Guide to incorporate numerical problems from the past 10 years of JEE exams related to the topic of motion

Extended Guide: Types of Motion with Past JEE Numerical Problems

1. Linear Motion

Key Concepts Tested in JEE:

1. Equations of motion under uniform and non-uniform acceleration.
2. Graphical interpretation of motion (displacement-time, velocity-time graphs).
3. Variable acceleration problems (integration-based approach).

Example 1: JEE Main 2019

Problem:
A particle starts from rest and moves with an acceleration a=2ta = 2t. Find the displacement of the particle in the first 3 seconds.

Solution:
Acceleration: a=2ta = 2t.
Velocity: v=∫a dt=∫2t dt=t2+C.v = \int a \, dt = \int 2t \, dt = t^2 + C.

Since v=0v = 0 at t=0t = 0, C=0C = 0, so v=t2v = t^2.
Displacement: s=∫v dt=∫t2 dt=t33+C.s = \int v \, dt = \int t^2 \, dt = \frac{t^3}{3} + C.

Again, s=0s = 0 at t=0t = 0, so C=0C = 0.
At t=3t = 3, s=333=9 ms = \frac{3^3}{3} = 9 \, \text{m}.

Example 2: JEE Advanced 2016

Problem:
A car accelerates from rest at a constant rate α\alpha for some time and then decelerates at a constant rate β\beta to come to rest. If the total time for the journey is TT, find the maximum velocity attained.

Solution:
Time for acceleration: t1t_1.
Time for deceleration: t2=T−t1t_2 = T – t_1.
Using symmetry of motion: vmax=αt1=β(T−t1).v_{\text{max}} = \alpha t_1 = \beta (T – t_1).

Simplifying: t1=βTα+β.t_1 = \frac{\beta T}{\alpha + \beta}.

Substitute t1t_1: vmax=α⋅βTα+β=αβTα+β.v_{\text{max}} = \alpha \cdot \frac{\beta T}{\alpha + \beta} = \frac{\alpha \beta T}{\alpha + \beta}.

2. Circular Motion

Key Concepts Tested in JEE:

1. Centripetal acceleration and force.
2. Banking of roads.
3. Non-uniform circular motion with tangential acceleration.

Example 3: JEE Main 2020

Problem:
A particle is moving in a circular path of radius 5 m5 \, \text{m} with a constant speed of 10 m/s10 \, \text{m/s}. Calculate the centripetal force acting on the particle if its mass is 2 kg2 \, \text{kg}.

Solution:
Centripetal force: Fc=mv2r=2(10)25=40 N.F_c = \frac{mv^2}{r} = \frac{2(10)^2}{5} = 40 \, \text{N}.

Example 4: JEE Advanced 2018

Problem:
A car of mass mm moves on a banked road of angle θ\theta with speed vv. The coefficient of friction is μ\mu. Derive the condition for the car to avoid slipping.

Solution:
For no slipping: μmg+mv2rcos⁡θ=mgsin⁡θ.\mu mg + \frac{mv^2}{r}\cos\theta = mg\sin\theta.

Rearranging: v2=r⋅g(sin⁡θ+μcos⁡θ)/(cos⁡θ−μsin⁡θ).v^2 = r \cdot g (\sin\theta + \mu \cos\theta) / (\cos\theta – \mu \sin\theta).

3. Simple Harmonic Motion (SHM)

Key Concepts Tested in JEE:

1. Energy conservation in SHM.
2. Equation of motion in terms of displacement, velocity, and acceleration.
3. Oscillations in combined systems (e.g., springs and pendulums).

Example 5: JEE Advanced 2015

Problem:
A block of mass 1 kg1 \, \text{kg} is attached to a spring of spring constant 200 N/m200 \, \text{N/m}. The block is displaced by 0.1 m0.1 \, \text{m} from its equilibrium position and released. Find the time period and maximum speed of the block.

Solution:
Time period: T=2πmk=2π1200=0.44 s.T = 2\pi\sqrt{\frac{m}{k}} = 2\pi\sqrt{\frac{1}{200}} = 0.44 \, \text{s}.

Maximum speed: vmax=ωA=km⋅A=2001⋅0.1=1.41 m/s.v_{\text{max}} = \omega A = \sqrt{\frac{k}{m}} \cdot A = \sqrt{\frac{200}{1}} \cdot 0.1 = 1.41 \, \text{m/s}.

Example 6: JEE Main 2017

Problem:
Two springs with spring constants k1=100 N/mk_1 = 100 \, \text{N/m} and k2=200 N/mk_2 = 200 \, \text{N/m} are connected in series. A block of mass 2 kg2 \, \text{kg} is attached. Find the effective time period of oscillation.

Solution:
Effective spring constant: 1keff=1k1+1k2=1100+1200=3200.\frac{1}{k_{\text{eff}}} = \frac{1}{k_1} + \frac{1}{k_2} = \frac{1}{100} + \frac{1}{200} = \frac{3}{200}. keff=2003.k_{\text{eff}} = \frac{200}{3}.

Time period: T=2πmkeff=2π2200/3=1.54 s.T = 2\pi\sqrt{\frac{m}{k_{\text{eff}}}} = 2\pi\sqrt{\frac{2}{200/3}} = 1.54 \, \text{s}.

4. Projectile Motion

Key Concepts Tested in JEE:

1. Maximum height, time of flight, and range of a projectile.
2. Projectile on an inclined plane.
3. Relative motion and velocity.

Example 7: JEE Main 2018

Problem:
A ball is thrown with velocity 20 m/s20 \, \text{m/s} at an angle of 45∘45^\circ. Calculate the range and maximum height.

Solution:
Horizontal range: R=u2sin⁡2θg=(20)2sin⁡90∘10=40 m.R = \frac{u^2\sin2\theta}{g} = \frac{(20)^2\sin90^\circ}{10} = 40 \, \text{m}.

Maximum height: H=u2sin⁡2θ2g=(20)2(sin⁡45∘)22(10)=10 m.H = \frac{u^2\sin^2\theta}{2g} = \frac{(20)^2(\sin45^\circ)^2}{2(10)} = 10 \, \text{m}.

Summary of Observations:

1. Graphical Interpretation: Many JEE questions use velocity-time and acceleration-time graphs to test analytical skills.
2. Mixed-Concept Problems: Advanced-level questions often combine multiple types of motion, such as projectile motion with relative velocity.
3. Energy-Based Approaches: Problems on SHM and circular motion frequently leverage energy conservation principles.
4. Complex Numerical Approaches: Advanced problems in JEE often require integration or differentiation for non-uniform motion.

By practicing these problems, aspirants can develop a robust understanding of motion and ensure their preparation aligns with JEE’s high standards.