}|\cos \theta_{Ai}]=[|\vec{v}_{Af}|\cos \theta_{Af}+|\vec{v}_{Bf}|\cos \theta_{Bf}]}$
4. Considering momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow \sum |\vec{p}_{i(y)}|=\sum |\vec{p}_{f(y)}|}$
$\mathbf\small{\Rightarrow [0+0]=[m|\vec{v}_{Af(y)}|+m|\vec{v}_{Bf(y)}|]}$
$\mathbf\small{\Rightarrow [0]=[|\vec{v}_{Af(y)}|+|\vec{v}_{Bf(y)}|]}$
$\mathbf\small{\Rightarrow [0]=[|\vec{v}_{Af}|\sin \theta_{Af}+|\vec{v}_{Bf}|\sin \theta_{Bf}]}$
5. So we have two equations:
(i) From (3) we have: $\mathbf\small{|\vec{v}_{Af}|\cos \theta_{Af}+|\vec{v}_{Bf}|\cos \theta_{Bf}=|\vec{v}_{Ai}|\cos \theta_{Ai}}$
(ii) From (4) we have: $\mathbf\small{|\vec{v}_{Af}|\sin \theta_{Af}+|\vec{v}_{Bf}|\sin \theta_{Bf}=0}$
6. But we have three unknowns. So we must have three equations
• For the third equation, we apply conservation of kinetic energy
• Total initial kinetic energy = Total final kinetic energy
• Since the masses are equal, we have: $\mathbf\small{|\vec{v_{Ai}}|^2+|\vec{v_{Bi}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow |\vec{v_{Ai}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
[∵ v_Bi = 0]
7. Let us write the three equations together:
(i) From (3) we have: $\mathbf\small{|\vec{v}_{Af}|\cos \theta_{Af}+|\vec{v}_{Bf}|\cos \theta_{Bf}=|\vec{v}_{Ai}|\cos \theta_{Ai}}$
(ii) From (4) we have: $\mathbf\small{|\vec{v}_{Af}|\sin \theta_{Af}+|\vec{v}_{Bf}|\sin \theta_{Bf}=0}$

(iii) From (6) we have: $\mathbf\small{|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2=|\vec{v_{Ai}}|^2}$

8. Let us write them in a 'easy to manipulate' form:
(i) $\mathbf\small{A\;\cos \theta_{A}+B\cos \theta_{B}=k_1}$
(ii) $\mathbf\small{A\;\sin \theta_{A}+B\sin \theta_{B}=0}$
(iii) $\mathbf\small{A^2+B^2=(k_1)^2}$ 
Where:
• $\mathbf\small{A=|\vec{v_{Af}}|}$. A quantity that we have to find
• $\mathbf\small{B=|\vec{v_{Bf}}|}$. A quantity that we have to find
• $\mathbf\small{\theta_{B}=\theta_{B_f}}$. A quantity that we have to find
• $\mathbf\small{\theta_{A}=\theta_{Af}}$. A constant, which can be calculated from the given data

• $\mathbf\small{k_1=|\vec{v}_{Ai}|\cos \theta_{Ai}}$ 
    ♦ But $\mathbf\small{\theta_{Ai}=0}$. So $\mathbf\small{\cos \theta_{Ai}}$ =1
    ♦ Thus we get: $\mathbf\small{k_1=|\vec{v}_{Ai}|\cos \theta_{Ai}=|\vec{v}_{Ai}|}$ 
    ♦ So k₁ is a constant, which can be calculated from the given data

■ Now we can write the steps for obtaining the 3 unknowns:
(iv) Squaring (i), we get:
$\mathbf\small{A^2\,\cos^2\theta_A+B^2\,\cos^2\theta_B+2AB\,\cos\theta_A\,\cos\theta_B=(k_1)^2}$
(v) Squaring (ii), we get:
$\mathbf\small{A^2\,\sin^2\theta_A+B^2\,\sin^2\theta_B+2AB\,\sin\theta_A\,\sin\theta_B=0}$
(vi) Adding (iv) and (v), we get:
$\mathbf\small{(A^2\,\cos^2\theta_A+A^2\,\sin^2\theta_A)+(A^2\,\sin^2\theta_B+B^2\,\sin^2\theta_B)+(2AB\,\cos\theta_A\,\cos\theta_B+2AB\,\sin\theta_A\,\sin\theta_B)=(k_1)^2+0}$
$\mathbf\small{\Rightarrow A^2(\cos^2\theta_A+\sin^2\theta_A)+B^2(\sin^2\theta_B+\sin^2\theta_B)+(2AB\,\cos\theta_A\,\cos\theta_B+2AB\,\sin\theta_A\,\sin\theta_B)=(k_1)^2+0}$
$\mathbf\small{\Rightarrow A^2+B^2+(2AB\,\cos\theta_A\,\cos\theta_B+2AB\,\sin\theta_A\,\sin\theta_B)=(k_1)^2}$
(vii) But from (iii), we have: $\mathbf\small{A^2+B^2=(k_1)^2}$
• Substituting this in (v), we get:
$\mathbf\small{(k_1)^2+(2AB\,\cos\theta_A\,\cos\theta_B+2AB\,\sin\theta_A\,\sin\theta_B)=(k_1)^2}$
$\mathbf\small{\Rightarrow 2AB\,\cos\theta_A\,\cos\theta_B+2AB\,\sin\theta_A\,\sin\theta_B=0}$
$\mathbf\small{\Rightarrow 2AB(\cos\theta_A\,\cos\theta_B+\,\sin\theta_A\,\sin\theta_B)=0}$
(viii) Using the identity $\mathbf\small{\cos(\theta_1-\theta_2)=\cos\theta_1\,\cos\theta_2+\sin\theta_1\,\sin\theta_2}$, we get:
$\mathbf\small{2AB\,\cos(\theta_A-\theta_B)=0}$
(ix) Putting back the values, we get:
• $\mathbf\small{2|\vec{v}_{Af}|\,|\vec{v}_{Bf}|\,\cos(\theta_{Af}-\theta_{Bf})=0}$
• $\mathbf\small{|\vec{v}_{Af}|}$ can be zero only if all the three conditions given below are satisfied:
    ♦ It is a one-dimensional collision (Details here)
    ♦ The two masses are equal
    ♦ Object B is stationary before collision
• In our present case, the second and third conditions are satisfied
• But it is not a one-dimensional collision. What we have is a two-dimensional collision
• So $\mathbf\small{|\vec{v}_{Af}|}$ cannot be zero  
• $\mathbf\small{|\vec{v}_{Bf}|}$ can be zero only if the mass of B is very large compared to mass of A. But in our present case, both masses are equal
• So $\mathbf\small{|\vec{v}_{Bf}|}$ cannot be zero
• Thus the only option is: $\mathbf\small{\cos(\theta_{Af}-\theta_{Bf})=0}$
• If $\mathbf\small{\cos(\theta_{Af}-\theta_{Bf})=0}$, then $\mathbf\small{(\theta_{Af}-\theta_{Bf})=90^\text{o}}$  
■ This relation $\mathbf\small{(\theta_{Af}-\theta_{Bf})=90^\text{o}}$ is our key
9. In fig.6.83(a) below, we see that, $\mathbf\small{(\theta_{Af}-\theta_{Bf})}$ is the 'angle between the two final vectors'

Fig.6.83

• If $\mathbf\small{(\theta_{Af}-\theta_{Bf})}$ is to be 90, the 'after collision' fig. will be as in (b)
• Here also, $\mathbf\small{(\theta_{Af}-\theta_{Bf})}$ is the 'angle between the two final vectors'

10. Another possibility is shown in fig.c

• After collision, the object 'B' is below the x-axis. It's final velocity vector makes an angle of $\mathbf\small{-\theta_{Bf}}$ with the x-axis  
• So $\mathbf\small{(\theta_{Af}-\theta_{Bf})}$ becomes: $\mathbf\small{[\theta_{Af}-(-\theta_{Bf})]=[theta_{Af}+\theta_{Bf}]}$ 
• Here also, $\mathbf\small{(\theta_{Af}-\theta_{Bf})}$ is the 'angle between the two final vectors'
■ Thus we can write: After a 2-dimensional collision, the two objects will scatter at 90^o to each other when the following two conditions are satisfied:
    ♦ The two objects have the same mass
    ♦ One of the objects is stationary before collision

Solved example 6.43
Two identical objects A and B collide on a smooth horizontal surface. B was originally at rest. A has an initial velocity of 6 ms^-1. After collision it scatters at an angle of 30^o to the original direction.
(a) What is the magnitude of the velocity of 'A' after the collision ?
(b) What is the magnitude and direction of the velocity of 'B' after the collision ?
Solution:
1. Fig.6.84 below shows the collision:

Fig.6.84

• In the fig.6.84, θ_Bf is shown below the initial direction. But the initial direction is considered as the x-axis
• So θ_Bf is below the x-axis. That means, θ_Bf is negative
• But we need not write '(-θ_Bf)' and do the calculations. We can do the calculations using '(+θ_Bf)'
• If in the final results, we get a negative value for 'θ_Bf', the positions in fig.6.83 will be justified
2. The table is shown below:

• All the cells in columns 1 to 5 can be filled up (using the given data) except (A,3), (B,3) and (B,5)
    ♦ (A,3) contains the unknown $\mathbf\small{|\vec{v_{Af}}|}$
    ♦ (B,3) contains the unknown $\mathbf\small{|\vec{v_{Bf}}|}$
    ♦ (B,5) contains the unknown $\mathbf\small{\theta_{Bf}}$
• So there are three unknowns. We need to form three equations
• Also note that, there are 3 'question marks' in the fig.6.84 above
• Most of the cells in columns 6 to 9 need to be calculated except (B,6), (A,7) and (B,7)
• Those 3 cells have zero values because:
    ♦ B has zero initial x-velocity
    ♦ A has zero initial y-velocity
    ♦ B has zero initial y-velocity
3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow [6m+0]=[m|\vec{v_{Af(x)}}|+m|\vec{v_{Bf(x)}}|]}$
$\mathbf\small{\Rightarrow [6]=[|\vec{v_{Af}}| \cos \theta_{Af}+|\vec{v_{Bf}}| \cos \theta_{Bf}]}$
$\mathbf\small{\Rightarrow [6]=[|\vec{v_{Af}}| \cos 30+|\vec{v_{Bf}}| \cos \theta_{Bf}]}$
$\mathbf\small{\Rightarrow [6]=[0.866|\vec{v_{Af}}|+|\vec{v_{Bf}}| \cos \theta_{Bf}]}$
4. Considering momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow [0]=[m|\vec{v_{Af(y)}}|+m|\vec{v_{Bf(y)}}|]}$
$\mathbf\small{\Rightarrow [0]=[|\vec{v_{Af}}| \sin \theta_{Af}+|\vec{v_{Bf}}| \sin \theta_{Bf}]}$
$\mathbf\small{\Rightarrow [0]=[|\vec{v_{Af}}| \sin 30+|\vec{v_{Bf}}| \sin \theta_{Bf}]}$
$\mathbf\small{\Rightarrow [0]=[0.5|\vec{v_{Af}}|+|\vec{v_{Bf}}| \sin \theta_{Bf}]}$
5. So we have two equations:
(i) From (3) we have: $\mathbf\small{0.866|\vec{v_{Af}}|+|\vec{v_{Bf}}| \cos \theta_{Bf}=6}$
(ii) From (4) we have: $\mathbf\small{0.5|\vec{v_{Af}}|+|\vec{v_{Bf}}| \sin \theta_{Bf}=0}$
6. But we have three unknowns. So we must have three equations
• For the third equation, we apply conservation of kinetic energy
Total initial kinetic energy = Total final kinetic energy
• Since the masses are equal, we have: $\mathbf\small{|\vec{v_{Ai}}|^2+|\vec{v_{Bi}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow |\vec{v_{Ai}}|^2+0=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow 6^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow |\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2=36}$
7. Let us write the three equations together:
(i) From (3) we have: $\mathbf\small{0.866|\vec{v_{Af}}|+|\vec{v_{Bf}}| \cos \theta_{Bf}=6}$
(ii) From (4) we have: $\mathbf\small{0.5|\vec{v_{Af}}|+|\vec{v_{Bf}}| \sin \theta_{Bf}=0}$
(iii) From (6) we have: $\mathbf\small{|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2=36}$

8. Now we apply the relation $\mathbf\small{\theta_{Af}-\theta_{Bf}=90^\text{o}}$
We get: $\mathbf\small{30-\theta_{Bf}=90}$
$\mathbf\small{\Rightarrow \theta_{Bf}=30-90=-60^\text{o}}$ 
• Note: We can apply this relation because, the following two conditions are satisfied:
    ♦ The two objects have the same mass
    ♦ One of the objects is stationary before collision
9. Substituting this value of $\mathbf\small{\theta_{Bf}}$ in (i), we get:
$\mathbf\small{0.866|\vec{v_{Af}}|+|\vec{v_{Bf}}| \cos (-60)=6}$
$\mathbf\small{\Rightarrow 0.866|\vec{v_{Af}}|+0.5|\vec{v_{Bf}}|=6}$
10. Substituting this value of $\mathbf\small{\theta_{Bf}}$ in (ii), we get:
$\mathbf\small{0.5|\vec{v_{Af}}|+|\vec{v_{Bf}}| \sin (-60)=0}$
$\mathbf\small{\Rightarrow 0.5|\vec{v_{Af}}|-0.866|\vec{v_{Bf}}|=0}$
11. Solving (9) and (10), we get:
$\mathbf\small{|\vec{v_{Af}}|}$ = 5.196 ms^-1
$\mathbf\small{|\vec{v_{Bf}}|}$ = 3
• Note: For solving, the following steps can be used:
    ♦ Multiply (9) by 0.866
    ♦ Multiply (10) by 0.5
    ♦ Add the results
12. Check:
• Let us substitute the velocities in (iii). We have:
$\mathbf\small{|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2=36}$
• The LHS will be: $\mathbf\small{5.196^2+3^2=(27+9)=36}$
• RHS = 36
• So the results obtained in (11) are correct

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We have completed this discussion on work, energy and power. In the next chapter, we will discuss about rotational motion

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