Class 11 Physics · Original EZT practice
Laws of Motion (Class 11 Chapter 4): summary and practice
Laws of Motion turns pushes and pulls into exact rules: inertia keeps a body steady, F = ma sizes up any acceleration, action-reaction pairs never cancel because they act on different bodies, and friction and circular motion fall out of the same three laws. All questions are original EZT tutor practice following the chapter's own section order.
6 original worked examples, 6 fresh self-check questions and 4 mixed questions, with explained answers. These are not copied textbook exercises, an official chapter index or a full answer key. No affiliation with NCERT is claimed.
Official sources for topic and edition context, checked 21 October 2026:
1. Inertia and Newton's first law
Aristotle's fallacy says motion needs a constant push; the chapter shows uniform motion needs no force at all. Inertia is a body's resistance to changes in its state of rest or uniform motion, and mass measures it. The first law says a body keeps its velocity unless a net external force acts.
2. Newton's second law and momentum
Momentum is p = mv. The second law says net force equals the rate of change of momentum, F = dp/dt, which becomes the everyday F = ma when mass is constant. A cricketer pulls the hands back while catching to stretch the time and shrink the force.
3. Newton's third law
For every action there is an equal and opposite reaction - but the two forces act on different bodies, so they never cancel each other. Walking pushes the ground back; the ground pushes you forward.
4. Friction (section 4.9.1)
Static friction adjusts up to a maximum f_s <= mu_s N; once sliding starts, kinetic friction is nearly constant at f_k = mu_k N, opposing relative motion. Friction is what lets us walk and lets cars corner.
5. Circular motion and solving problems
Uniform circular motion needs a net centripetal force F = mv^2/r towards the centre - friction or tension usually supplies it. The chapter's method: isolate the body, draw every force on a free-body diagram, resolve along axes, then apply F = ma to each direction.
Choose, solve, check, then try a fresh question.
Before calculating, read the method prompt and name the condition that makes it useful. Write your own attempt. Open the steps only when you need them, and use the check to catch a wrong sign, unit or interpretation. Finish with the fresh question before reading its answer.
Size a force with F = ma
Choose the method: Net force equals mass times acceleration.
EZT practice question
A 2 kg block accelerates at 3 m/s^2 on a smooth floor. What net force acts on it?
Show the worked solution
- F = ma.
- F = 2 x 3.
Answer: F = 6 N in the direction of the acceleration.
Check: Units work out: kg m/s^2 is exactly a newton.
Watch for this: Use the net force, not any single push - on a rough floor you must subtract friction first.
Now try it yourself
A 5 kg trolley speeds up at 4 m/s^2. Find the net force.
Check your answer
20 N.
F = ma = 5 x 4 = 20 N.
Tell mass from weight
Choose the method: Mass is inertia in kg; weight is the force W = mg.
EZT practice question
Find the weight of a 70 kg person, taking g = 9.8 m/s^2.
Show the worked solution
- W = mg.
- W = 70 x 9.8.
Answer: W = 686 N downwards.
Check: A 1 kg mass weighs 9.8 N, so 70 kg must weigh 70 times that.
Watch for this: Mass stays 70 kg everywhere; weight changes with g - do not quote kg as a force.
Now try it yourself
What is the weight of a 10 kg bag?
Check your answer
98 N.
W = mg = 10 x 9.8 = 98 N.
Slide a block against friction
Choose the method: Kinetic friction f_k = mu_k N, then net force = push - friction.
EZT practice question
A 10 kg block slides on a floor with mu_k = 0.3. Find the kinetic friction, and the acceleration under a 50 N push.
Show the worked solution
- N = mg = 10 x 9.8 = 98 N, so f_k = 0.3 x 98 = 29.4 N.
- Net force = 50 - 29.4 = 20.6 N.
- a = 20.6/10 = 2.06 m/s^2.
Answer: f_k = 29.4 N and a = 2.06 m/s^2.
Check: Without friction a would be 5 m/s^2 - friction must slow it, and 2.06 is less.
Watch for this: Friction uses the normal force N, not the push - on a flat floor N = mg.
Now try it yourself
The same block is pushed with only 20 N from rest. Does it slide?
Check your answer
No - it stays at rest.
Even the maximum static friction is about mu_s N, at least 29.4 N here, which beats the 20 N push.
Conserve momentum in a collision
Choose the method: Total momentum before equals total after when no net external force acts.
EZT practice question
A 2 kg lump moving at 3 m/s sticks to a 1 kg lump at rest. Find their common speed.
Show the worked solution
- Momentum before = 2 x 3 + 1 x 0 = 6 kg m/s.
- Combined mass = 3 kg, so v = 6/3.
Answer: v = 2 m/s in the original direction.
Check: Momentum after = 3 x 2 = 6 kg m/s - matches.
Watch for this: Do not average the speeds - momentum is mass times velocity, so masses matter.
Now try it yourself
A 3 kg ball at 2 m/s sticks to a 3 kg ball at rest. Common speed?
Check your answer
1 m/s.
Momentum before = 6 kg m/s over a combined 6 kg, giving 1 m/s.
Sort action-reaction pairs
Choose the method: The pair acts on two different bodies, so the forces never cancel.
EZT practice question
A book rests on a table. Name the third-law pairs involved and explain why the book does not fly off.
Show the worked solution
- Earth pulls the book down (weight); the book pulls Earth up - one pair.
- Table pushes the book up (normal force); the book pushes the table down - a second pair.
- Weight and normal force both act on the book, so they are NOT a third-law pair; they balance because the book is in equilibrium.
Answer: The pairs are book-Earth and book-table; weight and normal force merely balance on the same body.
Check: Each third-law partner acts on a different body - that is the test.
Watch for this: Never add action and reaction on one body to claim zero net force.
Now try it yourself
A bat hits a ball. Which force is larger, bat on ball or ball on bat?
Check your answer
Neither - they are equal and opposite.
The third law sizes them the same; the ball accelerates more because its mass is smaller.
Find the centripetal force
Choose the method: F = mv^2/r towards the centre.
EZT practice question
A 0.5 kg stone whirls at 4 m/s on a string of radius 2 m. What force pulls it towards the centre?
Show the worked solution
- F = mv^2/r.
- F = 0.5 x 16/2 = 0.5 x 8.
Answer: F = 4 N towards the centre, supplied by the string's tension.
Check: Doubling the speed would quadruple the force - the v^2 dependence is the signature.
Watch for this: There is no outward centrifugal force on the stone - the only real force points inward.
Now try it yourself
A 1 kg mass moves at 3 m/s in a circle of radius 3 m. Find the centripetal force.
Check your answer
3 N.
F = 1 x 9/3 = 3 N.
Try a mixed set
Choose a method or explanation before opening each answer.
1. What is inertia?
Show answer and reason
A body's resistance to any change in its state of rest or uniform motion.
Mass is the measure of inertia.
2. Newton's second law?
Show answer and reason
F = dp/dt, which is F = ma for constant mass.
Force changes momentum, not velocity directly.
3. Why don't action and reaction cancel?
Show answer and reason
They act on different bodies.
Only forces on the same body can balance.
4. Kinetic friction formula?
Show answer and reason
f_k = mu_k N.
Static friction instead adjusts up to mu_s N.
Try it yourself
Work these on paper before checking any notes. Answers are deliberately not shown here.
- A 5 kg trolley accelerates at 4 m/s^2. Find the net force on it.
- A 3 kg ball moving at 2 m/s sticks to a 3 kg ball at rest. Find their common speed.
- A net force of 20 N acts on a 4 kg block. Find its acceleration.
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Questions about this practice page
- What does Newton's first law say?
- A body at rest stays at rest, and a body in uniform straight-line motion keeps moving that way, unless a net external force acts on it. Uniform motion needs no force - that is Aristotle's fallacy overturned.
- Why do action-reaction pairs not cancel?
- Because the two forces act on two different bodies. Only forces acting on the same body can add to zero, so a pair never cancels itself out.
- What is friction in the laws of motion?
- A contact force that opposes relative motion (or its tendency) between surfaces. Static friction adjusts up to mu_s N; once sliding starts, kinetic friction is about mu_k N.
- Why is static friction written with <= ?
- Static friction is self-adjusting: it matches the applied push only up to its maximum mu_s N. Writing f_s <= mu_s N captures that it is usually less than the cap.
- Are these the NCERT exercise questions for Laws of Motion (Class 11)?
- No. EZT pages are original practice following the chapter's section order - use them to prepare for the textbook exercises, not as a substitute for your book.