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EAPCET Physics · Intermediate first year

Laws of Motion

Laws of Motion covers Newton’s three laws, momentum and impulse, friction, and the dynamics of circular motion.

Subject
Physics
Taught in
Intermediate first year
Physics in the paper
40 of 160 questions

Key formulas and results

  • F=dpdt=maF = \frac{dp}{dt} = ma
  • Impulse =F Δt=Δp= F\,\Delta t = \Delta p
  • Static friction f≤μsNf \le \mu_{s}N; kinetic friction f=μkNf = \mu_{k}N
  • Angle of repose: tan⁡θ=μs\tan\theta = \mu_{s}
  • Banking without friction: tan⁡θ=v2rg\tan\theta = \frac{v^{2}}{rg}
  • Apparent weight in a lift: m(g+a)m(g + a) when accelerating up, m(g−a)m(g - a) when accelerating down
  • Least force to drag a block on a rough floor (applied at tan⁡θ=μ\tan\theta = \mu): Fmin⁡=μmg1+μ2F_{\min} = \frac{\mu mg}{\sqrt{1 + \mu^{2}}}
  • Rough incline: sliding down a=g(sin⁡θ−μcos⁡θ)a = g(\sin\theta - \mu\cos\theta); moving up it retards at g(sin⁡θ+μcos⁡θ)g(\sin\theta + \mu\cos\theta)
  • Stopping distance of a block sliding on a rough floor: s=u22μgs = \frac{u^{2}}{2\mu g}
  • Conical pendulum: ω=gtan⁡θr\omega = \sqrt{\frac{g\tan\theta}{r}}, period T=2πlcos⁡θgT = 2\pi\sqrt{\frac{l\cos\theta}{g}}
  • Cyclist on a turn leans at tan⁡θ=v2rg\tan\theta = \frac{v^{2}}{rg}; maximum safe speed on a flat turn vmax⁡=μrgv_{\max} = \sqrt{\mu rg}
  • Well of death: minimum speed v=rg/μv = \sqrt{rg/\mu}
  • Force to hold a gun firing nn bullets of mass mm at speed vv in time tt: F=nmvtF = \frac{nmv}{t}; conveyor belt: F=vdmdtF = v\frac{dm}{dt}

What this chapter covers

  • Newton’s laws of motion
  • Momentum and impulse
  • Conservation of momentum
  • Friction
  • Connected bodies
  • Circular motion and banking

How it is usually asked

  • Bodies connected by strings and pulleys
  • Friction on horizontal and inclined surfaces
  • Banking of roads
  • Impulse and momentum

Understanding Laws of Motion

Newton’s first law says a body keeps its state of rest or uniform motion unless a net external force acts. The second law makes this quantitative: the net force equals the rate of change of momentum, which reduces to F = ma for constant mass. The third law says forces come in equal and opposite pairs, but these pairs act on different bodies, so they never cancel each other in one free-body diagram.

Friction opposes relative motion. Static friction adjusts itself up to a limit μₛN, while kinetic friction is roughly constant at μₖN. In circular motion some real force, such as friction, tension or a component of the normal reaction, must provide the centripetal force mv²/r; banking a road lets the normal reaction supply it.

EAPCET tests this chapter through blocks connected by strings and pulleys, inclines with friction, apparent weight in lifts, impulse and the banking of roads. The reliable method is always the same: isolate each body, draw every force acting on it, choose axes along and perpendicular to the motion, and write F = ma separately for each body before solving the equations together.

Worked example

EAPCET-style question

Two blocks of masses 4 kg and 6 kg are in contact on a smooth horizontal surface. A horizontal force of 20 N is applied to the 4 kg block, pushing both blocks. The contact force between the blocks is

  1. (1)12 N
  2. (2)8 N
  3. (3)20 N
  4. (4)10 N

Solution

  1. Treat both blocks as one system: a=Fm1+m2=2010=2 m/s2a = \frac{F}{m_{1} + m_{2}} = \frac{20}{10} = 2\ \text{m/s}^{2}.
  2. The only horizontal force on the 6 kg block is the contact force N from the 4 kg block.
  3. N=6×2=12 NN = 6 \times 2 = 12\ \text{N}.
  4. Check on the 4 kg block: 20−12=8 N=4×220 - 12 = 8\ \text{N} = 4 \times 2, which is consistent.

Answer: 12 N

Shortcuts and results to remember

  • Block on an incline: tan⁡θ>μ\tan\theta > \mu slides down; tan⁡θ=μ\tan\theta = \mu is just about to slide; tan⁡θ<μ\tan\theta < \mu stays at rest and static friction equals mgsin⁡θmg\sin\theta, not μN\mu N.
  • If the time to slide down a rough incline is nn times the time to go up the same distance, tan⁡θ=μn2+1n2−1\tan\theta = \mu\frac{n^{2} + 1}{n^{2} - 1}; for n=2n = 2, tan⁡θ=5μ3\tan\theta = \frac{5\mu}{3}.
  • Chain on a rough table: the largest fraction that can hang over the edge is μμ+1\frac{\mu}{\mu + 1}.
  • A block on a truck floor does not slip if the truck’s acceleration a≤μga \le \mu g; a block pressed against the truck’s front face does not fall if a≥g/μa \ge g/\mu.
  • Pulley constraint: for an ideal string, ∑(tension×velocity of each end)=0\sum(\text{tension} \times \text{velocity of each end}) = 0, which links the velocities and accelerations of the blocks.

More practice questions

Practice question 1

A 10 kg block rests on a rough incline of 37° with μ=0.8\mu = 0.8. The frictional force on it is (g=10 m/s2g = 10\ \text{m/s}^{2}, sin⁡37∘=3/5\sin 37^\circ = 3/5)

  1. (1)64 N
  2. (2)60 N
  3. (3)80 N
  4. (4)48 N

Solution

  1. Component of weight down the plane: mgsin⁡37∘=100×35=60 Nmg\sin 37^\circ = 100 \times \tfrac{3}{5} = 60\ \text{N}
  2. Normal force N=mgcos⁡37∘=80 NN = mg\cos 37^\circ = 80\ \text{N}, so limiting friction =0.8×80=64 N= 0.8 \times 80 = 64\ \text{N}
  3. Since 60 N<64 N60\ \text{N} < 64\ \text{N}, the block stays at rest
  4. Static friction balances the pull: f=60 Nf = 60\ \text{N} (not 64 N)

Answer: 60 N

Practice question 2

A 10 kg block on a rough floor (μ=0.6\mu = 0.6) is pulled by a 100 N force at 37° above the horizontal. Its acceleration is (g=10 m/s2g = 10\ \text{m/s}^{2})

  1. (1)2 m/s²
  2. (2)5.6 m/s²
  3. (3)8 m/s²
  4. (4)3.2 m/s²

Solution

  1. Horizontal part =100cos⁡37∘=80 N= 100\cos 37^\circ = 80\ \text{N}; vertical part =100sin⁡37∘=60 N= 100\sin 37^\circ = 60\ \text{N}
  2. Normal force N=100−60=40 NN = 100 - 60 = 40\ \text{N}
  3. Kinetic friction =0.6×40=24 N= 0.6 \times 40 = 24\ \text{N}
  4. a=80−2410=5.6 m/s2a = \frac{80 - 24}{10} = 5.6\ \text{m/s}^{2}

Answer: 5.6 m/s²

Common mistakes to avoid

  • Cancelling an action–reaction pair on the same body — the two forces act on different bodies.
  • Always writing friction as μN — static friction equals the applied force until the limiting value μₛN is reached.
  • Getting apparent weight in a lift wrong — it is m(g + a) when accelerating upwards and m(g − a) when accelerating downwards.
  • Adding a centrifugal force while working in the ground (inertial) frame — in that frame only real forces act, and their resultant provides mv²/r.

How to prepare it

Draw a free-body diagram for every body before writing any equation.

Once the formulas feel familiar, test them the way the exam does: practise Laws of Motion questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Physics chapters

See every chapter in the EAPCET Physics syllabus.

Practise Laws of Motion chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Laws of Motion part of the EAPCET Physics syllabus?
Yes. Laws of Motion is taught in Intermediate first year Physics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Laws of Motion for EAPCET?
Draw a free-body diagram for every body before writing any equation.
How many Physics questions are there in EAPCET?
40 of the 160 questions in the Engineering paper are Physics, spread across 30 chapters.
What mistakes do students make in Laws of Motion?
Cancelling an action–reaction pair on the same body — the two forces act on different bodies. Always writing friction as μN — static friction equals the applied force until the limiting value μₛN is reached. Getting apparent weight in a lift wrong — it is m(g + a) when accelerating upwards and m(g − a) when accelerating downwards. Adding a centrifugal force while working in the ground (inertial) frame — in that frame only real forces act, and their resultant provides mv²/r.