In the previous section we saw details about relative velocity in rectilinear motion. In this section we will see a special case.
In the next chapter, we will see motion in a plane.
Consider fig.3.55(a) below:
1. A 'platform on wheels' of very long length is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1.
• It is moving on a railway track
2. A bridge runs directly above and parallel to the railway track. On that bridge, a car is travelling towards the positive side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
• This 'particular velocity' is the velocity of the car w.r.t the platform. It is given by: VCP = (VC - VP)
4. Since both the car and the platform travels towards the positive side of the x axis, both are positive values
• So, if VC is greater than VP, we get a positive value for VCP
• That means, when viewed from M, the car appears to be moving towards the positive side of the x axis
5. If VC is less than VP, we get a negative value for VCP
• That means, when viewed from M, the car appears to be moving towards the negative side of the x axis
• That means, when viewed from M, the car appears to be moving further and further backwards
6. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the positive side of the x axis with a velocity of VC.
• The car will surely pass the obsever, whatever be the velocity of the car
Now consider fig(b):
1. As before, the platform is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1
2. The car is now moving on the platform
• It is travelling towards the positive side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
This 'particular velocity' is nothing but the velocity VC shown by the speedometer of the car
So this time we do not have to make a subtraction
4. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the positive side of the x axis with a velocity of (VC + VP)
• The car will surely pass the obsever, whatever be the velocity of the car
Consider fig.3.56(a) below:
1. A 'platform on wheels' of very long length is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1.
• It is moving on a railway track
2. A bridge runs directly above and parallel to the railway track. On that bridge, a car is travelling towards the negative side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
• This 'particular velocity' is the velocity of the car w.r.t the platform. It is given by:
VCP = (-VC -VP) = -(VC + VP)
• VC is given a negative sign because, the car is moving towards the negative side of the x axis
4. Since VCP is negtive, we can write:
When viewed from M, the car appears to be moving towards the negative side of the x axis
5. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the negative side of the x axis with a velocity of VC.
• The car will surely pass the obsever, whatever be the velocity of the car
Now consider fig(b):
1. As before, the platform is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1
2. The car is now moving on the platform
• It is travelling towards the negative side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
This 'particular velocity' is nothing but the velocity VC shown by the speedometer of the car
So this time we do not have to make a subtraction
4. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the negative side of the x axis with a velocity of (VC - VP)
• The car will move towards the right and away from the observer, if VC is less than VP.
♦ The car will never be able to pass the observer in this case
• The car will appear to be stationary, if VC is equal to VP
Now we will see a solved example:
Solved example 3.10
A jet airplane travelling at the speed of 500 $ km~h^{–1} $ ejects its products of combustion at the speed of 1500 $ km~h^{–1} $ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Solution:
1. Let us assume that, the jet travels towards the positive side of the x axis
2. Let $ V_J $ be the 'actual velocity' of the jet.
• That is., $ V_J $ is the velocity of the jet respect to the ground
• Then we can write: $ V_J $ = +500 $ km~h^{–1} $
3. The products of combustion (smoke) will be moving towards the negative side of the x axis
4. Let $ V_S $ be the 'actual velocity' of the smoke.
• Consider the following two velocities:
(i) 'velocity with which the smoke appears to move', when viewed from the jet
(ii) 'velocity with which the smoke of combustion appears to move', when viewed from the ground
• Surely, (i) is the 'actual velocity' of the smoke
• So we can write: $ V_S $ = -1500 $ km~h^{–1} $
5. Each particle of the gas emerges from the jet. But the jet is moving towards the right
• So the observer on the ground sees only a net velocity.
• We can write: Velocity of smoke with respect to ground = -1500 + 500 = -1000 $ km~h^{–1} $
• The negative sign indicates that, the smoke appears ot move with a velocity of 1000 $ km~h^{–1} $ towards the negative side of the x axis
 |
| Fig.3.55 |
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1.
• It is moving on a railway track
2. A bridge runs directly above and parallel to the railway track. On that bridge, a car is travelling towards the positive side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
• This 'particular velocity' is the velocity of the car w.r.t the platform. It is given by: VCP = (VC - VP)
4. Since both the car and the platform travels towards the positive side of the x axis, both are positive values
• So, if VC is greater than VP, we get a positive value for VCP
• That means, when viewed from M, the car appears to be moving towards the positive side of the x axis
5. If VC is less than VP, we get a negative value for VCP
• That means, when viewed from M, the car appears to be moving towards the negative side of the x axis
• That means, when viewed from M, the car appears to be moving further and further backwards
6. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the positive side of the x axis with a velocity of VC.
• The car will surely pass the obsever, whatever be the velocity of the car
Now consider fig(b):
1. As before, the platform is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1
2. The car is now moving on the platform
• It is travelling towards the positive side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
This 'particular velocity' is nothing but the velocity VC shown by the speedometer of the car
So this time we do not have to make a subtraction
4. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the positive side of the x axis with a velocity of (VC + VP)
• The car will surely pass the obsever, whatever be the velocity of the car
Consider fig.3.56(a) below:
 |
| Fig.3.56 |
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1.
• It is moving on a railway track
2. A bridge runs directly above and parallel to the railway track. On that bridge, a car is travelling towards the negative side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
• This 'particular velocity' is the velocity of the car w.r.t the platform. It is given by:
VCP = (-VC -VP) = -(VC + VP)
• VC is given a negative sign because, the car is moving towards the negative side of the x axis
4. Since VCP is negtive, we can write:
When viewed from M, the car appears to be moving towards the negative side of the x axis
5. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the negative side of the x axis with a velocity of VC.
• The car will surely pass the obsever, whatever be the velocity of the car
Now consider fig(b):
1. As before, the platform is moving towards the positive side of the x axis.
• The reading shown by the speedometer of the moving platform is VP ms-1
• So the platform is moving with a velocity of VP ms-1
2. The car is now moving on the platform
• It is travelling towards the negative side of the x axis
• The reading shown by the speedometer of the car is VC ms-1
• So the car is moving with a velocity of VC ms-1
3. A point M is marked on the platform
• When viewed from M, the car will appear to be moving with a 'particular velocity'
This 'particular velocity' is nothing but the velocity VC shown by the speedometer of the car
So this time we do not have to make a subtraction
4. Consider the point N marked on the ground
• A stationary observer at N will see the following:
♦ The platform moving towards the positive side of the x axis with a velocity of VP
♦ The car moving towards the negative side of the x axis with a velocity of (VC - VP)
• The car will move towards the right and away from the observer, if VC is less than VP.
♦ The car will never be able to pass the observer in this case
• The car will appear to be stationary, if VC is equal to VP
Now we will see a solved example:
Solved example 3.10
A jet airplane travelling at the speed of 500 $ km~h^{–1} $ ejects its products of combustion at the speed of 1500 $ km~h^{–1} $ relative to the jet plane. What is the speed of the latter with respect to an observer on the ground ?
Solution:
1. Let us assume that, the jet travels towards the positive side of the x axis
2. Let $ V_J $ be the 'actual velocity' of the jet.
• That is., $ V_J $ is the velocity of the jet respect to the ground
• Then we can write: $ V_J $ = +500 $ km~h^{–1} $
3. The products of combustion (smoke) will be moving towards the negative side of the x axis
4. Let $ V_S $ be the 'actual velocity' of the smoke.
• Consider the following two velocities:
(i) 'velocity with which the smoke appears to move', when viewed from the jet
(ii) 'velocity with which the smoke of combustion appears to move', when viewed from the ground
• Surely, (i) is the 'actual velocity' of the smoke
• So we can write: $ V_S $ = -1500 $ km~h^{–1} $
5. Each particle of the gas emerges from the jet. But the jet is moving towards the right
• So the observer on the ground sees only a net velocity.
• We can write: Velocity of smoke with respect to ground = -1500 + 500 = -1000 $ km~h^{–1} $
• The negative sign indicates that, the smoke appears ot move with a velocity of 1000 $ km~h^{–1} $ towards the negative side of the x axis
In the next chapter, we will see motion in a plane.
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