Derive Newton's first law from second law?

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According to Newton’s second law of motion, the rate of change of momentum of a body is directly proportional to the applied force and it takes place in the direction in which the force acts.

Consider a body of mass m moving with velocity v, then the linear momentum of the body is given by

p = mv

Now from, Newton’s second law,

F ∝ \(\frac{dp}{dt}\)

⇒ F = k \(\frac{dp}{dt}\)

Where, k is constant of proportionality.

As, p = mv

⇒ F = k \(\frac{d(mv)}{dt}\) = km \(\frac{dv}{dt}\)

But, \(\frac{dv}{dt}\) = a, acceleration of the body

The value of the constant of proportionality k is considered as 1 for simplicity.

By taking k = 1, we get

F = ma

According to Newton’s first law of motion, everybody continues in its state of rest or uniform motion in a particular direction until and unless an external force is applied to change that state.

Now, if net force, F = 0, then by Newton’s second law acceleration, a = 0.

This shows that, if there is no force acting on the body, its acceleration is zero i.e. the body will be in its state of rest or uniform motion. Hence, Newton’s first law is derived from the second law.

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The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

\( \frac{q}{4 \pi \epsilon_0 l^2} \) pointing along MA
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Derive Newton's first law from second law?

CBSE CLASS XII Related Questions

1.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

3.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

5.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Similar Physics Concepts

Comments

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Derive Newton's first law from second law?

CBSE CLASS XII Related Questions

1.A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

3.Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

5. If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.

Similar Physics Concepts

Comments

SUBSCRIBE TO OUR NEWS LETTER

1.
A light copper ring is freely suspended by a light string. A bar magnet is held horizontally with its length along the axis of the ring. The magnet is moved towards the ring with its N pole facing the loop. What will happen to the ring and its position? Explain.

2.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

3.
Two heaters rated as \((P_1,V)\) and \((P_2,V)\) are connected in series across a dc source of \(V/2\) volt. The power consumed by the combination will be –

4.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

5.
If Bohr’s quantization postulate (angular momentum \( = \frac{nh}{2\pi} \)) is a basic law of nature, it should be equally valid for the case of planetary motion also. Why, then, do we never speak of quantization of orbits of planets around the Sun? Explain.

6.
Photoemission of electrons occurs from a metal (\( \phi_0 = 1.96 \, \text{eV} \)) when light of frequency \( 6.4 \times 10^{14} \, \text{Hz} \) is incident on it. Calculate: Energy of a photon in the incident light, The maximum kinetic energy of the emitted electrons, and The stopping potential.