In the previous chapter, we saw Newton's First law of motion. We also saw what inertia is and that mass is a measure of inertia. In this section we will see Newton's Second law of motion.
We saw that objects in motion tend to continue in the state of motion. Let us see the effects that such moving objects cause on our surroundings. For that, we will do a simple experiment.
1. Fig.2.5(a) below shows a channel which is curved in shape.
2. A marble is placed at B. It is at rest.
3. Another marble is released from A
4. It rolls down through the channel and hits the one at B
5. When hit, the marble at B is displaced to C. Note the position of C
Now repeat the experiment:
1. Fig.2.5(b) above shows the arrangement. A marble is placed at B. It is at rest.
2. Another marble is released from A. This time A is at a higher level than the A in fig.2.5(a)
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C. This time C is at a higher level than the C in fig.2.5(a)
• That means B is hit harder this time.
• The same mass was released from A in both the experiments.
• But the heights of release were different.
• The higher level of A in the second experiment gave it more hitting power.
■ Let us analyse this:
1. When an object ‘falls down from a height’ or ‘rolls down from a height’, it acquires a certain velocity.
2. If the height of fall is more, that is., if the object falls from a greater height, the velocity that it acquires is more.
[This we will learn in the next chapter. At present all we need to know is that, in the second experiment, the marble from A was travelling with a greater velocity when it reached B]
3. So we can infer this:
Even if masses are the same, a greater velocity will give more hitting power.
An example is:
• Throw a tennis ball at a foot ball at a moderate speed. The foot ball will move some distance.
♦ Throw the tennis ball at the foot ball at a higher speed. The foot ball will move a greater distance.
Now let us do the rolling down of marbles in another way:
1. Fig.2.6(a) below shows a channel in a curved shape.
2. A marble is placed at B. It is at rest. Another marble is released from A. Note the position of A
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C
Now repeat the experiment:
1. Fig.2.6(b) above shows the arrangement. A marble is placed at B. It is at rest. Another marble is released from A. This time A is at the same level as the A in fig.2.6(a)
2. But this time a heavier marble is released from A than the one in fig.2.6(a)
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C. This time C is at a higher level than the C in fig.2.6(a)
• That means B is hit harder this time.
• The heights of release were the same in both the cases.
• But the masses were different.
• The greater mass from A in the second experiment gave it more hitting power.
■ Let us analyse this:
1. When an object ‘falls down from a height’ or ‘rolls down from a height’, it acquires a certain velocity.
2. We said earlier that: If the height of fall is more, that is., if the object falls from a greater height, the velocity that it acquires is more. [This we will learn in the next chapter. At that time we will learn this also:
• Even if the masses are different, two objects will acquire the same velocity if they are released from the same height]
3. In our experiments in fig.2.6, the marbles at A were released from the same heights in both fig.2.6(a) and (b). 4. So both of them hit B with the same velocity.
5. Even when hit by the same velocity, the marble at B moved a greater distance
6. This is because, it was hit by a greater mass.
7. So we can infer this:
Even if velocities are the same, a greater mass will give more hitting power.
An example:
• A car is at rest. It is hit by a cycle moving at 10 m s-1. The hit causes minor damages to the car.
♦ The same car is at rest. It is hit by a truck moving at 10 m s-1. It causes severe damages to the car.
■ So we see that both mass and velocity give special property for a moving object. Newton called this property as Momentum. It is denoted by the letter 'p'.
■ For an object in motion, it’s momentum is the product of it’s mass and it’s velocity. That is.,
If an object of mass m is moving with a velocity v, then for that object, p = mv
Now we can continue our discussion based on momentum:
1. Consider a car which would not start by itself. It has engine trouble. It has to be pushed up to the workshop. So two men are pushing it on a level road along a straight line.
2. They are pushing the car and the car is moving with the velocity u
3. After some time, they decide to move faster, so that they can reach the workshop sooner. For that, they have to increase the velocity.
4. For increasing the velocity, they have to put extra effort. That is., they have to push harder.
5. If they give an instantaneous hard push, the car will not increase it’s velocity. An instantaneous push is like giving one hard push quickly, and taking the hands off. Such a push will not increase the velocity.
6. They have to apply the extra effort for a prolonged time. This ‘prolonged time’ may be 5 or 10 seconds. But it is not instantaneous.
7. So they push harder for a few seconds and the car increases it’s velocity to v
8. So the momentum changes to mv. Earlier it was mu
9. As v is greater than u, the momentum mv is greater than mu. The increase in momentum is :
p2 - p1 =(mv – mu) = m(u-v)
• The pushing force is responsible for this increasing in momentum
10. Also we saw that the pushing force has to be applied for a prolonged time duration. Then only the increase in momentum can be achieved. So we see that, three quantities are related. The three quantities are:
• Change in momentum
• Pushing force
• Time for which the push was applied
■ There are no other quantities involved while pushing the car from velocity u to velocity v.
♦ We know what mass is and how to measure it
♦ We know what velocity is and how to measure it
• We know what is time and how to measure it
• We do not know what force is and how to measure it yet
12. So the two groups will be as shown below:
13. The group on the right have two known quantities. Let us try to combine them into a single quantity:
• If we divide the 'change of momentum' by the time t, we will get:
• This is 'the change in momentum per unit time'.
• But 'change in momentum per unit time' is the 'rate of change of momentum'.
• So the two groups can be written as:
14. So we have just two quantities. How are they related?
15. Based on his studies, Sir Isaac Newton found out that the 'rate of change of momentum' which we can achieve, is directly proportional to the force applied.
16. We have learned about direct proportions in our maths classes. Details here and here. So we can write:
• Force ∝ Rate of change of momentum
This is same as:
[The symbol '∝' denotes: 'Is proportional to']
17. In the above expression, (v-u)⁄t is 'rate of change of velocity', which is the acceleration 'a'.
18. So we can write: F ∝ ma
⇒ F = kma. Where k is the constant of proportionality
19. So our next task is to find 'k'
(i) If we carefully choose the unit for force, we will get a simple value for 'k'. Let us try:
(ii) Let us call the unit of force as 'newton'. In short form it is written as 'N'.
(iii) Let 1 N be the force required to produce an acceleration of 1 m s-2 on a mass of 1 kg
(iv) Now consider a mass of 1 kg moving with an acceleration of 1 m s-2.
(v) Then obviously, a force of 1 N is acting on that mass.
(vi) So we can write: 1 N = k × (1 kg) × (1 m s-2)
same as:
(vii) So the value of k is 1, and it's unit is Ns2⁄kg m.
20. With this value of 'k', we can find the force acting on any object which is subjected to an acceleration.
An example: A 2 kg mass is moving at an acceleration of 7 m s-2. What is the magnitude of the force which is causing this acceleration?
Solution:
We have: F = kma.
• Where k = 1 Ns2⁄kg m.
• m = 2 kg
• a = 7 m s-2.
■ So we can write:
Note that, in the final answer, the numerical part is 14 and the unit is 'N'. This is because, all the other units cancel out as shown below:
• kg in the numerator and kg in the denominator cancels out
• m in the numerator and m in the denominator cancels out
• s2 in the numerator and s-2 in the numerator cancels out
• All that remains is N
Another example:
A 7500 g mass is moving at an acceleration of 20 cm s-2. What is the magnitude of the force which is causing this acceleration?
Solution:
We have: F = kma.
• Where k = 1 Ns2⁄kg m.
• m = 7500 g = 7.5 kg
• a = 20 cm s-2 = 0.2 m s-2.
■ So we can write:
Note that, in the final answer, the numerical part is 1.5 and the unit is 'N'. This is because, all the other units cancel out as shown below:
• kg in the numerator and kg in the denominator cancels out
• m in the numerator and m in the denominator cancels out
• s2 in the numerator and s-2 in the numerator cancels out
• All that remains is N
• It is important to use proper units:
♦ Mass must be in kg
♦ Acceleration must be in m s-2.
• Then we will get the final answer in N
In the next section, we will see some practical applications of Newton's Second Law of motion.
We saw that objects in motion tend to continue in the state of motion. Let us see the effects that such moving objects cause on our surroundings. For that, we will do a simple experiment.
1. Fig.2.5(a) below shows a channel which is curved in shape.
 |
| Fig.2.5 |
3. Another marble is released from A
4. It rolls down through the channel and hits the one at B
5. When hit, the marble at B is displaced to C. Note the position of C
Now repeat the experiment:
1. Fig.2.5(b) above shows the arrangement. A marble is placed at B. It is at rest.
2. Another marble is released from A. This time A is at a higher level than the A in fig.2.5(a)
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C. This time C is at a higher level than the C in fig.2.5(a)
• That means B is hit harder this time.
• The same mass was released from A in both the experiments.
• But the heights of release were different.
• The higher level of A in the second experiment gave it more hitting power.
■ Let us analyse this:
1. When an object ‘falls down from a height’ or ‘rolls down from a height’, it acquires a certain velocity.
2. If the height of fall is more, that is., if the object falls from a greater height, the velocity that it acquires is more.
[This we will learn in the next chapter. At present all we need to know is that, in the second experiment, the marble from A was travelling with a greater velocity when it reached B]
3. So we can infer this:
Even if masses are the same, a greater velocity will give more hitting power.
An example is:
• Throw a tennis ball at a foot ball at a moderate speed. The foot ball will move some distance.
♦ Throw the tennis ball at the foot ball at a higher speed. The foot ball will move a greater distance.
Now let us do the rolling down of marbles in another way:
1. Fig.2.6(a) below shows a channel in a curved shape.
 |
| Fig.2.6 |
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C
Now repeat the experiment:
1. Fig.2.6(b) above shows the arrangement. A marble is placed at B. It is at rest. Another marble is released from A. This time A is at the same level as the A in fig.2.6(a)
2. But this time a heavier marble is released from A than the one in fig.2.6(a)
3. It rolls down through the channel and hits the one at B
4. When hit, the marble at B is displaced to C. Note the position of C. This time C is at a higher level than the C in fig.2.6(a)
• That means B is hit harder this time.
• The heights of release were the same in both the cases.
• But the masses were different.
• The greater mass from A in the second experiment gave it more hitting power.
■ Let us analyse this:
1. When an object ‘falls down from a height’ or ‘rolls down from a height’, it acquires a certain velocity.
2. We said earlier that: If the height of fall is more, that is., if the object falls from a greater height, the velocity that it acquires is more. [This we will learn in the next chapter. At that time we will learn this also:
• Even if the masses are different, two objects will acquire the same velocity if they are released from the same height]
3. In our experiments in fig.2.6, the marbles at A were released from the same heights in both fig.2.6(a) and (b). 4. So both of them hit B with the same velocity.
5. Even when hit by the same velocity, the marble at B moved a greater distance
6. This is because, it was hit by a greater mass.
7. So we can infer this:
Even if velocities are the same, a greater mass will give more hitting power.
An example:
• A car is at rest. It is hit by a cycle moving at 10 m s-1. The hit causes minor damages to the car.
♦ The same car is at rest. It is hit by a truck moving at 10 m s-1. It causes severe damages to the car.
■ For an object in motion, it’s momentum is the product of it’s mass and it’s velocity. That is.,
If an object of mass m is moving with a velocity v, then for that object, p = mv
Now we can continue our discussion based on momentum:
1. Consider a car which would not start by itself. It has engine trouble. It has to be pushed up to the workshop. So two men are pushing it on a level road along a straight line.
2. They are pushing the car and the car is moving with the velocity u
3. After some time, they decide to move faster, so that they can reach the workshop sooner. For that, they have to increase the velocity.
4. For increasing the velocity, they have to put extra effort. That is., they have to push harder.
5. If they give an instantaneous hard push, the car will not increase it’s velocity. An instantaneous push is like giving one hard push quickly, and taking the hands off. Such a push will not increase the velocity.
6. They have to apply the extra effort for a prolonged time. This ‘prolonged time’ may be 5 or 10 seconds. But it is not instantaneous.
7. So they push harder for a few seconds and the car increases it’s velocity to v
8. So the momentum changes to mv. Earlier it was mu
9. As v is greater than u, the momentum mv is greater than mu. The increase in momentum is :
p2 - p1 =(mv – mu) = m(u-v)
• The pushing force is responsible for this increasing in momentum
10. Also we saw that the pushing force has to be applied for a prolonged time duration. Then only the increase in momentum can be achieved. So we see that, three quantities are related. The three quantities are:
• Change in momentum
• Pushing force
• Time for which the push was applied
■ There are no other quantities involved while pushing the car from velocity u to velocity v.
11. Let us group the above three into 2 groups. One group will contain known quantities and the other group will contain unknown quantities.
• We know what 'change in momentum' [m(v – u)] is, and how to measure it. Because:♦ We know what mass is and how to measure it
♦ We know what velocity is and how to measure it
• We know what is time and how to measure it
• We do not know what force is and how to measure it yet
12. So the two groups will be as shown below:
• If we divide the 'change of momentum' by the time t, we will get:
• But 'change in momentum per unit time' is the 'rate of change of momentum'.
• So the two groups can be written as:
15. Based on his studies, Sir Isaac Newton found out that the 'rate of change of momentum' which we can achieve, is directly proportional to the force applied.
16. We have learned about direct proportions in our maths classes. Details here and here. So we can write:
• Force ∝ Rate of change of momentum
This is same as:
17. In the above expression, (v-u)⁄t is 'rate of change of velocity', which is the acceleration 'a'.
18. So we can write: F ∝ ma
⇒ F = kma. Where k is the constant of proportionality
19. So our next task is to find 'k'
(i) If we carefully choose the unit for force, we will get a simple value for 'k'. Let us try:
(ii) Let us call the unit of force as 'newton'. In short form it is written as 'N'.
(iii) Let 1 N be the force required to produce an acceleration of 1 m s-2 on a mass of 1 kg
(iv) Now consider a mass of 1 kg moving with an acceleration of 1 m s-2.
(v) Then obviously, a force of 1 N is acting on that mass.
(vi) So we can write: 1 N = k × (1 kg) × (1 m s-2)
same as:
(vii) So the value of k is 1, and it's unit is Ns2⁄kg m.
20. With this value of 'k', we can find the force acting on any object which is subjected to an acceleration.
An example: A 2 kg mass is moving at an acceleration of 7 m s-2. What is the magnitude of the force which is causing this acceleration?
Solution:
We have: F = kma.
• Where k = 1 Ns2⁄kg m.
• m = 2 kg
• a = 7 m s-2.
■ So we can write:
Note that, in the final answer, the numerical part is 14 and the unit is 'N'. This is because, all the other units cancel out as shown below:
• kg in the numerator and kg in the denominator cancels out
• m in the numerator and m in the denominator cancels out
• s2 in the numerator and s-2 in the numerator cancels out
• All that remains is N
Another example:
A 7500 g mass is moving at an acceleration of 20 cm s-2. What is the magnitude of the force which is causing this acceleration?
Solution:
We have: F = kma.
• Where k = 1 Ns2⁄kg m.
• m = 7500 g = 7.5 kg
• a = 20 cm s-2 = 0.2 m s-2.
■ So we can write:
Note that, in the final answer, the numerical part is 1.5 and the unit is 'N'. This is because, all the other units cancel out as shown below:
• kg in the numerator and kg in the denominator cancels out
• m in the numerator and m in the denominator cancels out
• s2 in the numerator and s-2 in the numerator cancels out
• All that remains is N
• From the above two examples it is clear that, We need not put the value of k as 1Ns2⁄kg m.
• We need to put the values of mass and acceleration only. That means we can calculate the force using the equation: F = ma• It is important to use proper units:
♦ Mass must be in kg
♦ Acceleration must be in m s-2.
• Then we will get the final answer in N
In the next section, we will see some practical applications of Newton's Second Law of motion.
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