In the previous section we saw some solved examples on Newton's Second law of motion. In this section we will see the Third law of motion.
Let us do an experiment:
1. Take a boiling tube and put a small amount of water in it.
2. Close it gently with a stop cork. Suspend the boiling tube using two strings as shown in the fig.2.8(a) below:
• Note that the boiling tube should not be in a vertical position. It should also be not in a horizontal position. It must be nearly horizontal. This is because, if it is horizontal, the water will flow up to the stop cork.
3. Heat the boiling tube with a burner until the water vapourizes.
4. When the water vapourizes, the vapour pressure inside the boiling tube increases, and after some time, the stop cork will pop off. This is shown in the fig.2.8(b).
5. When the cork pop off, the boiling tube will swing in the opposite direction.
6. We know why the cork pops off. It is because of the excess pressure inside the tube.
7. But we do not know why the boiling tube swing in the opposite direction. The reason is explained below:
8. The cork was initially at rest. When it pops off, it is moving with a certain velocity. That means, it's velocity increases from zero to a certain value. So it is clear that the cork is accelerated.
9. Let this acceleration of the cork be ac. Let the mass of the cork be mc. Then the force Fc responsible for the acceleration is given by Fc = mcac
10. Who applied this force on the cork?
Ans: We see no component other than the 'boiling tube with water'. It is the only component in contact with the cork. So the force Fc came from the boiling tube. To be precise, Fc came from the high pressure vapour inside the boiling tube.
11. So a force has been applied on the cork. According to Newtons third law of motion, the cork will apply an equal force on the boiling tube. This force causes the swinging of the boiling tube.
12. Also, this force acts in the opposite direction. So, if the force applied by the cork on the boiling tube is Ft, we can write: Fc = -Ft. This is same as:
mcac = -mbab Where mb and ab are the mass and acceleration of the boiling tube.
13. In the above equation, mc is smaller than mb.
14. So, to maintain the equality, ac will be larger than at ab. This is the reason why, the swinging speed of the boiling tube is smaller than the popping speed of the cork.
Another example:
1. Fill a balloon with air. Close the mouth of the balloon with your fingers.
2. Then suddenly release the fingers. We can see that the air rushes out. And at the same time, the balloon moves in the opposite direction.
3. Here the mass of air which rushed out was acted upon by a force.
♦ This is evident because, the velocity of air increased from zero to a certain value.
♦ This force is supplied by the high pressure of air in the balloon.
4. This force Fa is given by: Fa = maaa Where ma is the mass of air and aa is the acceleration of air.
5. The air will exert a same force on the balloon. Let this force acting on the balloon be Fb.
6. Then, based on the third law of motion, we can write Fa = -Fb.
7. If mb is the mass of the balloon and ab the acceleration of the balloon, then Fb = mbab.
8. Substituting this in (6) we get: maaa = -mbab.
9. Here, just like air, the balloon also has a low mass. That means, ma and mb are comparable. That is why, the balloon also moves with a high velocity, just like the air which rushes out in the opposite direction.
Another example:
1. When a bullet is fired from the gun, the bullet moves forward with a high acceleration. That means a force is acting on the bullet. Let this force be Fb.
2. When this force Fb acts on the bullet, the bullet exerts an equal and opposite force Fg on the gun.
3. So the gun moves backwards. This is called the recoil of the gun.
4. Then, based on the third law of motion, we can write Fb = -Fg.
⇒ mbab = -mgag. Here mg is much larger than mb. So ag will be much smaller than ab
One more example:
1. When a sailor jumps to the shore from a boat, sailor moves forward but the boat moves backwards.
2. The sailor exerts a force on the boat. The boat exerts an equal and opposite force on the sailor. That is why the boat moves backwards.
3. we can write Fs = -Fb ⇒ msas = -mbab.
4. Here both the sailor and boat moves with approximately equal accelerations. This is because, for a small boat, mb will be nearly equal to ms.
5. But if the boat is very large, mb will be very large. Then ab will be so small that, we won't be able to even notice the movement of the boat.
■ So a natural question arises:
If the forces are equal and opposite, why don't they cancel each other?
The answer can be given with the help of the following fig.2.9(a):
1. In the fig.2.9(a), equal and opposite forces are acting on the same object P. They will cancel out each other. Because of this, the object P will not have any movement.
2. In fig.2.9(b), the two forces are equal and opposite but they are acting on different objects. So the objects will move in opposite directions. The forces will not cancel each other.
• Ball A has a mass mA and Ball B has a mass mB
• Ball A is moving with a velocity uA and Ball B is moving with a velocity uB
2. Let uA be greater than uB. Then Ball A will soon reach up to Ball B and a collision will take place. This is shown in fig.b
3. Let during the collision, the two balls be in contact with each other for a time 't'
• During this time, Ball A will exert a force FAB on Ball B
• And Ball B will exert a force FBA on Ball A
4. Based on Newton's third law of motion, FAB = -FBA
5. After the collision,
• Let Ball A move with a velocity of vA
• Let Ball B move with a velocity of vB.
This is shown in fig.c. Note that, these velocity changes occur because, the balls are acted upon by forces FAB and FBA
6. Now we can begin the calculations:
• Momentum of Ball A before collision = mAuA
• Momentum of Ball A after collision = mAvA
• Time duration of the collision = t
• Therefore rate of change of momentum =
• But we know that rate of change of momentum is the force being applied on the object.
• So the above rate of change of moment which is experienced by Ball A is the force FBA applied by Ball B on Ball A So we can write:
7. Momentum of Ball B before collision = mBuB
• Momentum of Ball B after collision = mBvB
• Time duration of the collision = t
Therefore rate of change of momentum =
• But we know that rate of change of momentum is the force being applied on the object.
• So the above rate of change of moment which is experienced by Ball B is the force FAB applied by Ball A on Ball B So we can write:
8. From Newton's third law, the forces FAB and FBA must be equal and opposite. That is., FAB = -FBA
9. So we can write:
10. From this we get: mA(vA-uA) = mB(vB-uB)
⇒ mAvA - mAuA = -mBvB + mBuB
⇒ mAuA + mBuB = mAvA + mBvB .
11. The final result in the above step is important. Consider the terms on the left side:
• mAuA is the momentum of Ball A before collision
• mBuB is the momentum of Ball B before collision
■ So (mAuA + mBuB) is the total momentum available in the system before collision
Consider the terms on the right side:
• mAvA is the momentum of Ball A after collision
• mBvB is the momentum of Ball B after collision
■ So (mAvA + mBvB) is the total momentum available in the system after collision
12. From (10) and (11), we can write:
■ Total momentum available in the system before collision = Total momentum available in the system after collision
Note that, the only forces involved are the ones due to collision between the balls. There is no unbalanced force in the system.
As a result of the above collision experiment, we say that, the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum.
This statement can alternatively be given as: The total momentum of the two objects is unchanged or conserved by the collision.
Now we will see some solved examples:
Solved example 2.10
A bullet of mass 20 g is horizontally fired with a velocity 150 m s-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
Solution:
1. Masses:
• Given: Mass of bullet = mB = 20 g = 0.02 kg
• Given: Mass of pistol = mP = 2 kg
2. Initial velocities:
• The bullet accelerates from rest. So uB = 0
• The pistol accelerates from rest. So uP = 0
3. Final velocities:
• Given that velocity of bullet = vB= 150 m s-1.
• We have to find vP.
4. Total momentum before the pistol is fired = (mBuB +mPuP) = (0.02 × 0 + 2 × 0) = 0 kg m s-1
• Total momentum after the pistol is fired = (mBvB +mPvP) = (0.02 × 150 + 2 × vP) = (2vP+ 3) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the fire = Total momentum before the fire. So we can write:
2vP+ 3 = 0 ⇒ vP = - 3⁄2 = -1.5 m s-1.
(The negative sign indicates that, the direction of motion of the pistol is opposite to the direction of motion of the bullet)
Solved example 2.11
A girl of mass 40 kg jumps with a horizontal velocity of 5 m s-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.
Solution:
In this problem we have some special situations:
• The girl jumped with an initial velocity. So uG is not equal to Zero.
• After she lands on the cart, she and the cart moves together. So after the collision, the mass is equal to the sum of the masses
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of girl = mG = 40 kg
• Given: Mass of cart = mC = 3 kg
2. Final mass = mf = mG + mC = 40 + 3 = 43 kg
3. Initial velocities:
• The girl jumps with a velocity. Given uG = 5 m s-1
• The cart is initially at rest. So uC = 0
4. Final velocity:
• After collision, that is., after she lands on the cart, she and cart will move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mGuG + mCuC) = (40 × 5 + 3 × 0) = 200 kg m s-1.
• Total momentum after collision = mfvf = 43 × vf = 43vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
200 = 43vf ⇒ vf = 200⁄43 = 4.65 m s-1.
7. Note that the answer is positive. So the cart and girl will move together in the same direction in which the girl jumped on to the cart.
Solved example 2.12
Two hockey players of opposite teams, while trying to hit a hockey ball on the ground, collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s-1 while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s-1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.
Solution:
In this problem, we have a special situation:
• The players become entangled after the collision. That means, after the collision, the players move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of player A = mA = 60 kg
• Given: Mass of player B = mB = 55 kg
2. Final mass = mf = mA + mB = 60 + 55 = 115 kg
3. Initial velocities:
• Player A was moving with a velocity. Given uA = 5 m s-1
• Player B was moving with a velocity. Given uB = 6 m s-1
♦ But Player B was moving in a direction opposite to that of player A. So the velocity of player B has to be taken as negative. Thus we can write: uB = -6 m s-1
4. Final velocity:
• After collision, the two players move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (60 × 5 + 55 × -6) = 300 - 330 = -30 kg m s-1.
• Total momentum after collision = mfvf = 115 × vf = 115vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
-30 = 115vf ⇒vf = -30⁄115 = -0.26 m s-1.
7. Note that the answer is negative. So the entangled player moves together in the same direction in which player B was moving. Because, his initial velocity was taken as negative.
♦ Conservation of energy
♦ Conservation of angular momentum
These are considered to be fundamental laws in physics.
• These laws are based on observations and experiments. It is important to remember that a conservation law cannot be proved.
■ There are two types of results:
♦ We get results from experiments.
♦ The results can be predicted using the law.
• If both the results are same then we can say that 'The law is verified'.
♦ Of course, trial experiments has to be done several times to get confirmation
• Even if both results are same, in all trial experiments, we cannot say that 'The law is proved'
• On the other hand, if the results are different, in a single trial, we can say: 'The law is disproved'.
■ The law of conservation of momentum has been deduced from large number of observations and experiments.
• The results of all those observations and experiments are in confirmation with the results predicted by the law.
• This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realised so far, which contradicts this law.
• Several experiences of every-day life can be explained on the basis of the law of conservation of momentum.
In the next section, we will see some more solved examples.
Let us do an experiment:
1. Take a boiling tube and put a small amount of water in it.
2. Close it gently with a stop cork. Suspend the boiling tube using two strings as shown in the fig.2.8(a) below:
 |
| Fig.2.8 |
3. Heat the boiling tube with a burner until the water vapourizes.
4. When the water vapourizes, the vapour pressure inside the boiling tube increases, and after some time, the stop cork will pop off. This is shown in the fig.2.8(b).
5. When the cork pop off, the boiling tube will swing in the opposite direction.
6. We know why the cork pops off. It is because of the excess pressure inside the tube.
7. But we do not know why the boiling tube swing in the opposite direction. The reason is explained below:
8. The cork was initially at rest. When it pops off, it is moving with a certain velocity. That means, it's velocity increases from zero to a certain value. So it is clear that the cork is accelerated.
9. Let this acceleration of the cork be ac. Let the mass of the cork be mc. Then the force Fc responsible for the acceleration is given by Fc = mcac
10. Who applied this force on the cork?
Ans: We see no component other than the 'boiling tube with water'. It is the only component in contact with the cork. So the force Fc came from the boiling tube. To be precise, Fc came from the high pressure vapour inside the boiling tube.
11. So a force has been applied on the cork. According to Newtons third law of motion, the cork will apply an equal force on the boiling tube. This force causes the swinging of the boiling tube.
12. Also, this force acts in the opposite direction. So, if the force applied by the cork on the boiling tube is Ft, we can write: Fc = -Ft. This is same as:
mcac = -mbab Where mb and ab are the mass and acceleration of the boiling tube.
13. In the above equation, mc is smaller than mb.
14. So, to maintain the equality, ac will be larger than at ab. This is the reason why, the swinging speed of the boiling tube is smaller than the popping speed of the cork.
Another example:
1. Fill a balloon with air. Close the mouth of the balloon with your fingers.
2. Then suddenly release the fingers. We can see that the air rushes out. And at the same time, the balloon moves in the opposite direction.
3. Here the mass of air which rushed out was acted upon by a force.
♦ This is evident because, the velocity of air increased from zero to a certain value.
♦ This force is supplied by the high pressure of air in the balloon.
4. This force Fa is given by: Fa = maaa Where ma is the mass of air and aa is the acceleration of air.
5. The air will exert a same force on the balloon. Let this force acting on the balloon be Fb.
6. Then, based on the third law of motion, we can write Fa = -Fb.
7. If mb is the mass of the balloon and ab the acceleration of the balloon, then Fb = mbab.
8. Substituting this in (6) we get: maaa = -mbab.
9. Here, just like air, the balloon also has a low mass. That means, ma and mb are comparable. That is why, the balloon also moves with a high velocity, just like the air which rushes out in the opposite direction.
Another example:
1. When a bullet is fired from the gun, the bullet moves forward with a high acceleration. That means a force is acting on the bullet. Let this force be Fb.
2. When this force Fb acts on the bullet, the bullet exerts an equal and opposite force Fg on the gun.
3. So the gun moves backwards. This is called the recoil of the gun.
4. Then, based on the third law of motion, we can write Fb = -Fg.
⇒ mbab = -mgag. Here mg is much larger than mb. So ag will be much smaller than ab
One more example:
1. When a sailor jumps to the shore from a boat, sailor moves forward but the boat moves backwards.
2. The sailor exerts a force on the boat. The boat exerts an equal and opposite force on the sailor. That is why the boat moves backwards.
3. we can write Fs = -Fb ⇒ msas = -mbab.
4. Here both the sailor and boat moves with approximately equal accelerations. This is because, for a small boat, mb will be nearly equal to ms.
5. But if the boat is very large, mb will be very large. Then ab will be so small that, we won't be able to even notice the movement of the boat.
■ In all the above examples, we are seeing Newton's third law in action. It can be stated as:
For every action, there is an equal and opposite reaction.■ So a natural question arises:
If the forces are equal and opposite, why don't they cancel each other?
The answer can be given with the help of the following fig.2.9(a):
 |
| Fig.2.9 |
2. In fig.2.9(b), the two forces are equal and opposite but they are acting on different objects. So the objects will move in opposite directions. The forces will not cancel each other.
Conservation of Momentum
1. Consider two objects, say two balls A and B, moving along a straight line as shown in fig.2.10(a) below: |
| Fig.2.10 |
• Ball A is moving with a velocity uA and Ball B is moving with a velocity uB
2. Let uA be greater than uB. Then Ball A will soon reach up to Ball B and a collision will take place. This is shown in fig.b
3. Let during the collision, the two balls be in contact with each other for a time 't'
• During this time, Ball A will exert a force FAB on Ball B
• And Ball B will exert a force FBA on Ball A
4. Based on Newton's third law of motion, FAB = -FBA
5. After the collision,
• Let Ball A move with a velocity of vA
• Let Ball B move with a velocity of vB.
This is shown in fig.c. Note that, these velocity changes occur because, the balls are acted upon by forces FAB and FBA
6. Now we can begin the calculations:
• Momentum of Ball A before collision = mAuA
• Momentum of Ball A after collision = mAvA
• Time duration of the collision = t
• Therefore rate of change of momentum =
• So the above rate of change of moment which is experienced by Ball A is the force FBA applied by Ball B on Ball A So we can write:
• Momentum of Ball B after collision = mBvB
• Time duration of the collision = t
Therefore rate of change of momentum =
• So the above rate of change of moment which is experienced by Ball B is the force FAB applied by Ball A on Ball B So we can write:
9. So we can write:
⇒ mAvA - mAuA = -mBvB + mBuB
⇒ mAuA + mBuB = mAvA + mBvB .
11. The final result in the above step is important. Consider the terms on the left side:
• mAuA is the momentum of Ball A before collision
• mBuB is the momentum of Ball B before collision
■ So (mAuA + mBuB) is the total momentum available in the system before collision
Consider the terms on the right side:
• mAvA is the momentum of Ball A after collision
• mBvB is the momentum of Ball B after collision
■ So (mAvA + mBvB) is the total momentum available in the system after collision
12. From (10) and (11), we can write:
■ Total momentum available in the system before collision = Total momentum available in the system after collision
Note that, the only forces involved are the ones due to collision between the balls. There is no unbalanced force in the system.
As a result of the above collision experiment, we say that, the sum of momenta of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them. This is known as the law of conservation of momentum.
This statement can alternatively be given as: The total momentum of the two objects is unchanged or conserved by the collision.
Now we will see some solved examples:
Solved example 2.10
A bullet of mass 20 g is horizontally fired with a velocity 150 m s-1 from a pistol of mass 2 kg. What is the recoil velocity of the pistol?
Solution:
1. Masses:
• Given: Mass of bullet = mB = 20 g = 0.02 kg
• Given: Mass of pistol = mP = 2 kg
2. Initial velocities:
• The bullet accelerates from rest. So uB = 0
• The pistol accelerates from rest. So uP = 0
3. Final velocities:
• Given that velocity of bullet = vB= 150 m s-1.
• We have to find vP.
4. Total momentum before the pistol is fired = (mBuB +mPuP) = (0.02 × 0 + 2 × 0) = 0 kg m s-1
• Total momentum after the pistol is fired = (mBvB +mPvP) = (0.02 × 150 + 2 × vP) = (2vP+ 3) kg m s-1
5. According to the law of conservation of momentum, Total momentum after the fire = Total momentum before the fire. So we can write:
2vP+ 3 = 0 ⇒ vP = - 3⁄2 = -1.5 m s-1.
(The negative sign indicates that, the direction of motion of the pistol is opposite to the direction of motion of the bullet)
Solved example 2.11
A girl of mass 40 kg jumps with a horizontal velocity of 5 m s-1 onto a stationary cart with frictionless wheels. The mass of the cart is 3 kg. What is her velocity as the cart starts moving? Assume that there is no external unbalanced force working in the horizontal direction.
Solution:
In this problem we have some special situations:
• The girl jumped with an initial velocity. So uG is not equal to Zero.
• After she lands on the cart, she and the cart moves together. So after the collision, the mass is equal to the sum of the masses
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of girl = mG = 40 kg
• Given: Mass of cart = mC = 3 kg
2. Final mass = mf = mG + mC = 40 + 3 = 43 kg
3. Initial velocities:
• The girl jumps with a velocity. Given uG = 5 m s-1
• The cart is initially at rest. So uC = 0
4. Final velocity:
• After collision, that is., after she lands on the cart, she and cart will move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mGuG + mCuC) = (40 × 5 + 3 × 0) = 200 kg m s-1.
• Total momentum after collision = mfvf = 43 × vf = 43vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
200 = 43vf ⇒ vf = 200⁄43 = 4.65 m s-1.
7. Note that the answer is positive. So the cart and girl will move together in the same direction in which the girl jumped on to the cart.
Solved example 2.12
Two hockey players of opposite teams, while trying to hit a hockey ball on the ground, collide and immediately become entangled. One has a mass of 60 kg and was moving with a velocity 5.0 m s-1 while the other has a mass of 55 kg and was moving faster with a velocity 6.0 m s-1 towards the first player. In which direction and with what velocity will they move after they become entangled? Assume that the frictional force acting between the feet of the two players and ground is negligible.
Solution:
In this problem, we have a special situation:
• The players become entangled after the collision. That means, after the collision, the players move together as a single mass.
So we will write the steps as follows:
1. Initial Masses:
• Given: Mass of player A = mA = 60 kg
• Given: Mass of player B = mB = 55 kg
2. Final mass = mf = mA + mB = 60 + 55 = 115 kg
3. Initial velocities:
• Player A was moving with a velocity. Given uA = 5 m s-1
• Player B was moving with a velocity. Given uB = 6 m s-1
♦ But Player B was moving in a direction opposite to that of player A. So the velocity of player B has to be taken as negative. Thus we can write: uB = -6 m s-1
4. Final velocity:
• After collision, the two players move together. We have to find the final velocity vf with which they move together.
5. Total momentum before collision = (mAuA + mBuB) = (60 × 5 + 55 × -6) = 300 - 330 = -30 kg m s-1.
• Total momentum after collision = mfvf = 115 × vf = 115vf kg m s-1
6. According to the law of conservation of momentum, Total momentum after collision = Total momentum before collision. So we can write:
-30 = 115vf ⇒vf = -30⁄115 = -0.26 m s-1.
7. Note that the answer is negative. So the entangled player moves together in the same direction in which player B was moving. Because, his initial velocity was taken as negative.
Proof for Conservation laws
• Following are some of the conservation laws:
♦ Conservation of momentum♦ Conservation of energy
♦ Conservation of angular momentum
These are considered to be fundamental laws in physics.
• These laws are based on observations and experiments. It is important to remember that a conservation law cannot be proved.
■ There are two types of results:
♦ We get results from experiments.
♦ The results can be predicted using the law.
• If both the results are same then we can say that 'The law is verified'.
♦ Of course, trial experiments has to be done several times to get confirmation
• Even if both results are same, in all trial experiments, we cannot say that 'The law is proved'
• On the other hand, if the results are different, in a single trial, we can say: 'The law is disproved'.
■ The law of conservation of momentum has been deduced from large number of observations and experiments.
• The results of all those observations and experiments are in confirmation with the results predicted by the law.
• This law was formulated nearly three centuries ago. It is interesting to note that not a single situation has been realised so far, which contradicts this law.
• Several experiences of every-day life can be explained on the basis of the law of conservation of momentum.
In the next section, we will see some more solved examples.
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