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Derive newton second law and third law
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Solution
Newton's Second law of motion: Newton's Second law of motion states that the second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of force.
Newton’s Third law of motion: Newton's Third law of motion states that for every action there is an equal and opposite reaction.
Derivation of Newton's Second Law
Suppose an object of mass, m is moving along a straight line with an initial velocity, u.
It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F
throughout the time, t. The initial and final momentum of the object will be,
p1 = mu and p2 = mv respectively.
The change in momentum ∝ p2 – p1
The change in momentum∝ mv – mu
The change in momentum∝ m × (v – u).
The rate of change of momentum ∝ m × (v −u)/t
Or, the applied force,
F ∝m × (v −u)/t
F = km (v - u)/t
F = kma
Here a [a = (v – u)/t ] is the acceleration, which is the rate of change of velocity. The quantity, k is a constant of proportionality.The SI units of mass and acceleration are kg and m s-2 respectively. The unit of force is so chosen that the value of the constant, k becomes one. For this, one unit of force is
defined as the amount that produces an acceleration of 1 m s-2 in an object of 1 kg mass.
That is, 1 unit of force = k × (1 kg) × (1 m s-2).
Thus, the value of k becomes 1. and
F = ma which is the mathematical expression on the Newton's second law of motion.
newton's third law of motion from second law
Consider an isolated system of 2 bodies A & B.An isolated system means where there is no external force acting.Now let FAB be the force acting on B by A & FBA be the force acting on A by B.Now rate of change of momentum of A =dpA/dt and rate of change of momentum of B =dpB/dt
thus FAB=dpB/dt (i)......
FBA=dpA/dt (ii)........
Adding 1 and 2 we get FAB+FBA=dpB/dt+dpA/dt =d(pB+pA)/dt
But if no force is applied momentum will also be 0 becoz no velocity will be there.so rate of change of momentum will also be 0
thus d(pA+pB)/dt=0
Read more on Brainly.in - https://brainly.in/question/954432#readmore
Newton’s Third law of motion: Newton's Third law of motion states that for every action there is an equal and opposite reaction.
Derivation of Newton's Second Law
Suppose an object of mass, m is moving along a straight line with an initial velocity, u.
It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F
throughout the time, t. The initial and final momentum of the object will be,
p1 = mu and p2 = mv respectively.
The change in momentum ∝ p2 – p1
The change in momentum∝ mv – mu
The change in momentum∝ m × (v – u).
The rate of change of momentum ∝ m × (v −u)/t
Or, the applied force,
F ∝m × (v −u)/t
F = km (v - u)/t
F = kma
Here a [a = (v – u)/t ] is the acceleration, which is the rate of change of velocity. The quantity, k is a constant of proportionality.The SI units of mass and acceleration are kg and m s-2 respectively. The unit of force is so chosen that the value of the constant, k becomes one. For this, one unit of force is
defined as the amount that produces an acceleration of 1 m s-2 in an object of 1 kg mass.
That is, 1 unit of force = k × (1 kg) × (1 m s-2).
Thus, the value of k becomes 1. and
F = ma which is the mathematical expression on the Newton's second law of motion.
newton's third law of motion from second law
Consider an isolated system of 2 bodies A & B.An isolated system means where there is no external force acting.Now let FAB be the force acting on B by A & FBA be the force acting on A by B.Now rate of change of momentum of A =dpA/dt and rate of change of momentum of B =dpB/dt
thus FAB=dpB/dt (i)......
FBA=dpA/dt (ii)........
Adding 1 and 2 we get FAB+FBA=dpB/dt+dpA/dt =d(pB+pA)/dt
But if no force is applied momentum will also be 0 becoz no velocity will be there.so rate of change of momentum will also be 0
thus d(pA+pB)/dt=0
Read more on Brainly.in - https://brainly.in/question/954432#readmore
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