Stopping of Two Blocks – Complete Notes, Revision, Important Questions & Downloads
Stopping of Two Blocks covers the TOC subtopic Ratio of Stopping Distances. Two blocks of masses m₁ and m₂ are compressed against each other (spring or direct contact) and suddenly released. By conservation of momentum (total initial momentum = 0), they fly apart with velocities u₁ and u₂ satisfying m₁u₁ = m₂u₂. On the same rough surface (same μ), each block decelerates at μg. Stopping distance S ∝ u² ∝ 1/m². Therefore S₁/S₂ = (m₂/m₁)². NEET tests the ratio S₁/S₂ = (m₂/m₁)². Key physics: heavier block moves slower after release (less KE for same momentum) and stops in shorter distance.
NEET Weightage — Stopping of Two Blocks
Friction (Chapter 5)| NEET Year | Questions from this Topic | Bar | Marks |
|---|---|---|---|
| 2024 | 0 | 0 | |
| 2023 | 1 | 4 | |
| 2022 | 0 | 0 | |
| 2021 | 1 | 4 | |
| 2020 | 0 | 0 | |
| 2019 | 0 | 0 | |
| 6-Year Total (2019–2024) | 0–2 | 0–8 |
STOPPING DISTANCES: S = u²/(2μg). S₁ = u₁²/(2μg). S₂ = u₂²/(2μg). Ratio: S₁/S₂ = u₁²/u₂² = (u₁/u₂)² = (m₂/m₁)². INTERPRETATION: Heavier block (say m₁ > m₂) has smaller velocity (u₁ < u₂) → smaller stopping distance (S₁ < S₂). Lighter block travels further. The ratio of stopping distances equals the square of the inverse mass ratio.
STOPPING TIMES: t = u/(μg). t₁/t₂ = u₁/u₂ = m₂/m₁. (Linear ratio, not squared.) KE RATIO: KE₁/KE₂ = (½m₁u₁²)/(½m₂u₂²) = (m₁/m₂)×(u₁/u₂)² = (m₁/m₂)×(m₂/m₁)² = m₂/m₁. So lighter block has MORE kinetic energy; the ratio KE₁/KE₂ = m₂/m₁ (lighter block has KE proportional to inverse mass).
How to Solve the Two-Blocks-Released Problem
Step 1 — Apply conservation of momentum Both initially at rest: m₁u₁ = m₂u₂ (magnitude). Velocity ratio: u₁/u₂ = m₂/m₁.
Step 2 — Write stopping distances Deceleration same for both: a = μg. S₁ = u₁²/(2μg); S₂ = u₂²/(2μg).
Step 3 — Find the ratio S₁/S₂ = (u₁/u₂)² = (m₂/m₁)². Time ratio: t₁/t₂ = u₁/u₂ = m₂/m₁. KE ratio: KE₁/KE₂ = m₂/m₁.
Study Materials — Stopping of Two Blocks
PDF · Cheat Sheet · MCQ Set · PYQSubtopics — Stopping of Two Blocks
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Rapid Revision — Stopping of Two Blocks
Concept → Trap → Example1) Ratio of Stopping Distances — Two Blocks Released from Compression
Ratio of Stopping DistancesSETUP: Blocks of masses m₁ and m₂ are initially at rest, compressed together (e.g., a spring between them). When released, they fly apart in opposite directions. Rough floor with same friction coefficient μ for both. Both blocks eventually stop. Find stopping distances S₁ and S₂. STEP 1 — Momentum conservation: Before release: total momentum = 0. After release: m₁v₁ + m₂v₂ = 0 (vector). So m₁u₁ = m₂u₂ (magnitudes, opposite directions). Velocity ratio: u₁/u₂ = m₂/m₁. STEP 2 — Stopping distance: Each block decelerates entirely due to kinetic friction. Deceleration of each: a = μg (friction force = μmg, divided by mass m → μg, mass-independent). Using v²=u²-2aS with v=0: S = u²/(2μg). STEP 3 — Ratio: S₁/S₂ = (u₁²/2μg)/(u₂²/2μg) = (u₁/u₂)² = (m₂/m₁)². FINAL RESULT: S₁/S₂ = (m₂/m₁)². The ratio is the square of the INVERSE mass ratio. KEY ADDITIONAL RESULTS: Stopping time ratio: t₁/t₂ = u₁/(μg) ÷ u₂/(μg) = u₁/u₂ = m₂/m₁. KE ratio: KE₁/KE₂ = ½m₁u₁²/½m₂u₂² = (m₁/m₂)(u₁/u₂)² = (m₁/m₂)(m₂/m₁)² = m₂/m₁.
- PHYSICAL INSIGHT: When two blocks are released from compression, they share the same magnitude of momentum (p = m₁u₁ = m₂u₂). The lighter block moves faster (v ∝ 1/m for fixed p). Since stopping distance S ∝ v² ∝ 1/m², the lighter block travels MUCH farther. Specifically S₁/S₂ = (m₂/m₁)². If m₁ = m₂: S₁ = S₂ (both stop at same distance from release point — symmetric). If m₁ < m₂: S₁ > S₂ (lighter block travels farther).
- ENERGY: When two masses compressed towards each other and suddenly released then energy acquired by each block will be dissipated against friction and finally block comes to rest. The energy stored in the spring (or compression) splits between the two blocks. KE₁ = ½m₁u₁² = p²/(2m₁) where p = m₁u₁. KE₂ = p²/(2m₂). So KE₁/KE₂ = m₂/m₁ — lighter block gets more kinetic energy (and travels farther). Total KE = p²/2 × (1/m₁ + 1/m₂).
- NEET TRAP: The stopping distance ratio is (m₂/m₁)² (squared), not m₂/m₁ (linear). Students often write S₁/S₂ = m₂/m₁ (which is the time ratio, not the distance ratio). The squared ratio comes from S ∝ u² and u ∝ 1/m. Doubling the mass ratio doubles the velocity ratio but quadruples the distance ratio. Always remember: S₁/S₂ = (m₂/m₁)² and t₁/t₂ = m₂/m₁ are DIFFERENT.
US Curriculum Gaps — Stopping of Two Blocks
Topics in this section are in NEET but may be framed differently in US physics courses.Momentum-Friction Combined Problems in AP Physics 1
AP Physics 1 tests momentum conservation (explosions, collisions) and friction (stopping distance) in separate units. The combined problem — two blocks released from compression, then stopped by friction on a rough surface → compare stopping distances — is not a standard AP Physics 1 question type. NRI students need to practise the two-step derivation: momentum conservation first, then friction kinematics.
- AP Physics 1: momentum and friction tested separately; combined explosion-then-friction not standard
- NEET: S₁/S₂=(m₂/m₁)² combines both concepts — derive once, memorise for direct application
- Key sequence: (1) momentum conservation → u₁/u₂=m₂/m₁; (2) S∝u² → S₁/S₂=(u₁/u₂)²=(m₂/m₁)²
Energy Distribution in Explosions in University Physics
Halliday & Resnick covers internal explosions and energy distribution in the center-of-mass frame. The specific result KE₁/KE₂ = m₂/m₁ (lighter block gets more KE) is derivable but not always highlighted as a named result. NEET tests KE ratio questions paired with stopping distance in the same problem. NRI students should know that KE ∝ p²/(2m) — for equal momenta, KE is inversely proportional to mass.
- US university: energy in explosions covered; KE₁/KE₂=m₂/m₁ derivable but not a named formula
- NEET: three ratios from the same setup: S₁/S₂=(m₂/m₁)²; t₁/t₂=m₂/m₁; KE₁/KE₂=m₂/m₁
- Memory trick: distance uses m² (squared ratio); time and KE use m (linear ratio)
NEET-Style Practice — Stopping of Two Blocks
4 QuestionsPractice Problems — Stopping of Two Blocks
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Physics — Friction Revision Checklist
Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.
FAQ — Stopping of Two Blocks
Notes · Downloads · Revision · Important QuestionsWhat is the ratio of stopping distances for two blocks released from a compressed spring?
Why is the stopping distance ratio (m₂/m₁)² and not just m₂/m₁?
What is the KE ratio for the two blocks after release?
Does the coefficient of friction affect the stopping distance RATIO?
What if the two blocks are on surfaces with different coefficients of friction?
Which block stops first (shorter stopping time)?
Can I use energy methods to find stopping distance?
What is the physical interpretation of S₁/S₂ = (m₂/m₁)²?
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