Laws of Motion for NEET UG
Newton's three laws connect force and motion. With a free-body diagram and F = ma you can solve almost every mechanics problem in NEET.
PhysicsClass 1168 MCQs4 lessons13 flashcards4 memory tricks
What you'll learn
- Newton's laws and the free-body diagram
State the three laws and draw a free-body diagram to find acceleration. - Momentum and impulse
Use impulse = change in momentum, and conservation of momentum, to solve collision and recoil problems. - Friction
Tell static from kinetic friction, and solve problems on rough surfaces and inclines. - Circular motion on roads: banking
Find the safe speed on a level curve and on a banked road.
Key formulas
| Momentum | p = mv |
|---|---|
| Impulse | J = F × Δt = Δp |
| Conservation (no external force) | m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂ |
| Maximum static friction | f_s(max) = μ_s N |
| Kinetic friction | f_k = μ_k N |
| Angle of repose | tan θ = μ_s |
| Maximum speed on a level curve | v = √(μrg) |
| Speed needing no friction on a banked road | v = √(rg tan θ) |
Memory tricks
Newton's third law forces are equal, opposite and simultaneous, but they act on different bodies, so they never cancel each other.
F = Δp/Δt. Pulling the hands back increases the time over which the ball stops, so the force on the hands is smaller.
Static friction adjusts to equal the applied force, up to μₛN. Once the body slides, kinetic friction μₖN takes over, and μₖ < μₛ, so it is easier to keep a box moving than to start it.
Normal force N = m(g + a) when accelerating upwards, m(g − a) when accelerating downwards, and 0 in free fall (a = g).
Try 3 NEET-style questions
Q1. A force of 10 N acts on a body of mass 2 kg initially at rest. Its acceleration is
- 0.2 m/s²
- 5 m/s²
- 20 m/s²
- 12 m/s²
Show answer
Answer: (2) 5 m/s²
a = F/m = 10/2 = 5 m/s².
Q2. A force of 10 N acts at 60° above the horizontal on a 2 kg block resting on a smooth horizontal floor. The block's acceleration is
- 5 m/s²
- 2.5 m/s²
- 4.3 m/s²
- 10 m/s²
Show answer
Answer: (2) 2.5 m/s²
Only the horizontal component moves it: 10 cos 60° = 5 N, so a = 5/2 = 2.5 m/s².
Q3. A block of mass 5 kg rests on a horizontal surface with coefficient of kinetic friction 0.2. Taking g = 10 m/s², the kinetic friction on the block is
- 2 N
- 10 N
- 25 N
- 50 N
Show answer
Answer: (2) 10 N
Normal reaction N = mg = 50 N. Kinetic friction = μk N = 0.2 × 50 = 10 N.
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