Newton Law Of Motion – Discussions On The Second Law
Newton’s Second Law of Motion Notes for Class 11
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 the quantity of motion. The 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 respectively, 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.
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- 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 x unit of velocity
- CGS System: g · m · s-1
- SI kg · m · s-1
Dimension of momentum = dimension of mass x dimension of velocity = M x LT-1 = MLT-1
Mathematical Expressions For The Second Law: Let, m = 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, \(F \propto \frac{d \vec{p}}{d t} \text {, i.e., } \vec{F} \propto \frac{d}{d t}(m \vec{v})\)
Here, Two Different Situations May Arise:
1. The mass of the body is a constant, i.e., m = constant. Then the rate of change of momentum is,
⇒ \(\frac{d \vec{p}}{d t}=\frac{d}{d t}(m \vec{v})=m \frac{d \vec{v}}{d t} ; \text { hence, } \vec{F} \propto m \frac{d \vec{v}}{d t}\)
This is the most 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.
2. 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, \(\frac{d \vec{p}}{d t}=\frac{d}{d t}(m \vec{\nu})=\frac{d m}{d t} \vec{\nu}+m \frac{d \vec{\nu}}{d t}\)
The second law should be expressed as \(\vec{F} \propto\left(\frac{d m}{d t} \vec{v}+m \frac{d \vec{v}}{d t}\right)\)
Derivation of Newton’s Second Law of Motion
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 the acceleration of the body, a = \(\frac{v-u}{t}\).
Here, the rate of change of momentum of the body = \(\frac{m v-m u}{t}=m \frac{v-u}{t}=m a\)
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 as 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 ….(1)
This, essentially, is a vector equation; \(\vec{F}=m \vec{a}\)
i.e., force = mass x 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.
Key Concepts of Newton’s Second Law
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 standardized, 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 \(\vec{F}=m \vec{a}\) According To Calculus: Suppose the external constant force \(\vec{F}\) is applied on a body of constant mass m.
The acceleration of the body is, \(\vec{a}=\frac{d \vec{v}}{d t}\)
Therefore, according to Newton’s second law of motion, \(\vec{F}=m \frac{d \vec{v}}{d t}\)…(2)
Units Of Force In Different Systems: Using equation (1), units of force in different systems can be defined as:
Dimension Of Force: [m] = M, [a] = LT-2;
so, [F] = [ma] = MLT-2
Newton’s Second Law in Different Reference Frames
Relationship Between The Units Of Force: Relation between N and dyn 1 N = 1 kg x 1 m · s-2
= 1000 g x 100 cm · s-2
= 105 x 1g x 1 cm · s-2 = 105 dyn
Alternative Units Of Momentum: Force = \(\frac{\text { change in momentum }}{\text { time }}\)
So, momentum has the unit of force x time Units:
- CGS: dyn · s
- SI: N · s
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, the 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, \(\vec{F}=m \vec{a}\)
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 these three laws are valid only in these frames.
- When we speak of jet planes, rockets, etc., there are no doubts that 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 ≈ 7m0; 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 < 0.1 c, the mass of a body may safely be assumed to be a constant.
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Unit 3 Laws Of Motion Chapter 1 Newton Law Of Motion Second Law 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
A force of 100 dyn acts on a mass of 25 g for 5 s.
Given, u = 0, F = 100 dyn, t = 5 s and m = 25 g.
Now we know, that F = ma.
∴ 100 = 25 x <2 or, a = 4 cm · s-2
Therefore, v = u+at = 0 + 4×5 = 20cm · s-1
Example 2. A force acts on a mass of 16g for 3s and then ceases to act. In the next 3 seconds, the mass travels 81 cm. What was the magnitude of the force?
Solution:
Given
A force acts on a mass of 16g for 3s and then ceases to act. In the next 3 seconds, the mass travels 81 cm.
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 u, generated when the force was acting on the body.
s = vt or, v = \(\frac{s}{t}\) = \(\frac{81}{3}\) = 27 cm · s-1
If the acceleration on the body due to the force was a then v = u + at or, 27 = 0 + a x 3 or, a = 9 cm · s-2
∴ Applied force, F = ma – 16 x 9 = 144 dyn.
Example 3. A bullet of mass 50 g moving at 400 m s-1 penetrates a wall with an average force of 4 x 104 N. It comes out of the Other Side of the wall at 50 m · 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:
Given
A bullet of mass 50 g moving at 400 m s-1 penetrates a wall with an average force of 4 x 104 N. It comes out of the Other Side of the wall at 50 m · s-1.
In the case of the 1st bullet, m = 50 g = 5 x 10-3 kg
As the average force = 4 x 104 N,
a = average retardation = \(\frac{4 \times 10^4}{50 \times 10^{-3}} \mathrm{~m} \cdot \mathrm{s}^{-2}\)
u = initial velocity = 400 m · s-1
v = final velocity = 50 m · s-1
Let the thickness of the wall = s.
From v² = u²-2as, s = \(\frac{u^2-v^2}{2 a}=\frac{\left(400^2-50^2\right) \times 50 \times 10^{-3}}{2 \times 4 \times 10^4}=0.0984 \mathrm{~m}\)
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, \(m_0=\frac{2 F s}{u^2}=\frac{2 \times 4 \times 10^4 \times 0.0984}{400^2}=0.0492 \mathrm{~kg}\)
Newton Law Of Motion Second Law Problems Examples
Example 4. The force on a particle of mass 10g is (\(10 \hat{i}+5 \hat{j}\))N. If it starts from rest what would be its position at time t = 5s?
Solution:
Given
The force on a particle of mass 10g is (\(10 \hat{i}+5 \hat{j}\))N.
We have, Fx = 10N (given) [x component of force is 10]
∴ \(a_x=\frac{F_x}{m}=\frac{10}{0.01}=1000 \mathrm{~m} / \mathrm{s}^2\)
As this is a case of constant acceleration in x -direction,
x = \(u_x t+\frac{1}{2} a_x t^2=0+\frac{1}{2} \times 1000 \times(5)^2=12500 \mathrm{~m}\)
Similarly, \(a_y=\frac{F_y}{m}=\frac{5}{0.01}=500 \mathrm{~m} / \mathrm{s}^2\) and \(y=\frac{1}{2} a_y t^2=\frac{1}{2} \times 500 \times(5)^2=6250 \mathrm{~m}\)
Thus, the position of the particle at \(t=5 \mathrm{~s}\) is, \(\vec{r}=(12500 \hat{i}+6250 \hat{j}) \mathrm{m}\).
Example 5. Raindrops of radius I 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:
Given
Raindrops of radius I 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.
Here, r = 1 mm = 0.001 m = 10-3m
m = 4 mg = 4 x 10-6kg
s = 10-3m, v = 0, u = 30 m/s
∴ a = \(\frac{v^2-u^2}{2 s}=\frac{-30 \times 30}{2 \times 10^{-3}} \mathrm{~m} / \mathrm{s}^2\)
= -4.5 x 105 m/s2 (decelerating)
Taking the magnitude only, deceleration is 4.5 x 105 m/s²
∴ Force, F = 4 x 10-6 x 4.5 x 105 = 1.8 N.