In our present chapter, we are discussing about motions in two dimensions. In the previous section we completed a discussion on uniform circular motion. In this section, we will see relative velocity in two dimensions.
We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.
1. Consider two objects A and B shown in fig.4.41(a) below:
• They are moving in two different directions and with two different magnitudes
♦ A is moving with a velocity of $\mathbf\small{\vec {v_A}}$
♦ B is moving with a velocity of $\mathbf\small{\vec {v_B}}$
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [$\mathbf\small{\vec{v_B}-\vec{v_A}}$]
• That is., Relative velocity of B with respect to A = $\mathbf\small{\vec{v_{BA}}=\vec{v_B}-\vec{v_A}}$
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [$\mathbf\small{\vec{v_A}-\vec{v_B}}$]
• That is., Relative velocity of A with respect to B = $\mathbf\small{\vec{v_{AB}}=\vec{v_A}-\vec{v_B}}$
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method:
• We use the horizontal and vertical components. Fig.4.41(b) shows those components.
4. Consider the resultant $\mathbf\small{\vec{v_{BA}}}$ in case 1
♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Bx}}-\vec{v_{Ax}}}$
♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{By}}-\vec{v_{Ay}}}$
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant $\mathbf\small{\vec{v_{AB}}}$ in case 2
♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Ax}}-\vec{v_{Bx}}}$
♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{Ay}}-\vec{v_{By}}}$
Once the components are obtained, we can easily calculate the original
We saw 'relative velocity in the case of motion in one dimension' in a previous chapter. Our present discussion is based on that.
1. Consider two objects A and B shown in fig.4.41(a) below:
 |
| Fig.4.41 |
♦ A is moving with a velocity of $\mathbf\small{\vec {v_A}}$
♦ B is moving with a velocity of $\mathbf\small{\vec {v_B}}$
2. We can write two cases:
(i) Case 1: When viewed from A, The other object B will be moving with a velocity of [$\mathbf\small{\vec{v_B}-\vec{v_A}}$]
• That is., Relative velocity of B with respect to A = $\mathbf\small{\vec{v_{BA}}=\vec{v_B}-\vec{v_A}}$
(ii) Case 2: When viewed from B, The other object A will be moving with a velocity of [$\mathbf\small{\vec{v_A}-\vec{v_B}}$]
• That is., Relative velocity of A with respect to B = $\mathbf\small{\vec{v_{AB}}=\vec{v_A}-\vec{v_B}}$
3. So we need to do vector subtraction to find relative velocities. We already know the methods to do such subtractions. We have both graphical and analytical methods.
■ Here we will use analytical method:
• We use the horizontal and vertical components. Fig.4.41(b) shows those components.
4. Consider the resultant $\mathbf\small{\vec{v_{BA}}}$ in case 1
♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Bx}}-\vec{v_{Ax}}}$
♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{By}}-\vec{v_{Ay}}}$
Once the components are obtained, we can easily calculate the resultant
• Consider the resultant $\mathbf\small{\vec{v_{AB}}}$ in case 2
♦ The x component of this resultant will be: $\mathbf\small{\vec{v_{Ax}}-\vec{v_{Bx}}}$
♦ The y component of this resultant will be: $\mathbf\small{\vec{v_{Ay}}-\vec{v_{By}}}$
Once the components are obtained, we can easily calculate the original
Solved example 4.16
Rain is falling vertically with a speed of 35 ms-1. A woman rides a bicycle with a speed of 12 ms-1 in east to west direction. What is the direction in which she should hold her umbrella ?
Solution:
1. Let the rain fall be denoted by $\mathbf\small{\vec {v_R}}$
• Let the motion of bicycle be denoted by $\mathbf\small{\vec {v_B}}$
• The two vectors are shown in the fig.4.42(a) below:
2. We want to find 'how the velocity of the rain will appear, when viewed from the bicycle'
 |
| Fig.4.42 |
♦ That is., we want to find the velocity of the rain with respect to the bicycle.
♦ That is., we want to find $\mathbf\small{\vec {v_{RB}}}$
♦ That is., we want to find $\mathbf\small{\vec{v_R}-\vec{v_B}}$
3. For that, we need to find the following two:
(i) x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)
♦ It is given by: $\mathbf\small{\vec{v_{Rx}}-\vec{v_{Bx}}}$
(ii) y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$)
♦ It is given by: $\mathbf\small{\vec{v_{Ry}}-\vec{v_{By}}}$
4. Let us write the values:
• $\mathbf\small{\vec{v_{Rx}}}$ = 0 (∵ the rain is falling vertically and so has no horizontal component)
• $\mathbf\small{\vec{v_{Bx}}}$ = -12 $\mathbf\small{\hat{i}}$ (The travel is from left to right. So it is taken as negative)
• $\mathbf\small{\vec{v_{Ry}}}$ = -35 $\mathbf\small{\hat{j}}$ (The travel is from top to bottom. So it is taken as negative)
• $\mathbf\small{\vec{v_{By}}}$ = 0 (∵ the bicycle is travelling horizontally and so has no vertical component)
5. Substituting the values in 3(i), we get:
• x component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBx}}}$ = [0 - (-12 $\mathbf\small{\hat{i}}$)] = 12 $\mathbf\small{\hat{i}}$
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
Substituting the values in 3(ii), we get:
• y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBy}}}$ = [(-35 $\mathbf\small{\hat{i}}$ - 0)] = -35 $\mathbf\small{\hat{j}}$
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
• y component of ($\mathbf\small{\vec{v_R}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{RBy}}}$ = [(-35 $\mathbf\small{\hat{i}}$ - 0)] = -35 $\mathbf\small{\hat{j}}$
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
6. Now we can find the magnitude:
Magnitude of $\mathbf\small{\vec{v_R}-\vec{v_B}}$ = $\mathbf\small{|\vec{v_R}-\vec{v_B}|}$ = $\mathbf\small{|\vec{v_{RB}}|}$ = $\mathbf\small{\sqrt{12^2+(-35)^2}}$ = 37 ms-1.
7. Next we find the direction:
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v{RBy}}|}{|\vec{v{RBx}}|}=\frac{35}{12}}$
• Thus we get: θ = 71.076o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ)
• Thus we get: φ = (90-71.076) = 18.924o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle
Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
♦ E-W direction is the x axis
♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45o with the x axis.
• This is shown in fig.4.43(a) below:
3. Now we can write the details about the velocity vector of wind
• Let us denote it as $\mathbf\small{\vec {v_W}}$
• We can write:
♦ $\mathbf\small{|\vec {v_W}|}$ = 72 km/h
♦ $\mathbf\small{\vec {v_W}}$ makes an angle 45o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction
Magnitude of $\mathbf\small{\vec{v_R}-\vec{v_B}}$ = $\mathbf\small{|\vec{v_R}-\vec{v_B}|}$ = $\mathbf\small{|\vec{v_{RB}}|}$ = $\mathbf\small{\sqrt{12^2+(-35)^2}}$ = 37 ms-1.
7. Next we find the direction:
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v{RBy}}|}{|\vec{v{RBx}}|}=\frac{35}{12}}$
• Thus we get: θ = 71.076o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
8. But what we want is 'φ' which the resultant makes with the vertical
• It can be easily obtained because, in fig.c, we can see that φ = (90-θ)
• Thus we get: φ = (90-71.076) = 18.924o.
9. So we can write the result:
• When viewed from the bicycle, rain falls towards the cyclist at an angle of 18.924o with the vertical
• So the cyclist must hold the umbrella at an angle of 18.924o with the vertical,
10. We must note the difference between this example and example 4.1.
• In that example, the boy was stationary. The rain was falling at an angle because of the wind
• But here, the rain is falling vertically. But to the cyclist, it appears to be falling at an angle
Solved example 4.17
In a harbour, wind is blowing at a speed of 72 km/h and the flag on the mast of a boat anchored in the harbour flutters along the N-E direction. If the boat starts moving at a speed of 51 km/h to the north, what is the direction of the flag on the mast of the boat?
Solution:
1. The speed and direction of the wind is given:
• Speed is 72 km/h and direction is N-E
2. Let us set up the coordinate axes as follows:
♦ E-W direction is the x axis
♦ N-S direction is the y axis
• Then the N-E direction (in which the wind is blowing) will be inclined at 45o with the x axis.
• This is shown in fig.4.43(a) below:
 |
| Fig.4.43 |
• Let us denote it as $\mathbf\small{\vec {v_W}}$
• We can write:
♦ $\mathbf\small{|\vec {v_W}|}$ = 72 km/h
♦ $\mathbf\small{\vec {v_W}}$ makes an angle 45o with the x axis
• This is also shown in fig.a
4. This velocity of the wind will not change
• But when viewed from a moving boat, the wind will appear to have a different magnitude and different direction
• We want to find 'how the velocity of the wind will appear, when viewed from the moving boat'
♦ That is., we want to find the velocity of the wind with respect to the moving boat.
♦ That is., we want to find $\mathbf\small{\vec {v_{WB}}}$
♦ That is., we want to find $\mathbf\small{\vec{v_W}-\vec{v_B}}$
Where $\mathbf\small{\vec {v_B}}$ is the velocity of the boat
Where $\mathbf\small{\vec {v_B}}$ is the velocity of the boat
5. For that, we need to find the following two:
(i) x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)
♦ It is given by: $\mathbf\small{\vec{v_{Wx}}-\vec{v_{Bx}}}$
(ii) y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$)
♦ It is given by: $\mathbf\small{\vec{v_{Wy}}-\vec{v_{By}}}$
6. Let us write the values:
• $\mathbf\small{\vec{v_{Wx}}}$ = 72 cos 45 $\mathbf\small{\hat{i}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h
• $\mathbf\small{\vec{v_{Bx}}}$ = 0 (∵ the boat is travelling north and so has no component towards east)
• $\mathbf\small{\vec{v_{Wy}}}$ = 72 sin 45 $\mathbf\small{\hat{j}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h
• $\mathbf\small{\vec{v_{By}}}$ = 51 $\mathbf\small{\hat{j}}$
• $\mathbf\small{\vec{v_{Wy}}}$ = 72 sin 45 $\mathbf\small{\hat{j}}$ = 50.912 $\mathbf\small{\hat{i}}$ km/h
• $\mathbf\small{\vec{v_{By}}}$ = 51 $\mathbf\small{\hat{j}}$
7. Substituting the values in 5(i), we get:
• x component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBx}}}$ = [50.912 $\mathbf\small{\hat{i}}$ - 0] = 50.912 $\mathbf\small{\hat{i}}$
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
♦ We get a positive value. So this vector is directed towards the positive side of the x axis.
♦ This is shown in fig.b
Substituting the values in 5(ii), we get:
• y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBy}}}$ = [(72 sin 45 $\mathbf\small{\hat{j}}$ - 51 $\mathbf\small{\hat{j}}$)] = -0.0883
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v_{WBy}}|}{|\vec{v_{WBx}}|}=\frac{0.0883}{50.912}}$
• Thus we get: θ = 0.0997o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately
Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., $\mathbf\small{\vec{v_{RM}}}$ will slope towards the right
♦ Where $\mathbf\small{\vec{v_{RM}}}$ is the relative velocity of the rain with respect to the man
♦ This is shown in fig.4.44(a) below:
2. But in our present case, even when there is motion at 3 km/h, the rain appears to be exact vertical
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of $\mathbf\small{\vec{v_{R}}}$ in fig.b
3. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.b are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
4. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{Mx}}=3 \hat{i}}$
• Substituting in 4(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 4(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
5. Now consider the $\mathbf\small{\vec{v_{RM}}}$ in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, $\mathbf\small{\vec{v_{RMx}}}$ = 0
• From 4(i)b, we can write: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$ = 0
■ From this we get: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
This is one of the two equations which we will need to find $\mathbf\small{|\vec{v_{R}}|}$ and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the $\mathbf\small{\vec{v_{RM}}}$ is indeed at a slope. This is shown in fig.d
♦ Given that, it makes an angle of 45° with the vertical
♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) $\mathbf\small{\vec{v_{R}}}$ in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change
(ii) $\mathbf\small{\vec{v_{M}}}$ in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.d are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
8. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{Mx}}=6 \hat{i}}$
• Substituting in 8(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-6 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 8(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
9. Using the 'formula for direction of the resultant', we can write:
$\mathbf\small{\tan 45=\frac{|\vec{v_{RMy}}|}{|\vec{v_{RMx}}|}=\frac{||\vec{v_R}|\sin \theta| }{|(|\vec{v_R}|\cos \theta)-6|}}$
10. But in (5), we obtained: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
Substituting this in (9), we get: $\mathbf\small{\tan 45=\frac{||\vec{v_R}|\sin \theta| }{|(3)-6|}=\frac{||\vec{v_R}|\sin \theta| }{|(-3)|}=\frac{|\vec{v_R}|\sin \theta }{3}}$
11. But tan 45 = $\mathbf\small{\frac{1}{\sqrt{2}}}$
Thus we get: $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
12.So we have two results:
(i) $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
(ii) $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
Dividing (ii) by (i), we get: $\mathbf\small{\tan \theta =\frac{1}{\sqrt{2}}}$
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
$\mathbf\small{|\vec{v_{R}}|\cos 45=3}$
$\mathbf\small{\Rightarrow |\vec{v_{R}}|=\frac{3}{\frac{1}{\sqrt{2}}}=3\sqrt{2}=4.243\: km/h}$
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right
• y component of ($\mathbf\small{\vec{v_W}-\vec{v_B}}$) = $\mathbf\small{\vec{v_{WBy}}}$ = [(72 sin 45 $\mathbf\small{\hat{j}}$ - 51 $\mathbf\small{\hat{j}}$)] = -0.0883
♦ We get a negative value. So this vector is directed towards the negative side of the y axis
♦ This is also shown in fig.b
8. Now we can find the magnitude. But it is not asked in the question
9. But we do have to find the direction
• We have: $\mathbf\small{\tan \theta =\frac{|\vec{v_{WBy}}|}{|\vec{v_{WBx}}|}=\frac{0.0883}{50.912}}$
• Thus we get: θ = 0.0997o
■ We must note the following points 3 while using this formula:
(i) Magnitude of the y component is in the numerator
(ii) Magnitude of the x component is in the denominator
(iii) The angle 'θ' thus obtained will always be the angle between the resultant and the x component
• This 'θ' is marked in the fig.c
10. We see that, the relative velocity vector falls between the east and south directions
• But the deviation from east is only 0.0997o, which is a very small quantity
• So we can write:
■ The flag flutters in the east direction approximately
Solved example 4.18
When a man walks at a rate of 3 km/h, the rain appears to fall vertically. When he walks at the rate of 6 km/h, it appears to fall at an angle of 45° with the vertical. What is the original magnitude and direction of the rain?
Solution:
1. Consider the simple case when the rain falls in the exact vertical direction
■ For a man moving from right to left, the rain will obviously appear to be falling at a slope
• This slope will be towards the right.
• That is., $\mathbf\small{\vec{v_{RM}}}$ will slope towards the right
♦ Where $\mathbf\small{\vec{v_{RM}}}$ is the relative velocity of the rain with respect to the man
♦ This is shown in fig.4.44(a) below:
 |
| Fig.4.44 |
■ So it is clear that, the original rain is sloping towards the left. This is shown in fig.b
• We are asked to find this original magnitude and direction
• That is., we need to find the details of $\mathbf\small{\vec{v_{R}}}$ in fig.b
3. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.b are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
4. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{Mx}}=3 \hat{i}}$
• Substituting in 4(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.c
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 4(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
5. Now consider the $\mathbf\small{\vec{v_{RM}}}$ in fig.c
• It is perfect vertical. That means, it has no horizontal component
• That means, $\mathbf\small{\vec{v_{RMx}}}$ = 0
• From 4(i)b, we can write: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-3 \hat{i}}$ = 0
■ From this we get: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
This is one of the two equations which we will need to find $\mathbf\small{|\vec{v_{R}}|}$ and θ.
6. To get the other equation, we consider the motion at 6 km/h
• This time, the $\mathbf\small{\vec{v_{RM}}}$ is indeed at a slope. This is shown in fig.d
♦ Given that, it makes an angle of 45° with the vertical
♦ So it will make the same angle of 45° with the horizontal also
Note the two points:
(i) $\mathbf\small{\vec{v_{R}}}$ in fig.d is same as that in fig.c. This is because, the original speed and direction of rain does not change
(ii) $\mathbf\small{\vec{v_{M}}}$ in fig.d has the same direction as that in fig.c But the magnitude changed from 3 to 6 km/h
7. We know this:
• The two components of $\mathbf\small{\vec{v_{RM}}}$ in fig.d are:
(i) $\mathbf\small{\vec{v_{RMx}}}$
(ii) $\mathbf\small{\vec{v_{RMy}}}$
8. So we will first find those components:
(i) We have: $\mathbf\small{\vec{v_{RMx}}=\vec{v_{Rx}}-\vec{v_{Mx}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Rx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{Mx}}=6 \hat{i}}$
• Substituting in 8(i), we get: $\mathbf\small{\vec{v_{RMx}}=(|\vec{v_{R}}|\cos \theta) \hat{i}-6 \hat{i}}$
(ii) We have: $\mathbf\small{\vec{v_{RMy}}=\vec{v_{Ry}}-\vec{v_{My}}}$
• Let us write the values:
(a) $\mathbf\small{\vec{v_{Ry}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
♦ Where θ is the angle made by $\mathbf\small{\vec{v_{R}}}$ with the horizontal
♦ This is shown in fig.d
(b) $\mathbf\small{\vec{v_{My}}=0}$
• Substituting in 8(ii), we get: $\mathbf\small{\vec{v_{RMy}}=(|\vec{v_{R}}|\sin \theta) \hat{j}}$
9. Using the 'formula for direction of the resultant', we can write:
$\mathbf\small{\tan 45=\frac{|\vec{v_{RMy}}|}{|\vec{v_{RMx}}|}=\frac{||\vec{v_R}|\sin \theta| }{|(|\vec{v_R}|\cos \theta)-6|}}$
10. But in (5), we obtained: $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
Substituting this in (9), we get: $\mathbf\small{\tan 45=\frac{||\vec{v_R}|\sin \theta| }{|(3)-6|}=\frac{||\vec{v_R}|\sin \theta| }{|(-3)|}=\frac{|\vec{v_R}|\sin \theta }{3}}$
11. But tan 45 = $\mathbf\small{\frac{1}{\sqrt{2}}}$
Thus we get: $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
12.So we have two results:
(i) $\mathbf\small{|\vec{v_{R}}|\cos \theta=3}$
(ii) $\mathbf\small{|\vec{v_R}|\sin \theta =\frac{3}{\sqrt{2}}}$
Dividing (ii) by (i), we get: $\mathbf\small{\tan \theta =\frac{1}{\sqrt{2}}}$
Thus we get θ = 45°.
13. Substituting this value of θ in 12(i), we get:
$\mathbf\small{|\vec{v_{R}}|\cos 45=3}$
$\mathbf\small{\Rightarrow |\vec{v_{R}}|=\frac{3}{\frac{1}{\sqrt{2}}}=3\sqrt{2}=4.243\: km/h}$
14. So we can write:
• The rain is actually falling at an angle of 45° towards the left
• It has a magnitude of 4.243 km/h
• But when the man walks ar the rate of 6 km/h, the rain appears to be falling at an angle of 45° towards the right
In the next section, we will see a few more solved examples.
Copyright©2018 Higher Secondary Physics. blogspot.in - All Rights Reserved
No comments:
Post a Comment