Dynamics | Complete notes | Important Questions | Solutions | Class 11 (Physics) | Physics in Depth

Newton's Laws of Motion

Image source : Britannica

In 1687, Newton formulated his laws of motion and published it in his book Philosophiae Naturalis Principia Mathematica . Isaac Newton first understood the relation between a force and the acceleration and study of this relation is given a name , Newtonian Mechanics. Though, there are some restrictions to Newtonian Mechanics such as:

1. If the speeds of the interacting bodies are very large - an appreciable fraction of the speed of light- we must replace Newtonian mechanics with Einstein's special theory of relativity.
2. If the interacting bodies are on the scale of atomic structure (for example, electrons) we must replace Newtonian mechanics with quantum mechanics.

, it still applies to the motion of objects ranging in size from the very small (almost on the scale of atomic structure) to astronomical (galaxies and clusters of galaxies). Since, theory of relativity and quantum mechanics are beyond the syllabus of class 11, I will not discuss these on this post. However, if you like to take a brief insight to this topics, please drop comment, I might help you!

Image source : Wikipedia

Newton's First Law

It states that, "every body in this universe continues to be in its state of rest or of uniform motion in a straight line unless it is compelled by some external force to change the state. In other words, if the body is at rest, it stays at rest. If it is moving, it continues to move with the same velocity (same magnitude and same direction) unless we apply an external force.

external force Vs internal force?
A system consists of one or more bodies, and any force on the bodies inside the system from bodies outside the system is called an external force and internal forces are the forces between bodies inside the system.

Significance of Newton's first law of motion : Newton's first law expresses the idea of inertia . The inertia of a body is its tendency to remain at its original state (i.e., either in rest or in motion). Thus an object at rest begins to move only when it is pushed or pulled i.e., when a force acts on it. Similarly, if an object is moving in a straight line with constant velocity, it's direction will only change or it's motion will only be faster when a new force acts on it.

Newton's Second Law

It states that, "The net force on a body is equal to the product of the body's mass and its acceleration". We will define it in another way very soon!

F=ma

Significance of Newton's second law of motion : Newton's second law defines mass of a body as the characteristic that relates a force on the body to the resulting acceleration. The physical sensation of mass is felt when you try to accelerate a body, as in the case of kicking a football.

Newton's Third Law

It states that, "To every action there is always an equal and opposite reaction".

Significance of Newton's third law of motion : Consider a wood block on a table . So, the force \(\vec{F_{TB}}\) on the table from the wood block and a force \(\vec{F_{BT}}\) on the wood block from the table are the interacting forces in this this case. Then from Newton's third law of motion, \[F_{TB}=-F_{BT}\]

Examples of Newton's third law of motion

• Suppose that, the Earth pulls down on an apple with a force of magnitude 0.80 N. The apple must pull up on Earth with a force of magnitude 0.80 N, which we take to act at the center of Earth. These forces form a force pair. Eventhough they are same in magnitude, this forces produce different accelerations (compare masses of Earth and apple) when the apple is released. The acceleration of apple is about 9.8 m/s\(^2\) however the acceleration of earth is only about \(1.25 \times 10^{-25} m/s^2\).
• In the case of rocket, the downward force on the burning gases from the exhaust is equal to the upward force on the rocket which pushes the rocket in forward direction.

Here, a question may arise in your mind. Why don't the action and reaction forces cancel each other?
It is because they do not act on the same body. Imagine you are driving a car and suddenly your car is bumped to another car. Then from Newton's third law, your car exert a force on another car and the same magnitude of force act on your's car which slows down your car. The action-reaction forces here acts on different body i.e., one force (either action or reaction force) on your car and the other force (either action or reaction) on another car.

Linear Momentum

Linear momentum is the product of mass and velocity. \[\vec{p}=m\vec{v}\] It is measure of how much motion is contained by a body. It is a vector quantity and has the direction of velocity. It's unit is Kgm/s. The relation between force and linear momentum is revealed by the Newton's second law of motion.

Newton's second law of motion states that, "The rate of change of momentum of a body per unit time is directly proportional to the net external force applied on it and change takes place in the direction of force." \[F=\frac{dp}{dt}\] where p is the momentum.

Second law is the real law of motion
In order to prove it, we must prove that both the first law of motion and third law of motion follows from the Newton's second law of motion.
First law of motion : Consider no external force acts on the system , then F=ma reduces to ma=0. But m\(\neq0\), \(\therefore a=0\). We know that, \(a=\frac{dv}{dt}\). So, \(\frac{dv}{dt}=0 \implies v=\text{constant}\).
This proves the first law of motion since when no external force acts on the system, an object at rest stays in the rest and an object moving with uniform speed (v=constant) in a straight line will continue to do so.
Third law of motion : Consider two bodies A and B moving in a straight line collide for a time 't' (i.e., time of impact). As a result of collision, A will exert a force \(F_{BA}\) on B and B will exert an force \(F_{AB}\) on A which will change the linear momentum of each body.
Change in linear momentum of body A is, \[\Delta p=F_{AB}\times t\] and change in linear momentum of body B is, \[\Delta p'=F_{BA}\times t\] If no external force acts on the system of colliding bodies, there is no net change of momentum (i.e., the total momentum of the system remains constant). So, \[F_{AB}\times t+F_{BA}\times t=0\] \[\therefore F_{AB}=-F_{BA}\] This proves the Newton's third law of motion.

Impulse

The force acting for a short interval of time 't' on a body of mass 'm' is, \[F=\frac{mv-mu}{t}\] and , \[Ft=mv-mu=\text{change in momentum}\] The quantity Ft(force \(\times\) time) is known as the impulse of the force on the object. It's unit is newton second (Ns) or kgm/s.

• While catching a cricket ball, the players lowers his hands to increase the time of impact (i.e., t). Since, \(F\times t\)= change in momentum, as the change in momentum is same in both cases whether you lower or dont lower your hand, but when you lower your hand the time of impact increases which from the relation can be understood that the force will decrease.

Suppose sand is allowed to fall vertically at a steady rate of 100 g/s onto a horizontal conveyer belt moving at a steady velocity of 5 cm/s . The initial velocity of the sand is zero. The final horizontal velocity is 5 cm/s. Find the momentum change per second.
momentum change per second horizontally, \[\begin{align*} &=m(v-u)\\ &=0.1 \times 5 \times 10^{-2}\\ &=5\times 10^{-3} \hspace{0.1cm} N \end{align*}\] So, the momentum change per second is \(5 \times 10^{-3}\) N.
It can also be solved using calculus,
\[F=\frac{dp}{dt}=\frac{d(mv)}{dt} = v\frac{dm}{dt} \because v = \text{constant and } m = \text{changing with time}\] Now, use v = 5cm/s and \(\frac{dm}{dt}\)=100 to get the solution.

A ball of mass 0.2 kg falls from a height of 45 m. On striking the ground, it rebounds in 0.1 s with two-thirds of the velocity with which it struck the ground. Calculate (i) the momentum change on striking the ground and (ii) the force on the ball due to the impact.
Here, the velocity on impact v is, \[v=\sqrt{2gh} \because v^2=u^2+2gh, u=0\] Then, v = 30 m/s. Since, the ball rebounds with ⅔ of the velocity of impact, \[u'=-\frac{2}{3} \times 30\] Here, negative sign is due to our sign convention (the downward direction is taken to be positive while upward is taken to be negative). So, u'=20 m/s.
Thus, (i) the momentum change on striking the ground is, \[mv-mu'=0.2 \times (30+20)=10 \hspace{0.1cm} Ns\] and, (ii) force on the ball due to impact is, \[\frac{(mv-mu')}{t}=\frac{10}{0.1}= 100 \hspace{0.1cm} N\]

A cricket ball of mass 250 g collides with a bat with velocity 10 m/s and returns with same velocity within 0.01 s. What is the force acted on bat?
Here, mass of ball (m) = 0.25 kg, initial velocity (u) = 10 m/s, final velocity (v) = - 10 m/s (negative sign indicates the opposite direction) and time of impact (t) = 0.01 s.
Then, change in momentum = \(0.25 \times 20 = 5\) Ns. and thus, \[F=\frac{5}{0.01}=500 \hspace{0.1cm} N\] Jump to Mass Pulley system, apparent weight, conservation of linear momentum, etc.