Questions and Answers (Subjective)

Q1. State Newton's second law of motion and prove that it is the 'real' law of motion.

Answer: Newton's Second Law of Motion states that the rate of change of the linear momentum of a body is directly proportional to the net external force applied on the body, and this change takes place in the direction of the applied force. $$\vec{F} = \frac{d\vec{p}}{dt}$$

It is considered the 'real' law of motion because both the first and third laws can be derived from it.

1. First Law from Second Law:

The first law describes the motion of a body in the absence of a net external force. According to the second law, if $\vec{F}_{ext} = 0$, then $\frac{d\vec{p}}{dt} = 0$. This implies that the momentum $\vec{p}$ is constant. Since $\vec{p} = m\vec{v}$ and mass m is constant, the velocity $\vec{v}$ must also be constant. A constant velocity means the body is either at rest or in uniform motion, which is precisely the statement of the First Law.

2. Third Law from Second Law:

Consider an isolated system of two bodies, A and B, that interact with each other. Since the system is isolated, the net external force is zero, and the total momentum of the system is conserved. $$\vec{p}_A + \vec{p}_B = \text{constant}$$ Differentiating with respect to time: $$\frac{d}{dt}(\vec{p}_A + \vec{p}_B) = 0$$ $$\frac{d\vec{p}_A}{dt} + \frac{d\vec{p}_B}{dt} = 0$$ From the second law, $\frac{d\vec{p}_A}{dt}$ is the force on A due to B ($\vec{F}_{AB}$), and $\frac{d\vec{p}_B}{dt}$ is the force on B due to A ($\vec{F}_{BA}$). Therefore: $$\vec{F}_{AB} + \vec{F}_{BA} = 0 \implies \vec{F}_{AB} = -\vec{F}_{BA}$$ This is the statement of the Third Law.

Q2. Define static friction, limiting friction, and kinetic friction. Show their variation with the applied force using a graph.

Answer:

• Static Friction ($f_s$): The force of friction which comes into play between two bodies before one body actually starts moving over the other. It is a self-adjusting force that is always equal and opposite to the applied force, up to its maximum limit.
• Limiting Friction ($f_{s,max}$): The maximum value of static friction which comes into play when a body is just about to slide over the surface of another body. $f_{s,max} = \mu_s N$, where $\mu_s$ is the coefficient of static friction.
• Kinetic Friction ($f_k$): The force of friction which comes into play when a body is in a state of steady motion over the surface of another body. It is approximately constant and is given by $f_k = \mu_k N$, where $\mu_k$ is the coefficient of kinetic friction. Generally, $\mu_k < \mu_s$.

Graph: The graph shows that as the applied force increases, the static friction increases linearly with it, keeping the body at rest. When the applied force exceeds the limiting friction, the body starts to move, and the friction opposing the motion drops to the constant value of kinetic friction.

Q3. Why is centripetal force required for circular motion? Explain why banking of roads is necessary for vehicles taking a turn at high speeds.

Answer: Centripetal force is required for circular motion because an object moving in a circle is continuously changing its direction. A change in the direction of velocity constitutes an acceleration, even if the speed is constant. According to Newton's second law, a net force is required to produce an acceleration. For circular motion, this acceleration (centripetal acceleration, $a_c = v^2/r$) is always directed towards the center of the circle. Therefore, there must be a net force, the centripetal force, directed towards the center to cause this change in velocity.

Banking of Roads: When a car takes a turn on a flat (unbanked) road, the necessary centripetal force is provided solely by the force of static friction between the tires and the road. $$\frac{mv^2}{r} = f_s \le \mu_s N = \mu_s mg$$ This sets a maximum safe speed, $v_{max} = \sqrt{\mu_s g r}$. If the speed of the car exceeds this value, the frictional force is insufficient, and the car will skid.

To allow vehicles to turn at higher speeds safely, roads are banked. By tilting the road surface by an angle $\theta$, the normal force (N) from the road is also tilted. This tilted normal force has a horizontal component, $N\sin\theta$, which points towards the center of the turn. This horizontal component provides some or all of the required centripetal force. This reduces the reliance on friction, allowing for a much higher maximum safe speed.

Q4. Three blocks are connected as shown in the diagram on a horizontal frictionless table and pulled to the right with a force $F=60$ N. If $m_1 = 10$ kg, $m_2 = 20$ kg, and $m_3 = 30$ kg, find the tensions $T_1$ and $T_2$ in the strings.

Answer:

1. Find the acceleration of the entire system: The three blocks move together as a single system. The total mass is $M = m_1 + m_2 + m_3 = 10 + 20 + 30 = 60$ kg. The net external force is $F = 60$ N. Using Newton's Second Law for the system: $$a = \frac{F}{M} = \frac{60\ N}{60\ kg} = 1\ m/s^2$$

2. Find Tension $T_2$: Consider the FBD of mass $m_3$. The only horizontal force acting on it is the tension $T_2$ pulling it to the left. The question implies F is pulling $m_3$ to the right. Let's assume F pulls $m_1$ to the right. Then the string between $m_2$ and $m_3$ has tension $T_2$. The force equation for $m_3$ is $T_2 = m_3a$. $T_2 = 30 \times 1 = 30$ N.

3. Find Tension $T_1$: Consider the FBD of mass $m_2$. The force pulling it right is $T_1$, and the force pulling it left is $T_2$. The net force is $T_1 - T_2 = m_2a$. $T_1 - 30 = 20(1) \implies T_1 = 20 + 30 = 50$ N.

Let's re-read the standard diagram for this. Usually, F pulls the first block, $m_1$. Let's assume that. Then the force equation on $m_3$ is $T_2 = m_3 a = 30 \times 1 = 30$ N. The force equation on $m_2$ is $T_1 - T_2 = m_2 a \implies T_1 - 30 = 20 \times 1 \implies T_1 = 50$ N. Check with $m_1$: $F - T_1 = m_1 a \implies 60 - 50 = 10 \times 1 \implies 10 = 10$. It is consistent.

The tensions are $T_1 = 50\ N$ and $T_2 = 30\ N$.

Solved Examples from Laws of Motion

These examples are designed to cover various concepts from Newton’s laws:

Example 1: Friction on Inclined Plane

Q: A 2 kg block is placed on a 30° inclined plane with coefficient of static friction 0.5. Will it slide?

Solution: Force down the plane: $mg\sin\theta = 2 \cdot 9.8 \cdot 0.5 = 9.8$ N Normal force: $N = mg\cos\theta = 2 \cdot 9.8 \cdot 0.866 = 16.97$ N Max friction: $f_s = \mu_s N = 0.5 \cdot 16.97 = 8.48$ N Since 9.8 > 8.48, block will slide.

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Example 2: Pulley - Atwood Machine

Q: Masses 4 kg and 6 kg are suspended over a pulley. Find acceleration and tension.

Solution: Acceleration: $a = \frac{6 - 4}{6 + 4} \cdot 9.8 = 1.96$ m/s² Tension: $T = \frac{2 \cdot 6 \cdot 4}{6 + 4} \cdot 9.8 = 47.04$ N

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Example 3: Banking without Friction

Q: A car moves at 18 m/s on a road curve of radius 30 m. What should be the banking angle (no friction)?

Solution: $\tan\theta = \frac{v^2}{rg} = \frac{18^2}{30 \cdot 9.8} = \frac{324}{294} = 1.10$ $\theta = \tan^{-1}(1.10) ≈ 47.7°$

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Example 4: Apparent Weight in Lift

Q: A person of mass 60 kg stands in a lift accelerating upward at 2 m/s². What is his apparent weight?

Solution: Apparent weight $= N = m(g + a) = 60(9.8 + 2) = 708$ N

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Example 5: Force and Acceleration

Q: A force of 25 N is applied to a 5 kg block. What is the acceleration?

Solution: $a = F/m = 25/5 = 5$ m/s²

Q5. Derive the expression for tension and acceleration in an Atwood machine.

Answer: Let masses be $m_1$ and $m_2$ with $m_1 > m_2$.

• Net force = $(m_1 - m_2)g$
• Total mass = $m_1 + m_2$ So, acceleration: $$a = \frac{(m_1 - m_2)g}{m_1 + m_2}$$ Tension: $$T = \frac{2m_1 m_2}{m_1 + m_2} g$$

Q6. Explain the concept of apparent weight. How does it change in a lift?

Answer: Apparent weight is the normal force exerted on a body. In a lift:

• At rest or constant velocity: $N = mg$
• Accelerating upward: $N = m(g + a)$ → heavier
• Accelerating downward: $N = m(g - a)$ → lighter
• Free fall: $N = 0$ → weightless

Q7. A 10 kg box is pulled with a 50 N force at 30° to the horizontal. Friction coefficient is 0.3. Find acceleration.

Answer:

• Horizontal component: $F_x = 50\cos30° = 43.3\,N$
• Vertical component: $F_y = 50\sin30° = 25\,N$
• Normal: $N = mg - F_y = 98 - 25 = 73\,N$
• Friction: $f = 0.3 \times 73 = 21.9\,N$
• Net Force: $F = 43.3 - 21.9 = 21.4\,N$
• Acceleration: $a = F/m = 21.4/10 = 2.14\,m/s^2$