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 30o with it's original direction. Object 'B' moves in a direction which makes -30o 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:
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:
vAf = 2.598 ms-1
vBf = 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
• 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 30o 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:
• 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 90o with the East-West direction (the x-axis)
♦ So we have '90' in the cell (A,4)
♦ Velocity of 'B' makes an angle of 30o 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 56.33o with the x-axis
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 30o 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:
• In the fig.6.84, θBf is shown below the initial direction. But the initial direction is considered as the x-axis
• So θBf is below the x-axis. That means, θBf is negative
• But we need not write '(-θBf)' and do the calculations. We can do the calculations using '(+θBf)'
• If in the final results, we get a negative value for 'θBf', 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: $\mathbf\small{|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2=36}$
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 30o with it's original direction. Object 'B' moves in a direction which makes -30o 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 |
♦ (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:
vAf = 2.598 ms-1
vBf = 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
• 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 30o 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 |
2. The table is shown below:
• However, be careful while filling column 4
♦ Velocity of 'A' makes an angle of 90o with the East-West direction (the x-axis)
♦ So we have '90' in the cell (A,4)
♦ Velocity of 'B' makes an angle of 30o 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 56.33o with the x-axis
• 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 90o 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'
• Both the spheres have the same mass. So we can write: mA = mB = 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: $\mathbf\small{|\vec{v_{Ai}}|^2+|\vec{v_{Bi}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow |\vec{v_{Ai}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
[∵ vBi = 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: $\mathbf\small{|\vec{v}_{Af}|\sin \theta_{Af}+|\vec{v}_{Bf}|\sin \theta_{Bf}=0}$
■ 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 90o 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 |
• We have to prove that $\mathbf\small{(\theta_{Af}+\theta_{Bf})=90^\text{o}}$
2. The table is shown below:
• 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: $\mathbf\small{|\vec{v_{Ai}}|^2+|\vec{v_{Bi}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow |\vec{v_{Ai}}|^2=|\vec{v_{Af}}|^2+|\vec{v_{Bf}}|^2}$
[∵ vBi = 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
(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
♦ 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'
• 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'
• 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 90o 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
(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 |
• 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 90o 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 30o 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 |
• So θBf is below the x-axis. That means, θBf is negative
• But we need not write '(-θBf)' and do the calculations. We can do the calculations using '(+θBf)'
• If in the final results, we get a negative value for 'θBf', 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
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
We have completed this discussion on work, energy and power. In the next chapter, we will discuss about rotational motion
