PHALIN THAKKAR · LEARNING MATERIAL

Physics practice sheet

Begin with a diagram or graph. State the conditions before choosing an equation.

Foundation establishes the diagram, signs and calculation. Core practice combines ideas and conditions. Advanced extension adds parameters, stages, limiting cases or calculus. These are selected resources for JEE preparation, not a complete syllabus or an official difficulty classification.

Use g = 10 m/s² throughout. Start with the foundation lesson · Worked solutions

Foundation

P1. Resolve the force

Prerequisites: Force components and vertical equilibrium

A 4 kg block is pulled by a 20 N force at 30° above a smooth horizontal surface. Find Fx, Fy, N and ax. Draw and label the directions of all actual forces.

My attempt for P1

Hint 1

Resolve the pull before writing either equation.

Hint 2

While contact remains, vertical acceleration is zero.

Worked solution for P1

Foundation

P2. Include kinetic friction

Prerequisites: P1; kinetic friction

The same block is already sliding right. The coefficient of kinetic friction is 0.25. Find the friction force and horizontal acceleration. Explain why N is not 40 N.

My attempt for P2

Hint 1

Friction depends on the normal force.

Hint 2

Use the normal force from P1.

Worked solution for P2

Core practice

P3. Check whether motion can begin

Prerequisites: P1 and P2; static friction as an inequality

A 4 kg block is initially at rest. A variable force F acts at 30° above horizontal. The surface has μs = 0.50 and μk = 0.25.

(a) Derive the limiting force for the onset of rightward motion.

(b) At F = 12 N, find the actual static friction.

(d) Find the force at which the normal reaction first becomes zero.

My attempt for P3

Hint 1

Static friction can be smaller than μsN.

Hint 2

At limiting equilibrium, F cos30° = μs(mg − F sin30°).

Worked solution for P3

Core practice

P4. Two different optimal angles

Prerequisites: P3; trigonometric identities and optimisation

Use a 4 kg block.

(a) With μs = 0.50, find the pulling angle that minimises the force needed to initiate rightward motion. Find that minimum force.

(b) The block is already sliding right. With F = 20 N and μk = 0.25, find the angle that maximises its instantaneous horizontal acceleration.

Take 0° ≤ θ < 90° and verify that contact is retained at each optimum.

My attempt for P4

Hint 1

Rewrite cosθ + μ sinθ as a single shifted cosine.

Hint 2

Optimising the threshold force and optimising sliding acceleration use different coefficients.

Worked solution for P4

Core practice

P5. When the surface stops supporting the block

Prerequisites: P1; unilateral contact and Newton’s second law

A 2 kg block is initially on a smooth horizontal surface. A 30 N force is applied at angle θ above horizontal.

(a) Find the angle at which N first becomes zero.

(b) At θ = 60°, find ax, ay and N immediately after release.

My attempt for P5

Hint 1

A passive surface cannot pull the block downward.

Hint 2

After detachment, resolve the applied force and gravity without a normal force.

Worked solution for P5

Core practice

P6. Two blocks and a slipping condition

Prerequisites: P2 and P3; separate free body diagrams

A 1 kg block rests on a 3 kg block. The lower block slides right on a horizontal floor with μk = 0.20. The upper contact has μs = 0.50. Both blocks initially have the same rightward velocity.

A horizontal force F is applied to the lower block, with F ≥ 8 N.

(a) For F = 24 N, find their common acceleration and the friction on the upper block.

(b) Find the largest F for which they can continue moving together without relative slipping.

(c) If F = 32 N and the kinetic coefficient between the blocks is 0.30, find each block’s acceleration after relative sliding starts.

My attempt for P6

Hint 1

The floor supports the weight of both blocks.

Hint 2

The upper block is accelerated only by the friction at its lower face.

Worked solution for P6

Advanced extension

P7. A spring with friction

Prerequisites: Work and energy; spring force; static and kinetic friction

A 1 kg block is attached to a horizontal spring with k = 50 N/m. Let x = 0 be the spring’s relaxed position, with right positive. The block is released from rest at x = −0.40 m.

The surface has μs = 0.30 and μk = 0.20.

(a) Check that the block begins moving.

(b) Find its first turning point.

(d) Decide whether it remains at the first turning point.

(e) Find its eventual resting position for this ideal model.

My attempt for P7

Hint 1

Friction dissipates energy over the total distance travelled.

Hint 2

A turning point has zero speed; staying there also requires a static friction check.

Worked solution for P7

Advanced extension

P8. Energy and contact in a vertical loop

Prerequisites: Work and energy; radial acceleration; contact

A small particle of mass m is released from rest on a smooth ramp, at height H above a horizontal track. It then crosses a rough horizontal section of length L = 1.00 m with μk = 0.20 and enters the bottom of a smooth vertical circular loop of radius R = 0.50 m. It moves on the inside of the loop.

(a) Find the minimum H needed to complete the loop while maintaining contact.

(b) At that limiting H, find the normal reaction at the bottom.

(c) If H = 1.20 m, find the angle at which contact is lost and the speed then. Measure θ from the downward radius, so θ = 0 at the bottom.

Then select every correct statement:

A. The minimum release height is 1.45 m.

B. At that limiting height, the bottom reaction is 6mg.

C. At H = 1.20 m, the particle reverses along the track before losing contact.

D. At H = 1.20 m, contact is lost when cosθ = −2/3.

My attempt for P8

Hint 1

Reaching the top by energy alone does not ensure contact.

Hint 2

Combine the energy equation with N = m(v²/R + g cosθ).

Worked solution for P8

Attempts are not saved. Print your sheet to keep your work.

Continue with past JEE Advanced questions

These are independently written reasoning notes, not official worked solutions. Open the official papers for the original questions and diagrams. The 2026 selections agree with the published final answers.

2023 Paper 1 · Physics Q2: Polarisation and a prism

This combines Brewster’s angle, prism geometry and minimum deviation. For the blue ray, tan i = √3 gives i = 60° and Snell’s law gives r₁ = 30°. The stated deviation implies e − r₂ = 30°. Combining this with sin e = √3 sin r₂ gives r₂ = 30°, e = 60° and A = 60°.

n_{\rm red}=\frac{\sin[(60^\circ+30^\circ)/2]}{\sin30^\circ}=\sqrt2.

The incident blue light is polarized in the plane of incidence. Correct choices: A, C and D. A familiar polarisation idea now depends on several optical conditions.

Official 2023 Paper 1

2024 Paper 1 · Physics Q3: Constrained small oscillations

For a small angular displacement δ from the opposite point on the hoop, expand about stable equilibrium:

U=\frac{q^2}{8\pi\epsilon_0R\cos(\delta/2)}\approx U_0+\frac{q^2\delta^2}{64\pi\epsilon_0R}.

With kinetic energy ½mR²δ̇²,

\omega^2=\frac{q^2}{32\pi\epsilon_0mR^3}.

This gives option B. The deciding step is the potential expansion.

Official 2024 Paper 1

2025 Paper 1 · Physics Q5: Induction during partial entry

During partial entry into the magnetic region:

\frac{dv}{dt}=-Kv,\quad v=v_0e^{-Kt},\quad y=\frac{v_0}{K}(1-e^{-Kt}).

Complete entry requires v₀ > KL and occurs at

t=\frac1K\ln\frac{v_0}{v_0-KL}.

For v₀ = 3KL this gives ln(3/2)/K. Once fully inside, the net magnetic force is zero. During partial entry the velocity does not reach zero at a finite time. Correct choices: B and D.

Official 2025 Paper 1

2026 Paper 1 · Physics Q3: Loss of contact

While the rolling cylinder remains in contact with the corner,

v^2=v_0^2+\frac43gR(1-\cos\theta).

At detachment v² = gR cosθ. With v₀² = gR/3 this gives cosθ = 5/7 and v² = 5gR/7. Final answer: B. Energy and the contact condition must both be satisfied.

Official 2026 Paper 1 and final answers

The final answer document was verified as JEE (Advanced) 2026 Paper 1 on 10 October 2026. Its generic URL may be reused for a later year. Official paper archive.