Laws of Motion

CBSE Class 11 Physics â€¢ Chapter 04

Chapter Overview

This chapter covers the fundamental concepts of Laws of Motion. Ensure you understand the definitions and derivations thoroughly.

04.1 Newton's Laws of Motion

Inertia: Resistance to change in state of motion (Mass is measure of inertia).
Momentum ($p$): Quantity of motion. $\vec{p} = m\vec{v}$. Vector.
Impulse ($J$): Large force acting for short time. $\vec{J} = \vec{F}_{avg} \Delta t = \Delta \vec{p}$.

The Three Laws:

First Law (Law of Inertia): Object remains at rest or uniform motion unless acted upon by external force.
Second Law: Rate of change of momentum is proportional to applied force. $F \propto \frac{dp}{dt} \Rightarrow F = ma$.
Third Law: To every action, there is an equal and opposite reaction. $F_{AB} = -F_{BA}$.

04.7 Conservation of Momentum

Statement: If no external force acts on a system ($F_{ext}=0$), total momentum remains constant.

$\frac{dp}{dt} = 0 \Rightarrow p = \text{constant} \Rightarrow m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$.

Recoil of Gun:

$$ V_G = -\frac{m_b v_b}{M_G} $$

Negative sign indicates gun moves opposite to bullet.

Practice Problem 1

Q: A 50g bullet is fired from a 2kg gun with muzzle speed 400 m/s. Find recoil speed of gun.

Ans: $m_b = 0.05 \text{ kg}, v_b = 400 \text{ m/s}, M_G = 2 \text{ kg}$.
By Conservation of Momentum: $P_i = 0, P_f = m_b v_b + M_G V_G = 0$.
$V_G = -\frac{0.05 \times 400}{2} = -\frac{20}{2} = \mathbf{-10 \text{ m/s}}$.

04.8 Equilibrium & Friction

Equilibrium: Vector sum of all forces is zero. $\sum \vec{F} = 0$.
Static Friction ($f_s$): Self-adjusting force opposing impending motion. $f_s \le \mu_s N$.
Kinetic Friction ($f_k$): Opposes actual relative motion. $f_k = \mu_k N$. ($\mu_k < \mu_s$).

Figure 4.1: Forces on an Inclined Plane

04.1 Force and Inertia

Force: Push or pull that changes or tends to change state of rest or motion. Vector quantity.

Inertia: Inherent property of mass to resist change in state of motion.

Types:

Inertia of Rest: Tendency to remain at rest
Inertia of Motion: Tendency to keep moving
Inertia of Direction: Tendency to keep moving in same direction

Mass is measure of inertia - Greater mass â†’ Greater inertia

04.2 Newton's Laws of Motion

Newton's First Law (Law of Inertia)

Statement: A body continues in its state of rest or uniform motion in a straight line unless compelled by an external force.

Implications:

Defines inertial frame of reference
Force is needed only to change motion, not maintain it
Natural state is uniform motion (not rest)

Mathematical Form: If $\sum \vec{F} = 0$, then $\vec{v}$ = constant (or $\vec{a} = 0$)

Newton's Second Law

Statement: Rate of change of momentum is directly proportional to applied force and takes place in direction of force.

Derivation:

$\vec{F} \propto \frac{d\vec{p}}{dt}$

$\vec{F} = k\frac{d\vec{p}}{dt}$

where k = 1 (in SI units)

$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt}$

For constant mass:

$$ \vec{F} = m\frac{d\vec{v}}{dt} = m\vec{a} $$

SI Unit: Newton (N) = kgâ‹…m/sÂ²

Definition of 1 Newton: Force that produces acceleration of 1 m/sÂ² in 1 kg mass

Newton's Third Law

Statement: To every action, there is equal and opposite reaction.

Mathematical Form: $\vec{F}_{AB} = -\vec{F}_{BA}$

Important Points:

Action-reaction act on different bodies
They act simultaneously (not one after another)
Same magnitude, opposite direction, same type
They never cancel (act on different objects)

Worked Example 1

Q: A 5 kg block accelerates at 2 m/sÂ². Find net force.

Solution: $F = ma = 5 \times 2 = \mathbf{10 \text{ N}}$

04.3 Momentum and Impulse

Linear Momentum: $\vec{p} = m\vec{v}$

Vector quantity, SI unit: kgâ‹…m/s

Impulse: Change in momentum

$\vec{J} = \Delta \vec{p} = \vec{F} \cdot \Delta t$

$$ \vec{J} = \int \vec{F} dt $$

SI unit: Nâ‹…s = kgâ‹…m/s

Impulse-Momentum Theorem:

From $\vec{F} = \frac{d\vec{p}}{dt}$:

$\vec{F} dt = d\vec{p}$

Integrating: $\int_{t_1}^{t_2} \vec{F} dt = \int_{\vec{p_i}}^{\vec{p_f}} d\vec{p}$

$$ \vec{J} = \vec{p_f} - \vec{p_i} = \Delta \vec{p} $$

04.4 Conservation of Momentum

Statement: If no external force acts on a system, total momentum remains constant.

Derivation:

For two-body system: $\vec{F}_{12} = -\vec{F}_{21}$ (Newton's 3rd law)

$m_1\vec{a_1} = -m_2\vec{a_2}$

$m_1\frac{d\vec{v_1}}{dt} = -m_2\frac{d\vec{v_2}}{dt}$

$\frac{d}{dt}(m_1\vec{v_1} + m_2\vec{v_2}) = 0$

$$ m_1\vec{v_1} + m_2\vec{v_2} = \text{constant} $$

Worked Example 2 - Recoil

Q: Gun (5 kg) fires bullet (50 g) at 400 m/s. Find recoil speed.

Solution: Conservation of momentum:
$m_g v_g + m_b v_b = 0$ (initially at rest)
$5 \times v_g + 0.05 \times 400 = 0$
$v_g = \mathbf{-4 \text{ m/s}}$ (backward)

04.5 Friction

Friction: Force that opposes relative motion between surfaces in contact.

Types:

Static Friction (f_s): Prevents motion, 0 â‰¤ f_s â‰¤ Î¼_sN
Kinetic Friction (f_k): Opposes motion, f_k = Î¼_kN

where Î¼ = coefficient of friction, N = normal force

Note: Î¼_s > Î¼_k (harder to start than keep moving)

Worked Example 3

Q: Block (10 kg) on floor, Î¼_s = 0.5, Î¼_k = 0.3. Applied force = 40 N. Find friction and acceleration.

Solution:
N = mg = 10 Ã— 10 = 100 N
Max static friction = 0.5 Ã— 100 = 50 N
Since 40 N < 50 N, block doesn't move
Friction = 40 N (static), a = 0

04.6 Circular Motion Dynamics

Centripetal Force: Force towards center for circular motion

$$ F_c = \frac{mv^2}{r} = m\omega^2 r $$

Banking of Roads:

For safe turn without friction: $\tan\theta = \frac{v^2}{rg}$

Optimum speed: $v = \sqrt{rg\tan\theta}$

Worked Example 4

Q: Car (1000 kg) turns on banked road (r=50m, Î¸=30Â°). Find safe speed (g=10 m/sÂ²).

Solution:
$v = \sqrt{rg\tan\theta} = \sqrt{50 \times 10 \times \tan30Â°}$
$= \sqrt{500 \times 0.577} = \sqrt{288.5} \approx \mathbf{17 \text{ m/s}}$

Study Tip: Master the derivations in this chapter as they are frequently asked in exams. Practice numericals based on the formulas.