In the previous section we saw impulse. In this section we will see Newton's third law of motion.
We will write it in steps:
1. Consider a tug of war contest.
• One team pulls to the left while the other team pulls to the right.
2. 'Pulling' is in fact, 'applying a force'.
• If one of the team stops applying their force, the other team will not be able to apply their's
• That is., if one of the team stops applying their force, the other team's force will also disappear.
3. In the tug of war, we see 'pulling'.
• We see the same effect in 'pushing' also.
• When we push against a 'massive and stable wall', we will feel that someone is pushing us back with the same force
• This reverse force from the wall will disappear if we stop pushing
4. So we see that forces always occur in pairs.
• That is., When ever we apply a force on a body, there occurs another force.
• We can call the applied force as action and the resulting force as reaction
5. What is the magnitude of the reaction?
• We do not need to do any calculations for finding that magnitude
■ The 'magnitude of the reaction' will be exactly same as the 'magnitude of the action'
6. What about the direction?
■ The direction of the reaction is exactly opposite to the direction of the action
7. Newton's third law states that:
■ To every action, there is an equal and opposite reaction
Example 1:
1. Apply a compressive force on a spring
• Let the force applied on the spring by the hand be FSH
• Then FSH is the action
2. The spring will exert a reactive force on the hand
• This is the force exerted on the hand by the spring. We can call it FHS
■ According to Newton's third law, we can write FSH = -FHS .
Example 2:
1. Consider a stone falling from a height
• The earth is exerting a force on the stone. This force is due to gravity
2. It is an attractive force. That is., the stone is pulled towards the earth
• Let the force applied by the earth on the stone be FES
3. According to Newton’s third law, the stone will exert an equal and opposite force on the earth
• That means, the earth is pulled towards the stone
• Let this force be FSE
4. We can write: FES = -FSE
5. We see the stone moving towards the earth. That means, we can clearly see the effect of FES
• But we do not see the earth moving towards the stone. That means, we do not see the effect of FSE.
• Why is that so?
Ans: Since the two forces are equal and opposite, we can write:
FES = -FSE
• Applying Newton's second law, we get:
mE × aE = -mS × aS .
• Where:
♦ mE and mS are the masses of earth and stone respectively
♦ aE and aS are the accelerations of earth and stone respectively
• mE is very large compared to mE
• So, for the left side to become equal to right side, aE has to be extremely small
• It will be so small that, we will not be able to see the motion of earth towards the stone
We will write it in steps:
1. Consider a tug of war contest.
• One team pulls to the left while the other team pulls to the right.
2. 'Pulling' is in fact, 'applying a force'.
• If one of the team stops applying their force, the other team will not be able to apply their's
• That is., if one of the team stops applying their force, the other team's force will also disappear.
3. In the tug of war, we see 'pulling'.
• We see the same effect in 'pushing' also.
• When we push against a 'massive and stable wall', we will feel that someone is pushing us back with the same force
• This reverse force from the wall will disappear if we stop pushing
4. So we see that forces always occur in pairs.
• That is., When ever we apply a force on a body, there occurs another force.
• We can call the applied force as action and the resulting force as reaction
5. What is the magnitude of the reaction?
• We do not need to do any calculations for finding that magnitude
■ The 'magnitude of the reaction' will be exactly same as the 'magnitude of the action'
6. What about the direction?
■ The direction of the reaction is exactly opposite to the direction of the action
7. Newton's third law states that:
■ To every action, there is an equal and opposite reaction
Let us see some more examples:
1. Apply a compressive force on a spring
• Let the force applied on the spring by the hand be FSH
• Then FSH is the action
2. The spring will exert a reactive force on the hand
• This is the force exerted on the hand by the spring. We can call it FHS
■ According to Newton's third law, we can write FSH = -FHS .
Example 2:
1. Consider a stone falling from a height
• The earth is exerting a force on the stone. This force is due to gravity
2. It is an attractive force. That is., the stone is pulled towards the earth
• Let the force applied by the earth on the stone be FES
3. According to Newton’s third law, the stone will exert an equal and opposite force on the earth
• That means, the earth is pulled towards the stone
• Let this force be FSE
4. We can write: FES = -FSE
5. We see the stone moving towards the earth. That means, we can clearly see the effect of FES
• But we do not see the earth moving towards the stone. That means, we do not see the effect of FSE.
• Why is that so?
Ans: Since the two forces are equal and opposite, we can write:
FES = -FSE
• Applying Newton's second law, we get:
mE × aE = -mS × aS .
• Where:
♦ mE and mS are the masses of earth and stone respectively
♦ aE and aS are the accelerations of earth and stone respectively
• mE is very large compared to mE
• So, for the left side to become equal to right side, aE has to be extremely small
• It will be so small that, we will not be able to see the motion of earth towards the stone
Now we will see a practical application of the third law. We will write it in steps:
1. Consider a billiard ball striking against a rigid wall. See fig.5.16(a) below:
• The red line is a normal to the wall.
♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$.
• It travels along the red line. So we can write:
(i) The path of the ball is normal to the wall
(ii) The ball will hit the wall normally.
3. After hitting the wall, the ball rebounds
• The velocity after rebound is the same $\mathbf\small{\vec{u}}$. But in the reverse direction.
• The path after rebound is the same normal line shown in red color. This is shown in fig(b)
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=m\, \vec{u}}$
♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=-m\, \vec{u}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=(-m\, \vec{u}-m\, \vec{u})=-2m\, \vec{u}}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=m \times 0=0}$
1. Consider a billiard ball striking against a rigid wall. See fig.5.16(a) below:
 |
| Fig.5.16 |
♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$.
• It travels along the red line. So we can write:
(i) The path of the ball is normal to the wall
(ii) The ball will hit the wall normally.
3. After hitting the wall, the ball rebounds
• The velocity after rebound is the same $\mathbf\small{\vec{u}}$. But in the reverse direction.
• The path after rebound is the same normal line shown in red color. This is shown in fig(b)
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=m\, \vec{u}}$
♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=-m\, \vec{u}}$
■ So change in momentum in the x direction = $\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=(-m\, \vec{u}-m\, \vec{u})=-2m\, \vec{u}}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=m \times 0=0}$
(ii) Final momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y2}}=m \times \vec{v_{y2}}=m \times 0=0}$
■ So change in momentum in the y direction = $\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=(0-0)=0}$.
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of $\mathbf\small{\Delta \vec{p_x}}$ and $\mathbf\small{\Delta \vec{p_y}}$
• But $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_x}=-2m\, \vec{u}}$
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{-2m\, \vec{u}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{2m\, |\vec{u}|}$
(iv) It's direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=-2m\, \vec{u}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall
12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also
Let us write the steps:
1. Consider a billiard ball striking against a rigid wall. See fig.5.17(a) below:
• The red line is a normal to the wall.
♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$
• The direction of this $\mathbf\small{\vec{u}}$ makes an angle of 30o with the red line
• Also note that, the red line is drawn at the point where the ball hits the wall.
• So we can write:
The ball will hit the wall at an angle of 30o
3. After hitting the wall, the ball rebounds
• The velocity after rebound is $\mathbf\small{\vec{v}}$
• Let the path after rebound makes the same angle of 30o with the normal line shown in red color. This is shown in fig(b)
• Also let the magnitudes of the two velocities be the same. That is: $\mathbf\small{|\vec{u}|=|\vec{v}|}$
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=[m|\vec{u}|\cos\,30^o]\hat{i}}$
♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=[-m|\vec{v}|\cos\,30^o]\hat{i}}$
■ So change in momentum in the x direction =
$\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=\left([-m|\vec{v}|\cos\,30^o]\hat{i}-[m|\vec{u}|\cos\,30^o]\hat{i}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-|\vec{v}|-|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=[-m|\vec{u}|\sin\,30^o]\hat{j}}$
(ii) Final momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y2}}=m \times \vec{v_{y2}}=[-m|\vec{v}|\sin\,30^o]\hat{j}}$
• Both the above quantities are negative because, they are directed towards the negative side of the y axis
■ So change in momentum in the y direction =
$\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=\left([-m|\vec{v}|\sin\,30^o]\hat{j}-[-m|\vec{u}|\sin\,30^o]\hat{j}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{v}|--|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{u}|+|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}=0}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of $\mathbf\small{\Delta \vec{p_x}}$ and $\mathbf\small{\Delta \vec{p_y}}$
• But $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]}$
(iv) It's direction:
♦ The unit vector '$\mathbf\small{\hat{i}}$' indicates that, it acts parallel to the x axis
♦ The '-' sign indicates that it acts towards the negative side of the x axis
• So it's direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall
This is an interesting situation. We can write it as two points:
(i) First we made the ball to strike the wall normally
• Then we made the ball to strike the wall at an angle
(ii) In both cases, the force exerted by the wall on the ball acts along the normal to the wall
12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also
Let us write a summary in three points A, B and C:
A. A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the wall on the ball will be along the normal (at the point of contact) to the wall
• This force is directed 'away from the wall'
B. From (A), we get the direction of the force exerted by the ball on the wall. For that, we apply Newton's third law. We will write the steps again:
A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the ball on the wall will be along the normal (at the point of contact) to the wall
• This force is directed 'towards the wall'
C. We can obtain the 'magnitude of the force' exerted by the ball on the wall, if we have the following three items:
(i) Mass of the ball
(ii) Velocity of the ball
(iii) Time duration for which the ball is in contact with the wall
In the above discussion we saw this:
■ What ever be the direction in which the ball strikes the wall, the forces will be along the red line.
• What is so special about this red line?
Let us analyse:
1. When the ball hits the wall, there is a 'surface of contact'
• Since the ball is spherical, this surface of contact will be very small
• It will be like a 'point of contact'
2. If we draw a tangent (at the point of contact) to the sphere, that tangent will lie on the surface of the wall. Details about tangents can be seen here.
• So we can say that, the 'contact surface' is the surface of the wall itself
3. This is true for what ever direction the ball strikes the wall. It is shown in fig.5.18 below:
• In fig.5.18(a), the ball hits normally
♦ The 'contact surface' is a point
♦ The wall is tangential to that point
♦ So the red line is perpendicular to the contact surface
• In fig.5.18(b), the ball hits at an angle
♦ Here also, the 'contact surface' is a point
♦ The wall is tangential to that point
♦ So the red line is perpendicular to the contact surface
4. So we find that:
■ The forces act along the line perpendicular to the contact surface
5. So what is the speciality of the red line?
Answer can be written in two statements:
(i) The force always act along the line perpendicular to the contact surface
(ii) When a ball hits a wall at any angle, the red line (in figs.5.16 and 5.17) is the line perpendicular to the contact surface
• In the above example, the wall is a flat surface.
• How do we draw the red line when surfaces are not flat?
• We will see the case when spherical balls collide with each other:
1. In fig.5.19(a), two balls (of same size) A and B are moving with velocities VA and VB
2. Their paths intersect at P.
• After some time, let ball A reach P.
• If ball B also reach there at that same instant, a collision will take place.
• This is shown in fig.b.
3. In fig.c, a tangent is drawn at the point of contact
• We can say that, this tangent represents the surface of contact between the balls
• The line joining the 'centers of the balls' will be perpendicular to this surface
4. There will be two forces acting at the time of collision:
(i) FAB which is the force which A exerts on B
(ii) FBA which is the force which B exerts on A
5. By Newton's third law, FAB = - FBA
• Both the forces act along the line joining the centers of the balls. This is shown in the enlarged view in fig.d
■ So in this case, the 'line joining the ceters of the balls' is the 'red line'
So we have completed a basic discussion on Newton's third law
• The following points should be always kept in mind:
1. Forces always occur in pairs
• A 'pair' means two'
• So if we see FAB, we can be sure that, there will be FBA
♦ If FBA is not present, there will be no FAB
♦ If FAB is not present, there will be no FBA
2. The terms 'action' and 'reaction' may give the wrong impression that:
• Action occurs first, and reaction occurs later, as an 'effect' of the action
■ The fact is that, both occur at the same instant
3. The two forces act on two different bodies
• FAB acts on B
• FBA acts on A
■ So even though they are equal and opposite, they cannot cancel each other.
■ So change in momentum in the y direction = $\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=(0-0)=0}$.
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of $\mathbf\small{\Delta \vec{p_x}}$ and $\mathbf\small{\Delta \vec{p_y}}$
• But $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_x}=-2m\, \vec{u}}$
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{-2m\, \vec{u}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{2m\, |\vec{u}|}$
(iv) It's direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=-2m\, \vec{u}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall
12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also
In the above example, the ball strike the wall along a normal. What if it strikes at an angle?
1. Consider a billiard ball striking against a rigid wall. See fig.5.17(a) below:
 |
| Fig.5.17 |
♦ That means, the red line is perpendicular to the surface of the wall
2. The billiard ball has a linear motion. It’s velocity is $\mathbf\small{\vec{u}}$
• The direction of this $\mathbf\small{\vec{u}}$ makes an angle of 30o with the red line
• Also note that, the red line is drawn at the point where the ball hits the wall.
• So we can write:
The ball will hit the wall at an angle of 30o
3. After hitting the wall, the ball rebounds
• The velocity after rebound is $\mathbf\small{\vec{v}}$
• Let the path after rebound makes the same angle of 30o with the normal line shown in red color. This is shown in fig(b)
• Also let the magnitudes of the two velocities be the same. That is: $\mathbf\small{|\vec{u}|=|\vec{v}|}$
4. Now we can begin the calculations:
x direction:
(i) Initial momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x1}}=m \times \vec{v_{x1}}=[m|\vec{u}|\cos\,30^o]\hat{i}}$
♦ Where m is the mass of the ball
(ii) Final momentum of the ball in the x direction = $\mathbf\small{\vec{p_{x2}}=m \times \vec{v_{x2}}=[-m|\vec{v}|\cos\,30^o]\hat{i}}$
■ So change in momentum in the x direction =
$\mathbf\small{\Delta \vec{p_{x}}=(\vec{p_{x2}}-\vec{p_{x1}})=\left([-m|\vec{v}|\cos\,30^o]\hat{i}-[m|\vec{u}|\cos\,30^o]\hat{i}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-|\vec{v}|-|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
y direction:
(i) Initial momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y1}}=m \times \vec{v_{y1}}=[-m|\vec{u}|\sin\,30^o]\hat{j}}$
(ii) Final momentum of the ball in the y direction = $\mathbf\small{\vec{p_{y2}}=m \times \vec{v_{y2}}=[-m|\vec{v}|\sin\,30^o]\hat{j}}$
• Both the above quantities are negative because, they are directed towards the negative side of the y axis
■ So change in momentum in the y direction =
$\mathbf\small{\Delta \vec{p_{y}}=(\vec{p_{y2}}-\vec{p_{y1}})=\left([-m|\vec{v}|\sin\,30^o]\hat{j}-[-m|\vec{u}|\sin\,30^o]\hat{j}\right )}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{v}|--|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}}$
$\mathbf\small{\Rightarrow \Delta \vec{p_{y}}=\left[\left(-|\vec{u}|+|\vec{u}|\right )(m\sin\,30^o)\, \right ]\hat{j}=0}$
$\mathbf\small{(\because |\vec{u}|=|\vec{v}|)}$
5. Resultant change in momentum $\mathbf\small{\Delta \vec{p}}$ is the resultant of $\mathbf\small{\Delta \vec{p_x}}$ and $\mathbf\small{\Delta \vec{p_y}}$
• But $\mathbf\small{\Delta \vec{p_y}}$ = 0
■ So we get: $\mathbf\small{\Delta \vec{p}=\Delta \vec{p_{x}}=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
6. We can write the following four points:
(i) The ball suffers a change in momentum of $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
(ii) This 'change in momentum' is a vector quantity
(iii) It's magnitude is $\mathbf\small{\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]}$
(iv) It's direction:
♦ The unit vector '$\mathbf\small{\hat{i}}$' indicates that, it acts parallel to the x axis
♦ The '-' sign indicates that it acts towards the negative side of the x axis
• So it's direction is along the red line and towards the negative side of the x axis
7. The ball suffers a ‘change in momentum’ because, a force acted on it
• Obviously, this force was exerted by the wall.
♦ We can be sure about this because, there is no other object present in the vicinity
8. Let the force exerted by the wall on the ball be $\mathbf\small{\vec{F_{WB}}}$
• Let the time duration for which this force acted be Δt
• Consider the product: $\mathbf\small{\vec{F_{WB}}\times\Delta t}$
9. We know that this (product of force and time for which the force acts) is equal to the ‘change in momentum’
• So we can write $\mathbf\small{\vec{F_{WB}}\times\Delta t=\left[\left(-2|\vec{u}|\right )(m\cos\,30^o)\, \right ]\hat{i}}$
10. We have an 'equality of two vectors'
• The vector on the left is equal to the vector on the right
• So the two vectors have the same magnitude and same direction
11. That means, the direction of $\mathbf\small{\vec{F_{WB}}}$ is same as the direction of the ‘change in momentum’
• But we have seen that, the direction of the ‘change in momentum’ is along the red line and towards the negative side of the x axis
• That means the direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• So we obtain an important result:
■ The direction of the ‘force exerted by the wall on the ball’ is along the normal to the wall. Also it is directed away from the wall
This is an interesting situation. We can write it as two points:
(i) First we made the ball to strike the wall normally
• Then we made the ball to strike the wall at an angle
(ii) In both cases, the force exerted by the wall on the ball acts along the normal to the wall
12. Now we apply Newton's third law:
• We see that the wall applies a force of $\mathbf\small{\vec{F_{WB}}}$ on the wall
• According to the third law, the ball will apply a force whose magnitude is same as that of $\mathbf\small{\vec{F_{WB}}}$
(ii) Let the force applied by the ball on the wall be $\mathbf\small{\vec{F_{BW}}}$
• Then we can write: $\mathbf\small{|\vec{F_{WB}}|=|\vec{F_{BW}}|}$
13. Next we can consider the direction:
• We see that direction of $\mathbf\small{\vec{F_{WB}}}$ is along the red line and towards the negative side of the x axis
• Applying the third law, we get:
Direction of $\mathbf\small{\vec{F_{BW}}}$ is along the red line and towards the positive side of the x axis
14. If we are given the value of Δt, we can calculate the magnitude of $\mathbf\small{\vec{F_{WB}}}$
• All we have to do is, in (9), bring that Δt to the right side
• From that, we will get the magnitude of $\mathbf\small{\vec{F_{BW}}}$ also
Let us write a summary in three points A, B and C:
A. A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the wall on the ball will be along the normal (at the point of contact) to the wall
• This force is directed 'away from the wall'
B. From (A), we get the direction of the force exerted by the ball on the wall. For that, we apply Newton's third law. We will write the steps again:
A ball can strike a wall in two ways:
(i) It can strike the wall normally
(ii) It can strike the wall at an angle
• In both cases, the force exerted by the ball on the wall will be along the normal (at the point of contact) to the wall
• This force is directed 'towards the wall'
C. We can obtain the 'magnitude of the force' exerted by the ball on the wall, if we have the following three items:
(i) Mass of the ball
(ii) Velocity of the ball
(iii) Time duration for which the ball is in contact with the wall
In the above discussion we saw this:
■ What ever be the direction in which the ball strikes the wall, the forces will be along the red line.
• What is so special about this red line?
Let us analyse:
1. When the ball hits the wall, there is a 'surface of contact'
• Since the ball is spherical, this surface of contact will be very small
• It will be like a 'point of contact'
2. If we draw a tangent (at the point of contact) to the sphere, that tangent will lie on the surface of the wall. Details about tangents can be seen here.
• So we can say that, the 'contact surface' is the surface of the wall itself
3. This is true for what ever direction the ball strikes the wall. It is shown in fig.5.18 below:
 |
| Fig.5.18 |
♦ The 'contact surface' is a point
♦ The wall is tangential to that point
♦ So the red line is perpendicular to the contact surface
• In fig.5.18(b), the ball hits at an angle
♦ Here also, the 'contact surface' is a point
♦ The wall is tangential to that point
♦ So the red line is perpendicular to the contact surface
4. So we find that:
■ The forces act along the line perpendicular to the contact surface
5. So what is the speciality of the red line?
Answer can be written in two statements:
(i) The force always act along the line perpendicular to the contact surface
(ii) When a ball hits a wall at any angle, the red line (in figs.5.16 and 5.17) is the line perpendicular to the contact surface
• In the above example, the wall is a flat surface.
• How do we draw the red line when surfaces are not flat?
• We will see the case when spherical balls collide with each other:
1. In fig.5.19(a), two balls (of same size) A and B are moving with velocities VA and VB
 |
| Fig.5.19 |
• After some time, let ball A reach P.
• If ball B also reach there at that same instant, a collision will take place.
• This is shown in fig.b.
3. In fig.c, a tangent is drawn at the point of contact
• We can say that, this tangent represents the surface of contact between the balls
• The line joining the 'centers of the balls' will be perpendicular to this surface
4. There will be two forces acting at the time of collision:
(i) FAB which is the force which A exerts on B
(ii) FBA which is the force which B exerts on A
5. By Newton's third law, FAB = - FBA
• Both the forces act along the line joining the centers of the balls. This is shown in the enlarged view in fig.d
■ So in this case, the 'line joining the ceters of the balls' is the 'red line'
So we have completed a basic discussion on Newton's third law
• The following points should be always kept in mind:
1. Forces always occur in pairs
• A 'pair' means two'
• So if we see FAB, we can be sure that, there will be FBA
♦ If FBA is not present, there will be no FAB
♦ If FAB is not present, there will be no FBA
2. The terms 'action' and 'reaction' may give the wrong impression that:
• Action occurs first, and reaction occurs later, as an 'effect' of the action
■ The fact is that, both occur at the same instant
3. The two forces act on two different bodies
• FAB acts on B
• FBA acts on A
■ So even though they are equal and opposite, they cannot cancel each other.
We will see solved examples based on the third law, in a later section. In the next section, we will see Apparent weight in a lift
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