F = ma
F = μN
F = (m₁ + m₂)v/(m₁ + m₂)
a = Δv/Δt
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
F₁₂ = -F₂₁
T = 2π √(l/g)
W = F × s
P = Fv
K = (1/2)mv²
U = mgh
Impulse = Δp = FΔt

🪤 The 5 Mistakes That Cost Marks

Assuming friction is always kinetic, when ∈ fact it can be static or kinetic
Forgetting to consider the normal force ∈ problems involving inclined planes
Not using the correct equation for centripetal force: F = (mv²)/r
Incorrectly applying Newton's third law, forgetting that forces always come ∈ pairs
Not considering the system as a whole when applying conservation of momentum

✏️ 3 Solved PYQs

A block of mass 2 kg is moving with a velocity of 4 m/s on a frictionless surface. A force of 6 N is applied to the block ∈ the direction of motion. Calculate the acceleration of the block.
- F = ma, so a = F/m = 6/2 = 3 m/s²
A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. Calculate the force applied by the brakes.
- v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s²
- F = ma, so F = 1500 × -1.67 = -2505 N
A particle of mass 1 kg is attached to a string of length 1 m. The particle is rotating ∈ a circular path with a velocity of 2 m/s. Calculate the tension ∈ the string.
- F = (mv²)/r, so T = (1 × 2²)/1 = 4 N

🧠 The One Thing Most Students Get Wrong

Most students get the concept of pseudo force wrong. Pseudo force is a fictitious force that appears to act on an object when it is ∈ a non-inertial frame of reference. It is equal ∈ magnitude and opposite ∈ direction to the force that is actually acting on the object.

👁️ Ayush's Note

To solve problems ∈ laws of motion, always start by drawing a free body diagram. This will help you identify all the forces acting on the object and make it easier to apply Newton's laws.
Always consider the system as a whole when applying conservation of momentum.
Make sure to use the correct equation for centripetal force: F = (mv²)/r.
Pseudo force is a key concept ∈ laws of motion. Make sure you understand it well.

🔁 Last 5 Minutes Box

Last minute revision: go through all the formulas and make sure you understand the concepts of Newton's laws, friction, centripetal force, and pseudo force.
Check your notes and make sure you have all the formulas written down.
Go through the solved PYQs and make sure you understand the steps.
Take a few deep breaths and try to relax. You have prepared well for this exam.

📝 Practice MCQs

1. A block of mass 1 kg is moving with a velocity of 2 m/s on a frictionless surface. A force of 3 N is applied to the block ∈ the direction of motion. What is the acceleration of the block?

A) 1 m/s²

B) 2 m/s²

C) 3 m/s²

D) 4 m/s²

Answer: C) 3 m/s². Explanation: F = ma, so a = F/m = 3/1 = 3 m/s²

2. A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. What is the force applied by the brakes?

A) -1000 N

B) -1500 N

C) -2000 N

D) -2505 N

Answer: D) -2505 N. Explanation: v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s², F = ma, so F = 1500 × -1.67 = -2505 N

3. A particle of mass 1 kg is attached to a string of length 1 m. The particle is rotating ∈ a circular path with a velocity of 2 m/s. What is the tension ∈ the string?

A) 2 N

B) 3 N

C) 4 N

D) 5 N

Answer: C) 4 N. Explanation: F = (mv²)/r, so T = (1 × 2²)/1 = 4 N

4. A block of mass 2 kg is moving with a velocity of 4 m/s on a frictionless surface. A force of 6 N is applied to the block ∈ the direction of motion. What is the acceleration of the block?

A) 1 m/s²

B) 2 m/s²

C) 3 m/s²

D) 4 m/s²

Answer: B) 3 m/s². Explanation: F = ma, so a = F/m = 6/2 = 3 m/s²

5. A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. What is the distance traveled by the car before it comes to rest?

A) 50 m

B) 100 m

C) 150 m

D) 166.7 m

Answer: D) 166.7 m. Explanation: v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s², s = ut + (1/2)at² = 60 × (1000/3600) × 10 + (1/2) × -1.67 × 10² = 166.7 m

🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

🎬 Watch video explanations on YouTube →

This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides:

F = ma
F = μN
F = (m₁ + m₂)v/(m₁ + m₂)
a = Δv/Δt
v = u + at
s = ut + (1/2)at²
v² = u² + 2as
F₁₂ = -F₂₁
T = 2π √(l/g)
W = F × s
P = Fv
K = (1/2)mv²
U = mgh
Impulse = Δp = FΔt

🪤 The 5 Mistakes That Cost Marks

Assuming friction is always kinetic, when ∈ fact it can be static or kinetic
Forgetting to consider the normal force ∈ problems involving inclined planes
Not using the correct equation for centripetal force: F = (mv²)/r
Incorrectly applying Newton's third law, forgetting that forces always come ∈ pairs
Not considering the system as a whole when applying conservation of momentum

✏️ 3 Solved PYQs

A block of mass 2 kg is moving with a velocity of 4 m/s on a frictionless surface. A force of 6 N is applied to the block ∈ the direction of motion. Calculate the acceleration of the block.
- F = ma, so a = F/m = 6/2 = 3 m/s²
A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. Calculate the force applied by the brakes.
- v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s²
- F = ma, so F = 1500 × -1.67 = -2505 N
A particle of mass 1 kg is attached to a string of length 1 m. The particle is rotating ∈ a circular path with a velocity of 2 m/s. Calculate the tension ∈ the string.
- F = (mv²)/r, so T = (1 × 2²)/1 = 4 N

🧠 The One Thing Most Students Get Wrong

Most students get the concept of pseudo force wrong. Pseudo force is a fictitious force that appears to act on an object when it is ∈ a non-inertial frame of reference. It is equal ∈ magnitude and opposite ∈ direction to the force that is actually acting on the object.

👁️ Ayush's Note

To solve problems ∈ laws of motion, always start by drawing a free body diagram. This will help you identify all the forces acting on the object and make it easier to apply Newton's laws.
Always consider the system as a whole when applying conservation of momentum.
Make sure to use the correct equation for centripetal force: F = (mv²)/r.
Pseudo force is a key concept ∈ laws of motion. Make sure you understand it well.

🔁 Last 5 Minutes Box

Last minute revision: go through all the formulas and make sure you understand the concepts of Newton's laws, friction, centripetal force, and pseudo force.
Check your notes and make sure you have all the formulas written down.
Go through the solved PYQs and make sure you understand the steps.
Take a few deep breaths and try to relax. You have prepared well for this exam.

📝 Practice MCQs

1. A block of mass 1 kg is moving with a velocity of 2 m/s on a frictionless surface. A force of 3 N is applied to the block ∈ the direction of motion. What is the acceleration of the block?

A) 1 m/s²

B) 2 m/s²

C) 3 m/s²

D) 4 m/s²

Answer: C) 3 m/s². Explanation: F = ma, so a = F/m = 3/1 = 3 m/s²

2. A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. What is the force applied by the brakes?

A) -1000 N

B) -1500 N

C) -2000 N

D) -2505 N

Answer: D) -2505 N. Explanation: v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s², F = ma, so F = 1500 × -1.67 = -2505 N

3. A particle of mass 1 kg is attached to a string of length 1 m. The particle is rotating ∈ a circular path with a velocity of 2 m/s. What is the tension ∈ the string?

A) 2 N

B) 3 N

C) 4 N

D) 5 N

Answer: C) 4 N. Explanation: F = (mv²)/r, so T = (1 × 2²)/1 = 4 N

4. A block of mass 2 kg is moving with a velocity of 4 m/s on a frictionless surface. A force of 6 N is applied to the block ∈ the direction of motion. What is the acceleration of the block?

A) 1 m/s²

B) 2 m/s²

C) 3 m/s²

D) 4 m/s²

Answer: B) 3 m/s². Explanation: F = ma, so a = F/m = 6/2 = 3 m/s²

5. A car of mass 1500 kg is moving with a velocity of 60 km/h. The driver applies the brakes and the car comes to rest ∈ 10 s. What is the distance traveled by the car before it comes to rest?

A) 50 m

B) 100 m

C) 150 m

D) 166.7 m

Answer: D) 166.7 m. Explanation: v = u + at, so a = (v - u)/t = (0 - 60 × (1000/3600))/10 = -1.67 m/s², s = ut + (1/2)at² = 60 × (1000/3600) × 10 + (1/2) × -1.67 × 10² = 166.7 m

🚀 Ready to Ace Your Exam?

Put your knowledge to the test! Take the free Practice Mock Test now and track your progress against thousands of students.

🎬 Watch video explanations on YouTube →

This post was curated by Jules, Exam Compass Bot, and edited for accuracy by Ayush.

📚 Related Topics

Continue your revision with these related guides: