In the previous section, we completed a discussion on one-dimensional collision. In this section, we will see two-dimensional collision.
1. In the fig.6.75 below, four spheres A, B, P and Q rest on a smooth horizontal surface
♦ All the four spheres have the same mass
♦ All the four spheres have the same diameter
• The horizontal surface is shown in green color
2. Sphere P moves towards sphere Q with a velocity of vPi
• vPi is indicated by the cyan arrow
3. Sphere Q is initially at rest
• That is., vQi = 0
4. A white dashed line is drawn in alignment with vPi
• We see that, the white dashed line passes through the center of sphere Q
■ So we can say that, the collision between P and Q will be a one-dimensional collision
5. Now consider the sphere A
• It moves towards sphere B with a velocity of vAi
• vAi is indicated by the cyan arrow
6. Sphere B is initially at rest
• That is., vBi = 0
7. A white dashed line is drawn in alignment with vAi
• We see that, the white dashed line does not pass through the center of sphere B
■ So we can say that, the collision between A and B will not be a one-dimensional collision
8. The fig.6.75 that we saw above is a 3D view
• It is not convenient to draw 3D views every time
• So we usually draw 2D views.
9. 2D views are familiar to us. The spheres P and Q can be shown in 2D as in fig.6.76(a) below:
• This is the view that we will see when we look at the spheres from the side
• The green surface will appear as a thick green line
• The spheres will appear as circles
• The area of the green surface will not be visible when we look from the side
• For the one-dimensional collision between P and Q, this 2D view is sufficient
10. Next we look at the spheres A and B from the side. The 2D view will be as shown in fig.6.76(b)
■ Unfortunately, fig.6.76(b) does not convey the information that: 'Collision between A and B is not one-dimensional'. It looks as if, vAi passes through the center of sphere B
• So, if the collision is not one-dimensional, 'looking from the side' will not help
• We have to change our 'view point'
11. We have to look from 'above'
• The 2D view when looked from 'above' is shown in fig.6.77:
• Note that:
♦ In fig.6.76, where view is from 'side', the green surface appears as a 'thick green line'
♦ In fig.6.77, where view is from 'above', the green surface appears as a 'green rectangle'
• So in this type of 2D view, we can clearly see that the collision between A and B will not be one-dimensional
12. Now we can start the analysis of the collision
• In the fig.6.78 below, the conditions before and after collisions are shown separately
• The point of collision between the two spheres is shown by a red '❌' mark
13. After the collision, sphere A moves with a velocity of vAf
• This vAf makes a particular angle with the initial direction of A
• We will denote this angle as θAf
14. After the collision, sphere B moves with a velocity of vBf
• This vBf makes a particular angle with the initial direction of A
• We will denote this angle as θBf
15. The movements after the collision are in different directions
• So it is no longer a one-dimensional problem. We cannot describe the motions using the 'x-axis' alone
• We will need both x and y axes
■ The collision in fig.6.78 is a two-dimensional collision.
• So from now on, we will use vector notations for the analysis
16. In these types of problems, it is always convenient to consider the x-axis to lie along the 'initial direction'
• In the fig.6.78, the white dashed line indicates the 'initial direction'
• So we can write:
♦ $\mathbf\small{\vec{v_{Af}}}$ makes an angle of θAf with the x axis
♦ $\mathbf\small{\vec{v_{Bf}}}$ makes an angle of θBf with the x axis
17. The information obtained so far can be written in a tabular form as shown below:
The above table has only 5 columns. We will add additional columns as and when more information become available
18. So let us continue:
• We will apply the law of conservation of momentum
• But momentum is a vector quantity. So we will consider the x and y directions separately
• First we consider the x direction:
(i) Initial momentum of A in the x direction
$\mathbf\small{\vec{p_{Ai(x)}}=m_A \times \text{initial velocity of A in the x-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \cos (\theta_{Ai})]\hat{i}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \cos (0)]\hat{i}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times 1]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Ai}}|]\hat{i}}$
(ii) Initial momentum of B in the x direction
$\mathbf\small{\vec{p_{Bi(x)}}=m_B \times \text{initial velocity of B in the x-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bi}}|\times \cos (\theta_{Bi})]\hat{i}}$
$\mathbf\small{=m_B \times [0\times \cos (\theta_{Bi})]\hat{i}}$
= 0
• The above information about 'initial momentum in x direction' is entered as the sixth column as shown below:
• Note that, the 6th column can be filled up by using information from columns 1, 2 and 4
(cosine of the angle in column 4)
• So writing in the tabular form is indeed advantageous
19. Now we consider the y direction:
(i) Initial momentum of A in the y direction
$\mathbf\small{\vec{p_{Ai(y)}}=m_A \times \text{initial velocity of A in the y-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \sin (\theta_{Ai})]\hat{j}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \sin (0)]\hat{j}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times 0]\hat{j}}$
= 0
(ii) Initial momentum of B in the y direction
$\mathbf\small{\vec{p_{Bi(y)}}=m_B \times \text{initial velocity of B in the y-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bi}}|\times \sin (\theta_{Bi})]\hat{j}}$
$\mathbf\small{=m_B \times [0 \times \sin (\theta_{Bi})]\hat{j}}$
= 0
• The above information about 'initial momentum in y direction' is entered as the seventh column as shown below:
• Note that, the 7th column can be filled up by using information from columns 1, 2 and 4
(sine of the angle in column 4)
• So writing in the tabular form is indeed advantageous
20. Next we consider the final momentum:
• First we consider the x direction:
(i) Final momentum of A in the x direction
$\mathbf\small{\vec{p_{Af(x)}}=m_A \times \text{final velocity of A in the x-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
(ii) Final momentum of B in the x direction
$\mathbf\small{\vec{p_{Bf(x)}}=m_B \times \text{final velocity of B in the x-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
$\mathbf\small{=[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
• The above information about 'final momentum in x direction' is entered as the 8th column as shown below:
• Note that, the 8th column can be filled up by using information from columns 1, 3 and 5
(cosine of the angle in column 5)
21. Next we consider the y direction:
(i) Final momentum of A in the y direction
$\mathbf\small{\vec{p_{Af(y)}}=m_A \times \text{final velocity of A in the y-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
(ii) Final momentum of B in the y direction
$\mathbf\small{\vec{p_{Bf(y)}}=m_B \times \text{final velocity of B in the y-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
$\mathbf\small{=[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
• The above information about 'final momentum in x direction' is entered as the 9th column as shown below:
• Note that, the 9th column can be filled up by using information from columns 1, 3 and 5
(sine of the angle in column 5)
22. In our present case, this y component is negative because, it is directed towards the negative side of the x axis
• We will directly get this negative direction if we put 'negative value' for θBf
• For example, sin(-30) = -sin30 = -0.5
23. Now we apply the conservation of momentum in the x-direction. We have:
• Total initial momentum in the x-direction = Total final momentum in the x direction
That is., $\mathbf\small{\sum{|\vec{p_{i(x)}}|}=\sum{|\vec{p_{f(x)}}|}}$
$\mathbf\small{\Rightarrow|\vec{p_{iA(x)}}|+0=|\vec{p_{fA(x)}}|+|\vec{p_{fB(x)}}|}$
$\mathbf\small{\Rightarrow [m_A \times|\vec{v_{Ai}}|]=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]}$
• This equation can be easily formed from the table as follows:
Total from column 6 = Total from column 8
24. Next we apply the conservation of momentum in the y-direction. We have:
• Total initial momentum in the y-direction = Total final momentum in the y direction
That is., $\mathbf\small{\sum{|\vec{p_{i(y)}}|}=\sum{|\vec{p_{f(y)}}|}}$
$\mathbf\small{\Rightarrow0+0=|\vec{p_{fA(y)}}|+|\vec{p_{fB(y)}}|}$
$\mathbf\small{\Rightarrow 0=[m_A \times|\vec{v_{Af}}|\times \sin (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \sin (\theta_{Bf})]}$
• This equation can be easily formed from the table as follows:
Total from column 7 = Total from column 9
25. With that, we complete our calculations related to momentum. Next we consider energy
• Assuming an elastic collision, we can say: 'kinetic energy will be conserved'
• Energy is a scalar quantity. There is no need to consider x and y components
• So we can write:
$\mathbf\small{\frac{1}{2}m_A\;|\vec{v_{Ai}}|^2+\frac{1}{2}m_B\;|\vec{v_{Bi}}|^2=\frac{1}{2}m_A\;|\vec{v_{Af}}|^2+\frac{1}{2}m_BA\;|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow m_A\;|\vec{v_{Ai}}|^2+m_B\;|\vec{v_{Bi}}|^2=m_A\;|\vec{v_{Af}}|^2+m_BA\;|\vec{v_{Bf}}|^2}$
26. So we have three equations:
(i) From (23), we have: $\mathbf\small{[m_A \times|\vec{v_{Ai}}|]=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]}$
26. In the above 3 equations.
• Which are the 'known quantities' ?
• Which are the 'unknown quantities' ?
Let us analyse:
mA - Known quantity. Will be given in the question
mB - Known quantity. Will be given in the question
vAi - Known quantity. Will be given in the question
vBi - Known quantity. Will be given in the question
vAf - Unknown quantity. A final quantity which we have to determine
vBf - Unknown quantity. A final quantity which we have to determine
θAf - Unknown quantity. A final quantity which we have to determine
θBf - Unknown quantity. A final quantity which we have to determine
27. There are 4 unknowns. We have only 3 equations.
• So we use special 'angle measuring devices' to find one of the final angles θAf or θBf
• Then the 'number of unknowns' will become 3.
• We can easily find the other unknowns
28. Further, if both the final angles are known, the 'number of unknowns' will reduce to 2
• In such cases we need to form the first two equations only
Let us see some solved examples:
Solved example 6.39
Two objects A and B have the same mass of 0.1 kg. Object A has an initial speed of 5 ms-1. Object B is initially at rest. A collides with B and moves off with a speed of 2 ms-1 at an angle of 30o with the initial direction. What is the speed and direction of B?
Solution:
1. Fig.6.79 below shows the possible collision:
• In the above fig.6.79, 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.6.79 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 (B,3) and (B,5)
♦ (B,3) contains the unknown $\mathbf\small{|\vec{v_{Bf}}|}$
♦ (B,5) contains the unknown $\mathbf\small{\theta_{Bf}}$
• So there are only two unknowns. We need to form two equations only
• All the cells in columns 6 to 9 can be calculated except (B,8) and (B,9)
3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow 0.5=0.173+|\vec{P_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow 0.5=0.173+m_B\times|\vec{v_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow 0.5=0.173+0.1\times|\vec{v_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf(x)}}|=3.27}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|\times \cos \theta_{Bf}=3.27}$
4. Considering the momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow 0=0.1+|\vec{P_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow 0=0.1+m_B\times|\vec{v_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow 0=0.1+0.1\times|\vec{v_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf(y)}}|=-1}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|\times \sin \theta_{Bf}=-1}$
5. So we have two equations:
(i) From (3) we have: $\mathbf\small{|\vec{v_{Bf}}|\times \cos \theta_{Bf}=3.27}$
(ii) From (4) we have: $\mathbf\small{|\vec{v_{Bf}}|\times \sin \theta_{Bf}=-1}$
• Dividing 5(ii) by 5(i), we get:
$\mathbf\small{\frac{|\vec{v_{Bf}}|\times \sin \theta_{Bf}}{|\vec{v_{Bf}}|\times \cos \theta_{Bf}}=\frac{-1}{3.27}}$
$\mathbf\small{\Rightarrow \tan \theta_{Bf}=-\frac{1}{3.27}}$
$\mathbf\small{\Rightarrow \theta_{Bf}=-17^o}$
• θBf is negative. So the positions in fig.6.79 are justified
6. Substituting this value of θBf in 5(i), we get:
$\mathbf\small{|\vec{v_{Bf}}|\times \cos (-17)=3.27}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|=\frac{3.27}{\cos (-17)}=\frac{3.27}{0.9563}=3.42\,\,\text{ms}^{-1}}$
1. In the fig.6.75 below, four spheres A, B, P and Q rest on a smooth horizontal surface
♦ All the four spheres have the same mass
♦ All the four spheres have the same diameter
• The horizontal surface is shown in green color
2. Sphere P moves towards sphere Q with a velocity of vPi
• vPi is indicated by the cyan arrow
 |
| Fig.6.75 |
• That is., vQi = 0
4. A white dashed line is drawn in alignment with vPi
• We see that, the white dashed line passes through the center of sphere Q
■ So we can say that, the collision between P and Q will be a one-dimensional collision
5. Now consider the sphere A
• It moves towards sphere B with a velocity of vAi
• vAi is indicated by the cyan arrow
6. Sphere B is initially at rest
• That is., vBi = 0
7. A white dashed line is drawn in alignment with vAi
• We see that, the white dashed line does not pass through the center of sphere B
■ So we can say that, the collision between A and B will not be a one-dimensional collision
8. The fig.6.75 that we saw above is a 3D view
• It is not convenient to draw 3D views every time
• So we usually draw 2D views.
9. 2D views are familiar to us. The spheres P and Q can be shown in 2D as in fig.6.76(a) below:
 |
| Fig.6.76 |
• The green surface will appear as a thick green line
• The spheres will appear as circles
• The area of the green surface will not be visible when we look from the side
• For the one-dimensional collision between P and Q, this 2D view is sufficient
10. Next we look at the spheres A and B from the side. The 2D view will be as shown in fig.6.76(b)
■ Unfortunately, fig.6.76(b) does not convey the information that: 'Collision between A and B is not one-dimensional'. It looks as if, vAi passes through the center of sphere B
• So, if the collision is not one-dimensional, 'looking from the side' will not help
• We have to change our 'view point'
11. We have to look from 'above'
• The 2D view when looked from 'above' is shown in fig.6.77:
 |
| Fig.6.77 |
♦ In fig.6.76, where view is from 'side', the green surface appears as a 'thick green line'
♦ In fig.6.77, where view is from 'above', the green surface appears as a 'green rectangle'
• So in this type of 2D view, we can clearly see that the collision between A and B will not be one-dimensional
12. Now we can start the analysis of the collision
• In the fig.6.78 below, the conditions before and after collisions are shown separately
 |
| Fig.6.78 |
13. After the collision, sphere A moves with a velocity of vAf
• This vAf makes a particular angle with the initial direction of A
• We will denote this angle as θAf
14. After the collision, sphere B moves with a velocity of vBf
• This vBf makes a particular angle with the initial direction of A
• We will denote this angle as θBf
15. The movements after the collision are in different directions
• So it is no longer a one-dimensional problem. We cannot describe the motions using the 'x-axis' alone
• We will need both x and y axes
■ The collision in fig.6.78 is a two-dimensional collision.
• So from now on, we will use vector notations for the analysis
16. In these types of problems, it is always convenient to consider the x-axis to lie along the 'initial direction'
• In the fig.6.78, the white dashed line indicates the 'initial direction'
• So we can write:
♦ $\mathbf\small{\vec{v_{Af}}}$ makes an angle of θAf with the x axis
♦ $\mathbf\small{\vec{v_{Bf}}}$ makes an angle of θBf with the x axis
17. The information obtained so far can be written in a tabular form as shown below:
The above table has only 5 columns. We will add additional columns as and when more information become available
18. So let us continue:
• We will apply the law of conservation of momentum
• But momentum is a vector quantity. So we will consider the x and y directions separately
• First we consider the x direction:
(i) Initial momentum of A in the x direction
$\mathbf\small{\vec{p_{Ai(x)}}=m_A \times \text{initial velocity of A in the x-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \cos (\theta_{Ai})]\hat{i}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \cos (0)]\hat{i}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times 1]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Ai}}|]\hat{i}}$
(ii) Initial momentum of B in the x direction
$\mathbf\small{\vec{p_{Bi(x)}}=m_B \times \text{initial velocity of B in the x-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bi}}|\times \cos (\theta_{Bi})]\hat{i}}$
$\mathbf\small{=m_B \times [0\times \cos (\theta_{Bi})]\hat{i}}$
= 0
• The above information about 'initial momentum in x direction' is entered as the sixth column as shown below:
(cosine of the angle in column 4)
• So writing in the tabular form is indeed advantageous
19. Now we consider the y direction:
(i) Initial momentum of A in the y direction
$\mathbf\small{\vec{p_{Ai(y)}}=m_A \times \text{initial velocity of A in the y-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \sin (\theta_{Ai})]\hat{j}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times \sin (0)]\hat{j}}$
$\mathbf\small{=m_A \times [|\vec{v_{Ai}}|\times 0]\hat{j}}$
= 0
(ii) Initial momentum of B in the y direction
$\mathbf\small{\vec{p_{Bi(y)}}=m_B \times \text{initial velocity of B in the y-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bi}}|\times \sin (\theta_{Bi})]\hat{j}}$
$\mathbf\small{=m_B \times [0 \times \sin (\theta_{Bi})]\hat{j}}$
= 0
• The above information about 'initial momentum in y direction' is entered as the seventh column as shown below:
(sine of the angle in column 4)
• So writing in the tabular form is indeed advantageous
20. Next we consider the final momentum:
• First we consider the x direction:
(i) Final momentum of A in the x direction
$\mathbf\small{\vec{p_{Af(x)}}=m_A \times \text{final velocity of A in the x-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
(ii) Final momentum of B in the x direction
$\mathbf\small{\vec{p_{Bf(x)}}=m_B \times \text{final velocity of B in the x-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
$\mathbf\small{=[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
• The above information about 'final momentum in x direction' is entered as the 8th column as shown below:
(cosine of the angle in column 5)
21. Next we consider the y direction:
(i) Final momentum of A in the y direction
$\mathbf\small{\vec{p_{Af(y)}}=m_A \times \text{final velocity of A in the y-direction}}$
$\mathbf\small{=m_A \times [|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
$\mathbf\small{=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]\hat{i}}$
(ii) Final momentum of B in the y direction
$\mathbf\small{\vec{p_{Bf(y)}}=m_B \times \text{final velocity of B in the y-direction}}$
$\mathbf\small{=m_B \times [|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
$\mathbf\small{=[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]\hat{i}}$
• The above information about 'final momentum in x direction' is entered as the 9th column as shown below:
(sine of the angle in column 5)
22. In our present case, this y component is negative because, it is directed towards the negative side of the x axis
• We will directly get this negative direction if we put 'negative value' for θBf
• For example, sin(-30) = -sin30 = -0.5
23. Now we apply the conservation of momentum in the x-direction. We have:
• Total initial momentum in the x-direction = Total final momentum in the x direction
That is., $\mathbf\small{\sum{|\vec{p_{i(x)}}|}=\sum{|\vec{p_{f(x)}}|}}$
$\mathbf\small{\Rightarrow|\vec{p_{iA(x)}}|+0=|\vec{p_{fA(x)}}|+|\vec{p_{fB(x)}}|}$
$\mathbf\small{\Rightarrow [m_A \times|\vec{v_{Ai}}|]=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]}$
• This equation can be easily formed from the table as follows:
Total from column 6 = Total from column 8
24. Next we apply the conservation of momentum in the y-direction. We have:
• Total initial momentum in the y-direction = Total final momentum in the y direction
That is., $\mathbf\small{\sum{|\vec{p_{i(y)}}|}=\sum{|\vec{p_{f(y)}}|}}$
$\mathbf\small{\Rightarrow0+0=|\vec{p_{fA(y)}}|+|\vec{p_{fB(y)}}|}$
$\mathbf\small{\Rightarrow 0=[m_A \times|\vec{v_{Af}}|\times \sin (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \sin (\theta_{Bf})]}$
• This equation can be easily formed from the table as follows:
Total from column 7 = Total from column 9
25. With that, we complete our calculations related to momentum. Next we consider energy
• Assuming an elastic collision, we can say: 'kinetic energy will be conserved'
• Energy is a scalar quantity. There is no need to consider x and y components
• So we can write:
$\mathbf\small{\frac{1}{2}m_A\;|\vec{v_{Ai}}|^2+\frac{1}{2}m_B\;|\vec{v_{Bi}}|^2=\frac{1}{2}m_A\;|\vec{v_{Af}}|^2+\frac{1}{2}m_BA\;|\vec{v_{Bf}}|^2}$
$\mathbf\small{\Rightarrow m_A\;|\vec{v_{Ai}}|^2+m_B\;|\vec{v_{Bi}}|^2=m_A\;|\vec{v_{Af}}|^2+m_BA\;|\vec{v_{Bf}}|^2}$
26. So we have three equations:
(i) From (23), we have: $\mathbf\small{[m_A \times|\vec{v_{Ai}}|]=[m_A \times|\vec{v_{Af}}|\times \cos (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \cos (\theta_{Bf})]}$
(ii) From (24), we have: $\mathbf\small{0=[m_A \times|\vec{v_{Af}}|\times \sin (\theta_{Af})]+[m_B \times|\vec{v_{Bf}}|\times \sin (\theta_{Bf})]}$
(iii) From (25), we have: $\mathbf\small{m_A\;|\vec{v_{Ai}}|^2+m_B\;|\vec{v_{Bi}}|^2=m_A\;|\vec{v_{Af}}|^2+m_BA\;|\vec{v_{Bf}}|^2}$26. In the above 3 equations.
• Which are the 'known quantities' ?
• Which are the 'unknown quantities' ?
Let us analyse:
mA - Known quantity. Will be given in the question
mB - Known quantity. Will be given in the question
vAi - Known quantity. Will be given in the question
vBi - Known quantity. Will be given in the question
vAf - Unknown quantity. A final quantity which we have to determine
vBf - Unknown quantity. A final quantity which we have to determine
θAf - Unknown quantity. A final quantity which we have to determine
θBf - Unknown quantity. A final quantity which we have to determine
27. There are 4 unknowns. We have only 3 equations.
• So we use special 'angle measuring devices' to find one of the final angles θAf or θBf
• Then the 'number of unknowns' will become 3.
• We can easily find the other unknowns
28. Further, if both the final angles are known, the 'number of unknowns' will reduce to 2
• In such cases we need to form the first two equations only
Let us see some solved examples:
Solved example 6.39
Two objects A and B have the same mass of 0.1 kg. Object A has an initial speed of 5 ms-1. Object B is initially at rest. A collides with B and moves off with a speed of 2 ms-1 at an angle of 30o with the initial direction. What is the speed and direction of B?
Solution:
1. Fig.6.79 below shows the possible collision:
 |
| Fig.6.79 |
• 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.6.79 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 (B,3) and (B,5)
♦ (B,3) contains the unknown $\mathbf\small{|\vec{v_{Bf}}|}$
♦ (B,5) contains the unknown $\mathbf\small{\theta_{Bf}}$
• So there are only two unknowns. We need to form two equations only
• All the cells in columns 6 to 9 can be calculated except (B,8) and (B,9)
3. Considering momenta in the x-direction, we have:
• Sum in column 6 = Sum in column 8
$\mathbf\small{\Rightarrow 0.5=0.173+|\vec{P_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow 0.5=0.173+m_B\times|\vec{v_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow 0.5=0.173+0.1\times|\vec{v_{Bf(x)}}|}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf(x)}}|=3.27}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|\times \cos \theta_{Bf}=3.27}$
4. Considering the momenta in the y-direction, we have:
• Sum in column 7 = Sum in column 9
$\mathbf\small{\Rightarrow 0=0.1+|\vec{P_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow 0=0.1+m_B\times|\vec{v_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow 0=0.1+0.1\times|\vec{v_{Bf(y)}}|}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf(y)}}|=-1}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|\times \sin \theta_{Bf}=-1}$
5. So we have two equations:
(i) From (3) we have: $\mathbf\small{|\vec{v_{Bf}}|\times \cos \theta_{Bf}=3.27}$
(ii) From (4) we have: $\mathbf\small{|\vec{v_{Bf}}|\times \sin \theta_{Bf}=-1}$
• Dividing 5(ii) by 5(i), we get:
$\mathbf\small{\frac{|\vec{v_{Bf}}|\times \sin \theta_{Bf}}{|\vec{v_{Bf}}|\times \cos \theta_{Bf}}=\frac{-1}{3.27}}$
$\mathbf\small{\Rightarrow \tan \theta_{Bf}=-\frac{1}{3.27}}$
$\mathbf\small{\Rightarrow \theta_{Bf}=-17^o}$
• θBf is negative. So the positions in fig.6.79 are justified
6. Substituting this value of θBf in 5(i), we get:
$\mathbf\small{|\vec{v_{Bf}}|\times \cos (-17)=3.27}$
$\mathbf\small{\Rightarrow |\vec{v_{Bf}}|=\frac{3.27}{\cos (-17)}=\frac{3.27}{0.9563}=3.42\,\,\text{ms}^{-1}}$
We have completed this discussion on work, energy and power. In the next chapter, we will discuss about rotational motion
No comments:
Post a Comment