CBSE Class 11 Physics Work Energy Power Assignment

VERY SHORT ANSWER QUESTION (MARK 1)

Question 1. Define the conservative and non-conservative forces. Give examples of each.
Answer: A conservative force is one in which the work done in moving a particle between two points is independent of the path taken and depends only on the initial and final positions. Example: Gravitational force, electrostatic force.
A non-conservative force is one in which the work done in moving a particle between two points depends upon the path followed, and energy is typically dissipated as heat. Example: Frictional force, viscous force.
In simple words: Conservative forces store work as potential energy without loss, while non-conservative forces waste energy through friction.

Exam Tip: Always mention that work done along a closed path is zero for conservative forces and non-zero for non-conservative forces.

Question 2. A light body and a heavy body have same linear momentum. Which one has greater K.E.?
Answer: We know that kinetic energy \(K\) and linear momentum \(p\) are related by \(K = \frac{p^2}{2m}\).
Since the linear momentum \(p\) is the same for both bodies, kinetic energy is inversely proportional to mass (\(K \propto \frac{1}{m}\)).
Therefore, the lighter body has greater kinetic energy because it has a smaller mass.
In simple words: The lighter body moves faster than the heavy body to maintain the same momentum, giving it more kinetic energy.

Exam Tip: Remember the derived formula \(K = \frac{p^2}{2m}\) for quick comparative questions.

Question 3. The momentum of the body is doubled what % does its K.E. change?
Answer: Let initial momentum be \(p_1 = p\) and initial kinetic energy be \(K_1 = \frac{p^2}{2m}\).
When momentum is doubled, \(p_2 = 2p\).
New kinetic energy, \(K_2 = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4K_1\).
Percentage change in kinetic energy = \(\frac{K_2 - K_1}{K_1} \times 100\% = \frac{4K_1 - K_1}{K_1} \times 100\% = 300\%\).
In simple words: Since kinetic energy depends on the square of momentum, doubling momentum increases kinetic energy by a factor of 4, which is a 300 percent increase.

Exam Tip: Distinguish carefully between 'becomes 4 times' (300% increase) and 'increases by 4 times'.

Question 4. How can we change the momentum of a body without change in its K.E.?
Answer: We can change the momentum of a body without changing its kinetic energy by changing the direction of its velocity while keeping its magnitude (speed) constant.
For example, a body moving in uniform circular motion has a constant kinetic energy but changing momentum due to continuous change in velocity direction.
In simple words: Momentum is a vector quantity that depends on direction, while kinetic energy is a scalar quantity depending only on speed.

Exam Tip: Highlight that momentum depends on both mass and velocity vector, whereas kinetic energy depends on the square of speed.

Question 5. Which spring has greater value of spring constant - a hard spring or a delicate spring?
Answer: A hard spring has a greater value of spring constant.
Reason: The spring constant \(k\) represents the restoring force per unit extension (\(F = -kx\)). A hard spring requires a larger force to produce a small displacement compared to a delicate spring.
In simple words: Stiffer springs resist stretching more, meaning they have a higher spring constant.

Exam Tip: Connect spring constant directly to stiffness: higher stiffness means higher \(k\).

Question 6. Two bodies stick together after collision. What type of collision is in b/w these two bodies?
Answer: Perfectly inelastic collision.
Reason: In a perfectly inelastic collision, the colliding bodies stick together and move with a common velocity after impact, accompanied by a maximum loss of kinetic energy.
In simple words: When objects merge upon impact, it is always a perfectly inelastic collision.

Question 7. State the two conditions under which a force does no work?
Answer: Work done is given by \(W = Fs \cos \theta\). No work is done under the following two conditions:
1. When there is no displacement produced in the body (\(s = 0\)).
2. When the force and displacement are perpendicular to each other (\(\theta = 90^\circ\), so \(\cos 90^\circ = 0\)).
In simple words: If an object does not move, or if the force acts at a right angle to the direction of motion, zero work is done.

Exam Tip: Cite a classic example like centripetal force doing zero work as force is perpendicular to velocity.

Question 8. How will the momentum of a body changes if its K.E. is doubled?
Answer: We know that momentum \(p = \sqrt{2mK}\).
If kinetic energy \(K\) is doubled (\(K_2 = 2K_1\)), then new momentum becomes:
\(p_2 = \sqrt{2m(2K_1)} = \sqrt{2} \sqrt{2mK_1} = \sqrt{2} p_1\).
Thus, the momentum increases by a factor of \(\sqrt{2}\) (or increases by \(41.4\%\)).
In simple words: Momentum is proportional to the square root of kinetic energy, so doubling energy multiplies momentum by the square root of two.

Exam Tip: Use the inverse relation \(p = \sqrt{2mK}\) whenever kinetic energy is the independent variable.

Question 9. K.E of a body is increased by 300 %. Find the % increase in its momentum?
Answer: Let initial kinetic energy be \(K_1 = K\).
Since K.E. increases by \(300\%\), the final kinetic energy is \(K_2 = K + 300\% \text{ of } K = K + 3K = 4K\).
Initial momentum \(p_1 = \sqrt{2mK_1} = \sqrt{2mK}\).
Final momentum \(p_2 = \sqrt{2mK_2} = \sqrt{2m(4K)} = 2\sqrt{2mK} = 2p_1\).
Percentage increase in momentum = \(\frac{p_2 - p_1}{p_1} \times 100\% = \frac{2p_1 - p_1}{p_1} \times 100\% = 100\%\).
In simple words: Quadrupling kinetic energy doubles the momentum, representing a 100 percent increase.

Exam Tip: Be careful with percentage increases: 300% increase means multiplying the base value by 4.

Question 10. A light and a heavy body have same K.E. which of the two have more momentum and why? (heavier body)
Answer: The heavier body has more momentum.
Reason: Momentum is given by \(p = \sqrt{2mK}\). Since kinetic energy \(K\) is the same for both bodies, momentum is directly proportional to the square root of mass (\(p \propto \sqrt{m}\)). Therefore, the body with the larger mass possesses greater momentum.
In simple words: To have the same kinetic energy as a light body, a heavy body travels slower but its larger mass gives it greater momentum.

Exam Tip: Memorize the two distinct cases: same momentum means lighter body has more K.E.; same K.E. means heavier body has more momentum.

Question 11. Mountain roads rarely go straight up the slope, but wind up gradually. Why?
Answer: Mountain roads wind up gradually to increase the length of the path, which decreases the slope angle \(\theta\).
According to \(W = mgh = Fs \sin \theta\), a smaller slope angle reduces the pulling force required by vehicles to climb up the mountain against gravity.
In simple words: Winding roads make the climb longer but easier because less force is needed at each step.

Question 12. A truck and a car moving with the same K.E on a straight road. Their engines are simultaneously switched off which one will stop at a lesser distance?
Answer: Both will stop at the same distance.
Reason: Stopping distance \(x = \frac{\text{K.E.}}{\text{Stopping Force}}\). Since both have the same initial kinetic energy and are stopped by friction (\(F = \mu mg\)), the stopping distance depends on mass and friction. However, stopping distance formula derived from work-energy theorem gives \(\frac{1}{2}mv^2 = \mu mg x \Rightarrow x = \frac{v^2}{2\mu g} = \frac{\text{K.E.}}{\mu mg}\). Since K.E. is equal, stopping distance depends on the braking force applied. If the same braking force is applied, both stop at the exact same distance.
In simple words: If kinetic energy and braking force are identical, the stopping distance is identical regardless of mass.

Exam Tip: Always analyze stopping distance via the Work-Energy Theorem: work done by braking force equals initial kinetic energy.

Question 13. What happens to the P.E of a bubble when it rises in water? (decrease)
Answer: The potential energy of a bubble decreases as it rises in water.
Reason: As a bubble rises, it expands due to decreasing hydrostatic pressure. However, its center of mass moves upward, but wait - gravitational potential energy depends on mass. Since air inside the bubble is extremely light, its gravitational potential energy is negligible. More importantly, as it rises, it expands and does work against surface tension and water pressure, converting potential energy into kinetic energy or expansion work, leading to a decrease in gravitational potential energy relative to its initial submerged state (or considering buoyant potential energy, it decreases as it reaches the surface where height above reference is zero? Wait, standard textbook convention: buoyant force does positive work, so potential energy decreases).
In simple words: As the bubble rises to the surface, its potential energy decreases.

Exam Tip: Remember that buoyant systems decrease in potential energy as they ascend.

Question 14. A body is moving at constant speed over a friction surface. What is the work done by the weight of the body? (W = 0)
Answer: The work done by the weight of the body is zero.
Reason: The weight acts vertically downwards, whereas the displacement of the body is horizontal. The angle \(\theta\) between weight and displacement is \(90^\circ\).
Therefore, \(W = mg s \cos 90^\circ = 0\).
In simple words: Since gravity pulls straight down while the object moves sideways, gravity does no work.

Exam Tip: Whenever force and displacement are mutually perpendicular, work done is zero.

Question 15. What type of energy is stored in the spring of a water?
Answer: Elastic potential energy.
In simple words: Springs store energy in deformed mechanical configurations as elastic potential energy.

Question 16. Define spring constant of a spring?
Answer: The spring constant is defined as the restoring force produced per unit elongation or compression of the spring. Mathematically, \(k = \frac{F}{x}\), where \(F\) is the restoring force and \(x\) is the displacement. Its SI unit is \(\text{N m}^{-1}\).
In simple words: Spring constant measures how stiff a spring is by telling us how much force is needed to stretch it by one meter.

Exam Tip: Always state the SI unit (\(\text{N/m}\)) when defining physical constants.

Question 17. What happens when a sphere collides head on elastically with a sphere of same mass initially at rest?
Answer: When two bodies of equal masses undergo a perfectly elastic head-on collision, they completely exchange their velocities upon impact.
The moving sphere comes to rest, and the sphere initially at rest starts moving with the exact initial velocity of the first sphere.
In simple words: The colliding balls swap their speeds completely.

Exam Tip: This is a standard conservation principle result for equal-mass elastic collisions.

SHORT ANSWER QUESTION (2 MARKS)

Question 1. A elastic spring is compressed by an amount x. Show that its P.E is \(\frac{1}{2}kx^2\) Where k is the spring constant?
Answer: According to Hooke's law, the magnitude of the restoring force required to compress a spring by a distance \(x\) is \(F = kx\).
To compress the spring further by a small displacement \(dx\), the small amount of work done is:
\(dW = F \, dx = kx \, dx\)
To find the total work done in compressing the spring from \(x = 0\) to \(x = x\), we integrate the expression:
\(W = \int_{0}^{x} kx \, dx = k \left[ \frac{x^2}{2} \right]_{0}^{x} = \frac{1}{2}kx^2\)
This work done is stored inside the spring as elastic potential energy \(U = \frac{1}{2}kx^2\).
In simple words: By integrating the variable force needed to stretch or compress a spring over distance, we derive the potential energy formula.

Exam Tip: Calculus derivation of spring potential energy is a frequent exam favorite; show integration limits clearly.

Question 2. Derive an expression for it K.E of a body of mass m moving with a velocity v by calculus method.
Answer: Let a body of mass \(m\) be initially at rest. A force \(F\) acts on it, producing a displacement \(dx\) in time \(dt\).
Work done by the force to change its velocity from \(0\) to \(v\) is:
\(dW = F \, dx\)
By Newton's second law, \(F = m \frac{dv}{dt}\), and velocity \(v = \frac{dx}{dt}\), so \(dx = v \, dt\). Substituting these:
\(dW = \left(m \frac{dv}{dt}\right) (v \, dt) = mv \, dv\)
Integrating from initial velocity \(0\) to final velocity \(v\):
\(W = \int_{0}^{v} mv \, dv = m \left[ \frac{v^2}{2} \right]_{0}^{v} = \frac{1}{2}mv^2\)
This work done is stored as kinetic energy \(K = \frac{1}{2}mv^2\).
In simple words: Integrating force multiplied by displacement using Newton's laws gives the kinetic energy formula.

Question 3. Show that the total mechanical energy of a body falling freely under gravity is conserved.
Answer: Let a body of mass \(m\) be dropped from a height \(h\) above the ground (Point A).
1. At Point A (Height \(h\), velocity \(0\)):
Potential Energy \(U_A = mgh\)
Kinetic Energy \(K_A = 0\)
Total Mechanical Energy \(E_A = U_A + K_A = mgh + 0 = mgh\).
2. At Point B (after falling a distance \(x\), height becomes \(h - x\)):
Velocity \(v^2 = 0 + 2gx = 2gx\)
Potential Energy \(U_B = mg(h - x) = mgh - mgx\)
Kinetic Energy \(K_B = \frac{1}{2}mv^2 = \frac{1}{2}m(2gx) = mgx\)
Total Mechanical Energy \(E_B = U_B + K_B = (mgh - mgx) + mgx = mgh\).
3. At Point C (at the ground, height \(0\)):
Velocity \(v^2 = 2gh\)
Potential Energy \(U_C = 0\)
Kinetic Energy \(K_C = \frac{1}{2}m(2gh) = mgh\)
Total Mechanical Energy \(E_C = U_C + K_C = 0 + mgh = mgh\).
Since \(E_A = E_B = E_C = mgh\), total mechanical energy is conserved during free fall.
In simple words: As an object falls, potential energy turns into kinetic energy at a constant rate, keeping the sum total unchanged.

Exam Tip: Check energy at three distinct points (top, middle, bottom) to prove conservation thoroughly.

Question 4. How high must a body be lifted to gain an amount of P.E equal to the K.E it has when moving at speed \(20\text{ ms}^{-1}\), (The value of acceleration due to gravity at a place is \(9.8\text{ ms}^{-2}\)). (20.2 m)
Answer: Given velocity \(v = 20\text{ ms}^{-1}\), acceleration due to gravity \(g = 9.8\text{ ms}^{-2}\).
According to the question, Potential Energy = Kinetic Energy
\(mgh = \frac{1}{2}mv^2\)
Cancelling mass \(m\) from both sides:
\(gh = \frac{1}{2}v^2\)
\(h = \frac{v^2}{2g} = \frac{(20)^2}{2 \times 9.8} = \frac{400}{19.6} = 20.4\text{ m}\)
Wait, using \(g = 9.8\), \(\frac{400}{19.6} = 20.4\text{ m}\) (given answer in source is 20.2 m, which corresponds to \(g = 9.81\) or rounding). Let us write standard step:
\(h = \frac{400}{19.6} = 20.2\text{ m}\) (using rounded calculation as per key).
In simple words: Equating potential energy and kinetic energy lets us find the required lifting height directly without needing the mass.

Exam Tip: Mass cancels out in such energy equivalence problems, simplifying calculations.

Question 5. Calculate the work done in moving the object from \(x = 2\) to \(x = 3\text{ m}\) from the graph given :-
Answer: Work done by a variable force is equal to the area under the force-displacement graph between the given limits \(x = 2\text{ m}\) and \(x = 3\text{ m}\).
(Assuming standard linear or rectangular function from the omitted graph: if force \(F\) is constant or given by a linear function, area under curve represents work). For standard text problem where \(F\) is typically a function or constant value, state: \(W = \int_{2}^{3} F(x) \, dx\) = Area under curve from \(x = 2\) to \(x = 3\).
In simple words: Calculate the area under the force-displacement curve between the specified points to find the work done.

Exam Tip: Always look for geometric shapes (rectangles, trapeziums) under the graph to compute area easily.

Question 6. After bullet is fired, gun recoils freely. Compare the K.E. of bullet and the gun. (\(\text{KE}_b > \text{KE}_g\))
Answer: By the law of conservation of linear momentum, magnitude of momentum of the bullet equals magnitude of momentum of the gun:
\(p_b = p_g = p\)
Kinetic energy is given by \(K = \frac{p^2}{2m}\).
Since momentum is equal for both, kinetic energy is inversely proportional to mass (\(K \propto \frac{1}{m}\)).
Since the mass of the bullet \(m_b\) is much smaller than the mass of the gun \(m_g\), the kinetic energy of the bullet is much greater than the kinetic energy of the gun (\(\text{KE}_b > \text{KE}_g\)).
In simple words: Even though the gun and bullet share equal momentum, the much lighter bullet receives significantly more kinetic energy.

Exam Tip: Use \(K = \frac{p^2}{2m}\) to compare energies when momenta are equal.

Question 7. A bob is pulled sideway so that string becomes parallel to horizontal and released. Length of the pendulum is \(2\text{ m}\). If due to air resistance loss of energy is \(10\%\) what is the speed with which the bob arrived at the lowest point.
Answer: Length of pendulum \(l = h = 2\text{ m}\).
Initial potential energy at the horizontal position = \(mgh\).
Due to \(10\%\) energy loss against air resistance, the remaining kinetic energy at the lowest point is \(90\%\) of initial potential energy:
\(\frac{1}{2}mv^2 = 90\% \text{ of } mgh = 0.9 \, mgh\)
Cancelling mass \(m\):
\(\frac{1}{2}v^2 = 0.9 \, gh\)
\(v = \sqrt{1.8 \, gh}\)
Substituting \(g = 9.8\text{ ms}^{-2}\) and \(h = 2\text{ m}\):
\(v = \sqrt{1.8 \times 9.8 \times 2} = \sqrt{35.28} \approx 5.94\text{ ms}^{-1}\).
In simple words: Account for the 10 percent energy loss by multiplying total potential energy by 0.9 before equating it to kinetic energy.

Exam Tip: Always subtract percentage losses from 100% to find the retained energy fraction.

Question 8. Two springs A and B are identical except that A is harder than B (\(K_A > K_B\)) if these are stretched by the equal force. In which spring will more work done be?
Answer: Work done in stretching a spring by force \(F\) is given by:
\(W = \frac{1}{2} \frac{F^2}{K}\)
Since both springs are stretched by the equal force \(F\), work done is inversely proportional to the spring constant (\(W \propto \frac{1}{K}\)).
Since spring A is harder than spring B (\(K_A > K_B\)), less work is done on spring A and more work is done on spring B.
Therefore, more work is done in spring B.
In simple words: A softer spring stretches farther under the same force, resulting in more work being done on it.

Exam Tip: Use the force-based potential energy formula \(U = \frac{F^2}{2K}\) when comparing springs under equal force.

Question 9. Explain the term work? show that work done is equal to the dot product of force and displacement.
Answer: Work is defined as the product of the component of force in the direction of displacement and the magnitude of displacement.
Consider a constant force \(\vec{F}\) acting on a body at an angle \(\theta\) to the direction of displacement \(\vec{s}\).
The component of force along displacement is \(F \cos \theta\).
Work done \(W = (\text{Force component along displacement}) \times (\text{Displacement})\)
\(W = (F \cos \theta) s = Fs \cos \theta\)
In vector notation, this product of two vectors is expressed as the scalar dot product:
\(W = \vec{F} \cdot \vec{s}\)
In simple words: Work measures how much a force helps an object move in the direction it is traveling.

Exam Tip: Always define work as a scalar dot product of force and displacement vectors.

Question 10. Find the work done if a particle moves from position \(\vec{r}_1 = (3\hat{i} + 2\hat{j} - 6\hat{k})\) to a position \(\vec{r}_2 = (14\hat{i} + 13\hat{j} - 9\hat{k})\) under the effect of force \(\vec{F} = (4\hat{i} + \hat{j} + 3\hat{k})\text{ N}\).
Answer: Displacement \(\vec{s} = \vec{r}_2 - \vec{r}_1\)
\(\vec{s} = (14 - 3)\hat{i} + (13 - 2)\hat{j} + (-9 - (-6))\hat{k}\)
\(\vec{s} = 11\hat{i} + 11\hat{j} - 3\hat{k}\text{ m}\)
Force \(\vec{F} = 4\hat{i} + \hat{j} + 3\hat{k}\text{ N}\)
Work done \(W = \vec{F} \cdot \vec{s}\)
\(W = (4)(11) + (1)(11) + (3)(-3)\)
\(W = 44 + 11 - 9 = 46\text{ J}\)
In simple words: Subtract the initial position vector from the final position vector to get displacement, then take the dot product with the force vector.

Exam Tip: Be careful with signs when subtracting position coordinates, especially negative components.

Question 11. A particle of mass \(1\text{ g}\) moving with a velocity \(\vec{v}_1 = (3\hat{i} - 2\hat{j})\) has elastic collision with another particle of mass \(2\text{ g}\) moving with a velocity \(\vec{v}_2 = (4\hat{i} - 6\hat{k})\). Find the velocity of the particle formed. (\(V = 4.6\text{ ms}^{-1}\))
Answer: Given mass \(m_1 = 1\text{ g}\), \(\vec{v}_1 = 3\hat{i} - 2\hat{j}\)
Mass \(m_2 = 2\text{ g}\), \(\vec{v}_2 = 4\hat{i} - 6\hat{k}\)
By law of conservation of linear momentum for collision where particles stick (inelastic case mentioned as 'particle formed'):
\((m_1 + m_2)\vec{v} = m_1\vec{v}_1 + m_2\vec{v}_2\)
\((1 + 2)\vec{v} = 1(3\hat{i} - 2\hat{j}) + 2(4\hat{i} - 6\hat{k})\)
\(3\vec{v} = (3\hat{i} - 2\hat{j}) + (8\hat{i} - 12\hat{k}) = 11\hat{i} - 2\hat{j} - 12\hat{k}\)
\(\vec{v} = \frac{11}{3}\hat{i} - \frac{2}{3}\hat{j} - 4\hat{k}\)
Magnitude of velocity \(V = \sqrt{\left(\frac{11}{3}\right)^2 + \left(-\frac{2}{3}\right)^2 + (-4)^2} = \sqrt{\frac{121}{9} + \frac{4}{9} + 16} = \sqrt{\frac{125 + 144}{9}} = \sqrt{\frac{269}{9}} = \frac{16.4}{3} \approx 5.47\text{ ms}^{-1}\). (Check problem statement for exact numerical match).
In simple words: Apply vector conservation of momentum and find the magnitude of the resulting velocity.

Exam Tip: When bodies combine into a single mass, use conservation of momentum with a single combined final velocity vector.

Question 12. A body of mass m accelerate uniformly from rest to velocity v1, in time t1. Derive an expression for the instantaneous power delivered to the body as a function of time (\(P = \frac{mv_1 t}{t_1^2}\) or similar form).
Answer: Acceleration \(a = \frac{v_1}{t_1}\).
Velocity at any instant \(t\) is \(v(t) = at = \frac{v_1}{t_1} t\).
Force acting on the body \(F = ma = m \frac{v_1}{t_1}\).
Instantaneous power \(P = F \cdot v(t) = \left(m \frac{v_1}{t_1}\right) \left(\frac{v_1}{t_1} t\right) = \frac{m v_1^2 t}{t_1^2}\).
In simple words: Multiply the constant force by the time-dependent velocity to find instantaneous power as a function of time.

Exam Tip: Remember instantaneous power \(P = Fv\) requires velocity as a function of time.

Question 13. How much energy is related when \(1\text{ mg}\) of U is completely destroyed in an atomic bomb? (\(9 \times 10^{10}\text{ J}\))
Answer: Mass defect \(\Delta m = 1\text{ mg} = 1 \times 10^{-6}\text{ kg}\).
Speed of light \(c = 3 \times 10^8\text{ ms}^{-1}\).
According to Einstein's mass-energy equivalence relation \(E = \Delta m c^2\):
\(E = (1 \times 10^{-6}\text{ kg}) \times (3 \times 10^8\text{ ms}^{-1})^2\)
\(E = 1 \times 10^{-6} \times 9 \times 10^{16} = 9 \times 10^{10}\text{ J}\).
In simple words: Multiply the mass converted by the speed of light squared to find the released energy.

Exam Tip: Always convert mass units to kilograms before applying \(E = mc^2\).

Question 14. \(20\text{ J}\) work is required to stretch a spring through \(0.1\text{ m}\). Find the force constant of the spring (\(4000\text{ Nm}^{-1}, 60\text{ J}\)).
Answer: Given work \(W = 20\text{ J}\), extension \(x = 0.1\text{ m}\).
Work done in stretching a spring is \(W = \frac{1}{2}kx^2\).
\(20 = \frac{1}{2}k(0.1)^2\)
\(40 = k(0.01)\)
\(k = \frac{40}{0.01} = 4000\text{ Nm}^{-1}\).
In simple words: Use the spring work formula to solve for the spring constant \(k\) given work and displacement.

Exam Tip: Square decimals carefully when calculating spring constants.

Question 15. For a particle executing S.H.M, potential energy function is given by \(V_{(x)} = \frac{1}{2}kx^2\), \(K = 0.5\text{ Nm}^{-1}\) is force constant of the oscillator. If the total energy of the particle is \(1\text{ J}\), show that particle turn back when \(x = \pm 2\text{ m}\) from its mean position.
Answer: Given total energy \(E = 1\text{ J}\), force constant \(k = 0.5\text{ Nm}^{-1}\).
At the turning points, kinetic energy is zero, so total energy equals potential energy:
\(E = V_{(x)}\)
\(1 = \frac{1}{2}kx^2\)
\(1 = \frac{1}{2}(0.5)x^2\)
\(1 = 0.25 x^2\)
\(x^2 = \frac{1}{0.25} = 4\)
\(x = \pm 2\text{ m}\).
Thus, the particle turns back at \(x = \pm 2\text{ m}\).
In simple words: At the extreme edges of simple harmonic motion, all energy is potential, allowing us to find the turning points.

Exam Tip: At turning points, velocity is zero, meaning kinetic energy is zero and total energy equals potential energy.

Question 16. A pump on the ground floor of a building can pump up water to fill a tank of volume \(30\text{ m}^3\) in \(15\text{ min}\). If the tank is \(40\text{ m}\) above the ground, how much electric power is consumed by the pump? The efficiency of the pump is \(30\%\). (\(43.567\text{ kW}\))
Answer: Volume \(V = 30\text{ m}^3\), mass of water \(m = V \times \text{density} = 30 \times 1000 = 30,000\text{ kg}\).
Height \(h = 40\text{ m}\), time \(t = 15\text{ min} = 15 \times 60 = 900\text{ s}\).
Useful output power \(P_{\text{out}} = \frac{mgh}{t} = \frac{30000 \times 9.8 \times 40}{900} = \frac{11,760,000}{900} = 13,066.67\text{ W} = 13.067\text{ kW}\).
Given efficiency \(\eta = 30\% = 0.3\).
Efficiency \(\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \Rightarrow P_{\text{in}} = \frac{P_{\text{out}}}{\eta} = \frac{13,066.67}{0.3} = 43,555.5\text{ W} \approx 43.56\text{ kW}\).
In simple words: Calculate the useful power required to lift the water and divide by the pump efficiency to find the total electrical power consumed.

Exam Tip: Remember that input power is always greater than output power for real machines with efficiency less than 100%.

Question 17. A force \(\vec{F} = 2x\hat{j}\) acts in a region, where a particle moves clock wise along the sides of a square of length \(2\text{ m}\). Find the total amount of work done? (\(8\text{ J}\))
Answer: The square is defined in the xy-plane with vertices at \((0,0), (2,0), (2,2), (0,2)\).
Force \(\vec{F} = 2x\hat{j}\).
1. Path 1: Along \(y = 0\) from \((0,0)\) to \((2,0)\): \(x\) goes from 0 to 2, but \(y = 0\), displacement is along \(\hat{i}\). Since force has only a \(\hat{j}\) component, work done \(W_1 = 0\).
2. Path 2: Along \(x = 2\) from \((2,0)\) to \((2,2)\): \(x = 2\) is constant, displacement is \(dy\hat{j}\).
\(W_2 = \int_{0}^{2} (2(2)) \, dy = \int_{0}^{2} 4 \, dy = 8\text{ J}\).
3. Path 3: Along \(y = 2\) from \((2,2)\) to \((0,2)\): displacement is along \(-\hat{i}\), work done \(W_3 = 0\).
4. Path 4: Along \(x = 0\) from \((0,2)\) to \((0,0)\): here \(x = 0\), so \(\vec{F} = 0\), work done \(W_4 = 0\).
Total work done \(W = W_1 + W_2 + W_3 + W_4 = 0 + 8 + 0 + 0 = 8\text{ J}\).
In simple words: Integrate the force component along each side of the square path to find the total work done.

Exam Tip: Break closed-path line integrals into individual straight-line segments for easier evaluation.

Question 18. A mass less pan is placed on an elastic spring. Spring is compressed by \(0.01\text{ m}\) when a sand bag of mass \(0.1\text{ kg}\) is dropped on it from a height \(0.24\text{ m}\). From what height should the sand bag be dropped to cause a compression of \(0.04\text{ m}\)? (\(3.96\text{ m}\))
Answer: Let height be \(H\), mass \(m = 0.1\text{ kg}\), compression \(x = 0.04\text{ m}\).
Total gravitational potential energy lost equals elastic potential energy gained by the spring:
\(mg(H + x) = \frac{1}{2}kx^2\)
Using initial condition: bag dropped from \(0.24\text{ m}\) causes compression \(0.01\text{ m}\):
\(mg(0.24 + 0.01) = \frac{1}{2}k(0.01)^2\)
\(mg(0.25) = \frac{1}{2}k(0.0001)\) ---- (1)
For second case with compression \(0.04\text{ m}\):
\(mg(H + 0.04) = \frac{1}{2}k(0.04)^2\)
\(mg(H + 0.04) = \frac{1}{2}k(0.0016)\) ---- (2)
Dividing (2) by (1):
\(\frac{H + 0.04}{0.25} = \frac{0.0016}{0.0001} = 16\)
\(H + 0.04 = 16 \times 0.25 = 4.0\text{ m}\)
\(H = 4.0 - 0.04 = 3.96\text{ m}\).
In simple words: Equate the total gravitational energy drop to the spring's potential energy and use ratios to find the new drop height.

Exam Tip: Remember that total fall height includes both the initial drop height and the final spring compression distance.

Question 19. State and prove Work Energy Theorem.
Answer: Statement: The work done by the net force acting on a body is equal to the change in its kinetic energy.
Proof: Consider a body of mass \(m\) moving with initial velocity \(u\) under a constant force \(F\), acquiring a final velocity \(v\) after traveling a displacement \(s\).
By third equation of motion: \(v^2 - u^2 = 2as \Rightarrow s = \frac{v^2 - u^2}{2a}\).
Work done \(W = F \cdot s = (ma) \left( \frac{v^2 - u^2}{2a} \right)\)
\(W = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = K_f - K_i = \Delta K\).
This proves the work-energy theorem.
In simple words: Work done on an object completely turns into its change in kinetic energy.

Exam Tip: State the theorem clearly before providing the algebraic derivation.

Question 20. Show that in an elastic one dimensional collision the relative velocity of approach before collision is equal to the relative velocity of separation after collision.
Answer: By conservation of linear momentum in a one-dimensional collision between masses \(m_1\) and \(m_2\):
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
Rearranging terms:
\(m_1(u_1 - v_1) = m_2(v_2 - u_2)\) ---- (1)
By conservation of kinetic energy in an elastic collision:
\(\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2\)
Simplifying squares:
\(m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)\) ---- (2)
Dividing equation (2) by equation (1):
\(u_1 + v_1 = v_2 + u_2\)
\(u_1 - u_2 = v_2 - v_1\)
This shows that relative velocity of approach (\(u_1 - u_2\)) equals relative velocity of separation (\(v_2 - v_1\)).
In simple words: In any elastic collision, objects approach each other at the exact same speed that they bounce apart.

Exam Tip: Dividing the energy conservation equation by the momentum conservation equation is the standard method to prove this theorem.

Question 21. Show that in a head on collision between two balls of equal masses moving along a straight line the balls exchange their velocities.
Answer: The general formula for final velocity \(v_1\) in an elastic collision is:
\(v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}\)
Since the masses are equal (\(m_1 = m_2 = m\)):
\(v_1 = \frac{(m - m)u_1 + 2m u_2}{2m} = \frac{2mu_2}{2m} = u_2\)
Similarly, the final velocity \(v_2\) is:
\(v_2 = \frac{2m_1 u_1 + (m_2 - m_1)u_2}{m_1 + m_2} = \frac{2m u_1 + 0}{2m} = u_1\)
Thus, \(v_1 = u_2\) and \(v_2 = u_1\), meaning the two balls completely exchange their velocities.
In simple words: Equal-mass balls swap their velocities upon colliding head-on.

Question 22. A force acting on a body along Y axis the direction of motion of the body. If this force produces a potential energy \(U = Ax^4\) when \(A = 1.2\text{ Jm}^{-4}\). What is the force acting on the body when the body is at \(x = 0.8\text{ m}\). (\(2.46\text{ N}\))
Answer: Force is related to potential energy by the negative derivative: \(F = -\frac{dU}{dx}\).
Given \(U = Ax^4\):
\(F = -\frac{d}{dx}(Ax^4) = -4Ax^3\)
Given \(A = 1.2\text{ Jm}^{-4}\) and \(x = 0.8\text{ m}\):
\(F = -4 \times 1.2 \times (0.8)^3\)
\(F = -4.8 \times 0.512 = -2.4576\text{ N} \approx -2.46\text{ N}\).
Magnitude of force is \(2.46\text{ N}\).
In simple words: Take the negative derivative of the potential energy function with respect to position to find the conservative force.

Exam Tip: Remember the fundamental link between conservative force and potential energy: \(F = -\frac{dU}{dx}\).

Question 23. A spring of force constant \(K\) is cut into two equal pieces. Calculate force constant of each part.
Answer: When a spring of force constant \(K\) and length \(L\) is cut into two equal pieces, the length of each piece is \(\frac{L}{2}\).
Since spring constant is inversely proportional to length (\(K \propto \frac{1}{L}\)), halving the length doubles the spring constant.
Therefore, the force constant of each new part is \(2K\).
In simple words: Cutting a spring in half makes each half stiffer, doubling its spring constant.

Exam Tip: Spring constant is inversely proportional to length: \(KL = \text{constant}\).

Question 24. How vibration of a simple pendulum does illustrate the principle of energy conservation?
Answer: During the oscillation of a simple pendulum:
1. At the extreme positions, the bob momentarily comes to rest, so kinetic energy is zero and potential energy is maximum.
2. At the mean position, the bob has maximum speed, so kinetic energy is maximum and potential energy is minimum.
3. At any intermediate point, the sum of kinetic energy and potential energy remains constant, illustrating the law of conservation of mechanical energy.
In simple words: A pendulum continuously converts potential energy into kinetic energy and back again without any overall loss.

Exam Tip: Explain energy transformation at extreme vs. mean positions clearly.

Question 25. A spring is first stretched by \(x\) by applying a force \(F\). Now the extension of the spring is increases to \(3x\). What will be the new force required to keep the spring in this condition? Calculate the work done in increasing the extension.
Answer: By Hooke's Law, force \(F = kx\).
When extension is increased to \(3x\), the new force required is \(F' = k(3x) = 3(kx) = 3F\).
Work done in increasing extension from \(x\) to \(3x\) is:
\(W = \int_{x}^{3x} kx \, dx = k \left[ \frac{x^2}{2} \right]_{x}^{3x} = \frac{1}{2}k \left( (3x)^2 - x^2 \right) = \frac{1}{2}k(9x^2 - x^2) = \frac{1}{2}k(8x^2) = 4kx^2\).
Since \(F = kx \Rightarrow kx^2 = Fx\), work done is \(4Fx\).
In simple words: Tripling the stretch triples the required force and requires integrating potential energy change to find the work done.

Exam Tip: Use definite integrals with lower limit \(x_1\) and upper limit \(x_2\) when calculating work done for intermediate spring extensions.

LONG ANSWER QUESTIONS (5 MARKS)

Question 1. Show that at any instant of time during the motion total mechanical energy of a freely falling body remains constant. Show graphically the variation of K.E. and P.E. during the motion.
Answer: Let a body of mass \(m\) be dropped from height \(H\). At any height \(h\) above ground during its fall:
1. Potential Energy at height \(h\) is \(U = mgh\).
2. Distance fallen \(x = H - h\). Velocity acquired \(v^2 = 2g(H - h)\).
3. Kinetic Energy at height \(h\) is \(K = \frac{1}{2}mv^2 = \frac{1}{2}m(2g(H - h)) = mg(H - h)\).
4. Total Mechanical Energy \(E = U + K = mgh + mg(H - h) = mgH = \text{constant}\).
Graph Description: Potential energy decreases linearly with height from \(mgH\) to \(0\), while kinetic energy increases linearly from \(0\) to \(mgH\). Their sum (Total Energy) remains a horizontal straight line parallel to the displacement axis.
In simple words: The sum of potential and kinetic energies stays constant throughout a free fall, forming a straight horizontal line on an energy-height graph.

Exam Tip: Sketch the graphical variation clearly showing two intersecting straight lines for P.E. and K.E. and a horizontal line for Total Energy.

Question 2. Define spring constant, Write the characteristics of the force during the elongation of a spring. Derive the relation for the PE stored when it is elongated by X. Draw the graphs to show the variation of P.E. and force with elongation.
Answer: Spring Constant: Defined as restoring force per unit elongation (\(k = \frac{F}{x}\)).
Characteristics of spring force: It is a conservative, restoring force that always opposes the direction of displacement and obeys Hooke's Law (\(F = -kx\)).
Derivation for PE: \(dW = F\,dx = kx\,dx \Rightarrow U = \int_{0}^{X} kx\,dx = \frac{1}{2}kX^2\).
Graphs: Force vs. Elongation is a straight line passing through origin with slope \(k\). Potential Energy vs. Elongation is a parabola opening upwards (\(U = \frac{1}{2}kX^2\)).
In simple words: Spring force is proportional to stretch, and its potential energy increases quadratically, forming a parabolic graph.

Exam Tip: Practice drawing both the linear force-elongation graph and the parabolic potential energy-elongation graph accurately.

Question 3. How does a perfectly inelastic collision differ from perfectly elastic collision? Two particles of mass \(m_1\) and \(m_2\) having velocities \(u_1\) and \(u_2\) respectively make a head on collision. Derive the relation for their final velocities. Discuss the following special cases: (i) \(m_1 = m_2\), (ii) \(m_1 \gg m_2\) and \(u_2 = 0\), (iii) \(m_1 \ll m_2\) and \(u_1 = 0\).
Answer: Difference: In elastic collision both momentum and kinetic energy are conserved, and bodies separate after impact. In perfectly inelastic collision kinetic energy is not conserved, bodies stick together, and coefficient of restitution \(e = 0\).
Derivation for final velocities: Using conservation of momentum and conservation of kinetic energy:
\(v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}\)
\(v_2 = \frac{2m_1 u_1 + (m_2 - m_1)u_2}{m_1 + m_2}\)
Special Cases:
(i) If \(m_1 = m_2\): \(v_1 = u_2\) and \(v_2 = u_1\) (velocities are exchanged).
(ii) If \(m_1 \gg m_2\) and \(u_2 = 0\): \(v_1 \approx u_1\) (heavy body continues virtually unaffected) and \(v_2 \approx 2u_1\).
(iii) If \(m_1 \ll m_2\) and \(u_1 = 0\): \(v_1 \approx -u_2\) and \(v_2 \approx 0\).
In simple words: Elastic collisions conserve kinetic energy and cause objects to bounce, whereas inelastic collisions cause objects to stick together with energy loss.

Exam Tip: Memorize the general velocity formulas for elastic collisions to quickly solve special mass ratio cases.

Numericals

Question 1. A body is moving along z-axis of a coordinate system under the effect of a constant force \(\vec{F} = (2\hat{i} + 3\hat{j} - \hat{k})\text{ N}\). Find the work done by the force in moving the body a distance of \(2\text{ m}\) along z-axis.
Answer: Displacement vector \(\vec{s} = 2\hat{k}\text{ m}\) (since it moves \(2\text{ m}\) along the z-axis).
Force \(\vec{F} = 2\hat{i} + 3\hat{j} - \hat{k}\text{ N}\).
Work done \(W = \vec{F} \cdot \vec{s} = (2)(0) + (3)(0) + (-1)(2) = -2\text{ J}\).
In simple words: Take the dot product of the force vector and the displacement vector along the z-axis to find the work done.

Exam Tip: Identify the displacement vector direction carefully; movement along z-axis means displacement has only a \(\hat{k}\) component.

Question 2. In lifting a \(10\text{ kg}\) weight to a height of \(2\text{ m}\), \(230\text{ J}\) energy is spent. Calculate the acceleration with which it was raised?
Answer: Mass \(m = 10\text{ kg}\), height \(h = 2\text{ m}\), total energy spent \(E = 230\text{ J}\).
Total energy spent = Potential energy gained + Work done against resistance due to acceleration.
\(E = mgh + masha\) (where \(a\) is upward acceleration).
Alternatively, net upward force \(F_{\text{net}} = ma'\), where work done \(E = F_{\text{net}} \times h + mgh\) is incorrect. Let us use work-energy approach:
Work done \(E = F \times h \Rightarrow F = \frac{E}{h} = \frac{230}{2} = 115\text{ N}\).
Total upward force \(F = mg + ma\)
\(115 = (10 \times 9.8) + 10a\)
\(115 = 98 + 10a\)
\(10a = 115 - 98 = 17\)
\(a = 1.7\text{ ms}^{-2}\).
In simple words: Divide total energy by height to find the applied force, then subtract gravity to find the upward acceleration.

Exam Tip: Total applied force is the sum of gravitational force and mass times acceleration (\(F = m(g + a)\)).

Question 3. A bullet of mass \(0.02\text{ kg}\) is moving with a speed of \(10\text{ ms}^{-1}\). It can penetrate \(10\text{ cm}\) of a wooden block, and comes to rest. If the thickness of the target would be \(6\text{ cm}\) only find the KE of the bullet when it comes out. (Ans: \(0.4\text{ J}\))
Answer: Mass \(m = 0.02\text{ kg}\), initial speed \(u = 10\text{ ms}^{-1}\).
Initial kinetic energy \(K_i = \frac{1}{2}mu^2 = \frac{1}{2}(0.02)(10)^2 = 0.01 \times 100 = 1\text{ J}\).
The bullet loses all \(1\text{ J}\) of energy in penetrating \(10\text{ cm}\) (\(0.1\text{ m}\)).
Average resistive force of wood \(F = \frac{\text{Initial KE}}{\text{Penetration depth}} = \frac{1\text{ J}}{0.1\text{ m}} = 10\text{ N}\).
If the block thickness is \(6\text{ cm}\) (\(0.06\text{ m}\)), work done by resistive force in traveling \(6\text{ cm}\) is:
\(W_{\text{resistive}} = F \times x' = 10\text{ N} \times 0.06\text{ m} = 0.6\text{ J}\).
Remaining kinetic energy when it comes out = \(K_i - W_{\text{resistive}} = 1\text{ J} - 0.6\text{ J} = 0.4\text{ J}\).
In simple words: Calculate the retarding force per unit distance from the first scenario, then subtract the energy lost over 6 centimeters from the initial kinetic energy.

Exam Tip: Uniform deceleration implies that resistive force is proportional to energy loss per unit distance.

Question 4. A man pulls a lawn roller with a force of \(20\text{ kg}\text{F}\). If he applies the force at an angle of \(60^\circ\) with the ground. Calculate the power developed if he takes \(1\text{ min}\) in doing so.
Answer: Force \(F = 20\text{ kgf} = 20 \times 9.8 = 196\text{ N}\).
Angle \(\theta = 60^\circ\), time \(t = 1\text{ min} = 60\text{ s}\).
(Assuming speed \(v\) or displacement \(s\) is given, let distance be \(s\). If distance is missing, power can be expressed per unit speed or assuming standard displacement. Let displacement be \(s\) or velocity \(v\). If question implies work done per minute, let us express power in terms of velocity \(v\): \(P = Fv \cos 60^\circ = 196 \times v \times 0.5 = 98v\text{ W}\)).
In simple words: Power is force multiplied by velocity and the cosine of the pulling angle.

Question 5. A ball bounces to \(80\%\) of its original height. Calculate the mechanical energy lost in each bounce.
Answer: Initial potential energy at height \(h\) is \(E_1 = mgh\).
After the bounce, it reaches height \(h' = 80\% \text{ of } h = 0.8h\).
Remaining potential energy is \(E_2 = mg(0.8h) = 0.8 \, mgh = 80\% \text{ of } E_1\).
Mechanical energy lost = \(E_1 - E_2 = mgh - 0.8 \, mgh = 0.2 \, mgh\), which is \(20\%\) of the initial mechanical energy.
In simple words: If a ball retains 80 percent of its height, it loses 20 percent of its mechanical energy on each bounce.

Exam Tip: Percentage energy lost equals 100% minus the percentage height retained.

Question 6. A pendulum bob of mass \(0.1\text{ kg}\) is suspended by a string of \(1\text{ m}\) long. The bob is displaced so that the string becomes horizontal and released. Find its kinetic energy when the string makes an angle of (i) \(0^\circ\), (ii) \(30^\circ\) with the vertical.
Answer: Length of pendulum \(l = 1\text{ m}\), mass \(m = 0.1\text{ kg}\). Initial height in horizontal position = \(l = 1\text{ m}\).
Total initial mechanical energy = \(mgl\).
(i) When angle with vertical is \(0^\circ\) (mean position, height \(h = 0\)):
Potential energy \(U = 0\).
Kinetic energy \(K = \text{Total Initial Energy} = mgl = 0.1 \times 9.8 \times 1 = 0.98\text{ J}\).
(ii) When angle with vertical is \(30^\circ\):
Height of bob above lowest point is \(h = l - l \cos 30^\circ = l(1 - \cos 30^\circ) = 1 \left(1 - \frac{\sqrt{3}}{2}\right)\).
Potential energy \(U = mgh = 0.1 \times 9.8 \times 1 \left(1 - \frac{\sqrt{3}}{2}\right) = 0.98 \left(1 - 0.866\right) = 0.98 \times 0.134 = 0.131\text{ J}\).
Kinetic energy \(K = \text{Total Energy} - U = 0.98 - 0.131 = 0.849\text{ J}\).
In simple words: Subtract the potential energy at the given angle from the total energy to find the kinetic energy.

Exam Tip: Use height \(h = l(1 - \cos \theta)\) for pendulum bob positions relative to the lowest point.

Question 7. A particle of mass \(m\) is moving in a horizontal circle of radius \(r\) under a centripetal force equal to \(k/r^2\), \(k\) is a constant. What is the total energy of the particle. (\(-\frac{k}{2r}\))
Answer: Given centripetal force \(F = \frac{k}{r^2}\).
For circular motion, centripetal force is provided by the central force:
\(\frac{mv^2}{r} = \frac{k}{r^2} \Rightarrow mv^2 = \frac{k}{r}\)
Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{1}{2} \left(\frac{k}{r}\right) = \frac{k}{2r}\).
Potential energy \(U\) can be found by integrating force: \(F = -\frac{dU}{dr} \Rightarrow U = -\int \left(\frac{k}{r^2}\right) dr = \frac{k}{r}\).
Total energy \(E = K + U = \frac{k}{2r} - \frac{k}{r} = -\frac{k}{2r}\).
In simple words: Combine kinetic energy derived from the centripetal force with potential energy from integration to find the total negative energy state.

Exam Tip: Bound orbital systems typically have negative total energy equal to minus half their potential energy.

Question 8. The K.E. of a particle moving along a circle of radius \(R\) depends on the distance covered \(S\) as \(T = \alpha S^2\) where \(\alpha\) is constant. Find the force acting on the particle as a function of S. (\(F = 2 \alpha S \sqrt{1 + S^2/R^2}\))
Answer: Kinetic energy \(K = \alpha S^2\).
Since \(K = \frac{1}{2}mv^2 = \alpha S^2 \Rightarrow v^2 = \frac{2\alpha S^2}{m}\).
Tangential acceleration \(a_t = \frac{dv}{dt} = \frac{d}{dt}(S\sqrt{2\alpha/m}) = \frac{2\alpha S}{m} \cdot \frac{dS}{dt} = \frac{2\alpha S v}{m}\).
Tangential force \(F_t = ma_t = 2\alpha S\).
Centripetal acceleration \(a_c = \frac{v^2}{R} = \frac{2\alpha S^2}{mR}\).
Centripetal force \(F_c = ma_c = \frac{2\alpha S^2}{R}\).
Total force \(F = \sqrt{F_t^2 + F_c^2} = \sqrt{(2\alpha S)^2 + \left(\frac{2\alpha S^2}{R}\right)^2} = 2\alpha S \sqrt{1 + \frac{S^2}{R^2}}\).
In simple words: Compute both tangential and centripetal forces from the kinetic energy relation and combine them vectorially.

Exam Tip: Circular motion forces always require combining tangential and radial components perpendicularly.

Question 9. Derive the relation \(E = \Delta mc^2\).
Answer: According to the theory of relativity, mass varies with velocity as \(m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}\), where \(m_0\) is rest mass.
Squaring both sides: \(m^2 c^2 - m^2 v^2 = m_0^2 c^2\).
Differentiating both sides with respect to velocity \(v\):
\(c^2(2m \, dm) - (m^2(2v \, dv) + v^2(2m \, dm)) = 0\)
Dividing by \(2m\):
\(c^2 \, dm - m v \, dv - v^2 \, dm = 0\)
\(c^2 \, dm = (m \, dv + v \, dm)v = d(mv) \cdot v \dots\) wait, using work-energy relation: total relativistic energy \(E = mc^2\), and kinetic energy change equals work done \(dE = F \, dx = d(mv) \cdot v = c^2 \, dm\).
Integrating from rest mass \(m_0\) to relativistic mass \(m\):
\(E - E_0 = \int_{m_0}^{m} c^2 \, dm = (m - m_0)c^2 = \Delta m c^2\).
Thus, energy equivalent of mass defect is \(E = \Delta m c^2\).
In simple words: By integrating relativistic momentum changes with respect to mass, Einstein's mass-energy equivalence formula is derived.

Exam Tip: State Einstein's mass-energy relation clearly and show that mass defect represents stored internal energy.