Newton’s Laws of Motion

♦ Learn about : Newton’s first law of motion , Newton’s second law of motion & Numerical problems based on it ♦

Newton’s first law of motion :

Every body continues in its state of rest or of uniform motion in a straight line unless it is compelled by an externally impressed force to act otherwise.

The tendency of material bodies to maintain their velocity constant is known as inertia Mass is a measure of inertia. When an external force acts on a body, it changes its velocity.

Newton’s second law of motion :

The rate of change of momentum of a body is directly proportional to the applied force and acts in the direction of the force.

Momentum is the ‘quantity of motion’ present in a body. It is a vector quantity and for small enough velocities (v « c, the velocity of light), it is given by,

\displaystyle \vec{P} = m\vec{v} , where m is the mass of the body and v its velocity.

For a body ,

\displaystyle \frac{\vec{dp}}{dt} \varpropto \vec{F}

\displaystyle \frac{d (m\vec{v})}{dt} \varpropto \vec{F}

\displaystyle m\frac{\vec{dv}}{dt} \varpropto \vec{F}

\displaystyle \vec{F} = k m \frac{\vec{dv}}{dt}

where k is a constant. With proper choice of units, k = 1

Thus , \displaystyle \vec{F} = m \vec{a}

Illustration : A body of mass m = 1 kg falls from a height h = 20 m from the ground level
(a) What is the magnitude of the total change in momentum of the body before it strikes the ground ?
(b) What is the corresponding average force experienced by it?
(g = 10m/sec2).

Solution:
(a) Since the body falls from rest (u = 0) through a distance h before striking the ground, the speed v of the body is given by

v2 = u2 + 2 as ; putting a = g and s = h

we obtain v = √(2gh)

The magnitude of total change in momentum of the body

= ΔP = | mv – 0 | = mv

where v = √(2gh)

⇒ ΔP = m√(2gh)

ΔP = (1)√(2×10×20) kg m/sec

ΔP = 20 kg m/sec.

The average force experienced by the body = \displaystyle F_av = \frac{\Delta P}{\Delta t}

, where Δt = time of motion of the body = t(say)

We know ΔP = 20 kg m/sec.
Therefore we will have to find t using the given data. We know from kinematics that,

\displaystyle S = ut + \frac{1}{2}a t^2

\displaystyle h = \frac{1}{2}g t^2 , (Since , u = 0)

\displaystyle t = \sqrt {\frac{2h}{g} } , and \Delta P = m\sqrt{2gh}

\displaystyle F_av = \frac{\Delta P}{\Delta t}

Putting the general values of ΔP & t we obtain

Fav = mg

where mg is the weight (W) of the body and is directed vertically downwards. Therefore the body experiences a constant vertically downward force of magnitude mg.

Exercise :
A body of mass m = 2 kg changes its velocity v1 = (i + 2j + 3k) m/sec to v2 = (-2i + 2j – k) m/sec during 2 seconds.
(a) Find the magnitude of the change in momentum of the body .
(b) What is the magnitude of the average force experienced by the body ?

 Newton’s third law of motion

Every action has an equal and opposite reaction. This implies that if a force is applied by a body A on a body B, then a reaction force acts on A due to B.

According to the third law,
\displaystyle \vec{F_{AB}} = \vec{F_{BA}} , is they are equal in magnitude but opposite in direction.

This law is true for all forces – contact forces (such as friction)and non-contact forces (such as gravitation and electrostatics ) but does not apply to pseudo forces.

Illustration  : Two blocks of mass 2 kg and 4 kg are kept in contact with each other on a smooth horizontal surface. A horizontal force of 12 N is applied on the first block due to which the pair moves with a constant acceleration. Calculate the force acting between the blocks.

Solution : Let a be the common acceleration of the blocks,

R = Force between the two blocks

From FBD of Ist block

F – R = m1 a   . .. . (i)

From FBD of block 2,

R = m2 a     . . . .(ii)

From (i) & (ii),

F = (m1 + m2). a

a = F / (m1 + m2)

= 12/6 = 2 m/sec2

Force between blocks

= R = m2 a

= (4 ×2) N = 8 N.

Application of Newton’s laws of motion

Learn about : Application of laws of motion : Techniques & Approach

A separate diagram of the body is drawn showing all the different forces exerted by the bodies in the environment. This diagram is known as the free body diagram.

Application of Newton’s Laws to any system (consisting of one or more objects) can be done by following a systematic method.

We Recommend the following steps in the order given below :

(i) Draw the complete free body diagram (FBD), showing all the forces acting on each separate body.

(ii) Select proper coordinates for analysing the motion of each body.
Include any pseudo-forces within the FBD, if required.

(iii) If there are any constraints, write the proper constraint equations.

(iv) Apply Newton’s 2nd law of motion : F-> = ma-> for each body. This leads to a system of equations.

(v) Solve these equations:

(a) Identify the known and unknown quantities. Check that the no. of equations equals the number of unknowns.
(b) Check the equations using dimensional analysis.

(c) After solving, check the final solution using back substitution.

(vi) If the velocity (v ) or, position ( x) is required, proceed from a knowledge of the acceleration ( a->) as found from equations in step (v) and apply kinematics: dv/dt = a (known) and finally integrate.

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