Laws of Motion - NCERT Solutions

Chapter Overview

The Laws of Motion form the foundation of Classical Mechanics and are essential for understanding various physical phenomena. This chapter introduces Newton’s three laws of motion, which govern the relationship between forces and motion.

Key Concepts:

• Force: Interaction that causes acceleration
• Inertia: Resistance to change in state of motion
• Momentum: Product of mass and velocity
• Impulse: Change in momentum due to force
• Friction: Force opposing relative motion

NCERT Exercise Solutions

Exercise 5.1

Question: State the reason why a block placed on a table does not start moving on its own.

Solution: A block placed on a table does not start moving on its own because:

1. Balanced Forces: The gravitational force (weight) acting downward is balanced by the normal reaction force from the table acting upward.
2. No Unbalanced Force: Since there is no horizontal force acting on the block, there is no net force to cause acceleration.
3. Newton’s First Law: According to Newton’s first law of motion, an object at rest remains at rest unless acted upon by an unbalanced force.

Exercise 5.2

Question: A constant force acting on a body of mass 3.0 kg changes its speed from 2.0 m s⁻¹ to 3.5 m s⁻¹ in 25 s. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Solution: Given:

• Mass (m) = 3.0 kg
• Initial velocity (u) = 2.0 m/s
• Final velocity (v) = 3.5 m/s
• Time (t) = 25 s

Using Newton’s second law: F = ma

First, calculate acceleration (a): a = (v - u)/t = (3.5 - 2.0)/25 = 1.5/25 = 0.06 m/s²

Now, calculate force (F): F = ma = 3.0 × 0.06 = 0.18 N

Answer: The magnitude of the force is 0.18 N, and its direction is the same as the direction of motion.

Exercise 5.3

Question: A bullet of mass 10 g traveling horizontally with a velocity of 150 m s⁻¹ strikes a stationary wooden block and comes to rest in 0.03 s. Calculate the distance of penetration of the bullet into the block. Also calculate the magnitude of the force exerted by the wooden block on the bullet.

Solution: Given:

• Mass (m) = 10 g = 0.01 kg
• Initial velocity (u) = 150 m/s
• Final velocity (v) = 0 m/s
• Time (t) = 0.03 s

Step 1: Calculate acceleration Using v = u + at: 0 = 150 + a(0.03) a = -150/0.03 = -5000 m/s²

Step 2: Calculate distance penetrated Using v² = u² + 2as: 0 = (150)² + 2(-5000)s 0 = 22500 - 10000s 10000s = 22500 s = 2.25 m

Step 3: Calculate force F = ma = 0.01 × (-5000) = -50 N

Answer:

• Distance of penetration = 2.25 m
• Force exerted by the block = 50 N (opposite to the direction of motion)

Exercise 5.4

Question: A motorcar of mass 1200 kg is moving along a straight line with a uniform velocity of 90 km/h. Its velocity is slowed down to 18 km/h in 4 s by an unbalanced external force. Calculate the acceleration and change in momentum. Also calculate the magnitude of the force required.

Solution: Given:

• Mass (m) = 1200 kg
• Initial velocity (u) = 90 km/h = 25 m/s
• Final velocity (v) = 18 km/h = 5 m/s
• Time (t) = 4 s

Step 1: Convert velocities to m/s

• u = 90 × (1000/3600) = 25 m/s
• v = 18 × (1000/3600) = 5 m/s

Step 2: Calculate acceleration a = (v - u)/t = (5 - 25)/4 = -20/4 = -5 m/s²

Step 3: Calculate change in momentum Δp = m(v - u) = 1200 × (5 - 25) = 1200 × (-20) = -24000 kg·m/s

Step 4: Calculate force F = ma = 1200 × (-5) = -6000 N

Answer:

• Acceleration = -5 m/s² (deceleration)
• Change in momentum = -24000 kg·m/s
• Force required = 6000 N (opposite to direction of motion)

NCERT Highlights for JEE/NEET

🎯 Important Concepts for JEE:

1. Newton’s Second Law in Vector Form: $\vec{F} = \frac{d\vec{p}}{dt} = m\vec{a}$
2. Impulse-Momentum Theorem: $\vec{J} = \Delta\vec{p} = \int_{t_1}^{t_2} \vec{F}dt$
3. Conservation of Momentum: In absence of external forces, total momentum remains conserved
4. Friction Laws:
   • Static friction: $f_s \leq \mu_s N$
   • Kinetic friction: $f_k = \mu_k N$

🧬 Key Points for NEET:

1. Inertia: Directly proportional to mass
2. Newton’s Third Law: Action and reaction are equal and opposite, act on different bodies
3. Equilibrium Conditions: $\sum \vec{F} = 0$ and $\sum \vec{\tau} = 0$
4. Free Body Diagrams: Essential for problem-solving

In-Text Examples

Example 5.1

Question: A railway engine 1800 kg moving with a velocity of 8 km/h, collides with a stationary wagon of mass 2000 kg. After collision, they stick together and move with a common velocity. Find their common velocity after collision.

Solution: Given:

• Mass of engine (m₁) = 1800 kg
• Velocity of engine (u₁) = 8 km/h = 2.22 m/s
• Mass of wagon (m₂) = 2000 kg
• Velocity of wagon (u₂) = 0 m/s
• After collision: they stick together (perfectly inelastic collision)

Using conservation of momentum: $m_1u_1 + m_2u_2 = (m_1 + m_2)v$

$1800 × 2.22 + 2000 × 0 = (1800 + 2000)v$

$3996 = 3800v$

$v = 3996/3800 = 1.05$ m/s = 3.78 km/h

Answer: Common velocity = 3.78 km/h

JEE/NEET Practice Questions

Question 1 (JEE Level)

Statement: A block of mass m is placed on a rough inclined plane at angle θ. The coefficient of friction between the block and plane is μ. The minimum force required to move the block up the incline is:

Solution: For the block to move up the incline:

• Component of weight down the plane: mg sinθ
• Frictional force down the plane: μN = μmg cosθ
• Total force to overcome: mg sinθ + μmg cosθ

Answer: $F_{min} = mg(\sin\theta + \mu\cos\theta)$

Question 2 (NEET Level)

Statement: A person of mass 70 kg stands in a lift. The lift starts moving upward with uniform acceleration of 2 m/s². The apparent weight of the person is:

Solution: When lift accelerates upward:

• Real weight: W = mg = 70 × 9.8 = 686 N
• Additional force due to acceleration: F = ma = 70 × 2 = 140 N
• Apparent weight: $W_{apparent} = mg + ma = 686 + 140 = 826$ N

Answer: 826 N

Important Formulas

Common Mistakes to Avoid

1. Ignoring Direction: Force and acceleration are vector quantities
2. Wrong Free Body Diagram: Not including all forces acting on the body
3. Sign Conventions: Not maintaining consistent sign conventions
4. Friction Direction: Assuming friction always opposes motion without analyzing the situation

Success Tips

1. Draw Free Body Diagrams: Always start with a clear FBD
2. Apply Newton’s Laws Systematically: Follow a step-by-step approach
3. Check Units: Ensure all quantities are in SI units
4. Practice Different Scenarios: Cover various types of problems

Master Laws of Motion

Understanding Newton’s laws is crucial for success in JEE/NEET. Practice regularly and focus on conceptual clarity!

Remember: Newton’s laws are the foundation of mechanics. Master these concepts to excel in physics!

For additional practice, check out our Physics Practice Problems section.