Maharashtra Board Class 11 Physics Chapter 4 Laws of Motion Solutions

1. Choose The Correct Answer.

Question 1. Consider following pair of forces of equal magnitude and opposite directions: (P) Gravitational forces exerted on each other by two point masses separated by a distance. (Q) Couple of forces used to rotate a water tap. (R) Gravitational force and normal force experienced by an object kept on a table. For which of these pair/pairs the two forces do NOT cancel each other's translational effect?
(A) Only P
(B) Only P and Q
(C) Only R
(D) Only Q and R
Answer: (A) Only P
In simple words: Only in case P, the forces are an action-reaction pair acting on different objects, hence they do not cancel each other's translational effect on a single body. In Q and R, forces act on the same body or are not an action-reaction pair in the context of translational cancellation.

🎯 Exam Tip: Understanding Newton's third law and distinguishing between action-reaction pairs and balanced forces acting on a single object is crucial for scoring well.

Question 2. Consider following forces: (w) Force due to tension along a string, (x) Normal force given by a surface, (y) Force due to air resistance and (z) Buoyant force or upthrust given by a fluid. Which of these are electromagnetic forces?
(A) Only w, y and z
(B) Only w, x and y
(C) Only x, y and z
(D) All four.
Answer: (D) All four.
In simple words: All the listed forces- tension, normal force, air resistance, and buoyant force- originate from electromagnetic interactions between atoms and molecules. They are manifestations of the fundamental electromagnetic force.

🎯 Exam Tip: Recognize that many everyday forces, including contact forces and fluid forces, are fundamentally electromagnetic in origin, stemming from interatomic and intermolecular interactions.

Question 3. At a given instant three point masses m, 2m and 3m are equidistant from each other. Consider only the gravitational forces between them. Select correct statement/s for this instance only:
(A) Mass m experiences maximum force.
(B) Mass 2m experiences maximum force.
(C) Mass 3m experiences maximum force.
(D) All masses experience force of same magnitude.
Answer: (C) Mass 3m experiences maximum force.
In simple words: Gravitational force is proportional to the product of masses. The mass 3m, being the largest, will experience the strongest gravitational attraction from both m and 2m, resulting in the maximum net force.

🎯 Exam Tip: Remember Newton's Law of Gravitation, \(F = \frac{Gm_1 m_2}{r^2}\). The net force on a mass is the vector sum of forces from all other masses, and the largest mass will generally experience the greatest total force when interacting with other masses.

Question 4. The rough surface of a horizontal table offers a definite maximum opposing force to initiate the motion of a block along the table, which is proportional to the resultant normal force given by the table. Forces \(F_1\) and \(F_2\) act at the same angle \(\theta\) with the horizontal and both are just initiating the sliding motion of the block along the table. Force \(F_1\) is a pulling force while the force \(F_2\) is a pushing force. \(F_2 > F_1\), because
(A) Component of \(F_2\) adds up to weight to increase the normal reaction.
(B) Component of \(F_1\) adds up to weight to increase the normal reaction.
(C) Component of \(F_2\) adds up to the opposing force.
(D) Component of \(F_1\) adds up to the opposing force.
Answer: (A) Component of \(F_2\) adds up to weight to increase the normal reaction.
In simple words: When pushing down on the block (Force \(F_2\)), a vertical component of \(F_2\) adds to the block's weight, increasing the total downward force on the table. This leads to a greater normal reaction force, which in turn increases the maximum static friction, requiring a larger \(F_2\) to initiate motion compared to pulling up slightly (Force \(F_1\)).

🎯 Exam Tip: Always consider all components of forces, especially in situations involving friction. The normal force is crucial for determining friction and can be affected by the vertical components of applied forces.

Question 5. A mass 2m moving with some speed is directly approaching another mass m moving with double speed. After some time, they collide with coefficient of restitution 0.5. Ratio of their respective speeds after collision is
(A) 2/3
(B) 3/2
(C) 2
(D) 1/2
Answer: (B) 3/2
In simple words: This problem requires applying conservation of momentum and the definition of the coefficient of restitution to a collision scenario, then solving the resulting system of equations to find the ratio of final speeds.

🎯 Exam Tip: For collision problems, always list knowns and unknowns, then apply the conservation of linear momentum and the coefficient of restitution equation for inelastic collisions to derive the final velocities.

Question 6. A uniform rod of mass 2m is held horizontal by two sturdy, practically inextensible vertical strings tied at its ends. A boy of mass 3m hangs himself at one third length of the rod. Ratio of the tension in the string close to the boy to that in the other string is
(A) 2
(B) 1.5
(C) 4/3
(D) 5/3
Answer: (B) 1.5
In simple words: To solve this, apply the conditions for rotational equilibrium (net torque is zero) and translational equilibrium (net force is zero). Take torques about one end of the rod to find the tensions in the strings, then calculate their ratio.

🎯 Exam Tip: When dealing with equilibrium problems involving extended objects, ensure both net force and net torque are zero. Choosing the pivot point strategically can simplify torque calculations by eliminating unknown forces.

Question 7. Select WRONG statement about centre of mass:
(A) Centre of mass of a 'C' shaped uniform rod can never be a point on that rod.
(B) If the line of action of a force passes through the centre of mass, the moment of that force is zero.
(C) Centre of mass of our Earth is not at its geometrical centre.
(D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.
Answer: (D) While balancing an object on a pivot, the line of action of the gravitational force of the earth passes through the centre of mass of the object.
In simple words: The gravitational force effectively acts at the center of gravity. For balancing, the line of action of the gravitational force must pass through the pivot, which means the center of gravity (or center of mass for uniform gravity) must be directly above or below the pivot. Statement (D) is incorrect because for balance, the *line of action of gravity* must pass through the pivot, implying the *center of gravity* is at the pivot.

🎯 Exam Tip: Clearly differentiate between center of mass and center of gravity, although they are often the same for objects in a uniform gravitational field. Understand that a force applied through the center of mass/gravity produces no torque, crucial for equilibrium concepts.

Question 8. For which of the following objects will the centre of mass NOT be at their geometrical centre?
(I) An egg
(II) a cylindrical box full of rice
(III) a cubical box containing assorted sweets
(A) Only (I)
(B) Only (I) and (II)
(C) Only (III)
(D) All, (I), (II) and (III).
Answer: (D) All, (I), (II) and (III).
In simple words: The center of mass aligns with the geometrical center only for objects with uniform density and symmetrical shape. An egg has an asymmetrical mass distribution, and boxes filled with loose, assorted items (rice, sweets) will have an uneven mass distribution, shifting their center of mass away from the geometrical center.

🎯 Exam Tip: The center of mass depends on both the shape and the distribution of mass. Symmetrical objects with uniform density have their center of mass at the geometric center, but any non-uniformity or asymmetry will shift it.

2. Answer The Following Questions.

Question 1. In the following table, every entry on the left column can match with any number of entries on the right side. Pick up all those and write respectively against A, B, C and D.

Answer: 1. Force due to tension in string: Electromagnetic (EM) force, reaction force, non-conservative force. 2. Normal force: Electromagnetic (EM) force, non-conservative force. Reaction force 3. Frictional force: Electromagnetic (EM) force, reaction force, non-conservative force. 4. Resistive force offered by air or water for objects moving through it: Electromagnetic (EM) force, non-conservative force.
In simple words: All the listed forces-tension, normal, frictional, and resistive-are fundamentally electromagnetic in origin, arising from interatomic interactions. They are also reaction forces. Frictional and resistive forces are non-conservative as they dissipate energy, while tension and normal forces are generally non-conservative as well in real-world applications where energy is not fully recovered.

🎯 Exam Tip: Classify forces based on their fundamental nature (gravitational, electromagnetic, strong, weak) and their energy conservation properties (conservative vs. non-conservative). Understand that most contact forces are electromagnetic and non-conservative.

Question 2. In real life objects, never travel with uniform velocity, even on a horizontal surface, unless something is done? Why is it so? What is to be done?
Answer: 1. According to Newton's first law, for a body to achieve uniform velocity, the net force acting on it should be zero. 2. In real life, a body in motion is constantly being acted on by resistive or opposing force like friction, in the direction opposite to that of the motion. 3. To overcome these opposing forces, an additional external force is required. Thus, the net force is not maintained at zero, making it hard to achieve uniform velocity.
In simple words: Objects on a horizontal surface in real life experience resistive forces like friction. To maintain uniform velocity, these opposing forces must be continuously counteracted by an applied external force, otherwise, the net force won't be zero and the object will decelerate.

🎯 Exam Tip: Connect real-world observations to Newton's first law. Emphasize the role of resistive forces (like friction and air resistance) in preventing objects from maintaining uniform velocity without continuous external effort.

Question 3. For the study of any kind of motion, we never use Newton's first law of motion directly. Why should it be studied?
Answer: 1. Newton's first law shows an equivalence between the 'state of rest' and 'state of uniform motion along a straight line.' 2. Newton's first law of motion defines force as a physical quantity that brings about a change in 'state of rest' or 'state of uniform motion along a straight line' of a body. 3. Newton's first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line. Due to all these reasons, Newton's first law should be studied.
In simple words: Newton's first law, while not directly used for calculations, is foundational. It establishes the concept of inertia, defines force as what causes changes in motion (acceleration), and highlights the equivalence of rest and uniform motion, providing the basis for understanding force and motion.

🎯 Exam Tip: Explain the qualitative significance of Newton's first law: it defines inertia and provides a conceptual framework for understanding force as a cause of motion change, even if quantitative analysis primarily uses the second law.

Question 4. Are there any situations in which we cannot apply Newton's laws of motion? Is there any alternative for it?
Answer: 1. Limitation: Newton's laws of motion cannot be applied for objects moving in non-inertial (accelerated) frame of reference. Alternative solution: For non-inertial (accelerated) frame of reference, pseudo force needs to be considered along with all the other forces. 2. Limitation: Newton's laws of motion are applicable to point objects and rigid bodies. Alternative solution: Body needs to be approximated as a particle as the laws can be applied to individual particles in a rigid body and then summed up over the body. 3. Limitation: Newton's laws of motion cannot be applied for objects moving with speeds comparable to that of light. Alternative solution: Einstein's special theory of relativity has to be used. 4. Limitation: Newton's laws of motion cannot be applied for studying the behaviour and interactions of objects having atomic or molecular sizes. Alternative solution: Quantum mechanics has to be used.
In simple words: Newton's laws have limitations; they don't apply directly in accelerated frames (requiring pseudo forces), to objects moving near the speed of light (requiring special relativity), or to atomic/molecular scale phenomena (requiring quantum mechanics). They are also best suited for point masses or rigid bodies.

🎯 Exam Tip: Be aware of the domains where Newton's laws are valid. Highlighting their limitations and the appropriate alternative theories (relativity, quantum mechanics, pseudo forces) demonstrates a deeper understanding of physics principles.

Question 5. You are inside a closed capsule from where you are not able to see anything about the outside world. Suddenly you feel that you are pushed towards your right. Can you explain the possible cause (s)? Is it a feeling or a reality? Give at least one more situation like this.
Answer: 1. In a capsule, if we suddenly feel a push towards the right it is because the capsule is in motion and taking a turn towards the left. 2. The push towards the right is a feeling. In reality, when the capsule is beginning its turning motion towards the left, we continue in a straight line. 3. This happens because we try to maintain our direction of motion while the capsule takes a turn towards the left. 4. An external force is required to change our direction of motion. In accordance with one of the inferences from Newton's first law of motion, in the absence of any external force, we continue to move in a straight line at constant speed and feel the sudden push in the direction opposite to the motion of the capsule. 5. Example: While travelling by bus, when the bus takes a sudden turn we feel the push in the opposite direction.
In simple words: The feeling of being pushed to the right is due to inertia, not a real force pushing you. The capsule turns left, but your body, by inertia, tries to continue moving straight, relative to the turning capsule, it feels like a push to the right. This is similar to feeling pushed outwards when a bus takes a sharp turn.

🎯 Exam Tip: Clearly explain inertia and frames of reference. Distinguish between actual forces and pseudo-forces (like the sensation of being pushed) experienced in non-inertial frames, providing relevant examples.

Question 6. Among the four fundamental forces, only one force governs your daily life almost entirely. Justify the statement by stating that force.
Answer: 1. Electromagnetic force is the attractive and repulsive force between electrically charged particles. 2. Since electromagnetic force is much stronger than the gravitational force, it dominates all the phenomena on atomic and molecular scales. 3. Majority of the forces experienced in our daily life like friction, normal reaction, tension in strings, elastic forces, viscosity etc. are electromagnetic in nature. 4. The structure of atoms and molecules, the dynamics of chemical reactions etc. are governed by electromagnetic forces. Thus, out of the four fundamental forces, electromagnetic force governs our daily life almost entirely.
In simple words: The electromagnetic force governs almost all daily life phenomena because it's responsible for the interactions between atoms and molecules. Forces like friction, normal force, tension, and chemical reactions are all manifestations of this fundamental force, dominating over gravity at the macroscopic level and weak/strong nuclear forces which act at subatomic scales.

🎯 Exam Tip: Understand the relative strengths and ranges of the four fundamental forces. Explain how the electromagnetic force underpins common everyday phenomena, justifying its pervasive influence in our macroscopic world.

Question 7. Find the odd man out:
(i) Force responsible for a string to become taut on stretching
(ii) Weight of an object
(iii) The force due to which we can hold an object in hand.
Answer: Weight of an object. Reason: Weight of an object (force due to gravity) is a non-contact force while force responsible for a string to become taut (tension force) and force due to which we can hold an object in hand (normal force) are contact forces.
In simple words: Weight is the odd one out because it's a non-contact force (gravitational attraction), whereas tension in a string and the normal force from holding an object are both contact forces requiring direct physical interaction.

🎯 Exam Tip: Clearly differentiate between contact and non-contact forces. Remember that gravitational force (weight) acts over a distance, while forces like tension, friction, and normal force require physical contact between objects.

Question 8. You are sitting next to your friend on ground. Is there any gravitational force of attraction between you two? If so, why are you not coming together naturally? Is any force other than the gravitational force of the earth coming in picture?
Answer: 1. Yes, there exists a gravitational force between me and my friend sitting beside each other. 2. The gravitational force between any two objects is given by, \(F = \frac{Gm_1 m_2}{r^2}\) Where, G = universal gravitational constant, \(m_1\) and \(m_2\) = mass of the two objects, r = distance between centres of the two objects 3. Thus, me and my friend attract each other. But due to our small masses, we exert a force on each other, which is too small as compared to the gravitational force of the earth. Hence, me and my friend don't move towards each other. 4. Apart from gravitational force of the earth, there is the normal force and frictional force acting on both me and my friend. In Chapter 5, you will study about force of gravitation in detail.
In simple words: Yes, a gravitational force exists between you and your friend, but it's negligible because your masses are small. You don't move together due to the much stronger gravitational pull of the Earth and the presence of normal and frictional forces which are significantly larger.

🎯 Exam Tip: Explain that gravitational force is universal but weak for ordinary masses. Contrast the tiny gravitational attraction between individuals with the Earth's overwhelming gravitational pull and the contact forces (normal, friction) that prevent movement.

Question 9. Distinguish between:
(A) Real and pseudo forces,
(B) Conservative and non-conservative forces,
(C) Contact and non-contact forces,
(D) Inertial and non-inertial frames of reference.
Answer:
(A) Real and pseudo forces,

(B) Conservative and non-conservative forces,

(C) Contact and non-contact forces,

(D) Inertial and non-inertial frames of reference.

In simple words: This question requires understanding four key distinctions in forces and frames of reference: real forces are due to interactions, pseudo forces arise from accelerating frames; conservative forces allow mechanical energy conservation and path-independent work, unlike non-conservative forces; contact forces involve physical touch, non-contact forces act remotely; inertial frames are non-accelerating, non-inertial frames are accelerating.

🎯 Exam Tip: Mastering these distinctions is fundamental to understanding mechanics. Focus on the definitions, examples, and the implications of each type (e.g., energy conservation for conservative forces, appearance of pseudo forces in non-inertial frames).

Question 10. State the formula for calculating work done by a force. Are there any conditions or limitations in using it directly? If so, state those clearly. Is there any mathematical way out for it? Explain.
Answer: 1. Suppose a constant force \(\vec{F}\) acting on a body produces a displacement \(\vec{S}\) in the body along the positive X-direction. Then the work done by the force is given as, \(W = F \cdot s \cos \theta\) Where \(\theta\) is the angle between the applied force and displacement. 2. If displacement is in the direction of the force applied, \(\theta = 0^\circ\)
\(W = \vec{F} \cdot \vec{S}\) Conditions/limitations for application of work formula: 1. The formula for work done is applicable only if both force \(\vec{F}\) and displacement \(\vec{S}\) are constant and finite i.e., it cannot be applied when the force is variable. 2. The formula is not applicable in several real- life situations like lifting an object through several thousand kilometres since the gravitational force is not constant. It is not applicable to viscous forces like fluid resistance as they depend upon speed and thus are often not constant with time. 3. The method of integration has to be applied to find the work done by a variable force. Integral method to find work done by a variable force:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र (a) एक बल-विस्थापन ग्राफ को दर्शाता है। इसमें X-अक्ष पर विस्थापन (s) और Y-अक्ष पर बल (F) दिखाया गया है। वक्र A से B तक जाता है, और इसके नीचे एक छायांकित क्षेत्र है जो छोटे विस्थापन ds के लिए बल F को दर्शाता है, जिससे कार्य की गणना के लिए समाकलन विधि का उपयोग किया जाता है। 1. Let the force vary non-linearly in magnitude between the points A and B as shown in figure (a). 2. In order to calculate the total work done during the displacement from \(s_1\) to \(s_2\), we need to use integration. For integration, we need to divide the displacement into large numbers of infinitesimal (infinitely small) displacements. 3. Let at \(P_1\), the magnitude of force be \(F = P_1P_1'\). Due to this force, the body displaces through infinitesimally small displacement ds, in the direction of force. It moves from \(P_1\) to \(P_2\). \(\vec{ds} = P_1P_2\) 4. But direction of force and displacement are same, we have \(\vec{ds} = P_1P_2\) 5. \(\vec{ds}\) is so small that the force F is practically constant for the displacement. As the force is constant, the area of the strip \(\vec{F} \cdot \vec{ds}\) is the work done dW for this displacement. 6. Hence, small work done between \(P_1\) to \(P_2\) is dW and is given by
\(dW = \vec{F} \cdot \vec{ds} = P_1P_1' \times P_1P_2'\)
= Area of the strip \(P_1P_2P_2'P_1'\). 7. The total work done can be found out by dividing the portion AB into small strips like \(P_1P_2P_2'P_1'\) and taking sum of all the areas of the strips.
\( \implies W = \int_{s_{1}}^{s 2} \vec{F} \cdot d \vec{s}=\text { Area } A B B^{\prime} A^{\prime} \) 8. Method of integration is applicable if the exact way of variation in \(\vec{F}\) and \(\vec{s}\) is known and that function is integrable. 9. The work done by the non-linear variable force is represented by the area under the portion of force-displacement graph. 10. Similarly, in case of a linear variable force, the area under the curve from \(s_1\) to \(s_2\) (trapezium APQB) gives total work done W [figure (b)].
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र (b) एक बल-विस्थापन ग्राफ को दर्शाता है जहाँ बल F रैखिक रूप से विस्थापन s के साथ बदलता है। X-अक्ष पर विस्थापन (s) और Y-अक्ष पर बल (F) दिखाया गया है। वक्र A से B तक एक सीधी रेखा है, और \(s_1\) से \(s_2\) तक का क्षेत्र (एक समलंब APQB) इस रैखिक रूप से बदलते बल द्वारा किए गए कुल कार्य को दर्शाता है।
In simple words: Work done by a constant force is \(W = F \cdot s \cos \theta\). This formula is limited to constant forces; for variable forces, integration (\(W = \int \vec{F} \cdot d\vec{s}\)) is used to sum up infinitesimal work done over tiny displacements.

🎯 Exam Tip: Distinguish between work done by constant and variable forces. For variable forces, understand that work is the area under the force-displacement graph, which is calculated using integration, and be able to set up the integral correctly.

Question 11. Justify the statement, “Work and energy are the two sides of a coin".
Answer: 1. Work and energy both are scalar quantities. 2. Work and energy both have the same dimensions i.e., \([M^1L^2T^{-2}]\). 3. Work and energy both have the same units i.e., SI unit: joule and CGS unit: erg. 4. Energy refers to the total amount of work a body can do. 5. A body capable of doing more work possesses more energy and vice versa. 6. Work done on a body by a conservative force is equal to the change in its kinetic energy. Thus, work and energy are the two sides of the same
In simple words: Work and energy are intricately linked because they share the same units, dimensions, and are both scalar quantities. Energy is the capacity to do work, and work is the process of transferring energy. For example, work done by a conservative force equals the change in kinetic energy.

🎯 Exam Tip: To justify this statement, focus on the fundamental connections: energy is the capacity to do work, and work is the transfer of energy. Emphasize shared units, dimensions, and the work-energy theorem.

Question 12. From the terrace of a building of height H, you dropped a ball of mass m. It reached the ground with speed v. Is the relation \(mgH = \frac{1}{2}mv^2\) applicable exactly? If not, how can you account for the difference? Will the ball bounce to the same height from where it was dropped?
Answer: 1. Let the ball dropped from the terrace of a building of height h have mass m. During free fall, the ball is acted upon by gravity (accelerating conservative force). 2. While coming down, the work that is done is equal to the decrease in the potential energy. 3. This work done however is not entirely converted into kinetic energy but some part of it is used in overcoming the air resistance (retarding non-conservative force). This part of energy appears in some other forms such as heat, sound, etc. 4. Thus, in this case of an accelerating conservative force along with a retarding non-conservative force, the work-energy theorem is given as, Decrease in the gravitational P.E. = Increase in the kinetic energy + work done against non-conservative forces. 5. Thus, the relation \(mgH = \frac{1}{2}mv^2\) is not applicable when non-conservative forces are considered. The part of the energy converted to heat, sound etc also needs to be added to the equation, 6. The ball will not bounce to the same height from where it was dropped due to the loss in kinetic energy during the collision making it an inelastic collision.
In simple words: The relation \(mgH = \frac{1}{2}mv^2\) is not exactly applicable because air resistance (a non-conservative force) converts some potential energy into heat and sound. Therefore, the ball will not bounce back to the same height due to energy loss during the inelastic collision with the ground.

🎯 Exam Tip: For real-world energy conservation problems, always consider non-conservative forces like air resistance and friction. Explain that these forces cause a loss of mechanical energy, leading to phenomena like a ball not bouncing back to its initial height.

Question 13. State the law of conservation of linear momentum. It is a consequence of which law? Given an example from our daily life for conservation of momentum. Does it hold good during burst of a cracker?
Answer: 1. Statement: The total momentum of an isolated system is conserved during any interaction. 2. The law of conservation of linear momentum is a consequence of Newton's second law of motion, (in combination with Newton's third law) 3. Example: When a nail is driven into a wall by striking it with a hammer, the hammer is seen to rebound after striking the nail. This is because the hammer imparts a certain amount of momentum to the nail and the nail imparts an equal and opposite amount of momentum to the hammer. Linear momentum conservation during the burst of a cracker: - The law of conservation of linear momentum holds good during bursting of a cracker. - When a cracker is at rest before explosion, the linear momentum of the cracker is zero. - When cracker explodes into number of pieces, scattered in different directions, the vector sum of linear momentum of these pieces is also zero. This is as per the law of conservation of linear momentum.
In simple words: The law of conservation of linear momentum states that the total momentum of an isolated system remains constant. It's a consequence of Newton's second and third laws. An exploding cracker demonstrates this; its initial zero momentum is conserved, so the vector sum of the momenta of all scattered pieces after explosion is also zero.

🎯 Exam Tip: Clearly define an isolated system and state the conservation law. Connect it to Newton's laws and provide a real-world example (like an explosion or recoil) that demonstrates the vector nature of momentum conservation.

Question 14. Define coefficient of restitution and obtain its value for an elastic collision and a perfectly inelastic collision.
Answer: (i) For two colliding bodies, the negative of ratio of relative velocity of separation to relative velocity of approach is called the coefficient of restitution.
(ii) Consider an head-on collision of two bodies of masses \(m_1\) and \(m_2\) with respective initial velocities \(u_1\) and \(u_2\). As the collision is head-on, the colliding masses are along the same line before and after the collision. Relative velocity of approach is given as,
\(U_a = u_2 - u_1\)
Let \(v_1\) and \(v_2\) be their respective velocities after the collision. The relative velocity of recede (or separation) is then \(V_s = v_2 - v_1\)
Therefore, \(e = - \frac{V_s}{U_a} = - \frac{v_2 - v_1}{u_2 - u_1} = \frac{v_1 - v_2}{u_2 - u_1}\) ....(1)
(iii) For a head-on elastic collision, According to the principle of conservation of linear momentum,
Total initial momentum = Total final momentum
Therefore, \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\) ....(2)
\(m_1 (u_1 - v_1) = m_2 (v_2 - u_2)\) ....(3)
As the collision is elastic, total kinetic energy of the system is also conserved.
Therefore, \( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 = \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \) ....(4)
\(m_1 (u_1^2 - v_1^2) = m_2 (v_2^2 - u_2^2)\)
\(m_1 (u_1 + v_1) (u_1 - v_1) = m_2 (v_2 + u_2) (v_2 - u_2)\) ....(5)
Dividing equation (5) by equation (3), we get
\(u_1 + v_1 = u_2 + v_2\)
Therefore, \(u_2 - u_1 = v_1 - v_2\) ....(6)
Substituting this in equation (1),
\(e = \frac{v_1 - v_2}{u_2 - u_1} = 1\)
(iv) For a perfectly inelastic collision, the colliding bodies move jointly after the collision, i.e., \(v_1 = v_2\)
Therefore, \(v_1 - v_2 = 0\)
Substituting this in equation (1), \(e = 0\).
In simple words: Coefficient of restitution measures how "bouncy" a collision is. For a perfectly elastic collision, it's 1 (perfect bounce), and for a perfectly inelastic collision, it's 0 (objects stick together).

🎯 Exam Tip: Remember the definitions and values of 'e' for elastic and inelastic collisions as they are fundamental to solving collision problems and frequently asked in exams.

Question 15. Discuss the following as special cases of elastic collisions and obtain their exact or approximate final velocities in terms of their initial velocities.
(i) Colliding bodies are identical.
(ii) A very heavy object collides on a lighter object, initially at rest.
(iii) A very light object collides on a comparatively much massive object, initially at rest.
Answer: The final velocities after a head-on elastic collision is given as, \[v_1 = \left( \frac{m_1 - m_2}{m_1 + m_2} \right) u_1 + \left( \frac{2m_2}{m_1 + m_2} \right) u_2\] \[v_2 = \left( \frac{2m_1}{m_1 + m_2} \right) u_1 + \left( \frac{m_2 - m_1}{m_1 + m_2} \right) u_2\] (i) Colliding bodies are identical
If \(m_1 = m_2\), then \(v_1 = u_2\) and \(v_2 = u_1\)
Thus, objects will exchange their velocities after head-on elastic collision.
In simple words: When two identical objects collide elastically head-on, they simply swap their velocities.

🎯 Exam Tip: Understanding velocity exchange in identical elastic collisions simplifies many problems and highlights the conservation principles at play.

(ii) A very heavy object collides on a lighter object, initially at rest.
Let \(m_1\) be the mass of the heavier body and \(m_2\) be the mass of the lighter body i.e., \(m_1 \gg m_2\); lighter particle is at rest i.e., \(u_2 = 0\) then,
\(m_1 - m_2 \approx m_1\) and \(\frac{m_2}{m_1 + m_2} \approx 0\),
Therefore, \(v_1 \approx u_1\) and \(v_2 = 2u_1\)
i.e., the heavier colliding body is left unaffected and the lighter body which is struck, travels with double the speed of the massive striking body.
In simple words: If a very heavy object hits a light object at rest, the heavy object continues almost unchanged, and the light object bounces off with double the heavy object's speed.

🎯 Exam Tip: This case demonstrates that the momentum and energy transfer from a heavy object to a light one can result in a significant speed increase for the lighter object, which is crucial for understanding impact dynamics.

(iii) A very light object collides on a comparatively much massive object, initially at rest.
If \(m_1\) is mass of a light body and \(m_2\) is mass of heavy body i.e., \(m_1 \ll m_2\) and \(u_2 = 0\). Thus, \(m_1\) can be neglected.
Hence \(v_1 \approx -u_1\), and \(v_2 \approx 0\).
i.e., the tiny (lighter) object rebounds with same speed while the massive object is unaffected.
In simple words: When a very light object hits a massive object at rest, the light object bounces back with its initial speed, and the massive object remains almost stationary.

🎯 Exam Tip: This scenario illustrates why hitting a wall with a small ball barely moves the wall, but the ball rebounds sharply, emphasizing the effect of mass disparity in collisions.

Question 16. A bullet of mass \(m_1\) travelling with a velocity u strikes a stationary wooden block of mass \(m_2\) and gets embedded into it. Determine the expression for loss in the kinetic energy of the system. Is this violating the principle of conservation of energy? If not, how can you account for this loss?
Answer: 1. A bullet of mass \(m_1\) travelling with a velocity u, striking a stationary wooden block of mass \(m_2\) and getting embedded into it is a case of perfectly inelastic collision.
2. In a perfectly inelastic collision, although there is a loss in kinetic energy, the principle of conservation of energy is not violated as the total energy of the system is conserved.
Loss in the kinetic energy during a perfectly inelastic head-on collision:
1. Let two bodies A and B of masses \(m_1\) and \(m_2\) move with initial velocity \(\overrightarrow{u_1}\), and \(u_2\) respectively such that particle A collides head-on with particle B i.e., \(u_1 > u_2\).
2. If the collision is perfectly inelastic, the particles stick together and move with a common velocity \(\vec{V}\) after the collision along the same straight line.
loss in kinetic energy = total initial kinetic energy – total final kinetic energy,
3. By the law of conservation of momentum, \(m_1 u_1 + m_2 u_2 = (m_1 + m_2) v\)
Therefore, \(v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}\)
4. Loss of Kinetic energy, \[\Delta K.E. = \left( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \right) - \frac{1}{2} (m_1 + m_2) v^2\] \[= \left( \frac{1}{2} m_1 u_1^2 + \frac{1}{2} m_2 u_2^2 \right) - \frac{1}{2} (m_1 + m_2) \left( \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2} \right)^2\] \[= \frac{m_1^2 u_1^2 + m_1 m_2 u_2^2 + m_1 m_2 u_1^2 + m_2^2 u_2^2 - (m_1 u_1 + m_2 u_2)^2}{2(m_1 + m_2)}\] \[= \frac{m_1^2 u_1^2 + m_1 m_2 u_2^2 + m_1 m_2 u_1^2 + m_2^2 u_2^2 - (m_1^2 u_1^2 + m_2^2 u_2^2 + 2m_1 m_2 u_1 u_2)}{2(m_1 + m_2)}\] \[= \frac{m_1 m_2 u_2^2 + m_1 m_2 u_1^2 - 2m_1 m_2 u_1 u_2}{2(m_1 + m_2)}\] \[= \frac{m_1 m_2 (u_1 - u_2)^2}{2(m_1 + m_2)}\] 5. Both the masses and the term \((u_1 - u_2)^2\) are positive. Hence, there is always a loss in a perfectly inelastic collision. For a perfectly inelastic collision, as \(e = 0\), the loss is maximum.
In simple words: In a perfectly inelastic collision like a bullet embedding in a block, kinetic energy is lost, but total energy is still conserved; the "lost" kinetic energy is converted into other forms like heat, sound, or deformation.

🎯 Exam Tip: Always remember that while kinetic energy might not be conserved in inelastic collisions, the total mechanical energy is transformed, not destroyed, often into heat, sound, or deformation energy. Linear momentum, however, is always conserved in all types of collisions within an isolated system.

Question 17. One of the effects of a force is to change the momentum. Define the quantity related to this and explain it for a variable force. Usually when do we define it instead of using the force?
Answer: 1. Impulse is the quantity related to change in momentum.
2. Impulse is defined as the change of momentum of an object when the object is acted upon by a force for a given time interval.
Need to define impulse:
1. In cases when time for which an appreciable force acting on an object is extremely small, it becomes difficult to measure the force and time independently.
2. In such cases, however, the effect of the force i.e, the change in momentum due to the force is noticeable and can be measured.
3. For such cases, it is convenient to define impulse itself as a physical quantity.
4. Example: Hitting a ball with a bat, giving a kick to a foot-ball, hammering a nail, bouncing a ball from a hard surface, etc.
Impulse for a variable force:
1. Consider the collision between a bat and ball. The variation of the force as a function of time is shown below. The force axis is starting from zero.
2. From the graph, it can be seen that the force is zero before the impact. It rises to a maximum during the impact and decreases to zero after the impact.
3. The shaded area or the area under the curve of the force-time graph gives the product of force against the corresponding time (\(\Delta t\)) which is the impulse of the force.
Area of ABCDE = F. \(\Delta t\) = impulse of force
4. For a constant force, the area under the curve is a rectangle.
5. In case of a softer tennis ball, the collision time becomes larger and the maximum force becomes less keeping the area under curve of the (F-t) graph same.
Area of ABCDE = Area of PQRST
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक चर बल के कारण आवेग को दर्शाता है। इसमें एक बल-समय ग्राफ दिखाया गया है जहाँ बल (F) y-अक्ष पर और समय (t) x-अक्ष पर है। ग्राफ एक वक्र बनाता है और वक्र के नीचे छायांकित क्षेत्र (ABCDE) बल के आवेग का प्रतिनिधित्व करता है, जिसे J = fΔt द्वारा दर्शाया गया है।
In chapter 3, you have studied the concept of using area under the curve.
In simple words: Impulse is the measure of the effect of a force applied over a short time, calculated as the change in momentum. It's useful when the force itself varies or acts for a very brief duration, making the force-time graph's area a clear indicator.

🎯 Exam Tip: Impulse-momentum theorem is vital: Impulse = Change in Momentum. The area under a Force-Time graph always gives the impulse, regardless of whether the force is constant or variable.

Question 18. While rotating an object or while opening a door or a water tap we apply a force or forces. Under which conditions is this process easy for us? Why? Define the vector quantity concerned. How does it differ for a single force and for two opposite forces with different lines of action?
Answer: 1. Opening a door can be done with ease if the force applied is:
• proportional to the mass of the object
• far away from the axis of rotation and the direction of force is perpendicular to the line joining the axis of rotation with the point of application of force.
2. This is because, the rotational ability of a force depends not only upon the magnitude and direction of force but also on the point where the force acts with respect to the axis of rotation.
3. Rotating an object like a water tap can be done with ease if the two forces are equal in magnitude but opposite in direction are applied along different lines of action.
4. The ability of a force to produce rotational motion is measured by its turning effect called 'moment of force' or 'torque'.
5. However, a moment of couple or rotational effect of a couple is also called torque.
6. For differences in the two vector quantities.

In simple words: Opening a door is easiest when force is applied perpendicular to the door and farthest from the hinges, creating a large moment of force (torque). A couple, which is two equal and opposite forces on different lines, produces pure rotational motion, and its moment is independent of the axis of rotation.

🎯 Exam Tip: Distinguishing between the moment of a single force (torque) and the moment of a couple is crucial. A single force can cause both rotational and translational motion, while a couple causes only rotational motion, independent of the pivot point chosen on the rigid body.

Question 19. Why is the moment of a couple independent of the axis of rotation even if the axis is fixed?
Answer: 1. Consider a rectangular sheet free to rotate only about a fixed axis of rotation, perpendicular to the plane.
2. A couple of forces \(\vec{F}\) and \(-\vec{F}\) is acting on the sheet at two different locations.
3. Consider the torque of the couple as two torques due to individual forces causing rotation about the axis of rotation.
4. Case 1: The axis of rotation is between the lines of action of the two forces constituting the couple. Let x and y be the perpendicular distances of the axis of rotation from the forces \(\vec{F}\) and \(-\vec{F}\) respectively.
In this case, the pair of forces cause anticlockwise rotation. As a result, the direction of individual torques due to the two forces is the same.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक आयताकार शीट पर कार्य करने वाले बल युग्म को दर्शाता है जो एक निश्चित घूर्णन अक्ष के चारों ओर घूम सकती है। बल F और -F विपरीत दिशाओं में कार्य कर रहे हैं, जिनकी अक्ष से लंबवत दूरी क्रमशः x और y है, जिससे शीट में वामावर्त घूर्णन होता है।
5. Case 2: Lines of action of both the forces are on the same side of the axis of rotation. Let q and p be the perpendicular distances of the axis of rotation from the forces \(\vec{F}\) and \(-\vec{F}\)
In simple words: The moment of a couple is the product of one force and the perpendicular distance between the two forces. This distance is fixed regardless of where the axis of rotation is chosen, making the couple's moment independent of the axis.

🎯 Exam Tip: The key insight is that the net force of a couple is zero, but its net torque is non-zero and purely rotational. This characteristic makes it uniquely useful for generating rotational motion without translational acceleration, and its effect doesn't depend on the pivot point.

Question 20. Explain balancing or mechanical equilibrium. Linear velocity of a rotating fan as a whole is generally zero. Is it in mechanical equilibrium? Justify your answer.
Answer: 1. The state in which the momentum of a system is constant in the absence of an external unbalanced force is called mechanical equilibrium.
2. A particle is said to be in mechanical equilibrium, if no net force is acting upon it.
3. In case of a system of bodies to be in mechanical equilibrium, the net force acting on any part of the system should be zero i.e., the velocity or linear momentum of all parts of the system must be constant or zero. There should be no acceleration in any part of the system.
4. Mathematically, for a system in mechanical equilibrium, \(\sum \vec{F} = 0\).
5. In case of rotating fan, if linear velocity is zero, then the linear momentum is zero. That means there is no net force acting on the fan. Hence, the fan is in mechanical equilibrium.
In simple words: Mechanical equilibrium means a system has constant momentum (which can be zero) because the net external force on it is zero. A rotating fan at a constant speed, as a whole, has zero linear velocity and thus zero linear momentum, implying it is in mechanical equilibrium with respect to translation.

🎯 Exam Tip: Distinguish between translational and rotational equilibrium. For translational equilibrium, net force is zero. For rotational equilibrium, net torque is zero. A rotating fan can be in translational equilibrium even while rotating.

Question 21. Why do we need to know the centre of mass of an object? For which objects, its position may differ from that of the centre of gravity? Use \(g = 10 \text{ m s}^{-2}\), unless, otherwise stated.
Answer: 1. Centre of mass of an object allows us to apply Newton's laws of motion to finite objects (objects of measurable size) by considering these objects as point objects.
2. For objects in non-uniform gravitational field or whose size is comparable to that of the Earth (size at least few thousand km), the position of centre of mass will differ than that of centre of gravity.
In simple words: We need the center of mass to treat extended objects as point masses for simpler physics calculations. It can differ from the center of gravity if the gravitational field isn't uniform over the object, which usually happens only for very large objects.

🎯 Exam Tip: The center of mass is a geometric property, while the center of gravity depends on the gravitational field. They coincide in a uniform gravitational field, which is often the case in basic physics problems, but not always for large celestial bodies or non-uniform fields.

3. Solve the following problems.

Question 1. A truck of mass 5 ton is travelling on a horizontal road with 36 km \( \text{hr}^{-1} \) stops on travelling 1 km after its engine fails suddenly. What fraction of its weight is the frictional force exerted by the road? If we assume that the story repeats for a car of mass 1 ton i.e., can moving with same speed stops in similar distance same how much will the fraction be?
Answer: [Ans: \(\frac{1}{200}\) in the both]
Solution:
Given: \(m_{\text{truck}} = 5 \text{ ton} = 5000 \text{ kg}\),
\(m_{\text{car}} = 1 \text{ ton} = 1000 \text{ kg}\),
\(u = 36 \text{ km/hr} = 10 \text{ m/s}\),
\(v = 0 \text{ m/s}\), \(s = 1 \text{ km} = 1000 \text{ m}\)
To find: Ratio of force of friction to the weight of vehicle
Formulae:
i. \(v^2 = u^2 + 2as\)
ii. \(F = ma\)
Calculation: From formula (i),
\(2 \times a_{\text{truck}} \times s = v^2 - u^2\)
Therefore, \(2 \times a_{\text{truck}} \times 1000 = 0^2 - 10^2\)
Therefore, \(2000 a_{\text{truck}} = -100\)
Therefore, \(a_{\text{truck}} = -0.05 \text{ m/s}^2\)
Negative sign indicates that velocity is decreasing
From formula (ii),
\(F_{\text{truck}} = m_{\text{truck}} \times a_{\text{truck}} = 5000 \times 0.05\)
\( = 250 \text{ N}\)
Therefore, \(\frac{F_{\text{truck}}}{\text{weight}_{\text{truck}}} = \frac{250}{5000 \times 10} = \frac{1}{200}\)
From formula (i),
\(2 \times a_{\text{car}} \times s = v^2 - u^2\)
\(2 \times a_{\text{car}} \times 1000 = 0^2 - 10^2\)
\(2000 a_{\text{car}} = -100\)
\(a_{\text{car}} = -0.05 \text{ m/s}^2\)
From formula (ii),
\(F_{\text{car}} = m_{\text{car}} a_{\text{car}} = 1000 \times 0.05 = 50 \text{ N}\)
Therefore, \(\frac{F_{\text{car}}}{\text{weight}_{\text{car}}} = \frac{50}{1000 \times 10} = \frac{1}{200}\)
Answer: The frictional force acting on both the truck and the car is \(\frac{1}{200}\) of their weight.
In simple words: Both the truck and the car experience a frictional force that is 1/200th of their respective weights to stop within the same distance when their engines fail, showing that for the same initial speed and stopping distance, the fraction of frictional force to weight remains constant.

🎯 Exam Tip: This problem emphasizes that for the same initial speed and stopping distance, the ratio of frictional force to weight (which is proportional to the coefficient of friction) remains constant, irrespective of the object's mass. This is a common concept in kinematics and friction problems.

Question 2. A lighter object A and a heavier object B are initially at rest. Both are imparted the same linear momentum. Which will start with greater kinetic energy: A or B or both will start with the same energy?
Answer: [Ans: A]
Solution:
1. Let \(m_1\) and \(m_2\) be the masses of light object A and heavy object B and \(v_1\) and \(v_2\) be their respective velocities.
2. Since both are imparted with the same linear momentum,
\(m_1 v_1 = m_2 v_2\)
3. Kinetic energy of the lighter object A
\(K.E._A = \frac{1}{2} m_1 v_1^2\)
Kinetic energy of the heavier object B
\(K.E._B = \frac{1}{2} m_2 v_2^2\)
\(\frac{K.E._A}{K.E._B} = \frac{\frac{1}{2} m_1 v_1^2}{\frac{1}{2} m_2 v_2^2}\)
\(\frac{K.E._A}{K.E._B} = \frac{m_1 (m_2 v_2 / m_1)^2}{m_2 v_2^2} = \frac{m_2}{m_1}\) ....[ \(\because m_1 v_1 = m_2 v_2\)]
4. As \(m_1 < m_2\), therefore \(K.E._A > K.E._B\), i.e, the lighter body A has more kinetic energy.
In simple words: When two objects are given the same linear momentum, the lighter object will have greater kinetic energy because kinetic energy is inversely proportional to mass for a given momentum.

🎯 Exam Tip: Remember the relationship \(K.E. = \frac{p^2}{2m}\). This formula directly shows that for a constant momentum (p), kinetic energy is inversely proportional to mass, explaining why the lighter object has more KE.

Question 3. As I was standing on a weighing machine inside a lift it recorded 50 kg-wt. Suddenly for few seconds it recorded 45 kg-wt. What must have happened during that time? Explain with complete numerical analysis. [Ans: Lift must be coming down with acceleration \(\frac{g}{10} = 1 \text{ ms}^{-2}\)]
Answer: Solution:
The weight recorded by weighing machine is always apparent weight and a measure of reaction force acting on the person. As the apparent weight (45 kg-wt) in this case is less than actual weight (50 kg-wt) the lift must be accelerated downwards during that time.
Numerical Analysis
1. Weight on the weighing machine inside the lift is recorded as 50 kg-wt
Therefore, \(mg = 50 \text{ kg-wt}\)
ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक व्यक्ति को लिफ्ट के अंदर एक वेइंग मशीन पर खड़ा दिखाता है। इसमें व्यक्ति पर लगने वाले बल - नीचे की ओर mg (वजन) और ऊपर की ओर मशीन द्वारा दिया गया सामान्य प्रतिक्रिया बल R को दर्शाया गया है। लिफ्ट की दिशा और त्वरण a भी इंगित किया गया है।
2. This weight acts on the weighing machine which offers a reaction R given by the reading of the weighing machine
Therefore, \(R = 45 \text{ kg-wt} = \frac{9}{10} mg\)
3. The forces acting on person inside lift are as follows:
• Weight \(mg\) downward (exerted by the earth)
• Normal reaction (R) upward (exerted by the floor)
4. As, \(R < mg\), the net force is in downward direction and given as,
\(mg - R = ma\)
But \(R = \frac{9}{10} mg\).
Therefore, \(mg - \frac{9}{10} mg = ma\)
Therefore, \(\frac{mg}{10} = ma\)
Therefore, \(a = g/10\)
Therefore, \(a = 1 \text{ m/s}^2\) (g = \(10 \text{ m/s}^2\))
5. Therefore, the elevator must be accelerated downwards with an acceleration of \(1 \text{ m/s}^2\) at that time.
In simple words: When the apparent weight in a lift decreases, it means the lift is accelerating downwards. The numerical analysis confirms that the lift is accelerating downwards at \(g/10\), which is \(1 \text{ m/s}^2\).

🎯 Exam Tip: Apparent weight in a lift changes with acceleration. If apparent weight is less than actual weight, the lift is accelerating downwards. If it's more, it's accelerating upwards. If equal, it's either at rest or moving with constant velocity.

Question 4. Figure below shows a block of mass 35 kg resting on a table. The table is so rough that it offers a self-adjusting resistive force 10% of the weight of the block for its sliding motion along the table. A 20 kg-wt load is attached to the block and is passed over a pulley to hang freely on the left side. On the right side there is a 2 kg-wt pan attached to the block and hung freely. Weights of 1 kg-wt each, can be added to the pan. Minimum how many and maximum how many such weights can be added into the pan so that the block does not slide along the table? [Ans: Min 15, maximum 21].
Answer: ℹ️ चित्र व्याख्या (Diagram Explanation): यह चित्र एक मेज पर रखे 35 kg द्रव्यमान के ब्लॉक को दर्शाता है। ब्लॉक के बाईं ओर एक घिरनी के ऊपर से गुजरता हुआ 20 kg भार लटका हुआ है। दाईं ओर, एक और घिरनी के ऊपर से गुजरता हुआ 2 kg भार का एक पैन लटका हुआ है जिसमें अतिरिक्त 1 kg भार के वजन जोड़े जा सकते हैं। यह सेटअप ब्लॉक पर कार्य करने वाले विभिन्न बलों और उसके संतुलन को दर्शाता है।
Solution:
Frictional (resistive) force \(f = 10\% \text{ (weight)}\)
\(f = \frac{10}{100} \times 35 \times 10 = 35\text{ N}\)
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र (a) 20 kg द्रव्यमान के ब्लॉक का मुक्त पिंड आरेख (FBD) दर्शाता है, जिसमें नीचे की ओर वजन (mg = 200 N) और ऊपर की ओर तनाव (T1) बल कार्य कर रहा है। चित्र (b) मेज पर रखे 35 kg द्रव्यमान के ब्लॉक पर क्षैतिज रूप से कार्य करने वाले बलों को दर्शाता है: दाईं ओर तनाव (T1) और बाईं ओर तनाव (T2) तथा घर्षण बल (f) और त्वरण (m1a) भी दर्शाया गया है।
(i) Consider FBD for 20 kg-wt load. Initially, the block kept on the table is moving towards left, because of the movement of block of mass 20 kg in downward direction.
Thus, for block of mass 20 kg,
\(ma = mg - T_1\) .... (1)
Consider the forces acting on the block of mass 35 kg in horizontal direction only as shown in figure (b). Thus, the force equation for this block is, \(m_1 a = T_1 - T_2 - f\) .... (2)
To prevent the block from sliding across the table,
\(m_1 a = ma = 0\)
Therefore, \(T_1 = mg = 200 \text{ N}\) ....[From (1)]
\(T_1 = T_2 + f\) ....[From (2)]
Therefore, \(T_2 + f = 200\)
Therefore, \(T_2 = 200 - 35 = 165 \text{ N}\)
Thus, the total force acting on the block from right hand side should be \(165 \text{ N}\).
Therefore, Total mass = \(16.5 \text{ kg}\)
Therefore, Minimum weight to be added = \(16.5 - 2 = 14.5 \text{ kg}\)
\(\approx 15 \text{ weights of 1 kg each}\)
(ii) Now, considering motion of the block towards right, the force equations for the masses in the pan and the block of mass 35 kg can be determined from FBD shown
ℹ️ चित्र व्याख्या (Diagram Explanation): चित्र (c) मेज पर रखे 35 kg द्रव्यमान के ब्लॉक पर कार्य करने वाले क्षैतिज बलों को दर्शाता है, जिसमें बाईं ओर तनाव (T1) और दाईं ओर तनाव (T2) तथा घर्षण बल (f) और त्वरण (m1a) शामिल हैं। चित्र (d) लटके हुए ब्लॉक (m2) का मुक्त पिंड आरेख दर्शाता है, जिसमें नीचे की ओर वजन (m2g) और ऊपर की ओर तनाव (T2) बल तथा त्वरण (m2a) कार्य कर रहा है।
From figure (c)
\(m_1 a = T_2 - T_1 - f\) ....(iii)
From figure (d),
\(m_2 a = m_2 g - T_2\) ... .(iv)
To prevent the block of mass 35 kg from sliding across the table, \(m_1 a = m_2 a = 0\)
From equations (iii) and (iv),
\(T_2 = T_1 + f\)
\(T_2 = m_2 g\)
Therefore, \(m_2 g = 200 + 35 = 235 \text{ N}\)
Therefore, The maximum mass required to stop the sliding = \(23.5 - 2 = 21.5 \text{ kg} \approx 21 \text{ weights of 1 kg}\)
Answer: The minimum 15 weights and maximum 21 weights of 1 kg each are required to stop the block from sliding.
In simple words: To keep the 35 kg block from sliding, you need to add enough weight to the pan to balance the forces. The minimum 15 kg prevents sliding left, and the maximum 21 kg prevents sliding right, due to the 10% frictional force and the hanging 20 kg-wt load.

🎯 Exam Tip: For equilibrium problems involving friction, always analyze the forces in both potential directions of motion to determine the minimum and maximum conditions. Free-body diagrams are indispensable for correctly identifying all forces and setting up equations.

Question 5. Power is rate of doing work or the rate at which energy is supplied to the system. A constant force F is applied to a body of mass m. Power delivered by the force at time t from the start is proportional to
(a) t
(b) \(t^2\)
(c) \(\sqrt{t}\)
(d) \(t^0\)
Derive the expression for power in terms of F, m and t.
[Ans: \(p = \frac{F^2 t}{m}\) Therefore, \(p \propto t\)]
Answer: Solution:
Derivation for expression of power:
(i) A constant force F is applied to a body of mass (m) initially at rest (\(u = 0\)).
(ii) We have,
\(v = u + at\)
Therefore, \(v = 0 + at\)
Therefore, \(v = at\) .... (1)
(iii) Now, power is the rate of doing work,
Therefore, \(P = \frac{dW}{dt}\)
Therefore, \(P = F \cdot \frac{ds}{dt}\) ....[ \(\because dW = F \cdot ds\)]
(iv) But \(\frac{ds}{dt} = v\), the instantaneous velocity of the particle.
Therefore, \(P = F \cdot V\) ... (2)
(v) According to Newton's second law,
\(F = ma\) (3)
(vi) Substituting equations (1) and (3) in equation (2)
\(P = (ma) (at)\)
Therefore, \(P = ma^2 t\)
Therefore, \(P = \frac{m^2 a^2}{m} t\)
Therefore, \(P = \frac{F^2}{m} t\)
(vii) As F and m are constant, therefore, \(P \propto t\).
In simple words: When a constant force is applied to an object, its power delivery increases linearly with time because the object's velocity also increases linearly with time, and power is the product of force and velocity.

🎯 Exam Tip: Remember that power is \(P = Fv\) (force times velocity) and \(P = \frac{dW}{dt}\) (rate of doing work). For a constant force, velocity increases linearly with time (from \(v=at\)), leading to power being directly proportional to time.

Question 6. 40000 litre of oil of density 0.9 g cc is pumped from an oil tanker ship into a storage tank at 10 m higher level than the ship in half an hour. What should be the power of the pump? [Ans: 2 kW]
Answer: Solution:
\(h = 10 \text{ m}\), \(\rho = 0.9 \text{ g/cc} = 900 \text{ kg/m}^3\), \(g = 10 \text{ m/s}^2\),
\(V = 40000 \text{ litre} = 40000 \times 10^{-3} \text{ m}^3 = 40 \text{ m}^3\)
\(T = 30 \text{ min} = 1800 \text{ s}\)
To find: Power(P)
Formula: \(P = \frac{W}{t} = \frac{\rho g V h}{t}\)
Calculation: From formula,
\(P = \frac{10 \times 900 \times 10 \times 40}{1800}\)
Therefore, \(P = 2000 \text{ W}\)
Therefore, \(P = 2 \text{ kW}\)
Answer: The power of the pump is 2 kW.
In simple words: The pump needs to do work against gravity to lift 40000 liters of oil to a 10-meter height in 30 minutes. Calculating the potential energy gained by the oil and dividing by the time gives the required power, which is 2 kilowatts.

🎯 Exam Tip: Power problems often involve lifting objects against gravity. The work done is equal to the change in potential energy (\(W = mgh\)), and power is simply this work divided by the time taken. Ensure consistent units (SI units are preferred).

Question 7.
Ten identical masses (m each) are connected one below the other with 10 strings. Holding the topmost string, the system is accelerated upwards with acceleration g/2. What is the tension in the 6th string from the top (Topmost string being the first string)? [Ans: 6 mg]
Answer:
Solution:
The problem involves a system of masses accelerating upwards. We need to find the tension in a specific string (the 6th from the top).
Consider the 6th string from the top. The number of masses below the 6th string is 5. So, the total mass hanging below the 6th string is \(5m\).
ℹ️ चित्र व्याख्या (Diagram Explanation): Figure (a) illustrates 10 identical masses (m each) connected by strings in a vertical chain, with an upward acceleration \(a = g/2\). The 6th string from the top is identified, with 5 masses hanging below it. Figure (b) is a Free Body Diagram (FBD) for the point where the 6th string is attached to the 5 masses below it (total mass \(5m\)). The forces shown are tension T upwards and the total weight \(5mg\) downwards, with an upward acceleration \(a = g/2\).
The force equation for this system, applying Newton's second law (\(F_{net} = ma\)), considering upward direction as positive, is:
\( T - 5mg = 5ma \)
So, \( T = 5mg + 5ma \)
Given that the acceleration \(a = g/2\).
Substituting the value of 'a' into the equation for T:
\( T = 5mg + 5m \left(\frac{g}{2}\right) \)
\( T = 5mg + 2.5mg \)
\( T = 7.5mg \)
Thus, the tension in the 6th string is 7.5 mg.
In simple words: When a chain of masses accelerates upwards, the tension in any string segment must support the weight of all masses below it plus the force needed to accelerate them. By applying Newton's second law to the segment below the 6th string, we find the tension.

🎯 Exam Tip: Always draw a clear Free Body Diagram (FBD) for the specific section of the system you are analyzing, considering all forces and the direction of acceleration. Ensure all units are consistent.

[Note: The answer given above is modified considering the correct textual concepts.]
Question 8.
Two galaxies of masses 9 billion solar mass and 4 billion solar mass are 5 million light years apart. If, the Sun has to cross the line joining them, without being attracted by either of them, through what point it should pass? [Ans: 3 million light years from the 9 billion solar mass]
Answer:
Solution:
The Sun can cross the line joining the two galaxies without being attracted by either of them if it passes from a neutral point. A neutral point is a point on the line joining two objects where the effect of gravitational forces acting due to both objects is nullified.
Given that:
Mass of galaxy 1: \( m_1 = 9 \times 10^9 \text{ M_s} \) (where M_s is solar mass unit)
Mass of galaxy 2: \( m_2 = 4 \times 10^9 \text{ M_s} \)
Distance between galaxies: \( r = 5 \times 10^6 \text{ light years} \)
Let the neutral point be at a distance \(x\) from galaxy 1. If the Sun (mass \(M_{sun}\)) is present at that point, the gravitational forces due to \(m_1\) and \(m_2\) on the Sun must be equal and opposite.
\( \frac{G m_1 M_{sun}}{x^2} = \frac{G m_2 M_{sun}}{(r-x)^2} \)
\( \frac{m_1}{x^2} = \frac{m_2}{(r-x)^2} \)
Substituting the given values:
\( \frac{9 \times 10^9}{x^2} = \frac{4 \times 10^9}{(r-x)^2} \)
Taking square roots on both sides:
\( \frac{3}{x} = \frac{2}{r-x} \)
\( 3(r-x) = 2x \)
\( 3r - 3x = 2x \)
\( 3r = 5x \)
\( x = \frac{3r}{5} \)
Now, substitute the value of \(r\):
\( x = \frac{3 \times (5 \times 10^6 \text{ light years})}{5} \)
\( x = 3 \times 10^6 \text{ light years} \)
The Sun has to cross the line from a point at a distance 3 million light years from the galaxy of mass 9 billion solar mass.
In simple words: A neutral point is where the gravitational pull from two objects cancels out. By setting the gravitational forces from the two galaxies on the Sun equal and solving for the distance from one galaxy, we find the Sun's safe passage point.

🎯 Exam Tip: For problems involving gravitational neutral points, remember that the forces must be equal in magnitude and opposite in direction. The G and test mass terms cancel out, simplifying the calculation to a ratio of masses and distances.

Question 9.
While decreasing linearly from 5 N to 3 N, a force displaces an object from 3 m to 5 m. Calculate the work done by this force during this displacement. [Ans: 8 N]
Answer:
Solution:
For a variable force, work done is given by the area under the force-displacement graph. From the given data, the graph can be plotted as follows, forming a trapezium.
ℹ️ चित्र व्याख्या (Diagram Explanation): The graph shows Force (F) in Newtons on the Y-axis and Displacement (s) in meters on the X-axis. A line segment connects the point (3m, 5N) to (5m, 3N), representing the linearly decreasing force. The area under this line segment from s=3m to s=5m, and down to the x-axis, forms a trapezium with vertices at (3,5), (5,3), (5,0), and (3,0). The shaded region represents the work done.
To calculate the area, we can decompose the trapezium into a rectangle and a triangle.
Force at \(s_1 = 3 \text{ m}\) is \(F_1 = 5 \text{ N}\).
Force at \(s_2 = 5 \text{ m}\) is \(F_2 = 3 \text{ N}\).
Displacement \( \Delta s = s_2 - s_1 = 5 - 3 = 2 \text{ m} \).
Area of the trapezium (work done) can be calculated using the formula:
\( W = \frac{1}{2} (F_1 + F_2) \Delta s \)
Substituting the values:
\( W = \frac{1}{2} (5 \text{ N} + 3 \text{ N}) (2 \text{ m}) \)
\( W = \frac{1}{2} (8 \text{ N}) (2 \text{ m}) \)
\( W = 8 \text{ J} \)
Alternatively, using the decomposition as shown in the original solution's method:
Area of rectangle (base \(2 \text{ m}\), height \(3 \text{ N}\)) = \( (5-3) \times 3 = 2 \times 3 = 6 \text{ J} \)
Area of triangle (base \(2 \text{ m}\), height \(5-3=2 \text{ N}\)) = \( \frac{1}{2} \times (5-3) \times (5-3) = \frac{1}{2} \times 2 \times 2 = 2 \text{ J} \)
Total Work done = \( 6 \text{ J} + 2 \text{ J} = 8 \text{ J} \)
In simple words: Work done by a variable force is found by calculating the area under its force-displacement graph. When the force changes linearly, this area is a trapezoid, which can be easily calculated by its formula or by splitting it into a rectangle and a triangle.

🎯 Exam Tip: For force-displacement graphs where force is not constant, calculating the area under the curve is the standard method for finding work done. Recognize common geometric shapes like rectangles, triangles, or trapezoids to simplify calculations.

Question 10.
Variation of a force in a certain region is given by \( F = 6x^2 - 4x - 8 \). It displaces an object from \( x = 1 \text{ m} \) to \( x = 2 \text{ m} \) in this region. Calculate the amount of work done. [Ans: Zero]
Answer:
Solution:
For a variable force, the work done is calculated by integrating the force with respect to displacement.
Given force function: \( F = 6x^2 - 4x - 8 \)
Displacement limits: from \( x_1 = 1 \text{ m} \) to \( x_2 = 2 \text{ m} \)
The work done \( W \) is given by:
\[ W = \int_{x_1}^{x_2} F \, dx \]
\[ W = \int_{1}^{2} (6x^2 - 4x - 8) \, dx \]
Split the integral:
\[ W = \int_{1}^{2} 6x^2 \, dx - \int_{1}^{2} 4x \, dx - \int_{1}^{2} 8 \, dx \]
Perform the integration:
\[ W = \left[ \frac{6x^3}{3} \right]_{1}^{2} - \left[ \frac{4x^2}{2} \right]_{1}^{2} - [8x]_{1}^{2} \]
\[ W = [2x^3]_{1}^{2} - [2x^2]_{1}^{2} - [8x]_{1}^{2} \]
Evaluate at the limits:
\[ W = (2(2^3) - 2(1^3)) - (2(2^2) - 2(1^2)) - (8(2) - 8(1)) \]
\[ W = (2 \times 8 - 2 \times 1) - (2 \times 4 - 2 \times 1) - (16 - 8) \]
\[ W = (16 - 2) - (8 - 2) - 8 \]
\[ W = 14 - 6 - 8 \]
\[ W = 14 - 14 \]
\[ W = 0 \text{ J} \]
The work done is zero.
In simple words: When the force is a function of position, we use integration to calculate the total work done. By integrating the given force function over the specified displacement limits, we find the net work, which in this case turns out to be zero.

🎯 Exam Tip: Remember to correctly apply the power rule for integration (\(\int x^n dx = \frac{x^{n+1}}{n+1}\)) and evaluate the definite integral by subtracting the lower limit value from the upper limit value.

Question 11.
A ball of mass 100 g dropped on the ground from 5 m bounces repeatedly. During every bounce 64% of the potential energy is converted into kinetic energy. Calculate the following:
(a) Coefficient of restitution.
(b) Speed with which the ball comes up from the ground after third bounce.
(c) Impulse given by the ball to the ground during this bounce.
(d) Average force exerted by the ground if this impact lasts for 250 ms.
(e) Average pressure exerted by the ball on the ground during this impact if contact area of the ball is 0.5 cm².
[Ans: 0.8, 5.12 m/s, 1.152N s, 4.608 N, 9.216 × 10⁴ N/m²]
Answer:
Solution:
Given mass \(m = 100 \text{ g} = 0.1 \text{ kg}\). Dropped from height \(h = 5 \text{ m}\). Assume \(g = 10 \text{ m/s}^2\).
Initial velocity just before first impact (dropping from rest):
\( u = \sqrt{2gh} = \sqrt{2 \times 10 \times 5} = \sqrt{100} = 10 \text{ m/s} \)

(a) Coefficient of restitution (e):
For every bounce, 64% of the initial energy is converted into kinetic energy. This means the ratio of final kinetic energy (\(KE_f\)) to initial kinetic energy (\(KE_i\)) is 0.64.
The coefficient of restitution \(e\) is related to the ratio of kinetic energies by \( e^2 = \frac{KE_f}{KE_i} \).
\[ e^2 = \frac{64}{100} \]
\[ e = \sqrt{\frac{64}{100}} = \frac{8}{10} = 0.8 \]

(b) Speed with which the ball comes up from the ground after third bounce:
Let \(u\) be the speed just before the first bounce.
Speed after first bounce (upwards): \( v_1 = eu \)
Speed before second bounce (downwards): \( u_2 = v_1 = eu \)
Speed after second bounce (upwards): \( v_2 = e u_2 = e(eu) = e^2 u \)
Speed before third bounce (downwards): \( u_3 = v_2 = e^2 u \)
Speed after third bounce (upwards): \( v_3 = e u_3 = e(e^2 u) = e^3 u \)
Now, substitute the values of \(e\) and \(u\):
\( v_3 = (0.8)^3 \times 10 \text{ m/s} \)
\( v_3 = 0.512 \times 10 \text{ m/s} \)
\( v_3 = 5.12 \text{ m/s} \)

(c) Impulse given by the ball to the ground during this bounce (the first bounce is implied, but the context of (b) and (c) together often refers to the *last* calculated bounce, i.e., third bounce in this case, based on the `Ans` for 5.12 m/s). The problem asks for "Impulse given by the ball to the ground during *this* bounce", which usually refers to the third bounce after calculating its speed. Let's assume the question implies the third bounce, as suggested by the given answer. Speed before third bounce (downwards): \( u_3 = e^2 u = (0.8)^2 \times 10 = 0.64 \times 10 = 6.4 \text{ m/s} \)
Speed after third bounce (upwards): \( v_3 = e^3 u = 5.12 \text{ m/s} \)
Impulse \( J = \Delta p = m(v_3 - (-u_3)) = m(v_3 + u_3) \) (considering upwards as positive velocity for \(v_3\), and downwards as negative for \(u_3\)).
The provided solution uses \(J = m v_3 - m v_2\) for the third bounce. This assumes \(v_2\) is the velocity *before* the third bounce, and \(v_3\) is *after*. Let's stick to the solution's notation of \(v_2\) as velocity *before* the third bounce (which is \(u_3\) in my notation), and \(v_3\) as velocity *after* the third bounce. So, \(v_{before} = -e^2u\) (downwards), \(v_{after} = e^3u\) (upwards). Impulse \( J = \Delta p = m(v_{after} - v_{before}) \)
\( J = m(e^3 u - (-e^2 u)) \)
\( J = m(e^3 u + e^2 u) \)
\( J = m e^2 u (e+1) \)
Substituting values: \(m = 0.1 \text{ kg}\), \(e = 0.8\), \(u = 10 \text{ m/s}\)
\( J = 0.1 \times (0.8)^2 \times 10 \times (0.8+1) \)
\( J = 0.1 \times 0.64 \times 10 \times 1.8 \)
\( J = 0.64 \times 1.8 \)
\( J = 1.152 \text{ Ns} \)

(d) Average force exerted by the ground if this impact lasts for 250 ms:
Time of impact \( \Delta t = 250 \text{ ms} = 250 \times 10^{-3} \text{ s} = 0.250 \text{ s} \)
Average force \( F_{avg} = \frac{J}{\Delta t} \)
\( F_{avg} = \frac{1.152 \text{ Ns}}{0.250 \text{ s}} \)
\( F_{avg} = 4.608 \text{ N} \)

(e) Average pressure exerted by the ball on the ground during this impact if contact area of the ball is 0.5 cm²:
Contact area \( A = 0.5 \text{ cm}^2 = 0.5 \times 10^{-4} \text{ m}^2 \)
Average pressure \( P_{avg} = \frac{F_{avg}}{A} \)
\( P_{avg} = \frac{4.608 \text{ N}}{0.5 \times 10^{-4} \text{ m}^2} \)
\( P_{avg} = 9.216 \times 10^4 \text{ N/m}^2 \)
In simple words: This problem analyzes a bouncing ball using principles of energy conservation and impulse. The coefficient of restitution tells us how much kinetic energy is retained after a bounce, which helps calculate subsequent bounce speeds. Impulse is the change in momentum during impact, and from it, average force and pressure can be found.

🎯 Exam Tip: Remember that kinetic energy is proportional to the square of velocity, so \(e^2\) is the ratio of KE after to before collision. Impulse is \(m\Delta v\), ensuring to account for the change in direction of velocity (e.g., \(v_{final} - (-v_{initial})\)). Convert all units to SI before calculation.

Question 12.
A spring ball of mass 0.5 kg is dropped from some height. On falling freely for 10 s, it explodes into two fragments of mass ratio 1:2. The lighter fragment continues to travel downwards with speed of 60 m/s. Calculate the kinetic energy supplied during explosion. [Ans: 200 J]
Answer:
Solution:
Given:
Total mass of the ball: \( M = 0.5 \text{ kg} \)
Mass ratio of fragments: \( m_1 : m_2 = 1 : 2 \)
So, \( m_1 = \frac{1}{3} M = \frac{1}{3} \times 0.5 \text{ kg} = \frac{0.5}{3} \text{ kg} \)
And \( m_2 = \frac{2}{3} M = \frac{2}{3} \times 0.5 \text{ kg} = \frac{1}{3} \text{ kg} \)
Initial state: Ball falls freely for 10 s. Assume \(g = 10 \text{ m/s}^2\).
Velocity just before explosion (\(u\)):
\( v = u_{initial} + gt = 0 + 10 \times 10 = 100 \text{ m/s} \)
So, \( u = 100 \text{ m/s} \). This is the velocity of the combined mass just before explosion.
After explosion, the two fragments move with velocities \(V_1\) and \(V_2\).
Lighter fragment (\(m_1\)) continues downwards with speed \(V_1 = 60 \text{ m/s}\). Since it's downwards, we'll keep the velocity as positive as we assumed downward as positive for \(u\).
Applying the law of conservation of linear momentum for the explosion:
Total momentum before explosion = Total momentum after explosion
\( M u = m_1 V_1 + m_2 V_2 \)
\( 0.5 \times 100 = \left(\frac{0.5}{3}\right) \times 60 + \left(\frac{1}{3}\right) V_2 \)
\( 50 = 10 + \frac{1}{3} V_2 \)
\( 40 = \frac{1}{3} V_2 \)
\( V_2 = 120 \text{ m/s} \)
This is the speed of the heavier fragment (\(m_2\)) also downwards.

Kinetic energy supplied during explosion (\(\Delta K.E.\)):
This is the difference between the total kinetic energy after explosion and the total kinetic energy before explosion.
\( \Delta K.E. = (K.E_{after}) - (K.E_{before}) \)
\( \Delta K.E. = \left( \frac{1}{2} m_1 V_1^2 + \frac{1}{2} m_2 V_2^2 \right) - \left( \frac{1}{2} M u^2 \right) \)
\( \Delta K.E. = \left( \frac{1}{2} \times \frac{0.5}{3} \times 60^2 + \frac{1}{2} \times \frac{1}{3} \times 120^2 \right) - \left( \frac{1}{2} \times 0.5 \times 100^2 \right) \)
\( \Delta K.E. = \left( \frac{1}{2} \times \frac{0.5}{3} \times 3600 + \frac{1}{2} \times \frac{1}{3} \times 14400 \right) - \left( \frac{1}{2} \times 0.5 \times 10000 \right) \)
\( \Delta K.E. = (0.5 \times 600 + 0.5 \times 4800) - 2500 \)
\( \Delta K.E. = (300 + 2400) - 2500 \)
\( \Delta K.E. = 2700 - 2500 \)
\( \Delta K.E. = 200 \text{ J} \)
The kinetic energy supplied during explosion is 200 J.
In simple words: This problem involves an explosion, which conserves momentum. We first find the ball's velocity before exploding, then use momentum conservation to find the velocity of the second fragment after the explosion. The energy supplied by the explosion is the difference between the total kinetic energy of the fragments and the kinetic energy of the original ball.

🎯 Exam Tip: For explosion problems, momentum is always conserved (if no external forces). The kinetic energy changes, with the increase representing energy supplied by the explosion. Ensure correct calculation of initial and final kinetic energies for all involved parts.

Question 13.
A marble of mass 2m travelling at 6 cm/s is directly followed by another marble of mass m moving with double speed. After collision, the heavier one travels with the average initial speed of the two. Calculate the coefficient of restitution. [Ans: 0.5]
Answer:
Solution:
Given:
Mass of marble 1 (heavier): \( m_1 = 2m \)
Mass of marble 2 (lighter): \( m_2 = m \)
Initial velocity of marble 1: \( u_1 = 6 \text{ cm/s} \)
Initial velocity of marble 2: \( u_2 = 2u_1 = 2 \times 6 = 12 \text{ cm/s} \)
The heavier marble (\(m_1\)) is travelling slower than the lighter marble (\(m_2\)) but is "directly followed" by it, implying \(u_2 > u_1\), so collision will occur. After collision, the heavier one travels with the average initial speed of the two.
Final velocity of marble 1 (heavier): \( V_1 = \frac{u_1 + u_2}{2} = \frac{6 + 12}{2} = \frac{18}{2} = 9 \text{ cm/s} \)
To find: Coefficient of restitution (\(e\)).

First, apply the law of conservation of linear momentum to find \(V_2\) (final velocity of marble 2):
\( m_1 u_1 + m_2 u_2 = m_1 V_1 + m_2 V_2 \)
\( (2m)(6) + (m)(12) = (2m)(9) + m V_2 \)
\( 12m + 12m = 18m + m V_2 \)
\( 24m = 18m + m V_2 \)
\( m V_2 = 6m \)
\( V_2 = 6 \text{ cm/s} \)

Now, calculate the coefficient of restitution (\(e\)). The coefficient of restitution is defined as the negative ratio of the relative velocity of separation to the relative velocity of approach.
\( e = -\frac{V_2 - V_1}{u_2 - u_1} \)
Alternatively, \( e = \frac{V_1 - V_2}{u_2 - u_1} \)
Using the second form:
\( e = \frac{9 - 6}{12 - 6} \)
\( e = \frac{3}{6} \)
\( e = 0.5 \)
The coefficient of restitution is 0.5.
In simple words: This problem uses conservation of momentum to find the final velocity of one object after a collision, given the final velocity of the other. Once all velocities are known, the coefficient of restitution is calculated from the ratio of relative separation velocity to relative approach velocity.

🎯 Exam Tip: Always state the conservation laws you are applying (momentum, energy). For coefficient of restitution, pay close attention to the signs and order of velocities in the formula to correctly represent relative motion.

Question 14.
A 2 m long wooden plank of mass 20 kg is pivoted (supported from below) at 0.5 m from either end. A person of mass 40 kg starts walking from one of these pivots to the farther end. How far can the person walk before the plank topples? [Ans: 1.25 m]
Answer:
Solution:
Given:
Length of plank \(L = 2 \text{ m}\). Mass of plank \(M_{plank} = 20 \text{ kg}\).
Pivots P1 and P2 are located 0.5 m from each end.
Let the left end be at \(x=0\). Then P1 is at \(x=0.5 \text{ m}\).
The right end is at \(x=2 \text{ m}\). Then P2 is at \(x=2-0.5 = 1.5 \text{ m}\).
The center of mass (CM) of the uniform plank is at its center, i.e., \(x=1 \text{ m}\).
Mass of person \(M_{person} = 40 \text{ kg}\).

"A person starts walking from one of these pivots to the farther end." Let's assume the person starts walking from P2 (\(x=1.5 \text{ m}\)) towards the left end (the farther end in terms of walking distance available). The plank will tend to topple about pivot P1 (\(x=0.5 \text{ m}\)).
ℹ️ चित्र व्याख्या (Diagram Explanation): The diagrams show a 2m plank with two supports, P1 and P2, placed 0.5m from each end. The plank's mass (20kg) is effectively at its center (1m mark). A person (40kg) walks on the plank. The second diagram specifically illustrates the scenario where the plank is on the verge of toppling about P1. The plank's weight creates a clockwise moment about P1, while the person walking to the left of P1 creates an anti-clockwise moment. P2 acts as an additional support which will lift off when toppling occurs about P1.

For the plank to be in equilibrium (just before toppling about P1), the clockwise moment due to the plank's weight about P1 must be balanced by the anti-clockwise moment due to the person's weight about P1.
Distance of plank's CM from P1: \(1 \text{ m} - 0.5 \text{ m} = 0.5 \text{ m}\) (to the right of P1).
Let \(d\) be the distance the person walks to the left of P1.
Moment due to plank's weight about P1: \( M_{plank} g \times 0.5 \text{ m} = 20g \times 0.5 \)
Moment due to person's weight about P1: \( M_{person} g \times d = 40g \times d \)
At the point of toppling, these moments balance:
\( 40g \times d = 20g \times 0.5 \)
\( 40d = 10 \)
\( d = \frac{10}{40} = 0.25 \text{ m} \)
This \(d\) is the distance the person walks past P1 (to the left of P1).
The person started at P2 (\(x=1.5 \text{ m}\)) and walked left.
The distance from P2 to P1 is \(1.5 \text{ m} - 0.5 \text{ m} = 1 \text{ m}\).
Total distance walked by the person from P2 is \(1 \text{ m} + d\)
Total distance walked = \(1 \text{ m} + 0.25 \text{ m} = 1.25 \text{ m}\).
Hence, the total distance walked by the person is 1.25 m.
In simple words: To find how far a person can walk on a plank before it topples, we balance the moments (torques) around the pivot point. The person's moment to one side must equal the plank's moment to the other side relative to the pivot.

🎯 Exam Tip: When dealing with toppling problems, identify the pivot point (the edge about which rotation occurs) and ensure all moments (torques) are calculated relative to this point. Sum of clockwise moments must equal sum of anti-clockwise moments for equilibrium.

Question 15.
A 2 m long ladder of mass 10 kg is kept against a wall such that its base is 1.2 m away from the wall. The wall is smooth but the ground is rough. Roughness of the ground is such that it offers a maximum horizontal resistive force (for sliding motion) half that of normal reaction at the point of contact. A monkey of mass 20 kg starts climbing the ladder. How far can it climb along the ladder? How much is the horizontal reaction at the wall? [Ans: 1.5 m, 15 N]
Answer:
Solution:
Given:
Length of ladder \(AC = L = 2 \text{ m}\). Mass of ladder \(M_{ladder} = 10 \text{ kg}\).
Distance of base from wall \(BC = 1.2 \text{ m}\).
Using Pythagoras theorem for triangle ABC:
\( AB = \sqrt{AC^2 - BC^2} = \sqrt{2^2 - 1.2^2} = \sqrt{4 - 1.44} = \sqrt{2.56} = 1.6 \text{ m} \).
Weight of ladder: \( W_1 = M_{ladder} g = 10 \text{ kg} \times 10 \text{ m/s}^2 = 100 \text{ N} \). The CM of the ladder is at \(L/2 = 1 \text{ m}\) from C. Horizontal distance of ladder CM from wall is \(BC/2 = 0.6 \text{ m}\).
Mass of monkey: \(M_{monkey} = 20 \text{ kg}\). Weight of monkey: \(W_2 = M_{monkey} g = 20 \text{ kg} \times 10 \text{ m/s}^2 = 200 \text{ N}\).
Wall is smooth, so it exerts only a horizontal normal force (reaction H) at A.
Ground is rough, so it exerts a normal force (N) and a frictional force (F) at C.
Condition for friction: Maximum friction \(F = N/2\).

ℹ️ चित्र व्याख्या (Diagram Explanation): The diagram shows a ladder AC leaning against a vertical wall (AB). The base C is on the rough ground, and the top A is against the smooth wall. The length of the ladder is 2m, and its base is 1.2m from the wall. The ladder's weight (\(W_1=100N\)) acts at its center. A monkey (\(W_2=200N\)) is climbing the ladder. Horizontal force H is from the wall at A. Vertical normal force N and horizontal friction F are from the ground at C.

Equations of equilibrium:
1. Vertical equilibrium (Sum of vertical forces = 0):
\( N - W_1 - W_2 = 0 \)
\( N = W_1 + W_2 = 100 \text{ N} + 200 \text{ N} = 300 \text{ N} \)
2. Horizontal equilibrium (Sum of horizontal forces = 0):
\( H - F = 0 \implies H = F \)
Since \(F = N/2\):
\( H = \frac{N}{2} = \frac{300 \text{ N}}{2} = 150 \text{ N} \)
The horizontal reaction at the wall is 150 N.

3. Rotational equilibrium (Sum of moments about C = 0):
Let \(x\) be the horizontal distance of the monkey from the base C.
Moment due to horizontal reaction at wall H: \( H \times AB = H \times 1.6 \text{ m} \) (counter-clockwise moment)
Moment due to ladder's weight \(W_1\): \( W_1 \times (\frac{BC}{2}) = 100 \text{ N} \times \frac{1.2}{2} = 100 \times 0.6 = 60 \text{ N m} \) (clockwise moment)
Moment due to monkey's weight \(W_2\): \( W_2 \times x \) (clockwise moment)
\( H \times 1.6 - W_1 \times 0.6 - W_2 \times x = 0 \)
\( 150 \times 1.6 - 100 \times 0.6 - 200 \times x = 0 \)
\( 240 - 60 - 200x = 0 \)
\( 180 - 200x = 0 \)
\( 200x = 180 \)
\( x = \frac{180}{200} = 0.9 \text{ m} \)
This \(x\) is the horizontal distance of the monkey from the base C.
Now, we need to find how far the monkey climbs along the ladder. Let this distance be \(L_{monkey}\).
Using similar triangles (monkey's position and ladder):
\( \frac{L_{monkey}}{AC} = \frac{\text{horizontal distance from C (x)}}{BC} \)
\( \frac{L_{monkey}}{2} = \frac{0.9}{1.2} \)
\( L_{monkey} = 2 \times \frac{0.9}{1.2} = 2 \times \frac{9}{12} = 2 \times \frac{3}{4} = 1.5 \text{ m} \)
The monkey can climb up to 1.5 m along the ladder, and the horizontal reaction at the wall is 150 N.
In simple words: This problem involves static equilibrium. We apply the conditions for translational equilibrium (sum of forces is zero) and rotational equilibrium (sum of moments is zero) to find the unknown forces and distances. Friction and normal forces at the base, and normal force from the wall, must balance the weights.

🎯 Exam Tip: For problems involving inclined objects like ladders, resolve forces into horizontal and vertical components. Always take moments about a point where unknown forces act to eliminate them from the moment equation (e.g., about the base C to eliminate N and F). Remember to relate distances along the ladder to horizontal/vertical distances using similar triangles or trigonometry.

Question 16.
Four uniform solid cubes of edges 10 cm, 20 cm, 30 cm and 40 cm are kept on the ground, touching each other in order. Locate centre of mass of their system. [Ans: 65 cm, 17.7 cm]
Answer:
Solution:
Let the density of the material of the cubes be \( \rho \). The mass of a cube is proportional to its volume, i.e., \( m = \rho \times (\text{edge})^3 \).
ℹ️ चित्र व्याख्या (Diagram Explanation): The diagram depicts four solid cubes of increasing edge lengths (10cm, 20cm, 30cm, 40cm) arranged side-by-side on the ground. The smallest cube (10cm edge) is placed with its bottom-left corner at the origin (0,0) of the Cartesian coordinate system. Subsequent cubes are placed adjacent to the previous one along the x-axis, creating a stair-like profile. The x and y axes are marked in cm, illustrating the positions and sizes of the cubes.

Let's define the masses and their center of mass coordinates:
Cube 1 (edge \(a_1 = 10 \text{ cm}\)):
Mass \( m_1 = \rho a_1^3 = \rho (10)^3 = 1000\rho \)
CM position \( (x_1, y_1) = (a_1/2, a_1/2) = (5, 5) \text{ cm} \)

Cube 2 (edge \(a_2 = 20 \text{ cm}\)), placed next to Cube 1:
Mass \( m_2 = \rho a_2^3 = \rho (20)^3 = 8000\rho \)
CM position \( (x_2, y_2) = (a_1 + a_2/2, a_2/2) = (10 + 20/2, 20/2) = (20, 10) \text{ cm} \)

Cube 3 (edge \(a_3 = 30 \text{ cm}\)), placed next to Cube 2:
Mass \( m_3 = \rho a_3^3 = \rho (30)^3 = 27000\rho \)
CM position \( (x_3, y_3) = (a_1 + a_2 + a_3/2, a_3/2) = (10 + 20 + 30/2, 30/2) = (45, 15) \text{ cm} \)

Cube 4 (edge \(a_4 = 40 \text{ cm}\)), placed next to Cube 3:
Mass \( m_4 = \rho a_4^3 = \rho (40)^3 = 64000\rho \)
CM position \( (x_4, y_4) = (a_1 + a_2 + a_3 + a_4/2, a_4/2) = (10 + 20 + 30 + 40/2, 40/2) = (80, 20) \text{ cm} \)

Total mass of the system \( M_{total} = m_1 + m_2 + m_3 + m_4 \)
\( M_{total} = (1000 + 8000 + 27000 + 64000)\rho = 100000\rho \)

X-coordinate of the center of mass (\(X_{cm}\)):
\[ X_{cm} = \frac{\sum m_i x_i}{M_{total}} \]
\[ X_{cm} = \frac{(1000\rho \times 5) + (8000\rho \times 20) + (27000\rho \times 45) + (64000\rho \times 80)}{100000\rho} \]
\[ X_{cm} = \frac{\rho(5000 + 160000 + 1215000 + 5120000)}{\rho(1000 + 8000 + 27000 + 64000)} \]
\[ X_{cm} = \frac{6500000}{100000} = 65 \text{ cm} \]

Y-coordinate of the center of mass (\(Y_{cm}\)):
\[ Y_{cm} = \frac{\sum m_i y_i}{M_{total}} \]
\[ Y_{cm} = \frac{(1000\rho \times 5) + (8000\rho \times 10) + (27000\rho \times 15) + (64000\rho \times 20)}{100000\rho} \]
\[ Y_{cm} = \frac{\rho(5000 + 80000 + 405000 + 1280000)}{\rho(1000 + 8000 + 27000 + 64000)} \]
\[ Y_{cm} = \frac{1770000}{100000} = 17.7 \text{ cm} \]
The center of mass of the system is located at point (65 cm, 17.7 cm).
In simple words: To find the center of mass of a system of multiple objects, we treat each object as a point mass located at its own center of mass. We then calculate the weighted average of the x-coordinates and y-coordinates of these individual centers of mass, with the weights being the masses of the objects.

🎯 Exam Tip: For composite objects, break the system into simpler shapes and find the CM of each component. Ensure consistent coordinate system and correct mass calculation (e.g., mass proportional to volume for uniform density). Sum of masses goes in the denominator for the weighted average CM formula.

Question 17.
A uniform solid sphere of radius R has a hole of radius R/2 drilled inside it. One end of the hole is at the centre of the sphere while the other is at the boundary. Locate centre of mass of the remaining sphere. [Ans: -R/14 ]
Answer:
Solution:
We can find the center of mass of the remaining sphere by considering it as a complete solid sphere with a negative mass hole subtracted from it.
Let the center of the large sphere be the origin O (\(0,0,0\)).
Mass of the complete solid sphere of radius R: \( M = \frac{4}{3} \pi R^3 \rho \), where \( \rho \) is the density.
Its center of mass is at \(r_1 = 0\).

ℹ️ चित्र व्याख्या (Diagram Explanation): The diagram shows a large solid sphere of radius R with its center at O. A smaller sphere, representing the drilled hole, has a radius of R/2. The hole is positioned such that one end is at O and the other is at the boundary of the large sphere. This means the center of the smaller sphere (D) is at a distance of R/2 from O. The diagram visually represents the subtraction method for finding the center of mass.

The hole is a sphere of radius \(R' = R/2\). Its center is at \(D\), at a distance \(R/2\) from \(O\). Let's assume this is along the positive x-axis, so its position vector is \(r_2 = R/2\).
Mass of the drilled hole (sphere): \( M' = \frac{4}{3} \pi (R/2)^3 \rho \)
\( M' = \frac{4}{3} \pi \frac{R^3}{8} \rho = \frac{1}{8} \left( \frac{4}{3} \pi R^3 \rho \right) \)
So, \( M' = \frac{M}{8} \)

The center of mass of the remaining sphere \(r_{cm}\) is given by the formula for a composite system (where one part is removed, hence its mass is treated as negative):
\[ r_{cm} = \frac{M r_1 - M' r_2}{M - M'} \]
Substitute the values:
\[ r_{cm} = \frac{M \times 0 - \frac{M}{8} \times \frac{R}{2}}{M - \frac{M}{8}} \]
\[ r_{cm} = \frac{-\frac{MR}{16}}{\frac{7M}{8}} \]
\[ r_{cm} = -\frac{MR}{16} \times \frac{8}{7M} \]
\[ r_{cm} = -\frac{R}{14} \]
The negative sign indicates that the center of mass of the remaining sphere is on the left side of the origin (opposite to the direction of the hole's center from the origin).
In simple words: To find the center of mass of a sphere with a hole, we treat the hole as a region with negative mass within the larger sphere. We calculate the center of mass using the principle of superposition, where the total mass is the sphere's mass minus the hole's mass, and moments are similarly subtracted.

🎯 Exam Tip: For problems involving objects with removed parts, use the "negative mass" concept. Treat the removed part as having negative mass and being located at its own center of mass. This allows you to apply the standard center of mass formula to the composite system.

Question 18.
In the following table, every item on the left side can match with any number of items on the right hand side. Select all those.
 

Answer:
(a) Elastic collision: (ii), (iii), (vi)
(b) Inelastic collision: (i), (v)
(c) Perfectly inelastic collision: (iv)
(d) Head on collision: (iii), (vi)
In simple words: Collisions are categorized based on whether kinetic energy is conserved (elastic), partially lost (inelastic), or completely lost with objects sticking together (perfectly inelastic). Head-on collisions occur when objects move along the same line before and after impact.

🎯 Exam Tip: Understand the definitions of each collision type and common examples. Elastic collisions conserve both momentum and kinetic energy, while inelastic collisions conserve momentum but not kinetic energy. Perfectly inelastic collisions are a special type of inelastic collision where objects stick together.

Can You Recall? (Textbook Page No. 47)

Question 1.
What are different types of motions?
Answer:
The various types of motion are linear, uniform linear, non-uniform linear, oscillatory, circular, periodic and random motion.
In simple words: Motion describes how an object's position changes over time, and it can be categorized based on its path, speed, and whether it repeats or is unpredictable.

🎯 Exam Tip: Be able to define and provide examples for each type of motion, such as a car on a straight road (linear), a pendulum (oscillatory), or planets around the sun (periodic).

Question 2.
What do you mean by kinematical equations and what are they?
Answer:
A set of three equations which analyses rectilinear motion of uniformly accelerated body and helps to predict the position of body are called as kinematical equations.
1. Equation for velocity-time relation: \( v = u + at \)
2. Equation for position-time relation: \( s = ut + \frac{1}{2} at^2 \)
3. Position-velocity relation: \( v^2 = u^2 + 2as \)
In simple words: Kinematical equations are a set of three mathematical formulas that describe the motion of objects moving in a straight line with constant acceleration, allowing us to find things like final velocity, displacement, or time.

🎯 Exam Tip: Memorize the three kinematic equations and understand when each is most appropriate to use based on the known and unknown variables in a problem (e.g., if time is not given, use the third equation).

Can You Tell? (Textbook Page No. 48)

Question 1.
Was Aristotle correct? If correct, explain his statement with an illustration.
Answer:
Aristotle was not correct in stating that an external force is required to keep a body in uniform motion.
In simple words: Aristotle believed that objects naturally stopped moving unless a continuous force kept them going, which is incorrect because objects in motion will stay in motion unless acted upon by an unbalanced force (like friction).

🎯 Exam Tip: Understand the historical progression of ideas about motion. Aristotle's view contrasts with Newton's first law (inertia), which states that an object's natural state is constant velocity (or rest) if no net force acts on it.

Question 2.
If wrong, give the correct modified version of his statement.
Answer:
For an uninterrupted motion of a body, an additional external force is required for overcoming opposing/resistive forces.
In simple words: To maintain uninterrupted motion in real-world scenarios, an external force is needed to counteract resistive forces like friction or air resistance, which would otherwise slow the object down.

🎯 Exam Tip: Differentiate between ideal situations (no friction) and real-world scenarios. In an ideal world, no force is needed for uniform motion. In reality, a force is often required to overcome friction and maintain constant velocity.

Can You Tell? (Textbook Page No. 48)

Question 1.
What is then special about Newton's first law if it is derivable from Newton's second law?
Answer:
1. Newton's first law shows an equivalence between the 'state of rest' and 'state of uniform motion along a straight line.'
2. Newton's first law of motion defines force as a physical quantity that brings about a change in 'state of rest' or 'state of uniform motion along a straight line' of a body.
3. Newton's first law of motion defines inertia as a fundamental property of every physical object by which the object resists any change in its state of rest or of uniform motion along a straight line.
Due to all these reasons, Newton's first law should be studied.
In simple words: Even though Newton's first law can be derived from the second, it is crucial because it defines inertia, identifies force as the cause of changes in motion, and establishes the equivalence between rest and uniform motion.

🎯 Exam Tip: Focus on the conceptual significance of Newton's first law: its definition of inertia, the role of force in changing motion, and establishing a zero net force as the condition for uniform velocity.