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CBSE • कक्षा XI • Physics • अध्याय 4
अनुमानित समय: 45 Mins
प्रगति: अध्ययनरत

गति के नियम (Laws of Motion)

In Class 11 Physics, "Laws of Motion" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

🚀 Have You Ever Wondered?

Why do rocket engines fire roaring exhaust gases downward to blast into the silence of space, or how does a racing car bank along a high-speed curve w...

Why do rocket engines fire roaring exhaust gases downward to blast into the silence of space, or how does a racing car bank along a high-speed curve without skidding off the asphalt? Newton's laws of motion and friction govern dynamic equilibrium.

यह अध्याय क्यों महत्वपूर्ण है

In Class 11 Physics, "Laws of Motion" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

अध्ययन से पूर्व (आवश्यक ज्ञान)

  • Inertia and Newton's three laws from Class 9.
  • Momentum and force.
  • Frictional force.

इस अध्याय के लक्ष्य

  • Explain Newton's First Law (Inertia), Second Law ($F = \frac{dp}{dt} = ma$), and Third Law ($F_{AB} = -F_{BA}$).
  • Define Impulse ($J = F \Delta t = \Delta p$) and state the Law of Conservation of Linear Momentum.
  • Construct Free Body Diagrams (FBD) for connected bodies and pulleys.
  • Analyze Friction: Static friction ($f_s \le \mu_s N$), Limiting friction, and Kinetic friction ($f_k = \mu_k N$).
  • Analyze Circular Dynamics: Centripetal force and optimum banking angle of curved roads ($\tan\theta = \frac{v^2}{rg}$).

अध्याय रूपरेखा एवं प्रगति

1 1. Newton's Second Law & Momentum
2 2. Conservation of Momentum & FBDs
3 3. Friction & Banking of Roads

सम्पूर्ण सैद्धांतिक एवं वैचारिक अध्ययन

1. Newton's Second Law & Momentum

Linear Momentum is $\mathbf{p = mv}$. Newton's Second Law states that the rate of change of momentum is directly proportional to applied force: $$\mathbf{\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\vec{a}} \quad (\text{when mass } m \text{ is constant})$$ Impulse ($J$): $\mathbf{\vec{J} = \int \vec{F}\, dt = \Delta\vec{p}}$ (cricketer pulling hands back to cushion a catch).

2. Conservation of Momentum & FBDs

In an isolated system with zero net external force ($\vec{F}_{ext} = 0$), total momentum is conserved: $\mathbf{\vec{p}_{initial} = \vec{p}_{final}}$ (recoil of a gun: $v_{gun} = -\frac{m_{bullet}}{M_{gun}} v_{bullet}$). Free Body Diagrams (FBD) isolate bodies to sum normal forces, gravity, and tensions.

3. Friction & Banking of Roads

  • Friction: Self-adjusting static friction $f_s \le \mu_s N$. Once motion starts, kinetic friction $f_k = \mu_k N$ operates (where $\mu_k < \mu_s$).
  • Banking of Curved Roads: To prevent vehicles from skidding at high speeds, curves are banked at angle $\theta$ such that horizontal normal component supplies centripetal force: $$\mathbf{\tan\theta = \frac{v^2}{rg}} \implies \mathbf{v_{\text{optimum}} = \sqrt{rg\tan\theta}}$$

चित्रात्मक व्याख्या एवं मॉडल

Laws of Motion Master Matrix

Conceptual framework, core mechanisms, and analytical relationships
Academic Architecture

1. Newton's Second Law & Momentum • 2. Conservation of Momentum & FBDs

अध्याय का सार संक्षेप एवं 10 मुख्य निष्कर्ष

मुख्य बिंदु 1
Newton's Second Law: $F = dp/dt$; genuine fundamental law from which first and third can be derived.
मुख्य बिंदु 2
Impulse-Momentum Theorem: Area under force-time curve equals total momentum transfer.
मुख्य बिंदु 3
Free Body Diagram: Engineering schematic isolating all forces acting on a mechanical component.
मुख्य बिंदु 4
Friction Hierarchy: Static friction is self-adjusting; limiting friction is maximum; kinetic is lower.
मुख्य बिंदु 5
Optimum Banking: Tilting curved roadways so normal reaction provides required centripetal acceleration.

स्व-मूल्यांकन अभ्यास (Check Your Understanding)

मूल वैचारिक स्पष्टता की जांच के लिए नैदानिक प्रश्न। पहले स्वयं हल करें, फिर उत्तर देखें।

1
A bullet of mass 0.04 kg moving with a speed of $90\text{ m/s}$ enters a heavy wooden block and is stopped after a distance of $60\text{ cm}$. What is the average resistive force exerted by the block on the bullet?
उत्तर एवं व्याख्या देखें
उत्तर: $u = 90\text{ m/s}, v = 0, s = 0.6\text{ m}$. $v^2 = u^2 + 2as \implies 0 = 90^2 + 2a(0.6) \implies a = -\frac{8100}{1.2} = -6750\text{ m/s}^2$. Resistive force $F = ma = 0.04 \times 6750 = 270\text{ Newtons}$.
270 N.
2
Explain why a cricket fielder pulls his hands backwards while catching a fast-moving cricket ball.
उत्तर एवं व्याख्या देखें
उत्तर: By pulling hands back, the fielder increases the time interval $\Delta t$ over which the ball's momentum is reduced to zero. Since $F = \frac{\Delta p}{\Delta t}$, increasing time drastically reduces the impact force on the hands.
Increases time of impact, reducing stopping force.
3
A constant retarding force of $50\text{ N}$ is applied to a body of mass $20\text{ kg}$ moving initially with a speed of $15\text{ m/s}$. How long does the body take to stop?
उत्तर एवं व्याख्या देखें
उत्तर: Acceleration $a = -F/m = -50/20 = -2.5\text{ m/s}^2$. $v = u + at \implies 0 = 15 - 2.5t \implies t = \frac{15}{2.5} = 6\text{ seconds}$.
6 seconds.
4
State the laws of limiting friction.
उत्तर एवं व्याख्या देखें
उत्तर: (1) The magnitude of limiting friction is directly proportional to the normal reaction ($f_s = \mu_s N$), (2) The direction of friction opposes impending motion, (3) Friction is independent of apparent surface contact area, (4) Friction depends on the nature and roughness of materials in contact.
Proportional to normal reaction; opposes motion; independent of area.
5
Why is it easier to pull a heavy roller than to push it?
उत्तर एवं व्याख्या देखें
उत्तर: When pushing, the downward component of force increases the normal reaction ($N = mg + F\sin\theta$), increasing friction; when pulling, the upward component decreases the normal reaction ($N = mg - F\sin\theta$), significantly reducing friction.
Pulling decreases normal reaction and friction; pushing increases it.
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गति के नियम (Laws of Motion) में कोई संदेह या प्रश्न है? हमारे AI अध्ययन मित्र से तुरंत समझें।