Newton’s Second Law of Motion

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Newton’s Second Law of Motion

Newton’s Second Law of Motion helps us understand how an object’s movement changes when a force is applied to it. Simply put, it tells us that the force acting on an object depends on how quickly its momentum changes, and this change always happens in the direction of the applied force.When the object’s mass stays constant, the law is often written as F = ma, where F is the force, m is the mass, and a is the acceleration. This equation is a cornerstone of physics—it shows how force, mass, and acceleration are all connected, and it plays a key role in understanding how things move in the real world, from a falling apple to a speeding car.

1.0Statements of Newton’s Second Law of Motion

The rate of change of momentum of an object is directly proportional to the applied external force, and this change takes place in the direction of the applied force.

or

According to this law, "rate of change of momentum of any system is directly proportional to the applied external force".

$F_{n e t} = \frac{d p}{d t}$

2.0Derivation of Newton’s Second Law of Motion

According to this law, "rate of change of momentum of any system is directly proportional to the applied external force".

$F_{net} \propto \frac{d p}{d t} \Rightarrow F = k \frac{d p}{d t} [In S.I. units k = 1]$

$F_{net} = \frac{d p}{d t}$

The change in momentum occurs in the direction of the exerted force.

$F_{net} = \frac{d p}{d t} = \frac{d ( m v )}{d t} = m \frac{d v}{d t} + v \frac{d m}{d t}$

If m is constant $(\frac{d m}{d t} = 0)$ $\Rightarrow F = m \frac{d v}{d t} = m a$

$F_{net} = \frac{d ( m v )}{d t} = m \frac{d v}{d t} + v \frac{d m}{d t}$

If $v$ is constant $(\frac{d v}{d t} = 0)$

$F_{net} = v \frac{d m}{d t}$ variable mass system e.g.conveyor belt,rocket

The slope of the momentum-time graph is equal to the force on the particle at that instant.

$F = \frac{d p}{d t} = tan θ = slope$

Derivation of Newton’s Second Law of Motion - slope of the momentum-time graph

3.0Variable Mass Problem(Rocket Propulsion)

m=Mass of rocket

$v_{re l}$ =Velocity of exhaust gases w.r.t rocket

$\frac{d m}{d t}$ =Rate of burning of fuel

Variable Mass Problem(Rocket Propulsion)

Case-1 : If rocket is accelerating upwards, then

$v_{re l} ∣ \frac{d m}{d t} ∣ - m g = ma$

acceleration of the rocket, $a = \frac{v _{re l}}{m} ∣ \frac{d m}{d t} ∣ - m g$

Case-2 : If rocket is moving with constant velocity, then a = 0

$v_{re l} ∣ \frac{d m}{d t} ∣ = m g$

4.0Units of Force

Absolute units of Force

The force which produces a unit acceleration in a body of unit mass

Gravitational Unit of Force

The force which produces acceleration equal to 'g'(acceleration due to gravity ) in a body of unit mass

SI Unit-Newton

$1 N = 1 k g \times 1 m / s^{2} = 1 k g m / s^{2}$

SI Unit-Kilogram Weight(kg wt) or kilogram force(kg f)

$1 k g wt = 1 k g f = 9.8 N$

CGS Unit-Dyne

$1 d y n e = 1 g \times 1 c m / s^{2} = 1 g c m / s^{2}$

CGS Unit-gram weight(g wt) or gram force(g f)

$1 g wt = 1 g f = 980 d y n e$

Note: The absolute unit of force remains constant throughout the universe, while the gravitational unit varies with location because it depends on the local acceleration due to gravity (g).

$1 N = 1 0^{5} d y n e$

5.0Component Form of Newton’s Second Law

$F = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k}$

$p = p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k}$

$a = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}$

$F = \frac{d p}{d t} = m a$

$F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = \frac{d}{d t} (p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k})$

$F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = m (a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k})$

$F_{x} = \frac{d p _{x}}{d t} = m a_{x}$ , $F_{y} = \frac{d p _{y}}{d t} = m a_{y}$ , $F_{z} = \frac{d p _{z}}{d t} = m a_{z}$

6.0Concepts of Inertial Mass

Suppose a force F acting independently on two masses m1and m2 produces accelerations $a_{1}$ and $a_{2}$ in them. By using Newton’s second Law,

$F = m_{1} a_{1} = m_{2} a_{2} or \frac{a _{1}}{a _{2}} = \frac{m _{2}}{m _{1}}$

If $m_{2} < m_{1}$ , then $a_{2} > a_{1}$ a given force produces a larger acceleration in a lighter mass than that in a heavier mass.
Mass of the body gives a measure of the resistance or opposition to a given force that tends to change its velocity. That is why, $m = \frac{F}{a}$ is called inertial mass of the body.

7.0Newton Second Law is The Real Law of Motion

First Law is contained in the second law

According to Newton’s second Law, the force acting on a body is given by, F=ma
In the absence of any external force, F=0 or ma=0. As m ≠ 0 therefore a=0
Thus, when no force is applied, there is no acceleration. In the absence of any external force, a body at rest remains at rest, and a body in uniform motion continues to move with the same velocity along a straight line. This is essentially Newton’s First Law of Motion. Therefore, the First Law is actually a special case of the Second Law.

Third Law is Contained in the Second Law

An isolated system consisting of two bodies, A and B. Suppose two bodies interact mutually with one another.
Let $F_{B A}$ be the force (action) exerted by A on B. Let $\frac{d p _{B}}{d t}$ be the resulting change of momentum of B.
Let $F_{A B}$ be the force (reaction) exerted by B on A. Let $\frac{d p _{A}}{d t}$ be the resulting change of momentum of A. According to Newton’s second Law

$F_{B A} = \frac{d p _{B}}{d t} and F_{A B} = \frac{d p _{A}}{d t}$

$F_{B A} + F_{A B} = \frac{d p _{B}}{d t} + \frac{d p _{A}}{d t} = \frac{d}{d t} (p_{B} + p_{A})$

In the absence of external force, the rate of change of momentum must be zero, $\frac{d}{d t} (p_{B} + p_{A}) must be zero.$

$F_{B A} + F_{A B} = 0$

$F_{B A} = - F_{A B}$

Action=-Reaction

This is nothing but Newton’s Third Law of motion, hence the third law of motion contained in the second law.

Related Video: Newton's Laws of Motion

8.0Sample Questions on Newton’s Second Law of Motion

Q-1.A ball of 0.20 kg hits a wall with a velocity of 25 m/s at an inclination of 45° with horizontal. if the ball rebounds at 90° to the direction of incidence with the same speed, calculate the magnitude of change in momentum of the ball.

Practice questions on newton's second law of motion: A ball of 0.20 kg hits a wall with a velocity of 25 m/s at an inclination of 45° with horizontal. if the ball rebounds at 90° to the direction of incidence with the same speed, calculate the magnitude of change in momentum of the ball.

Solution:

Initial momentum $p_{i} = m \times v_{i} = m \times v (cos 4 5^{\circ}) \hat{i} + m v sin 4 5^{\circ} (- \hat{j})$

Final momentum $p_{f} = m \times v_{f} = m \times v cos 4 5^{\circ} (- \hat{i}) + m v sin 4 5^{\circ} (- \hat{j})$

Change in momentum= $(- m v cos 4 5^{\circ}) - (m v cos 4 5^{\circ}) = - 2 m v cos 4 5^{\circ}$

$∣ Δ p ∣ = 2 m v cos 4 5^{0} = 2 \times 0.2 \times \times 25 \times \frac{1}{2} = 52 N - s$

Q-2.A cricket ball weighing 150 grams approaches a batsman at a speed of 12 m/s. The batsman strikes the ball, reversing its direction and sending it back at a speed of 20 m/s. If the contact time between the bat and the ball is 0.01 seconds, calculate:

The change in momentum of the ball.
The force exerted by the bat on the ball.

Solution:

According to given problem change in momentum of the ball

$Δ p = p_{f} - p_{i} = m (v - u) = 150 \times 1 0^{- 3} [20 - (- 12)] = 4.8 N - s$

force in case of change in momentum= $\frac{Δ p}{Δ t} = \frac{4.80}{0.01} = 480 N$

Q-3.A force $F = (6 \hat{i} - 8 \hat{j} + 10 \hat{k})$ produces an acceleration of $2 m / s^{2}$ in a body. Calculate the mass of the body.

Solution:

Acceleration $a = \frac{F}{m} \Rightarrow m = \frac{∣ F ∣}{a} = \frac{6 ^{2} + 8 ^{2} + 1 0 ^{2}}{2} = 10 k g$

Q-4.A force of 50 N acts on a block in the direction as shown in figure. The block is of mass 5kg, resting on a smooth horizontal surface. Find out the acceleration of the block.

Solution:

Acceleration of the block, $50 s in 60° = \frac{50 3}{2}$

acceleration of the block,

$a = \frac{co m p o n e n t o f f orce in t h e d i rec t i o n o f a cce l er a t i o n}{ma ss} = \frac{50 3}{2} \times \frac{1}{5} = 53 m / s^{2}$

Q-5.A 200 g ball is moving at a speed of 10 m/s when a player catches it. If the player brings the ball to rest in 0.20 seconds, what is the magnitude of the average force exerted on the player's hand during the catch?

Solution:

$F_{a vg} = \frac{Δ p}{Δ t} = \frac{200 \times 1 0 ^{- 3} \times 10}{0.20} = 10 N$

Newton’s Second Law of Motion

Newton’s Second Law of Motion helps us understand how an object’s movement changes when a force is applied to it. Simply put, it tells us that the force acting on an object depends on how quickly its momentum changes, and this change always happens in the direction of the applied force.When the object’s mass stays constant, the law is often written as F = ma, where F is the force, m is the mass, and a is the acceleration. This equation is a cornerstone of physics—it shows how force, mass, and acceleration are all connected, and it plays a key role in understanding how things move in the real world, from a falling apple to a speeding car.

1.0Statements of Newton’s Second Law of Motion

The rate of change of momentum of an object is directly proportional to the applied external force, and this change takes place in the direction of the applied force.

or

According to this law, "rate of change of momentum of any system is directly proportional to the applied external force".

$F_{n e t} = \frac{d p}{d t}$

2.0Derivation of Newton’s Second Law of Motion

According to this law, "rate of change of momentum of any system is directly proportional to the applied external force".

$F_{net} \propto \frac{d p}{d t} \Rightarrow F = k \frac{d p}{d t} [In S.I. units k = 1]$

$F_{net} = \frac{d p}{d t}$

The change in momentum occurs in the direction of the exerted force.

$F_{net} = \frac{d p}{d t} = \frac{d ( m v )}{d t} = m \frac{d v}{d t} + v \frac{d m}{d t}$

If m is constant $(\frac{d m}{d t} = 0)$ $\Rightarrow F = m \frac{d v}{d t} = m a$

$F_{net} = \frac{d ( m v )}{d t} = m \frac{d v}{d t} + v \frac{d m}{d t}$

If $v$ is constant $(\frac{d v}{d t} = 0)$

$F_{net} = v \frac{d m}{d t}$ variable mass system e.g.conveyor belt,rocket

The slope of the momentum-time graph is equal to the force on the particle at that instant.

$F = \frac{d p}{d t} = tan θ = slope$

3.0Variable Mass Problem(Rocket Propulsion)

m=Mass of rocket

$v_{re l}$ =Velocity of exhaust gases w.r.t rocket

$\frac{d m}{d t}$ =Rate of burning of fuel

Case-1 : If rocket is accelerating upwards, then

$v_{re l} ∣ \frac{d m}{d t} ∣ - m g = ma$

acceleration of the rocket, $a = \frac{v _{re l}}{m} ∣ \frac{d m}{d t} ∣ - m g$

Case-2 : If rocket is moving with constant velocity, then a = 0

$v_{re l} ∣ \frac{d m}{d t} ∣ = m g$

4.0Units of Force

Absolute units of Force

The force which produces a unit acceleration in a body of unit mass

Gravitational Unit of Force

The force which produces acceleration equal to 'g'(acceleration due to gravity ) in a body of unit mass

SI Unit-Newton

$1 N = 1 k g \times 1 m / s^{2} = 1 k g m / s^{2}$

SI Unit-Kilogram Weight(kg wt) or kilogram force(kg f)

$1 k g wt = 1 k g f = 9.8 N$

CGS Unit-Dyne

$1 d y n e = 1 g \times 1 c m / s^{2} = 1 g c m / s^{2}$

CGS Unit-gram weight(g wt) or gram force(g f)

$1 g wt = 1 g f = 980 d y n e$

Note: The absolute unit of force remains constant throughout the universe, while the gravitational unit varies with location because it depends on the local acceleration due to gravity (g).

$1 N = 1 0^{5} d y n e$

5.0Component Form of Newton’s Second Law

$F = F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k}$

$p = p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k}$

$a = a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k}$

$F = \frac{d p}{d t} = m a$

$F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = \frac{d}{d t} (p_{x} \hat{i} + p_{y} \hat{j} + p_{z} \hat{k})$

$F_{x} \hat{i} + F_{y} \hat{j} + F_{z} \hat{k} = m (a_{x} \hat{i} + a_{y} \hat{j} + a_{z} \hat{k})$

$F_{x} = \frac{d p _{x}}{d t} = m a_{x}$ , $F_{y} = \frac{d p _{y}}{d t} = m a_{y}$ , $F_{z} = \frac{d p _{z}}{d t} = m a_{z}$

6.0Concepts of Inertial Mass

Suppose a force F acting independently on two masses m1and m2 produces accelerations $a_{1}$ and $a_{2}$ in them. By using Newton’s second Law,

$F = m_{1} a_{1} = m_{2} a_{2} or \frac{a _{1}}{a _{2}} = \frac{m _{2}}{m _{1}}$

If $m_{2} < m_{1}$ , then $a_{2} > a_{1}$ a given force produces a larger acceleration in a lighter mass than that in a heavier mass.
Mass of the body gives a measure of the resistance or opposition to a given force that tends to change its velocity. That is why, $m = \frac{F}{a}$ is called inertial mass of the body.

7.0Newton Second Law is The Real Law of Motion

First Law is contained in the second law

According to Newton’s second Law, the force acting on a body is given by, F=ma
In the absence of any external force, F=0 or ma=0. As m ≠ 0 therefore a=0
Thus, when no force is applied, there is no acceleration. In the absence of any external force, a body at rest remains at rest, and a body in uniform motion continues to move with the same velocity along a straight line. This is essentially Newton’s First Law of Motion. Therefore, the First Law is actually a special case of the Second Law.

Third Law is Contained in the Second Law

An isolated system consisting of two bodies, A and B. Suppose two bodies interact mutually with one another.
Let $F_{B A}$ be the force (action) exerted by A on B. Let $\frac{d p _{B}}{d t}$ be the resulting change of momentum of B.
Let $F_{A B}$ be the force (reaction) exerted by B on A. Let $\frac{d p _{A}}{d t}$ be the resulting change of momentum of A. According to Newton’s second Law

$F_{B A} = \frac{d p _{B}}{d t} and F_{A B} = \frac{d p _{A}}{d t}$

$F_{B A} + F_{A B} = \frac{d p _{B}}{d t} + \frac{d p _{A}}{d t} = \frac{d}{d t} (p_{B} + p_{A})$

In the absence of external force, the rate of change of momentum must be zero, $\frac{d}{d t} (p_{B} + p_{A}) must be zero.$

$F_{B A} + F_{A B} = 0$

$F_{B A} = - F_{A B}$

Action=-Reaction

This is nothing but Newton’s Third Law of motion, hence the third law of motion contained in the second law.

Related Video: Newton's Laws of Motion

8.0Sample Questions on Newton’s Second Law of Motion

Q-1.A ball of 0.20 kg hits a wall with a velocity of 25 m/s at an inclination of 45° with horizontal. if the ball rebounds at 90° to the direction of incidence with the same speed, calculate the magnitude of change in momentum of the ball.

Solution:

Initial momentum $p_{i} = m \times v_{i} = m \times v (cos 4 5^{\circ}) \hat{i} + m v sin 4 5^{\circ} (- \hat{j})$

Final momentum $p_{f} = m \times v_{f} = m \times v cos 4 5^{\circ} (- \hat{i}) + m v sin 4 5^{\circ} (- \hat{j})$

Change in momentum= $(- m v cos 4 5^{\circ}) - (m v cos 4 5^{\circ}) = - 2 m v cos 4 5^{\circ}$

$∣ Δ p ∣ = 2 m v cos 4 5^{0} = 2 \times 0.2 \times \times 25 \times \frac{1}{2} = 52 N - s$

Q-2.A cricket ball weighing 150 grams approaches a batsman at a speed of 12 m/s. The batsman strikes the ball, reversing its direction and sending it back at a speed of 20 m/s. If the contact time between the bat and the ball is 0.01 seconds, calculate:

The change in momentum of the ball.
The force exerted by the bat on the ball.

Solution:

According to given problem change in momentum of the ball

$Δ p = p_{f} - p_{i} = m (v - u) = 150 \times 1 0^{- 3} [20 - (- 12)] = 4.8 N - s$

force in case of change in momentum= $\frac{Δ p}{Δ t} = \frac{4.80}{0.01} = 480 N$

Q-3.A force $F = (6 \hat{i} - 8 \hat{j} + 10 \hat{k})$ produces an acceleration of $2 m / s^{2}$ in a body. Calculate the mass of the body.

Solution:

Acceleration $a = \frac{F}{m} \Rightarrow m = \frac{∣ F ∣}{a} = \frac{6 ^{2} + 8 ^{2} + 1 0 ^{2}}{2} = 10 k g$

Q-4.A force of 50 N acts on a block in the direction as shown in figure. The block is of mass 5kg, resting on a smooth horizontal surface. Find out the acceleration of the block.

Solution:

Acceleration of the block, $50 s in 60° = \frac{50 3}{2}$

acceleration of the block,

$a = \frac{co m p o n e n t o f f orce in t h e d i rec t i o n o f a cce l er a t i o n}{ma ss} = \frac{50 3}{2} \times \frac{1}{5} = 53 m / s^{2}$

Q-5.A 200 g ball is moving at a speed of 10 m/s when a player catches it. If the player brings the ball to rest in 0.20 seconds, what is the magnitude of the average force exerted on the player's hand during the catch?

Solution:

$F_{a vg} = \frac{Δ p}{Δ t} = \frac{200 \times 1 0 ^{- 3} \times 10}{0.20} = 10 N$

Join ALLEN!

Join ALLEN!

Newton’s Second Law of Motion

1.0Statements of Newton’s Second Law of Motion

2.0Derivation of Newton’s Second Law of Motion

3.0Variable Mass Problem(Rocket Propulsion)

4.0Units of Force

5.0Component Form of Newton’s Second Law

6.0Concepts of Inertial Mass

7.0Newton Second Law is The Real Law of Motion

8.0Sample Questions on Newton’s Second Law of Motion

Table of Contents

Newton’s Second Law of Motion

1.0Statements of Newton’s Second Law of Motion

2.0Derivation of Newton’s Second Law of Motion

3.0Variable Mass Problem(Rocket Propulsion)

4.0Units of Force

5.0Component Form of Newton’s Second Law

6.0Concepts of Inertial Mass

7.0Newton Second Law is The Real Law of Motion

8.0Sample Questions on Newton’s Second Law of Motion

Table of Contents