Motion of Blocks in Contact

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NEET Physics — Newton's Laws of Motion

Motion of Blocks in Contact – Complete Notes, Revision, Important Questions & Downloads

Motion of Blocks in Contact examines how two or three blocks pushed together by a horizontal force F accelerate as a single system. NEET tests: (1) identifying the contact force at a specific interface in a 2- or 3-block system; (2) recognising how the contact force changes when F is applied from the opposite end; (3) comparing contact forces at different interfaces using R = (mass on free side)/(total mass) × F. For two blocks (masses m₁ and m₂, smooth surface, horizontal push F on m₁): acceleration a = F/(m₁+m₂). The contact force (reaction) between the blocks equals R = m₂F/(m₁+m₂). For three blocks: acceleration = F/(m₁+m₂+m₃); contact force between last two blocks = m₃F/(m₁+m₂+m₃). This topic is foundational for multi-body Newton's-law problems in NEET.

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Contact Force Between BlocksNewton's Laws Ch.4a = F/(m₁+m₂)

Expected QuestionsQ

1–2

Blocks in contact appear regularly in NEET — typically identifying the contact force (reaction force) between adjacent blocks. Questions range from computing the reaction force given total force and masses, to comparing reactions at different contact points in a three-block system, to the inclined-plane version of blocks in contact.

Time Required⏱

45 min

15 min for the two-block system: treat as single body for acceleration, then isolate one block for contact force. 15 min for the three-block system: two contact forces, one from m₁ pushing m₂+m₃ and one from m₁+m₂ pushing m₃. 15 min for NEET practice: direction of force reversed, rough surfaces, blocks on inclines.

Difficulty⚡

Easy

The system-first approach (find acceleration treating all blocks as one mass) is straightforward. Contact forces require isolating the 'forward' blocks relative to the contact point. Common error: confusing which side of the contact you must isolate, leading to wrong formula. Another trap: thinking the contact force equals F or equals zero (neither is correct).

NRI USA Curriculum GapUS

Low

AP Physics 1 covers blocks in contact and contact forces as part of Newton's Third Law and multi-body problems. The NEET emphasis is on rapid formula application and comparing contact forces at different points in 3+ block systems. The concept is fully covered in US physics but NEET demands calculation speed.

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NEET Weightage — Motion of Blocks in Contact

Newton's Laws of Motion (Chapter 4)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
2022	0	0 Q	0
2021	1	1 Q	4
2020	0	0 Q	0
2019	0	0 Q	0
6-Year Total (2019–2024)	1–2		4–8

Two-block system (m₁ pushed by F, m₂ in front): a = F/(m₁+m₂). Contact force (reaction on m₂ from m₁) = R = m₂ × a = m₂F/(m₁+m₂). Also equals force exerted by m₂ on m₁ (Newton's Third Law).
Three-block system (F on m₁, then m₂, then m₃): a = F/(m₁+m₂+m₃). Contact between m₁ and m₂ = R₁ = (m₂+m₃)F/(m₁+m₂+m₃). Contact between m₂ and m₃ = R₂ = m₃F/(m₁+m₂+m₃). Always: R₁ > R₂ — contact forces decrease from applied force end to free end.
Direction matters: if F pushes from the right (on m₃), the formulas flip. Contact between m₃ and m₂: R = m₁F/(m₁+m₂+m₃). The block mass that matters is always the mass on the side AWAY from the applied force.

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0.3

Avg Questions / Year

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Total Marks (6 yrs)

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Direct

Pattern

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Easy

Difficulty

How to Prepare Motion of Blocks in Contact for NEET

Treat the system as one body first, then isolate for contact forces Step 1 (System): Net force on all blocks together = F (contact forces are internal). Total mass = m₁+m₂ (or +m₃). a = F/(total mass). This gives the common acceleration. Step 2 (Isolate): To find the contact force between blocks A and B, isolate all blocks on one side of the contact. The contact force on that sub-system = mass of sub-system × a. Which side to isolate: choose the side that does NOT have the applied force F on it (it has only the contact force acting on it externally). That contact force = (mass of sub-system) × a.

Apply the direction of push correctly in three-block systems In a three-block system, if F acts on the leftmost block m₁: Reaction at (m₁-m₂ contact) = (m₂+m₃) × a — because m₂ and m₃ together are the 'ahead' sub-system relative to this contact. Reaction at (m₂-m₃ contact) = m₃ × a. The reaction decreases because fewer masses are 'ahead'. Numerical check: R₁ = (m₂+m₃)/(m₁+m₂+m₃) × F; R₂ = m₃/(m₁+m₂+m₃) × F. R₁ > R₂ always (since m₂+m₃ > m₃).

Recognise blocks in contact on rough surfaces and on inclines Rough surface: the friction force on the entire system = μ(m₁+m₂)g. Adjust net force: a = [F − μ(m₁+m₂)g]/(m₁+m₂). Contact force: R = m₂a + μm₂g (friction now acts on m₂ as well). The friction on m₂ is μm₂g (opposing motion). On incline: component of gravity along slope replaces part of the force balance. The system approach still works — just include gravity terms.

Study Materials — Motion of Blocks in Contact

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Full Notes

Two-block system: acceleration and contact force derivation. Three-block system: two contact forces. Rough surface modification. Vertical blocks-in-contact (stacked blocks pushed by vertical force). Direction reversal scenarios.

Single topic2 pagesDerivation + Application

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Formula Sheet

a = F/(m₁+m₂). R = m₂F/(m₁+m₂). Three-block: R₁ = (m₂+m₃)F/ΣM, R₂ = m₃F/ΣM. Rule: R = (mass ahead) × a = (mass ahead)/(total mass) × F.

4 formulas1 pageQuick reference

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MCQ Practice

15 questions: 2-block contact force, 3-block contact forces, rough surface correction, direction reversal, contact force on incline, vertical stack problems.

15 MCQsAll variantsSolved

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PYQ

Year-tagged NEET questions on blocks in contact, contact force computation, and three-body contact problems.

4+ year-tagged Qs2015–2024Solved

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2-Column Table

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Rapid Revision — Motion of Blocks in Contact

Concept → Trap → Example

1) Two Blocks in Contact — System and Contact Force

Core

Blocks m₁ and m₂ on smooth horizontal surface. Force F applied on m₁, pushing m₂. Since blocks move together (same acceleration): F = (m₁+m₂)a → a = F/(m₁+m₂). Contact force R: isolate m₂ alone. Only force on m₂ horizontally = R (from m₁). Newton's 2nd law for m₂: R = m₂a = m₂F/(m₁+m₂). Alternative: isolate m₁. Forces on m₁: F (applied, forward) and R (reaction from m₂, backward). m₁a = F − R → R = F − m₁a = F − m₁F/(m₁+m₂) = m₂F/(m₁+m₂) ✓. Key insight: R is always less than F; it acts on m₂ forward (from m₁) and on m₁ backward (from m₂) — Newton's Third Law pair.

The contact force R depends on the mass being pushed (m₂) and the total mass. For fixed F: R increases as m₂ increases (more mass to push → more contact force). R = 0 if m₂ = 0 (nothing ahead → no contact force needed). R = F if m₁ = 0 (all mass ahead → contact force equals applied force). A useful sanity check: R is bounded between 0 and F.
If F is applied on m₂ instead (pushing m₁): same acceleration a = F/(m₁+m₂). But contact force is now: isolate m₁ (m₁ has no applied force, only contact force). R = m₁a = m₁F/(m₁+m₂). Now R depends on m₁ (the block on the side away from F). Swapping direction changes which mass appears in the contact force formula. NEET often tests this asymmetry.
Newton's Third Law perspective: the contact force at the m₁-m₂ interface is a Newton's Third Law pair. m₁ pushes m₂ with force R forward; m₂ pushes m₁ with force R backward. Both forces are equal in magnitude. The net force on the system from these contact forces is zero (internal forces cancel) — consistent with system acceleration = F/(m₁+m₂).

Example (NEET-style)Blocks of 4 kg and 6 kg on smooth surface, F = 20 N applied on the 4 kg block. (a) Find acceleration. (b) Find contact force. (a) a = F/(m₁+m₂) = 20/(4+6) = 2 m/s². (b) Isolate 6 kg block: R = m₂ × a = 6 × 2 = 12 N. Verify: isolate 4 kg: F − R = m₁a → 20 − 12 = 4×2 = 8 ✓. Now if F is applied on the 6 kg block instead: a = 2 m/s² (same). Contact force = m₁a = 4 × 2 = 8 N. The contact force is smaller now because fewer mass is 'ahead' of the contact (only the 4 kg block, not the 6 kg block).

2) Three Blocks in Contact — Two Contact Forces

High Priority

Three blocks m₁, m₂, m₃ on smooth surface. F applied on m₁. Acceleration a = F/(m₁+m₂+m₃). Contact force R₁ (between m₁ and m₂): isolate m₂+m₃ sub-system. Only external force on them = R₁ (forward). (m₂+m₃)a = R₁ → R₁ = (m₂+m₃)a = (m₂+m₃)F/(m₁+m₂+m₃). Contact force R₂ (between m₂ and m₃): isolate m₃ alone. Only external force = R₂. m₃a = R₂ → R₂ = m₃F/(m₁+m₂+m₃). Since m₂+m₃ > m₃: R₁ > R₂. Contact forces decrease from the applied force end to the free end.

If F is applied on m₃ instead (reverse direction): a = F/(m₁+m₂+m₃) (same). R₁ (contact between m₁ and m₂): isolate m₁ alone (no applied force on it). R₁ = m₁F/(m₁+m₂+m₃). R₂ (contact between m₂ and m₃): isolate m₁+m₂. R₂ = (m₁+m₂)F/(m₁+m₂+m₃). Now R₂ > R₁ — contact forces are larger near the applied force and smaller away from it. Regardless of direction: the block on the 'free side' of the contact determines the contact force.
NEET three-block mnemonics: For force applied on LEFT end → contact force AT position P = (total mass on the right of P)/(total mass) × F. For force applied on RIGHT end → contact force AT position P = (total mass on the left of P)/(total mass) × F. In both cases, use the mass on the side AWAY from the applied force to compute the contact force at that interface.
Three-block verification: R₁ + (m₁×a) vs F for m₁: F − R₁ = m₁a → F − [(m₂+m₃)/(m₁+m₂+m₃)]F = m₁F/(m₁+m₂+m₃) = m₁a ✓. For m₂: R₁ − R₂ = m₂a → [(m₂+m₃)−m₃]/(total) × F = m₂/(total) × F = m₂a ✓. All consistent.

Example (NEET-style)Three blocks: m₁ = 2 kg, m₂ = 3 kg, m₃ = 5 kg on smooth surface, F = 30 N on m₁. (a) acceleration; (b) contact force R₁ (between m₁ and m₂); (c) contact force R₂ (between m₂ and m₃). (a) a = 30/(2+3+5) = 3 m/s². (b) R₁ = (m₂+m₃)a = (3+5)×3 = 24 N. (c) R₂ = m₃×a = 5×3 = 15 N. Verify for m₁: F − R₁ = 30 − 24 = 6 = m₁a = 2×3 ✓. Verify for m₂: R₁ − R₂ = 24 − 15 = 9 = m₂a = 3×3 ✓. Verify for m₃: R₂ = 15 = m₃a = 5×3 ✓. Forces in system: applied 30 N → contact 24 N at first joint → contact 15 N at second joint → drives 5 kg block at 3 m/s².

3) Blocks in Contact on Rough Surface and Vertical Stack

Application

Rough horizontal surface (coefficient μ): Friction acts on both blocks opposing motion. System: NET force = F − μ(m₁+m₂)g. Acceleration a = [F − μ(m₁+m₂)g]/(m₁+m₂). Contact force: isolate m₂. Forces on m₂: R (forward from m₁) and friction f₂ = μm₂g (backward). R − μm₂g = m₂a → R = m₂(a + μg) = m₂[F/(m₁+m₂)]. Note: same result as smooth case for contact force using system approach (friction on m₂ is accounted for by the reduced acceleration). Alternatively: R = m₂a + μm₂g separately. Both consistent.

Vertical blocks (horizontal force pushing two blocks against a vertical wall or horizontally stacked between two walls): the analysis is identical but gravity and normal forces change. For two blocks stacked vertically (m₁ on top, horizontal push F on both or on the wall side): the contact force between them acts through the horizontal direction. Vertical equilibrium: N_wall = F, friction from wall supports weight if locked. This is a constraint problem.
Blocks on incline: two blocks m₁ and m₂ on a smooth incline, m₁ pushing m₂ up the incline. 'Applied force' equivalent = difference in gravity components. If a push F acts up the incline on the system: a = [F − (m₁+m₂)g sinθ]/(m₁+m₂). Contact force R = m₂(a + g sinθ) = m₂F/(m₁+m₂). Frictionless incline: same contact force formula as flat surface — only the effective driving force changes.
Blocks in decelerating system: if brakes are applied (deceleration a on system): contact force between front block (m₁) and rear block (m₂) = m₁ × a (front block must be decelerated by the contact force from the rear). This is the seatbelt/head-rest physics principle — the front block pushes backward on the rear (decelerating it) while the rear pushes backward on its support.

Example (NEET-style)Two blocks, m₁ = 5 kg and m₂ = 3 kg, on a rough surface (μ = 0.2), F = 16 N applied on m₁. (g = 10 m/s²) (a) Find acceleration. (b) Find contact force. (a) Net force = F − μ(m₁+m₂)g = 16 − 0.2×8×10 = 16 − 16 = 0 N. So a = 0 m/s². (b) With a = 0: contact force R = m₂×a + μm₂g = 0 + 0.2×3×10 = 6 N. This makes sense: the contact force from m₁ on m₂ just overcomes the friction on m₂ to keep it at constant velocity (zero acceleration). For m₁: F − R − μm₁g = 5×0 → 16 − 6 − 0.2×5×10 = 16 − 6 − 10 = 0 ✓.

US Curriculum Gaps — Motion of Blocks in Contact

Topics in this section are tested in NEET but organised differently in standard US physics courses.

Contact Force Symmetry Asymmetry (AP Physics 1 Gap)

AP Physics 1 covers Newton's Third Law and two-body problems, including the concept that contact forces are equal-and-opposite pairs. However, NEET specifically emphasises the asymmetry in contact force VALUE based on APPLICATION direction: applying F on m₁ gives contact force R = m₂F/(m₁+m₂), while applying F on m₂ gives R = m₁F/(m₁+m₂). This direction-dependent formula is a NEET exam pattern that requires the student to identify which mass is 'ahead' of the contact point. AP Physics 1 students often set up FBDs correctly but may not think of the shortcut formula and direction dependency.

NEET: force applied on left block — contact force = (right mass)/(total mass) × F
NEET: force applied on right block — contact force = (left mass)/(total mass) × F
AP Physics 1: FBD approach is standard but the pattern formula is not explicitly taught

Three-Body Contact Force at Specific Interface (AP Gap)

AP Physics 1 introduces three-body systems but focuses on acceleration and individual forces. NEET tests three-block contact force at a SPECIFIC interface (e.g., 'find force between block 2 and block 3 in this 3-block system'). The formula R = (mass on free side × total force) / total mass is a NEET computation shortcut not explicitly taught in AP Physics 1. NEET also tests the ratio R₁:R₂ in three-block systems, requiring students to compare mass fractions. This computational pattern appears in roughly 15% of NEET Newton's Laws questions.

NEET: three blocks, identify contact force at one specific interface
NEET: compare contact forces at two different interfaces — ratios
AP Physics 1: three-body problems covered but interface-specific contact force not a standard drill

NEET-Style Practice Questions — Motion of Blocks in Contact

4 Questions

1Three blocks of masses 2 kg, 3 kg, and 5 kg are placed on a smooth surface in contact. A horizontal force F = 20 N is applied on the 2 kg block. What is the contact force between the 3 kg and 5 kg blocks?Three-Block Contact

10 N

12 N

15 N

8 N

Total mass = 2 + 3 + 5 = 10 kg. Acceleration a = 20/10 = 2 m/s². Contact force between 3 kg and 5 kg blocks: isolate 5 kg block (free end). Only force on it = R₂. R₂ = 5 × 2 = 10 N. Alternatively: R₂ = m₃/(total) × F = (5/10) × 20 = 10 N. The contact force at the 2kg-3kg interface R₁ = (3+5)/10 × 20 = 16 N. Contact forces: applied 20 N → 16 N at first joint → 10 N at second joint. The force 'dissipates' at each contact point — each block absorbs part of the driving force to accelerate itself.

2Two blocks of masses 8 kg (block A) and 2 kg (block B) are in contact on a smooth surface. A horizontal force of 40 N is applied on A. Now the same force is applied on B. Find the ratio of contact forces (F_A applied : F_B applied):Direction Symmetry

1 : 4

4 : 1

1 : 1

2 : 1

Acceleration is the same in both cases: a = 40/(8+2) = 4 m/s². Case F on A: isolate B. R = m_B × a = 2 × 4 = 8 N. Case F on B: isolate A. R = m_A × a = 8 × 4 = 32 N. Ratio: 8 : 32 = 1 : 4. When F is applied on the heavier block (8 kg), the contact force is smaller (8 N) because only the lighter block (2 kg) needs to be pushed through the contact. When F is applied on the lighter block (2 kg), the contact force is larger (32 N) because the heavier block (8 kg) must be pushed through the contact. Key takeaway: the contact force = (mass on FREE side of contact) × a.

3Two blocks of masses 3 kg and 7 kg are in contact on a rough horizontal surface (μ = 0.4). A horizontal force F = 50 N is applied on the 3 kg block. Find the contact force between the blocks. (g = 10 m/s²)Rough Surface

35 N

28 N

35 N

42 N

Total friction = μ(m₁+m₂)g = 0.4×10×10 = 40 N. Net force = 50 − 40 = 10 N. Acceleration a = 10/(3+7) = 1 m/s². Contact force: isolate 7 kg block. Forces on 7 kg: R (forward from 3 kg block) and friction f₂ = μm₂g = 0.4×7×10 = 28 N (backward). R − 28 = 7×1 → R = 35 N. Verify for 3 kg: F − R − f₁ = m₁a → 50 − 35 − 0.4×3×10 = 50 − 35 − 12 = 3 = 3×1 ✓. Trap: students who forget friction on each block and use R = m₂ × a = 7 × 1 = 7 N (wrong — ignores friction on the 7 kg block itself).

4In a three-block system (masses 1 kg, 2 kg, 3 kg from left to right) on a smooth surface, same force F is first applied on 1 kg block (left end), then on 3 kg block (right end). The contact force between the middle (2 kg) and rightmost (3 kg) blocks in each case is, respectively:Comparative

F/2 and 3F/6 = F/2 (same)

F/2 and F/3 (Case 2 smaller)

3F/6 and F/2

Case 1: F/2; Case 2: F/2 — both equal

Total mass M = 1+2+3 = 6 kg. Acceleration a = F/6 in both cases. Case 1 (F on left, on 1 kg): contact force between 2 kg and 3 kg = mass on free side (3 kg) × a = 3 × F/6 = F/2. Case 2 (F on right, on 3 kg): free side of 2kg-3kg interface is now 2 kg AND 1 kg → mass on free side = 1+2 = 3 kg… wait. Free side is away from applied force. F on 3 kg (right): at the 2kg-3kg interface, the free side is 1kg + 2kg = 3 kg. R₂ = 3 × F/6 = F/2. Contact between 1 kg and 2 kg: free side = 1 kg. R₁ = 1 × F/6 = F/6. So in both cases, contact force between 2 kg and 3 kg = F/2. This might seem counter-intuitive — the symmetry arises because the mass on the 'free side' of the 2-3 interface is 3 kg in Case 1 (only the 3 kg block) and 1+2 = 3 kg in Case 2 (both remaining blocks). Coincidentally equal. Answer: both = F/2.

Practice Problems — Motion of Blocks in Contact

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1Four blocks of masses 1 kg, 2 kg, 3 kg, and 4 kg are placed in contact on a smooth horizontal surface. A horizontal force F = 20 N is applied on the 1 kg block (leftmost). Find: (a) acceleration, (b) contact force between 2 kg and 3 kg blocks, (c) contact force between 3 kg and 4 kg blocks.

(a) 2 m/s², (b) 14 N, (c) 8 N

(a) 2 m/s², (b) 12 N, (c) 6 N

(a) 2 m/s², (b) 7 N, (c) 4 N

(a) 1 m/s², (b) 14 N, (c) 8 N

👁 Reveal Answer

(a) Total mass = 1+2+3+4 = 10 kg. a = 20/10 = 2 m/s². (b) Contact force between 2 kg and 3 kg: mass on free side (right of this interface) = 3+4 = 7 kg. R = 7 × 2 = 14 N. Alternatively: R = (3+4)/10 × 20 = 14 N. (c) Contact force between 3 kg and 4 kg: mass on free side = 4 kg. R = 4 × 2 = 8 N. Verify: F = 20 N drives system. At each contact, force reduces by the amount absorbed by each block to accelerate it. End-block (4 kg) receives 8 N → makes sense. Sequence: 20 N → 18 N → 14 N → 8 N → 0 N at free end of 4 kg. Contact at 1-2 interface = (2+3+4)/10 × 20 = 18 N.

2Two blocks A (5 kg) and B (3 kg) are in contact on a smooth inclined plane (θ = 30°). A force F = 40 N is applied up the incline on A. B is on the far side (free end). Find: (a) acceleration of the system, (b) contact force between A and B. (g = 10 m/s², sin30° = 0.5)

(a) 2 m/s², (b) 9 N

(a) 1 m/s², (b) 6.5 N

(a) 2 m/s², (b) 6 N

(a) 3 m/s², (b) 9 N

👁 Reveal Answer

(a) Net force along incline = F − (m_A + m_B)g sinθ = 40 − 8×10×0.5 = 40 − 40 = 0 N. So a = 0 m/s². The system is in equilibrium — force F just balances gravity along incline. (b) Contact force with a = 0: isolate B. Forces on B along incline: R (up incline from A) and m_B g sinθ = 3×10×0.5 = 15 N (down incline). R − 15 = 3×0 → R = 15 N. Verify for A: F − R − m_A g sinθ = m_A × 0 → 40 − 15 − 5×10×0.5 = 40 − 15 − 25 = 0 ✓. The contact force (15 N) equals the weight component of B along the slope — A is preventing B from sliding down.

3Block A (6 kg) is on a smooth surface. Block B (4 kg) is placed on top of A. A horizontal force F = 25 N is applied on A alone. The coefficient of friction between A and B is μ = 0.3. Find: (a) do the blocks move together or slip? (b) if together, find acceleration and friction on B. If they slip, find acceleration of each. (g = 10 m/s²)

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N — but this exceeds μm_B g = 12 N, so they do NOT slip

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N (within limit 12 N — no slip)

(a) Slip; a_A = 3.43 m/s², a_B = 3 m/s²

👁 Reveal Answer

Test if together: assume a = F/(m_A+m_B) = 25/10 = 2.5 m/s². Force needed on B to achieve a = 2.5 m/s²: F_B = m_B × a = 4 × 2.5 = 10 N (this must come entirely from friction since force is only on A). Maximum static friction on B: μ × m_B × g = 0.3 × 4 × 10 = 12 N. Since required friction (10 N) < maximum friction (12 N), the blocks DO move together. Acceleration = 2.5 m/s². Friction on B = 10 N (forward, from A's surface on B). Friction on A from B = 10 N backward (Newton's Third Law). So the blocks move as a unit with the friction as the 'contact force' that drags B along. If we had required friction > 12 N: blocks would slip → solve separately: a_B = μg = 3 m/s²; a_A = (F − μm_B g)/m_A = (25−12)/6 ≈ 2.17 m/s².

4In a 3-block system on a smooth surface: m₁ = 1 kg, m₂ = 2 kg, m₃ = 3 kg, with F = 12 N applied on m₁. Find each block's acceleration and the forces at each contact. Verify force balance for m₂.

a = 2 m/s²; R₁₂ = 10 N; R₂₃ = 6 N

a = 2 m/s²; R₁₂ = 5 N; R₂₃ = 6 N

a = 2 m/s²; R₁₂ = 10 N; R₂₃ = 2 N

a = 2 m/s²; R₁₂ = 4 N; R₂₃ = 6 N

👁 Reveal Answer

a = F/(m₁+m₂+m₃) = 12/6 = 2 m/s². R₁₂ (between m₁ and m₂) = (m₂+m₃) × a = (2+3) × 2 = 10 N. R₂₃ (between m₂ and m₃) = m₃ × a = 3 × 2 = 6 N. Force balance on m₂ (receives R₁₂ forward and gives R₂₃ forward): m₂ × a = R₁₂ − R₂₃ → 2 × 2 = 10 − 6 = 4 ✓. The m₂ block receives 10 N from its left and passes 6 N to m₃ to its right; the difference (4 N) accelerates m₂ itself. This hierarchical force reduction is the central concept of blocks in contact.

Physics — Newton's Laws of Motion Revision Checklist

Check off chapters as you revise

Use this section for quick chapter tracking before mocks, part tests, and final NEET revision.

System acceleration formula: For N blocks on smooth surface with force F on one end: a = F/(sum of all masses). This applies regardless of how force is distributed. Write total mass quickly (add all given masses). Then a = F/total mass. Key: contact forces are INTERNAL to the system — they do not appear in the system equation.

Contact force isolation rule: To find contact force at any interface: isolate all blocks on the SIDE WITHOUT the applied force F. The only external force on them is the contact force R. So R = (mass of isolated sub-system) × a. Quick formula: R = (mass on free side / total mass) × F. Apply this for both two-block and three-block systems.

Rough surface extension — include friction in sub-system: On rough surface, friction acts on each block separately. When isolating a sub-system to find contact force: include friction on all blocks in the sub-system. R − μm_sub × g = m_sub × a → R = m_sub × (a + μg) = m_sub × [F/(total mass)]. Note: this equals m_sub × F / total mass — same as smooth case formula (friction is already embedded in the acceleration).

FAQ — Motion of Blocks in Contact

Notes · Downloads · Revision · Important Questions

Why do blocks in contact have the same acceleration?

They have the same acceleration because they are in contact with each other and move together as a rigid system — they cannot pass through each other, so they must have the same velocity and acceleration at all times (no relative motion). This is the constraint: the blocks are touching, not attached by a string. As long as the contact force is compressive (pushing, not pulling), they stay together and share the same acceleration. If the contact force would need to become tensile (pulling), the blocks separate and must be analyzed independently.

Why does the contact force change when the direction of F is reversed?

Because the contact force must provide the net force only on the sub-system on the 'free' side of the contact. When F is on the left (on m₁), the contact at the m₁-m₂ interface must push m₂ (and anything beyond) — so R = (mass beyond interface) × a. When F is reversed to the right (on m₂), the contact at the same interface must push m₁ — so R = m₁ × a. Different masses on the free side → different contact forces. The acceleration a = F/(m₁+m₂) remains the same, but the required contact force changes based on what mass needs to be accelerated through that interface.

Can the contact force ever exceed the applied force F?

No, the contact force is always less than or equal to F. The contact force equals (mass on free side)/(total mass) × F. Since the mass on the free side is always less than or equal to the total mass (it's a fraction), the contact force ≤ F. It equals F only in the degenerate case where the applied-force block has zero mass (m₁ = 0), making the free side = total mass. Physically, the applied force F is 'shared' among all blocks to accelerate them — the contact force at any internal interface only needs to drive the sub-system ahead of it.

What happens if the blocks are in contact but one has a different coefficient of friction than the other?

Each block experiences its own friction force: f₁ = μ₁m₁g and f₂ = μ₂m₂g. System acceleration: a = [F − μ₁m₁g − μ₂m₂g]/(m₁+m₂). Contact force: isolate m₂. On m₂: R − μ₂m₂g = m₂a → R = m₂a + μ₂m₂g = m₂(a + μ₂g). Different friction coefficients mean the individual friction forces are different but the approach is the same — compute system acceleration, then isolate a sub-system.

In the three-block system, which contact force is larger?

The contact force closer to the applied force is always larger. If F is applied on m₁ (left): R₁ (at m₁-m₂ interface) = (m₂+m₃)/(total) × F and R₂ (at m₂-m₃ interface) = m₃/(total) × F. Since m₂+m₃ > m₃: R₁ > R₂ always. Physically, the contact closer to F must push more mass (both m₂ and m₃), while the contact farther from F only needs to push m₃. The force 'decreases' as it passes through each block — each block 'uses up' part of the net force to accelerate itself.

What is the condition for blocks in contact to separate (not stay together)?

Blocks can only PUSH each other, not pull. If the required contact force becomes negative (tensile — pulling) in the analysis, the blocks separate. For blocks in contact, if the applied force reduces suddenly or changes direction so that the block in front would need to decelerate slower than the block behind, the contact force would become tensile — at that point, the blocks lose contact. Mathematically: contact force R = (mass ahead) × a. If acceleration becomes negative (deceleration) and the 'ahead' block can decelerate on its own (e.g., due to its own friction) faster than the 'behind' block — they separate.

How is the contact force concept applied in car crash safety physics?

In a car collision: the car decelerates rapidly. The driver (mass m₁) and passenger (mass m₂) inside form a 'blocks in contact' system with the car seat and seatbelt. During deceleration a: the contact force from the seatbelt on the driver = m₁ × a. This is why crash deceleration values (often 20–50g) produce enormous forces on occupants — the contact force scales with both mass and deceleration. Airbags increase the stopping time, reducing a, which reduces the contact force and hence the injury force on the driver. This is Newton's Second Law applied to multi-body systems.

Can we apply the blocks-in-contact analysis to more than three blocks?

Yes. For n blocks with masses m₁, m₂, …, mₙ on a smooth surface, force F on m₁: acceleration a = F/(m₁+m₂+…+mₙ). Contact force at the interface between the k-th and (k+1)-th block from the applied-force end: R_k = (m_{k+1} + m_{k+2} + … + mₙ) × a = (sum of remaining masses) / (total mass) × F. The pattern holds for any number of blocks. NEET has tested four-block systems. The key is always: identify the 'free side' mass.

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Newton's Laws of Motion

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Friction

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Motion of Blocks in Contact

NEET > Physics > Laws of Motion > Newton's Laws of Motion > Motion of Blocks in Contact

Unit Progress

NEET Physics — Newton's Laws of Motion

Motion of Blocks in Contact – Complete Notes, Revision, Important Questions & Downloads

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Contact Force Between BlocksNewton's Laws Ch.4a = F/(m₁+m₂)

Expected QuestionsQ

1–2

Time Required⏱

45 min

Difficulty⚡

Easy

NRI USA Curriculum GapUS

Low

0Subtopics

4+Practice Questions

4Free Downloads

45 minPrep Time

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NEET Weightage — Motion of Blocks in Contact

Newton's Laws of Motion (Chapter 4)

NEET Year	Questions from this Topic	Bar	Marks
2024	0	0 Q	0
2023	1	1 Q	4
2022	0	0 Q	0
2021	1	1 Q	4
2020	0	0 Q	0
2019	0	0 Q	0
6-Year Total (2019–2024)	1–2		4–8

📊

0.3

Avg Questions / Year

🎯

Total Marks (6 yrs)

📈

Direct

Pattern

⚠️

Easy

Difficulty

How to Prepare Motion of Blocks in Contact for NEET

Study Materials — Motion of Blocks in Contact

PDF · Cheat Sheet · MCQ Set · PYQ

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Full Notes

Single topic2 pagesDerivation + Application

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Formula Sheet

a = F/(m₁+m₂). R = m₂F/(m₁+m₂). Three-block: R₁ = (m₂+m₃)F/ΣM, R₂ = m₃F/ΣM. Rule: R = (mass ahead) × a = (mass ahead)/(total mass) × F.

4 formulas1 pageQuick reference

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MCQ Practice

15 questions: 2-block contact force, 3-block contact forces, rough surface correction, direction reversal, contact force on incline, vertical stack problems.

15 MCQsAll variantsSolved

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📒

PYQ

Year-tagged NEET questions on blocks in contact, contact force computation, and three-body contact problems.

4+ year-tagged Qs2015–2024Solved

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2-Column Table

Column A	Column B

Rapid Revision — Motion of Blocks in Contact

Concept → Trap → Example

1) Two Blocks in Contact — System and Contact Force

Core

The contact force R depends on the mass being pushed (m₂) and the total mass. For fixed F: R increases as m₂ increases (more mass to push → more contact force). R = 0 if m₂ = 0 (nothing ahead → no contact force needed). R = F if m₁ = 0 (all mass ahead → contact force equals applied force). A useful sanity check: R is bounded between 0 and F.
If F is applied on m₂ instead (pushing m₁): same acceleration a = F/(m₁+m₂). But contact force is now: isolate m₁ (m₁ has no applied force, only contact force). R = m₁a = m₁F/(m₁+m₂). Now R depends on m₁ (the block on the side away from F). Swapping direction changes which mass appears in the contact force formula. NEET often tests this asymmetry.
Newton's Third Law perspective: the contact force at the m₁-m₂ interface is a Newton's Third Law pair. m₁ pushes m₂ with force R forward; m₂ pushes m₁ with force R backward. Both forces are equal in magnitude. The net force on the system from these contact forces is zero (internal forces cancel) — consistent with system acceleration = F/(m₁+m₂).

2) Three Blocks in Contact — Two Contact Forces

High Priority

If F is applied on m₃ instead (reverse direction): a = F/(m₁+m₂+m₃) (same). R₁ (contact between m₁ and m₂): isolate m₁ alone (no applied force on it). R₁ = m₁F/(m₁+m₂+m₃). R₂ (contact between m₂ and m₃): isolate m₁+m₂. R₂ = (m₁+m₂)F/(m₁+m₂+m₃). Now R₂ > R₁ — contact forces are larger near the applied force and smaller away from it. Regardless of direction: the block on the 'free side' of the contact determines the contact force.
NEET three-block mnemonics: For force applied on LEFT end → contact force AT position P = (total mass on the right of P)/(total mass) × F. For force applied on RIGHT end → contact force AT position P = (total mass on the left of P)/(total mass) × F. In both cases, use the mass on the side AWAY from the applied force to compute the contact force at that interface.
Three-block verification: R₁ + (m₁×a) vs F for m₁: F − R₁ = m₁a → F − [(m₂+m₃)/(m₁+m₂+m₃)]F = m₁F/(m₁+m₂+m₃) = m₁a ✓. For m₂: R₁ − R₂ = m₂a → [(m₂+m₃)−m₃]/(total) × F = m₂/(total) × F = m₂a ✓. All consistent.

3) Blocks in Contact on Rough Surface and Vertical Stack

Application

Vertical blocks (horizontal force pushing two blocks against a vertical wall or horizontally stacked between two walls): the analysis is identical but gravity and normal forces change. For two blocks stacked vertically (m₁ on top, horizontal push F on both or on the wall side): the contact force between them acts through the horizontal direction. Vertical equilibrium: N_wall = F, friction from wall supports weight if locked. This is a constraint problem.
Blocks on incline: two blocks m₁ and m₂ on a smooth incline, m₁ pushing m₂ up the incline. 'Applied force' equivalent = difference in gravity components. If a push F acts up the incline on the system: a = [F − (m₁+m₂)g sinθ]/(m₁+m₂). Contact force R = m₂(a + g sinθ) = m₂F/(m₁+m₂). Frictionless incline: same contact force formula as flat surface — only the effective driving force changes.
Blocks in decelerating system: if brakes are applied (deceleration a on system): contact force between front block (m₁) and rear block (m₂) = m₁ × a (front block must be decelerated by the contact force from the rear). This is the seatbelt/head-rest physics principle — the front block pushes backward on the rear (decelerating it) while the rear pushes backward on its support.

US Curriculum Gaps — Motion of Blocks in Contact

Topics in this section are tested in NEET but organised differently in standard US physics courses.

Contact Force Symmetry Asymmetry (AP Physics 1 Gap)

NEET: force applied on left block — contact force = (right mass)/(total mass) × F
NEET: force applied on right block — contact force = (left mass)/(total mass) × F
AP Physics 1: FBD approach is standard but the pattern formula is not explicitly taught

Three-Body Contact Force at Specific Interface (AP Gap)

NEET: three blocks, identify contact force at one specific interface
NEET: compare contact forces at two different interfaces — ratios
AP Physics 1: three-body problems covered but interface-specific contact force not a standard drill

NEET-Style Practice Questions — Motion of Blocks in Contact

4 Questions

10 N

12 N

15 N

8 N

1 : 4

4 : 1

1 : 1

2 : 1

35 N

28 N

35 N

42 N

F/2 and 3F/6 = F/2 (same)

F/2 and F/3 (Case 2 smaller)

3F/6 and F/2

Case 1: F/2; Case 2: F/2 — both equal

Practice Problems — Motion of Blocks in Contact

Click "Reveal Answer" after attempting

(a) 2 m/s², (b) 14 N, (c) 8 N

(a) 2 m/s², (b) 12 N, (c) 6 N

(a) 2 m/s², (b) 7 N, (c) 4 N

(a) 1 m/s², (b) 14 N, (c) 8 N

👁 Reveal Answer

(a) 2 m/s², (b) 9 N

(a) 1 m/s², (b) 6.5 N

(a) 2 m/s², (b) 6 N

(a) 3 m/s², (b) 9 N

👁 Reveal Answer

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N — but this exceeds μm_B g = 12 N, so they do NOT slip

(a) Together; (b) a = 2.5 m/s², friction on B = 10 N (within limit 12 N — no slip)

(a) Slip; a_A = 3.43 m/s², a_B = 3 m/s²

👁 Reveal Answer

a = 2 m/s²; R₁₂ = 10 N; R₂₃ = 6 N

a = 2 m/s²; R₁₂ = 5 N; R₂₃ = 6 N

a = 2 m/s²; R₁₂ = 10 N; R₂₃ = 2 N

a = 2 m/s²; R₁₂ = 4 N; R₂₃ = 6 N

👁 Reveal Answer

Physics — Newton's Laws of Motion Revision Checklist

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FAQ — Motion of Blocks in Contact

Notes · Downloads · Revision · Important Questions

Why do blocks in contact have the same acceleration?

Why does the contact force change when the direction of F is reversed?

Can the contact force ever exceed the applied force F?

What happens if the blocks are in contact but one has a different coefficient of friction than the other?

In the three-block system, which contact force is larger?

What is the condition for blocks in contact to separate (not stay together)?

How is the contact force concept applied in car crash safety physics?

Can we apply the blocks-in-contact analysis to more than three blocks?

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