ANALYSIS OF MOTION FOR JEE 2025 ASPIRANTS

Exhaustive Analysis of Types of Motion for JEE 2025 Aspirants

The concept of motion forms the foundation of Physics for both CBSE and JEE. However, JEE goes beyond CBSE-level understanding by emphasizing problem-solving, critical reasoning, and real-world applications. Here’s an extended and in-depth explanation of types of motion with advanced problem-solving approaches, focusing on concepts relevant for JEE 2025.

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1. Linear Motion (Rectilinear Motion)

1.1 Uniform Motion

• Motion in a straight line with constant velocity.
• Displacement (ss) is proportional to time (tt): s=vts = vt.

1.2 Non-Uniform Motion

• Velocity changes with time due to acceleration (aa).
• Governed by Equations of Motion:
  1. v=u+atv = u + at
  2. s=ut+12at2s = ut + \frac{1}{2}at^2
  3. v2=u2+2asv^2 = u^2 + 2as

Advanced Tips:

• For JEE problems, always check whether acceleration is constant. If not, integrate: v=∫a(t) dtands=∫v(t) dtv = \int a(t) \, dt \quad \text{and} \quad s = \int v(t) \, dt

Sample JEE Numerical:

A car starts from rest and accelerates at a rate of 2 m/s22 \, \text{m/s}^2. It then decelerates uniformly to rest in 10 seconds over a distance of 100 m. Find the maximum velocity and acceleration during deceleration.

Solution:
For the first phase:
Using s=ut+12at2s = ut + \frac{1}{2}at^2:

100=0+12(2)(t2)⇒t2=100⇒t=10 s100 = 0 + \frac{1}{2}(2)(t^2) \Rightarrow t^2 = 100 \Rightarrow t = 10 \, \text{s}

Maximum velocity:

v=u+at=0+(2)(10)=20 m/sv = u + at = 0 + (2)(10) = 20 \, \text{m/s}

For deceleration:

v2=u2+2as⇒0=202+2a(100)⇒a=−2 m/s2v^2 = u^2 + 2as \quad \Rightarrow 0 = 20^2 + 2a(100) \Rightarrow a = -2 \, \text{m/s}^2

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2. Circular Motion

2.1 Uniform Circular Motion (UCM)

• Speed is constant, but velocity changes due to changing direction.
• Centripetal Force: Fc=mv2rF_c = \frac{mv^2}{r}
• Angular Velocity (ω\omega): ω=ΔθΔt\omega = \frac{\Delta \theta}{\Delta t}
• Relation Between Linear and Angular Quantities:
  1. v=rωv = r\omega
  2. ac=rω2a_c = r\omega^2

2.2 Non-Uniform Circular Motion

• Involves angular acceleration (α\alpha), leading to tangential acceleration (at=rαa_t = r\alpha).

Advanced Problem-Solving for JEE:

• Analyze the vector nature of centripetal and tangential accelerations.
• For complex problems involving banked curves, use: tan⁡θ=v2rg\tan\theta = \frac{v^2}{rg}

Sample JEE Numerical:

A car of mass 1000 kg1000 \, \text{kg} moves around a circular track of radius 100 m100 \, \text{m} with a constant speed of 36 km/h36 \, \text{km/h}. Find the centripetal force.

Solution:
Convert speed:

v=36 km/h=10 m/sv = 36 \, \text{km/h} = 10 \, \text{m/s}

Centripetal force:

Fc=mv2r=1000(10)2100=1000 NF_c = \frac{mv^2}{r} = \frac{1000(10)^2}{100} = 1000 \, \text{N}

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3. Periodic Motion

3.1 Simple Harmonic Motion (SHM)

• A specific type of periodic motion where acceleration is proportional to displacement and directed towards the mean position.
• Equation of motion: a=−ω2xa = -\omega^2x
• General Solution: x(t)=Asin⁡(ωt+ϕ)x(t) = A\sin(\omega t + \phi) Where:
  • AA: Amplitude
  • ω\omega: Angular frequency (ω=km)(\omega = \sqrt{\frac{k}{m}})
  • ϕ\phi: Phase constant

Energy in SHM:

• Total Energy (EE): 12kA2\frac{1}{2}kA^2
• Kinetic Energy (KEKE): 12k(A2−x2)\frac{1}{2}k(A^2 – x^2)
• Potential Energy (PEPE): 12kx2\frac{1}{2}kx^2

Sample JEE Numerical:

A block of mass 2 kg2 \, \text{kg} attached to a spring of spring constant 200 N/m200 \, \text{N/m} oscillates with an amplitude of 0.1 m0.1 \, \text{m}. Find the time period and total energy.

Solution:
Time period:

T=2πmk=2π2200=0.628 sT = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{2}{200}} = 0.628 \, \text{s}

Total energy:

E=12kA2=12(200)(0.1)2=1 JE = \frac{1}{2}kA^2 = \frac{1}{2}(200)(0.1)^2 = 1 \, \text{J}

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4. Projectile Motion

Key Parameters:

1. Horizontal Range (RR):
   R=u2sin⁡2θgR = \frac{u^2\sin2\theta}{g}
2. Time of Flight (TT):
   T=2usin⁡θgT = \frac{2u\sin\theta}{g}
3. Maximum Height (HH):
   H=u2sin⁡2θ2gH = \frac{u^2\sin^2\theta}{2g}

Advanced JEE Tip:

• For inclined plane problems, resolve components of velocity and acceleration along the incline.

Sample JEE Numerical:

A ball is projected with a velocity of 20 m/s20 \, \text{m/s} at an angle of 30∘30^\circ. Find the range and maximum height.

Solution:
Horizontal range:

R=u2sin⁡2θg=(20)2sin⁡60∘10=34.6 mR = \frac{u^2\sin2\theta}{g} = \frac{(20)^2\sin60^\circ}{10} = 34.6 \, \text{m}

Maximum height:

H=u2sin⁡2θ2g=(20)2(sin⁡30∘)22(10)=5 mH = \frac{u^2\sin^2\theta}{2g} = \frac{(20)^2(\sin30^\circ)^2}{2(10)} = 5 \, \text{m}

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Key Preparation Tips for JEE:

1. Master Fundamental Concepts:
   • Focus on vector analysis and calculus applications in motion.
   • Study graphs of motion, especially v−tv-t and s−ts-t curves.
2. Practice Problem-Solving:
   • Solve previous 10 years’ JEE Advanced and Main question papers.
   • Focus on mixed-concept problems (e.g., combining circular motion with SHM).
3. Use Resources:
   • Refer to standard books like H.C. Verma, D.C. Pandey, and I.E. Irodov.
   • Leverage online tools like video lectures and mock tests.
4. Mock Test Strategy:
   • Allocate time efficiently between theoretical and numerical questions.
   • Analyze mistakes to identify weak areas.

By following this exhaustive guide, you can ensure a deep understanding of motion and enhance your problem-solving skills to ace JEE 2025!

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