Solve Laws of Motion Numericals Fast: F=ma, Momentum and Friction Shortcuts for NEET & JEE

Quick Answer: Every laws of motion numerical in NEET and JEE reduces to three formulas: F = ma, p = mv (impulse = change in momentum), and f = μN. Solve in 3 steps — draw a free-body diagram, resolve forces along the motion axis, apply F = ma per axis. The top shortcut: for connected bodies, treat the whole system as one mass to get acceleration first, then find tension separately.

Newton’s Second Law: F = ma and Common Problem Patterns

Newton’s second law states that the net force on a body equals mass × acceleration: F_net = ma. Acceleration is always in the direction of the net force, not necessarily the direction of motion. Three patterns dominate exams:

• Block on a horizontal surface: If applied force F acts and friction opposes it, a = (F − μmg)/m.
• Lift problems: Apparent weight N = m(g + a) when accelerating up, N = m(g − a) when accelerating down, N = mg at constant velocity or rest.
• Block on a pulley (single body): a = net driving force ÷ total mass of the moving system.

Example: A 5 kg block in a lift accelerating upward at 2 m/s² (g = 10): N = 5(10 + 2) = 60 N. The reading on a spring balance inside the lift shows exactly this.

Momentum Numericals: p = mv and Impulse = Change in Momentum

Momentum p = mv (unit: kg·m/s). The impulse-momentum theorem states F × t = mv − mu. Use it whenever force acts for a short time — collisions, hits, recoil:

• Collision-style: A 0.15 kg ball at 20 m/s is stopped in 0.02 s. F = m(Δv)/t = 0.15 × 20 / 0.02 = 150 N.
• Recoil: Gun of mass M fires bullet of mass m with velocity v. Set total momentum conserved (initially zero): recoil velocity V = mv/M (opposite direction).

In NEET/JEE, always check whether the problem says “comes to rest” (v = 0) or “bounces back” (v reverses sign — momentum change is 2mv, a classic trap).

Friction Numericals: Static vs Kinetic and the f = μN Shortcut

Friction force f = μN, where N is the normal reaction, not always mg. Decision rule:

Key shortcut: if μ_s > tanθ for a block on an incline, the block stays at rest — no calculation of acceleration needed. This one-liner eliminates 2 of 4 options instantly.

Inclined Plane Problems: Breaking Weight into Components

Resolve weight mg into two perpendicular components:

• Along the incline (driving force): mg sinθ
• Perpendicular (balanced by normal): mg cosθ, so N = mg cosθ

With friction (block sliding down): a = g(sinθ − μ cosθ). Block pushed up the incline: friction reverses and a = g(sinθ + μ cosθ) acting down the slope.

NEET-style example: A block slides down a 30° incline with μ = 0.2, g = 10. a = 10(0.5 − 0.2 × 0.866) = 10(0.5 − 0.173) ≈ 3.27 m/s².

Sign-Convention Traps That Cost Marks

1. Friction direction on inclines: Friction always opposes relative motion (or tendency), not the direction of motion of the block. Going up the incline, friction acts down.
2. Lift acceleration: “Lift accelerates down at a > g” means the body loses contact (N = 0) — apparent weight is negative, physically impossible, so N = 0.
3. Negative acceleration: If you take right as positive and the body decelerates, keep a negative in F = ma — don’t flip the force sign manually as well, or you double-count.
4. Bounce problems: Momentum change is mv − (−mu) = m(v + u), not m(v − u).

Connected Bodies and Pulleys: The System Approach Shortcut

For two blocks (m₁ on table, m₂ hanging over a pulley):

Step 1 — system approach: a = driving force ÷ total mass = m₂g / (m₁ + m₂).

Step 2 — isolate one body: For m₂: m₂g − T = m₂a, so T = m₂(g − a).

Example: m₁ = 3 kg on frictionless table, m₂ = 2 kg, g = 10. a = 20/5 = 4 m/s²; T = 2(10 − 4) = 12 N. Note T is always less than m₂g when the system accelerates — a quick sanity check.

Speed Shortcuts and Formula Tricks for MCQ Accuracy

• Unit check: Force in newtons requires kg × m/s². If mass is in grams, convert first.
• Limiting cases: Put μ = 0 or θ = 0 in your answer — it should reduce to the standard frictionless result. If not, your expression is wrong.
• Dimension elimination: Options with wrong dimensions (e.g., m/s for a force) are out immediately.
• tanθ test: Block at rest on incline ⇔ μ_s ≥ tanθ — no full solution needed.
• Impulse = area under F–t graph — read it geometrically instead of integrating.

Worked Example Set: 5 Exam-Style Numericals Solved Step-by-Step

1. A 2 kg block is pushed with 20 N on a rough floor (μ = 0.5, g = 10). Find a.
f = 0.5 × 2 × 10 = 10 N; a = (20 − 10)/2 = 5 m/s²

2. A 0.05 kg bullet moving at 200 m/s embeds in a 1.95 kg block at rest. Find common velocity.
Conservation: 0.05 × 200 = 2 × v → v = 5 m/s

3. A 60 kg person stands in a lift accelerating down at 3 m/s² (g = 10). Apparent weight?
N = 60(10 − 3) = 420 N

4. Block at rest on 45° incline, μ_s = 0.8, g = 10. Does it slide?
tan45° = 1 > μ_s = 0.8, so yes, it slides with a = 10(0.7071 − 0.8 × 0.7071) ≈ 1.41 m/s²

5. A cricket ball of 0.16 kg hits a bat at 15 m/s and leaves at 20 m/s in the opposite direction. Impulse?
J = m(v + u) = 0.16 × 35 = 5.6 kg·m/s

Common Mistakes in Laws of Motion Numericals

• g = 10 vs 9.8: Use the value stated in the question. If unspecified, JEE Main expects g = 10 m/s² in most integer-type questions; NEET usually states it. Never mix (use 10 in f, 9.8 in weight).
• Using f = μmg on an incline instead of μmg cosθ.
• Ignoring that normal force changes in a lift or under an applied vertical force.
• Treating tension as equal to hanging weight in an accelerating system (it isn’t — T = m(g − a)).
• Forgetting that static friction adjusts up to μ_sN — it is not always at maximum.

Practice Questions with Answer Key

1. A 10 kg box accelerates at 3 m/s² on a frictionless floor. Net force?
2. A 1000 kg car brakes from 20 m/s to rest in 5 s. Average retarding force?
3. Block on incline, θ = 30°, μ = 0. Sliding acceleration (g = 10)?
4. A 70 kg person in a lift accelerating up at 2 m/s² reads what on the scale (g = 10)?
5. Minimum force to start a 5 kg block moving, μ_s = 0.4, g = 10?
6. Two blocks 4 kg and 6 kg connected over a frictionless pulley (Atwood machine). Acceleration (g = 10)?
7. A 0.02 kg bullet recoils a 4 kg gun at what speed if muzzle velocity is 300 m/s?
8. Angle of repose for μ_s = 0.75 is closest to?
9. A 2 kg block slides down 30° incline, μ_k = 0.3, g = 10. Acceleration?
10. Impulse needed to stop a 0.5 kg ball moving at 12 m/s?

Answer Key: 1. 30 N  2. 4000 N  3. 5 m/s²  4. 840 N  5. 20 N  6. 2 m/s²  7. 1.5 m/s  8. ≈ 36.9°  9. 2.4 m/s²  10. 6 kg·m/s

Frequently Asked Questions

Q: What is the fastest way to solve laws of motion numericals?

Draw a free-body diagram, write F = ma along each axis, and check units before marking the answer. For multi-body systems, apply the system approach first to get acceleration in one line.

Q: Why is friction on an incline μmg cosθ and not μmg?

Because the normal reaction on an incline equals the perpendicular component of weight, mg cosθ — the surface doesn’t support the full weight, only the component pressing into it.

Q: How does acceleration in a lift change apparent weight?

Accelerating upward: N = m(g + a) — you feel heavier. Accelerating downward: N = m(g − a) — you feel lighter. At constant velocity, N = mg. This is the classic exam trap.

Q: What is the impulse-momentum theorem?

Impulse (F × t) equals the change in momentum (mv − mu). It is the go-to relation for collision, bounce and recoil problems, and equals the area under the force-time graph.

Q: Should I use g = 10 or 9.8 in NEET/JEE numericals?

Follow the value given in the question. If none is specified, use 9.8 m/s² unless the options clearly indicate rounding to 10 — many JEE integer-answer questions assume g = 10 for clean numbers.

Related reading

• Motion Graphs: Reading a Journey Like a Sentence
• SI Units and Measurement Conversions: Prefixes, Dimensions and Common Exam Traps for NEET, JEE and SSC

Related explainers

Motion Graphs: Reading a Journey Like a Sentence

SI Units and Measurement Conversions: Prefixes, Dimensions and Common Exam Traps for NEET, JEE and SSC

Acceleration and the Three Golden Equations: The Motion Toolkit