Contents
- 1 What is Mathematical Expression of Newton’s Second Law?
- 1.1 Discussions On The First Law
- 1.2 Inertia of a Body
- 1.3 A few examples of inertia of rest:
- 1.4 Experimental demonstration of inertia of rest:
- 1.5 A few examples of inertia of motion:
- 1.6 Force
- 1.7 Inertial and Non-Inertial Frames of Reference
- 1.8 Discussions on The Second Law
- 1.9 Mathematical Expression for the Second Law
- 1.10 Relationship between the units of force:
- 1.11 Alternative units of momentum:
- 1.12 Numerical Examples
From the study of subatomic particles to the laws of motion, Physics Topics offer insights into the workings of the world around us.
What is Mathematical Expression of Newton’s Second Law?
In 1687, Sir Isaac Newton, in his book Principia, published three laws of motion related to all the moving objects in the universe. These are known as Newton’s laws of motion. These laws cannot be proved using any old concept of physics. Yet, their validity has been confirmed through successful explanations and predictions on numerous events in nature. Hence, the laws founded a new branch of physics called kinetics.
The laws are:
1st Law: Every body continues to be in its state of rest or of uniform motion in a straight line unless the body is compelled to change its state by a net external force.
2nd Law: The rate of change of momentum of a body is proportional to the impressed force and takes place in the direction in which the force acts.
3rd Law: To every action there is an equal and opposite reaction.
These laws involve the new concepts of
- inertia,
- force and
- momentum.
The study of these laws is the subject of this chapter and is called Newtonian mechanics.
It should be mentioned, however, that Newtonian mechanics does not apply to all situations.
- If the interacting bodies are moving with the speed of light (or an appreciable fraction of the speed of light), Newtonian mechanics should be replaced by the Einstein’s special theory of relativity.
- If the interacting bodies are of atomic dimensions (e.g., electrons in an atom), Newtonian mechanics should be replaced by quantum mechanics.
In short, Newtonian mechanics may be treated as a special case of these two more comprehensive theories.
Discussions On The First Law
From the first law, we come to know about
- inertia of a body and
- definition of force.
Inertia of a Body
The first law leads to the idea that the natural tendency of a body at rest, is to remain at rest; or, the natural tendency of a moving body is to continue to be in its state of uniform motion. In both the cases, the body cannot change its state by itself. The natural tendency of a body to resist any change in its state of rest or of motion is called the inertia of the body.
Definition: The property of a body which enables it to continue in its state of rest or of uniform motion is called inertia.
Inertia can be either
- inertia of rest or
- inertia of motion.
Inertia of rest: The tendency of a stationary body to remain at rest forever, is called the inertia of rest.
A few examples of inertia of rest:
i) When a vehicle suddenly starts moving, a passenger sitting or standing inside it lean backward. The passenger’s body was at rest when the vehicle was at rest. But when the vehicle starts moving, the feet being in contact with the vehicle are set into motion. However, the upper part of the body tries to maintain its state of rest and hence moves in the direction opposite to the direction of motion of the vehicle.
ii) From a vertical stack of books, if any book from the middle is rapidly pulled out, the remaining books of the stack do not topple over due to inertia.
iii) If we make a pile of carrom pieces and hit the bottom piece hard enough with a striker, it will move away, but the rest of the pile will remain at the original position. The lowest piece moves because of the force exerted by the striker on it. However the rest of the pile remains at its place due to inertia of rest.
iv) When a tree is vigorously shaken, the branches of the tree are in motion but the leaves tend to continue in their state of rest due to inertia of rest. As a result, leaves get separated from the branches of the tree and fall down.
Experimental demonstration of inertia of rest:
A postcard is kept on top of a glass. A coin is placed on the postcard [Fig.], just above the mouth of the glass. When the postcard is given a sudden horizontal flick of a finger, it flies off sideways and the coin, due to its inertia of rest, drops into the glass.
Inertia of motion The tendency of a moving body to maintain its motion in a straight line at a constant velocity is called inertia of motion.
A few examples of inertia of motion:
i) When a moving vehicle suddenly applies brakes, passengers lean forward or fall. They were in motion with the vehicle; but when it suddenly decelerates, their lower limbs being in contact with the vehicle slow down, but the upper parts of their bodies continue to move forward.
ii) Blades of an electrical fan continue to rotate for some time due to its inertia of motion even after the fan is switched off. Blades slow down and finally stop because of air resistance and other damping forces,
iii) A bicycle continues its motion for sometime, even after the cyclist ceases pedalling.
iv) Athletes run some distance before taking a long jump. The inertia of motion helps them to cover a longer distance.
v) A stone, tied to the end of a string, is rotated in a circular path [Fig.]. If the string snaps suddenly, due to inertia of motion the stone flies off tangentially with the velocity at that instant. Since, the direction of velocity in a circular motion, at any point, is along the tangent at that point, the stone follows that tangential path.
vi) For a simple pendulum, the lowest position of the bob at A is the equilibrium position [Fig.]. When the bob is taken to B and released, the earth’s pull makes the bob move towards A. Due to inertia of motion it does not stop at A, and continues up to C. Again, due to the pull of the earth it starts moving towards A. Thus, the pendulum oscillates about its mean position till air resistance brings it to rest.
Experimental demonstration of inertia of motion: A heavy metallic sphere A is placed on a plank B fitted with wheels [Fig.], The ball and the plank together, are set in motion. If the plank stops on colliding with an obstruction C, the sphere A, due to its inertia of motion, cannot stop and topples across the obstruction C.
Force
Newton’s first law of motion states that, the state of rest or of uniform motion of a body cannot be altered without the application of a force. In certain cases, there exists opposing forces like friction. Then, the applied force should be greater than the opposing forces to bring about a change in the state of motion of the body. For example, a heavy boulder cannot be moved, or a moving train cannot be stopped by the force applied by a single man. This discussion leads to the definition of force.
Definition: Force is an external influence which changes or tends to change the state, of rest or of uniform motion of a body.
Force is a vector quantity. It has both magnitude and direc-tion. In addition, the line of action of the force is also important. For example, we consider the sphere R on a horizontal floor having its centre at O [Fig.]. Let force F1 along BC or force F2 along OA, act on R.
While the force F2 will cause a rectilinear motion of R, F1 will set up a rotation.
Inertial and Non-Inertial Frames of Reference
When the motion of a body is considered on or near the surface of the earth, the earth’s surface is taken as the fixed frame of reference. A stone lying on the surface of the earth remains in that state of rest unless an external force is applied on it. Also a spherical object, lying on the floor of a train moving with a uniform velocity, remains at rest in the absence of an external force.
Hence, for a frame of reference, stationary or in uniform motion, Newton’s first law of motion holds good. Such sta-tionary or uniformly moving frames are called inertial frames of reference. In an inertial frame of reference, all the three laws of Newton are valid.
On the other hand, suppose the train, with the spherical object on its floor, accelerates suddenly. The object will also move with an acceleration but in the direction opposite to that of the train. Though no external force is applied on the spherical object, it will change its state of rest.
Hence, the first law of motion is not applicable in a train moving with an acceleration. A frame of reference undergoing acceleration, is called a non-inertial frame of reference. Newton’s laws of motion are not applicable in non-inertial frames of reference.
The fact that acceleration is produced even in the absence of an external force in a non-inertial frame of reference, can be explained by considering a fictitious or pseudo force. It implies that in a non-inertial frame, when any real external force is absent, a fictitious force changes the object’s state of rest or of uniform motion. Centrifugal force, in circular motion, is one of the most common pseudo forces, encountered in our daily lives [for details see the chapter Circular Motion].
In context it may be mentioned that the force-acceleration relation, i.e., F = ma may be applicable even in a non-inertial frame of reference, if the pseudo forces are properly taken into account.
Discussions on The Second Law
Momentum
Definition: The dynamical property arising from the combined effect of mass and velocity of a moving body is called its momentum.
Newton described momentum as quantity of motion. Momentum of a body is the product of its mass and velocity. So, if the mass and the velocity of a body are m and v respec-tively, then its momentum = mv. Thus a body’s momentum depends on both its mass and velocity. Mass is a scalar while velocity is a vector quantity. So momentum is also a vector quantity. It has both magnitude and direction. The direction of momentum is the same as the direction of velocity.
Concept of momentum: Suppose, two identical trucks, one loaded and the other empty, are moving with the same velocity. In this case, the momentum of the loaded truck is greater than that of the other because of its greater mass. To stop within the same interval of time, the loaded truck requires more force than the empty one.
Alternatively, let us consider two trucks of the same mass and the first one is moving with a higher velocity—so its momentum is also higher. Here, again the same principle will apply. To stop the two trucks within the same interval of time, a greater force has to be applied against the first one. Thus, the motion of a body is not described by its velocity only—it is described by the combination of mass and velocity i.e., the momentum of the body.
Unit and dimension of momentum: The unit of momentum = unit of mass × unit of velocity
Dimension of momentum
= dimension of mass × dimension of velocity = M × LT-1 = MLT-1
Mathematical Expression for the Second Law
Let, in = mass of a body; \(\vec{v}\) = its velocity; then momentum, \(\vec{p}\) = m\(\vec{v}\).
And, \(\vec{F}\) = net external force acting on the body.
Newton’s second law of motion states that,
F ∝ \(\frac{d \vec{p}}{d t}\), i.e., F ∝ \(\frac{d}{d t}\)(m\(\vec{v}\))
Here, two different situations may arise:
i) The mass of the body is a constant, i.e., m = constant. Then the rate of change of momentum is,
\(\frac{d \vec{p}}{d t}\) = \(\frac{d}{d t}(m \vec{v})\) = \(m \frac{d \vec{v}}{d t}\) ; hence, \(\vec{F}\) ∝ \(m \frac{d \vec{v}}{d t}\)
This is the mos common situation in nature. In our daily life, most of the moving bodies we observe do not change their masses with time; external forces produce changes in their velocities only.
ii) The mass of the body changes with time, i.e., m ≠ constant. Jet planes, rockets, etc. are familiar examples. Due to combustion and release of fuel, these objects change their velocities as well as their masses with time. Here,
\(\frac{d \vec{p}}{d t}\) = \(\frac{d}{d t}(m \vec{v})\) = \(\frac{d m}{d t} \vec{v}\) + \(m \frac{d \vec{v}}{d t}\)
The second law should be expressed as
\(\vec{F}\) ∝ (\(\frac{d m}{d t} \vec{v}\) + m\(\frac{d \vec{v}}{d t}\))
Derivation of the equation F = ma: Suppose,
u = initial velocity of a moving body
F= an external net force acting on this body for a time t
v = final velocity attained due to the action of the force \(\vec{F}\).
Using the relation v = u + at, we get
acceleration of the body, a = \(\frac{v-u}{t}\)
Here, the rate of change of momentum of the body
= \(\frac{m v-m u}{t}\) = m\(\frac{v-u}{t}\) = ma
From Newton’s second law,
F ∝ ma, or F = kma [k = constant]
This is a situation where we are free to define a unit of force. By convention, it is defined in such a way that the proportionality constant becomes unity, i.e., k = 1 . For m = 1 and a = 1, if we put F = 1, then k = 1.
Then we get the equation,
F = ma
This, essentially, is a vector equation; \(\vec{F}\) = m\(\vec{a}\),
i.e., force = mass × acceleration
The equation (1) is called the law of motion of a body of constant mass; this is the very basis of the study of mechanics
\(\vec{F}\) = m\(\vec{a}\) can be broken to three component equations, one for each axis of xyz cartesian coordinate system:
Fx = max, Fy = may, Fz = maz
This means that if a force is not parallel to the velocity of the body, i.e., makes an angle with it, it changes only the component of velocity along the direction of force. The component of velocity normal to the force remains unchanged.
Definition of unit force: The force, acting on a unit mass and producing a unit acceleration, is called a unit force.
Once this definition of the unit of force is standardised, Newton’s second law can be stated as: the rate of change of momentum of a body is equal to the impressed force.
The form of \(\overrightarrow{\boldsymbol{F}}\) = m\(\overrightarrow{\boldsymbol{a}}\) according to calculus:
Suppose the external constant force F is applied on a body of constant mass m.
The acceleration of the body is,
\(\overrightarrow{\boldsymbol{a}}\) = \(\frac{d \vec{v}}{d t}\)
Therefore, according to Newton’s second law of motion,
\(\overrightarrow{\boldsymbol{a}}\) = m\(\frac{d \vec{v}}{d t}\) ……. (2)
Units of force in different systems : Using equation (1), units of force in different systems can be defined as
Unit | Definition | |
CGS System | dyne (dyn) | The force that acts on a mass of 1 g and produces an acceleration of 1 cm ᐧ s-2 is called 1 dyne. So, 1 dyn = 1 g × 1 cm ᐧ s-2 . |
SI | Newton (N) | The force that acts on a mass of 1 kg and produces an acceleration of 1 m ᐧ s-2 is called 1 newton. 1N = 1 kg × 1 m ᐧ s-2. |
Dimension offeree: [m] = M, [a] = LT-2;
so, [F] = [ma] = MLT-2
Relationship between the units of force:
Relation 1 N = 1 kg × 1 m ᐧ s-2
between = 1000 g × 100 cm ᐧ s-2
N and dyn = 105 × 1 g × 1 cm ᐧ s-2 = 105 dyn
Alternative units of momentum:
Force \(=\frac{\text { change in momentum }}{\text { time }}\)
So, momentum has the unit of force × time:
These are identical to the units mentioned in Sec. 1.3.1.
Direction of force: Newton’s second law of motion not only helps to find the magnitude of a force, but also provides its direction. According to the law,
direction of force = direction of acceleration [∵ mass is a scalar quantity]
In the equation F = ma, both the sides represent vectors and hence it is a vector equation,
\(\overrightarrow{\boldsymbol{F}}\) = m\(\overrightarrow{\boldsymbol{a}}\)
Any decrease in momentum in a particular direction signifies that a force is acting and the acceleration is taking place in the opposite direction. This is a case of retardation.
It is to be noted that a body of constant mass will accelerate only as long as a force acts on it. i.e., from the equation F = ma, a = 0 when F = 0. As soon as F becomes 0, the velocity of the body ceases to change any further. Conversely, the acceleration of a body is always associated with a force acting on it. That is, if a ≠ 0, F ≠ 0.
To establish the first law from the second law of motion:
According to the second law,
F = ma or, F = \(\frac{m(v-u)}{t}\)
When there is no external force, F = 0.
∴ 0 = m(v – u)
As m ≠ 0, v – u = 0 or, v = u
This is the state of uniform motion in the absence of an external force.
Also if u = 0, then v = u = 0, i.e., the body remains in its state of rest in the absence of an external force.
Thus, the second law includes the first law. Yet, Newton mentioned the first law separately to establish the idea of inertial frames of reference—all this three laws are valid only in these frames.
When we speak of jet planes, rockets, etc., there are no doubts objects of variable mass and the second law should be applied on them accordingly. However, there is a more fundamental nature of variation of mass, namely, the relativistic increase of mass of a body with its velocity. Einstein’s special theory of relativity establishes that, even if a body does not gain or lose any amount of matter contained in it, it shows a significant increase of mass when its velocity reaches very near to that of light. As per this theory, the mass of a body moving with velocity v is
m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Where, m0 = mass of the body at rest, and c = velocity of light in vacuum.
Calculations, with this formula, show that, when v = 99% of c, m ≈ 7 m0; when v = 99.5% of c, m ≈ 10m0 and so on. Newton’s second law is also valid in this range—we have only to be careful about the variation of mass of a body.
However, in the low-velocity range, say for v < 0.1c, the mass of a body may safely be assumed to be a constant.
Numerical Examples
Example 1.
A force of 100 dyn acts on a mass of 25 g for 5 s. Find the velocity attained.
Solution:
Given, u = 0, F = 100 dyn, t = 5 s and m = 25 g.
Now we know, F = ma.
∴ 100 = 25 × a or, a = 4 cm ᐧ s-2
Therefore, v = u + at = 0 + 4 × 5 = 20 cm ᐧ s-1
Example 2.
A force acts on a mass of 16 g for 3 s and then ceases to act. In the next 3 seconds, the mass travels 81 cm. What was the magnitude of the force?
Solution:
While covering the distance of 81 cm, there was no external force on the body and therefore it must have covered this distance with a uniform velocity v, generated when the force was acting on the body.
s = vt or, v = \(\frac{s}{t}\) = \(\frac{81}{3}\) = 27 cm ᐧ s-1
If the acceleration on the body due to the force was a then
v = u + at or, 27 = 0 + a × 3 or, a = 9 cm ᐧ s-2
∴ Applied force, F = ma = 16 × 9 = 144 dyn.
Example 3.
A bullet of mass 50 g moving at 400 m ᐧ s-1 penetrates a wall with an average force of 4 × 104 N. It comes out of the other side of the wall at 50 m ᐧ s-1. Find the thickness of the wall. Another bullet of comparatively lower mass, moving with the same velocity cannot penetrate the same wall. What can be the maximum mass of the bullet?
Solution:
In the case of the 1st bullet,
m = 50 g = 5 × 10-3 kg
As the average force = 4 × 104 N,
a = average retardation = \(\frac{4 \times 10^4}{50 \times 10^{-3}}\) m ᐧ s-2
u = initial velocity = 400 m ᐧ s-1
v = final velocity = 50 m ᐧ s-1
Let the thickness of the wall = 5.
From v2 = u2 – 2as,
s = \(\frac{u^2-v^2}{2 a}\) = \(\frac{\left(400^2-50^2\right) \times 50 \times 10^{-3}}{2 \times 4 \times 10^4}\) = 0.0984 m
The second bullet cannot penetrate the wall; hence its final velocity should be 0, i.e., v = 0 .
As it cannot cover the distance s, the maximum possible mass m0 corresponds to s = 0.0984 m.
Average retardation, a = \(\frac{F}{m_0}\) = \(\frac{u^2-v^2}{2 s}\) = \(\frac{u^2}{2 s}\)
or, m0 = \(\frac{2 F s}{u^2}\) = \(\frac{2 \times 4 \times 10^4 \times 0.0984}{400^2}\) = 0.0492 kg
Example 4.
The force on a particle of mass 10g is (10\(\hat{\boldsymbol{i}}\) + 5\(\hat{\boldsymbol{j}}\))N. If it starts from rest what would be its position at time t = 5s?
Solution:
We have, Fx = 10N (given)
[∵ x component of force is 10]
∴ ax = \(\frac{F_x}{m}\) = \(\frac{10}{0.01}\) = 100 m/s2
As this is a case of constant acceleration in x -direction,
x = uxt + \(\frac{1}{2} a_x t^2\) = 0 + \(\frac{1}{2}\) × 1000 × (5)2 = 12500 m
Similarly ay = \(\frac{F_y}{m}\) = \(\frac{5}{0.01}\) = 500 m/s2
and y = \(\frac{1}{2} a_y t^2\) = \(\frac{1}{2}\) × 500 × (5)2 = 6250 m
Thus, the position of the particle at t = 5 s is, \(\vec{r}\) = (12500\(\hat{i}\) + 6250\(\hat{j}\))m.
Example 5.
Raindrops of radius 1 mm and mass 4 mg are falling with a speed of 30 m/s on the head of a bald person. The drops splash on the head and come to rest. Assuming equivalently that the drops cover a distance equal to their radii on the head, estimate the force exerted by each drop on the head.
Solution:
Here, r = 1 mm = 0.001 m = 10-3m
m = 4 mg = 4 × 10-6kg
s = 10-3m, v = 0, u = 30 m/s
∴ a = \(\frac{v^2-u^2}{2 s}\) = \(\frac{-30 \times 30}{2 \times 10^{-3}}\)m/s2
= -4.45 × 105m/s2 (decelerating)
Taking the magnitude only, deceleration is 4.5 × 105m/s2
∴ Force, F = 4 × 10-6 × 4.5 × 105 = 1.8 N.