Laws of Motion Class 11 NCERT Physics: Comprehensive Study Guide

Understanding the Foundations: Inertia, Momentum, and Force

Dynamics begins with Galileo's law of inertia, which debunked the ancient Aristotelian idea that a constant force is required to keep a body in uniform motion. Newton formalized this into his First Law of Motion: a body remains in its state of rest or uniform motion unless acted upon by an external net force.

To quantify motion, Newton introduced Linear Momentum ($\vec{p}$), defined as the product of mass ($m$) and velocity ($\vec{v}$). Newton's Second Law states that the rate of change of momentum is directly proportional to the applied external force and takes place in the direction of the force:
$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$
For a system with constant mass, this simplifies to the familiar relation $\vec{F} = m\vec{a}$.

Newton's Third Law emphasizes that forces always occur in matched action-reaction pairs: to every action, there is an equal and opposite reaction. It is vital to note that action and reaction act on two different bodies, which is why they never cancel each other out.

Essential Terminology

Inertia

The inherent property of an object to resist any change in its state of rest or uniform motion. It is quantitatively measured by the mass of the body.

Impulse

The measure of the total effect of a large force acting over a very short time interval. Mathematically, $\text{Impulse} = \vec{F} \times \Delta t = \Delta \vec{p}$ (change in momentum).

Static Friction

A self-adjusting contact force that resists the tendency of relative motion between two surfaces in contact. Its maximum value is called limiting friction ($f_{s,\text{max}} = \mu_s N$).

Banking of Roads

The phenomenon of raising the outer edge of a curved road over the inner edge to provide the necessary centripetal force to vehicles, reducing dependency on friction.

Step-by-Step Guide: Drawing Free Body Diagrams (FBD)

1. Isolate the Object — Choose the specific body under consideration and draw it separately, isolated from its surroundings.
2. Identify All Forces — Identify and draw all external forces acting on the body. This includes non-contact forces (like gravity, $mg$, pointing vertically downward) and contact forces (like normal reaction $N$, tension $T$, and friction $f$).
3. Choose Coordinate Axes — Establish a coordinate system. Align one axis along the direction of actual or potential acceleration to make the mathematical equations simpler to solve.
4. Apply Newton's Second Law — Resolve all vectors into components along your chosen axes. Set up equations: $\Sigma F_x = m a_x$ and $\Sigma F_y = m a_y$. Solve these simultaneous equations to find unknown quantities.

Common Exam Traps & Conceptual Clarifications

1. Normal Force is NOT always equal to $mg$: On an inclined plane of angle $\theta$, the normal force is $mg \cos\theta$. If you are pulling or pushing a block at an angle, the vertical component of your force changes the normal reaction. Always resolve forces vertically to calculate $N$ before applying the friction formula $f = \mu N$!
2. The Self-Adjusting Nature of Static Friction: Static friction is not always equal to $\mu_s N$. That is only its maximum (limiting) value. If a pulling force of $5\text{ N}$ is applied to a block and it doesn't move, the static friction is exactly $5\text{ N}$, even if $\mu_s N$ computes to $20\text{ N}$.
3. Action-Reaction Pair Mistake: Gravity pulling an object down and the normal force pushing it up are NOT an action-reaction pair. They act on the same body. An action-reaction pair always acts on different bodies (e.g., earth pulling the block down vs. the block pulling the earth up).

Practice Questions with Solutions

• Q: A constant force acting on a body of mass $3.0\text{ kg}$ changes its speed from $2.0\text{ m/s}$ to $3.5\text{ m/s}$ in $25\text{ s}$. The direction of motion of the body remains unchanged. What is the magnitude and direction of the force? A: Step 1: Write down the given values. Mass ($m$) = $3.0\text{ kg}$ Initial velocity ($u$) = $2.0\text{ m/s}$ Final velocity ($v$) = $3.5\text{ m/s}$ Time ($t$) = $25\text{ s}$ Step 2: Calculate the acceleration ($a$) using the equation of motion $v = u + at$: $a = \frac{v - u}{t} = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06\text{ m/s}^2$ Step 3: Apply Newton's second law to find force ($F$): $F = m \cdot a = 3.0\text{ kg} \times 0.06\text{ m/s}^2 = 0.18\text{ N}$ Since the speed increases in the direction of motion, the force acts in the direction of motion. Final answer: The magnitude of the force is $0.18\text{ N}$ in the direction of motion.
• Q: A man of mass $70\text{ kg}$ stands on a weighing scale in a lift which is moving (a) upwards with a uniform acceleration of $5\text{ m/s}^2$, and (b) downwards with a uniform acceleration of $5\text{ m/s}^2$. What are the readings on the scale in each case? (Take $g = 10\text{ m/s}^2$) A: Step 1: Understand that the weighing scale measures the normal reaction ($R$). Step 2: Case (a) - Lift moving upwards with acceleration $a = 5\text{ m/s}^2$. Using the equation of motion: $R - mg = ma \implies R = m(g + a)$ $R = 70(10 + 5) = 70 \times 15 = 1050\text{ N}$ Step 3: Case (b) - Lift moving downwards with acceleration $a = 5\text{ m/s}^2$. Using the equation of motion: $mg - R = ma \implies R = m(g - a)$ $R = 70(10 - 5) = 70 \times 5 = 350\text{ N}$ Final answer: (a) $1050\text{ N}$ (or $105\text{ kg}$ reading), (b) $350\text{ N}$ (or $35\text{ kg}$ reading).
• Q: A block of mass $4\text{ kg}$ rests on a rough horizontal plane. The coefficient of static friction between the block and the plane is $0.4$. If a horizontal force of $12\text{ N}$ is applied to the block, find the force of friction acting on the block. A: Step 1: Calculate the normal reaction force ($N$). For a horizontal plane, $N = mg = 4\text{ kg} \times 10\text{ m/s}^2 = 40\text{ N}$. Step 2: Calculate the maximum limiting static friction ($f_{s,\text{max}}$): $f_{s,\text{max}} = \mu_s N = 0.4 \times 40 = 16\text{ N}$ Step 3: Compare the applied horizontal force ($F = 12\text{ N}$) with $f_{s,\text{max}}$. Since the applied force ($12\text{ N}$) is less than the limiting friction ($16\text{ N}$), the block will not move. Step 4: Realize that static friction is self-adjusting. It matches the applied force up to its maximum limit. Therefore, the static friction force acting on the block is equal to the applied force. Final answer: The force of friction is $12\text{ N}$.
• Q: Two masses $m_1 = 5\text{ kg}$ and $m_2 = 10\text{ kg}$ connected by an inextensible string pass over a frictionless pulley. Find the acceleration of the system and the tension in the string when released. (Take $g = 10\text{ m/s}^2$) A: Step 1: Write down the equations of motion for both masses. Since $m_2 > m_1$, $m_2$ moves downwards with acceleration $a$ and $m_1$ moves upwards with acceleration $a$. For $m_1$: $T - m_1 g = m_1 a \implies T - 50 = 5a$ For $m_2$: $m_2 g - T = m_2 a \implies 100 - T = 10a$ Step 2: Add both equations to eliminate Tension ($T$): $(T - 50) + (100 - T) = 5a + 10a \implies 50 = 15a \implies a = \frac{50}{15} = 3.33\text{ m/s}^2$ Step 3: Substitute the value of $a$ into the equation for $m_1$ to find $T$: $T = m_1 g + m_1 a = 50 + 5(3.33) = 50 + 16.67 = 66.67\text{ N}$ Final answer: Acceleration is $3.33\text{ m/s}^2$ and Tension is $66.67\text{ N}$.

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