Laws of Motion

This chapter introduces the fundamental principles governing motion and its causes, formulated as Newton's three laws of motion. It moves beyond the description of motion (kinematics) to explain why objects move the way they do. The core concepts you will learn are force, inertia, linear momentum, and impulse. You will study different types of forces, including weight, normal reaction, tension, and friction.

The main skill to master is the application of these laws to solve problems. This is done by drawing free-body diagrams (FBDs), which are schematic representations of an object isolated from its surroundings, showing all the external forces acting on it. Using FBDs, you will set up and solve equations of motion for a wide range of situations, from a single block on an inclined plane to complex systems of multiple bodies connected by strings and pulleys.

Most students find the initial setup of a problem to be the hardest part, specifically identifying all forces correctly and resolving them into components. Problems involving friction, where you must distinguish between static and kinetic conditions, and those with constraints relating the motion of different parts of a system, require careful practice. A firm grasp of Kinematics and vector algebra from previous chapters is essential for success here.

Formulas for Laws of Motion

Newton's Second Law of Motion: $\vec{F}_{net} = \frac{d\vec{p}}{dt}$ The net external force is the rate of change of linear momentum.
Newton's Second Law (for constant mass): $\vec{F}_{net} = m\vec{a}$ The most common form, where m is mass and a is acceleration.
Linear Momentum: $\vec{p} = m\vec{v}$ p is momentum, m is mass, and v is velocity.
Impulse-Momentum Theorem: $\vec{J} = \int \vec{F} dt = \Delta \vec{p}$ Impulse J equals the total change in linear momentum Δp.
Conservation of Linear Momentum: $\text{If } \vec{F}_{ext} = 0, \text{ then } \vec{p}_{initial} = \vec{p}_{final}$ For an isolated system, the total momentum is conserved.
Limiting Static Friction: $f_{s, max} = \mu_s N$ The maximum force of static friction, where μs is the coefficient of static friction.
Kinetic Friction: $f_k = \mu_k N$ The force of kinetic friction, where μk is the coefficient of kinetic friction.
Centripetal Force: $F_c = \frac{mv^2}{r} = m\omega^2r$ The net force towards the center required for uniform circular motion.

One of these questions, worked through

JEE Advanced 2019

A block of mass $2M$ is attached to a massless spring with spring-constant $k$ . This block is connected to two other blocks of masses $M$ and $2M$ using two massless pulleys and strings. The acceleration of the blocks are $a_1$ , $a_2$ and $a_3$ as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is $x_0$ . Which of the following option(s) is/are correct? [ $g$ is the acceleration due to gravity. Neglect friction]

Laws of Motion A block of mass

2M

is attached to a massless spring with spring-constant

k

. This block is connected to two other blocks of masses

M

and

2M

using two massless pulleys and strings. The acceleration of the blocks are

a_1

a_2

and

a_3

as shown in the figure. The system is released from rest with the spring in its unstretched state. The maximum extension of the spring is

x_0

. Which of the following option(s) is/are correct? [

g

is the acceleration due to gravity. Neglect friction]

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Step 1: Establish the string constraint relation.

Let $x$ be the displacement of the block of mass $2M$ (attached to the spring) to the right. As this block moves right by $x$ , the movable pulley $P_2$ moves down by $x$ . Let $y_M$ and $y_{2M}$ be the downward displacements of the blocks of mass $M$ and $2M$ respectively.

The length of the string passing over $P_2$ is constant, which gives:

y_M + y_{2M} = 2x

Differentiating twice with respect to time:

a_2 + a_3 = 2a_1

Rearranging this relation:

a_2 - a_1 = a_1 - a_3

This shows that Option 4 is correct.

Step 2: Analyze the forces and tensions.

Let $T$ be the tension in the string connecting the blocks of mass $M$ and $2M$ . Since the pulley $P_2$ is massless, the tension $T_u$ in the upper string connected to the block of mass $2M$ on the table is:

T_u = 2T

For the block of mass $2M$ on the table, the equation of motion is:

2M a_1 = T_u - kx

2M a_1 = 2T - kx

Step 3: Determine the tension $T$ using the frame of the movable pulley.

In the accelerating frame of pulley $P_2$ (which has downward acceleration $a_1$ ), the effective acceleration due to gravity is:

g_{\text{eff}} = g - a_1

The equations of motion for the hanging blocks of mass $M$ and $2M$ relative to the pulley are:

M(g - a_1) - T = M a_{\text{rel}}

2M(g - a_1) - T = -2M a_{\text{rel}}

Multiplying the first equation by 2 and adding it to the second equation to eliminate $a_{\text{rel}}$ :

4M(g - a_1) - 3T = 0

T = \frac{4M}{3}(g - a_1)

Step 4: Find the acceleration $a_1$ as a function of $x$ .

Substitute the expression for $T$ into the equation of motion for the block on the table:

2M a_1 = \frac{8M}{3}(g - a_1) - kx

\frac{14M}{3} a_1 = \frac{8Mg}{3} - kx

a_1 = \frac{4g}{7} - \frac{3k}{14M}x

Step 5: Determine the maximum extension $x_0$ .

Rewrite the equation of motion in the standard form of simple harmonic motion:

a_1 = -\frac{3k}{14M}\left(x - \frac{8Mg}{3k}\right)

This represents simple harmonic motion with:

\omega^2 = \frac{3k}{14M}

x_{\text{eq}} = \frac{8Mg}{3k}

Since the system is released from rest at $x = 0$ , the amplitude of oscillation is $A = x_{\text{eq}}$ . The maximum extension $x_0$ is:

x_0 = 2x_{\text{eq}} = \frac{16Mg}{3k}

Since $\frac{16Mg}{3k} \neq \frac{4Mg}{k}$ , Option 2 is incorrect.

Step 6: Evaluate the acceleration at $x = \frac{x_0}{4}$ .

Substitute $x = \frac{x_0}{4} = \frac{4Mg}{3k}$ into the expression for $a_1$ :

a_1 = \frac{4g}{7} - \frac{3k}{14M}\left(\frac{4Mg}{3k}\right)

a_1 = \frac{2g}{7}

Since $\frac{2g}{7} \neq \frac{3g}{10}$ , Option 1 is incorrect.

Step 7: Evaluate the speed at $x = \frac{x_0}{2}$ .

At $x = \frac{x_0}{2} = x_{\text{eq}}$ , the block is at its equilibrium position, where its speed $v$ is maximum:

v = \omega x_{\text{eq}}

v = \sqrt{\frac{3k}{14M}} \cdot \frac{8Mg}{3k}

v = g\sqrt{\frac{32M}{21k}}

Since $g\sqrt{\frac{32M}{21k}} \neq 3g\sqrt{\frac{M}{5k}}$ , Option 3 is incorrect.

Answer: The only correct option is $a_2 - a_1 = a_1 - a_3$ .

Open this question on its own page →

Past-paper questions on Laws of Motion

JEE Advanced 40 questions