Step 1: Given Variables & Physical Constants

• Volume of tank ($V$) = $30 \, m^3$
• Time taken ($t$) = $15 \, min = 15 \times 60 = 900 \, s$
• Height of tank ($h$) = $40 \, m$
• Efficiency ($\eta$) = $30\% = 0.3$
• Density of water ($\rho$) = $10^3 \, kg/m^3$
• Acceleration due to gravity ($g$) = $9.8 \, m/s^2$

Step 2: Calculating Mass of Water

The mass ($m$) of the water pumped is determined by the volume and the density of water [By definition: $m = \rho V$].

$m = 10^3 \, kg/m^3 \times 30 \, m^3 = 30,000 \, kg$

Step 3: Calculating Useful Work Done (Output Energy)

The work done ($W_{out}$) against gravity to raise the water to height $h$ is defined by potential energy [Per the Work-Energy Theorem]:

$W_{out} = mgh$

$W_{out} = 30,000 \, kg \times 9.8 \, m/s^2 \times 40 \, m$

$W_{out} = 11,760,000 \, J = 1.176 \times 10^7 \, J$

Step 4: Calculating Useful Power Output

Power is the rate at which work is performed ($P = W/t$):

$P_{out} = \frac{1.176 \times 10^7 \, J}{900 \, s}$

$P_{out} = 13,066.67 \, W$

Step 5: Calculating Input Electric Power

Efficiency is the ratio of useful power output to total power input [Per the Law of Conservation of Energy]:

$\eta = \frac{P_{out}}{P_{in}} \implies P_{in} = \frac{P_{out}}{\eta}$

$P_{in} = \frac{13,066.67 \, W}{0.3}$

$P_{in} = 43,555.57 \, W \approx 43.56 \, kW$

Visual Representation of Power Conversion

Final Solution: The total electric power consumed by the pump is $43.56 \, kW$.

Given Variables & Initial Setup

• Mass of body ($m$) = $2\text{ kg}$
• Initial velocity ($u$) = $0\text{ m/s}$
• Applied horizontal force ($F_a$) = $7\text{ N}$
• Coefficient of kinetic friction ($\mu_k$) = $0.1$
• Time interval ($t$) = $10\text{ s}$
• Acceleration due to gravity ($g$) = $9.8\text{ m/s}^2$

Step 1: Calculating Forces and Acceleration

The frictional force ($f_k$) acting on the body is given by:

$f_k = \mu_k \cdot N = \mu_k \cdot mg$ [By the definition of Kinetic Friction]

$f_k = 0.1 \times 2\text{ kg} \times 9.8\text{ m/s}^2 = 1.96\text{ N}$

The net force ($F_{net}$) acting on the body is:

$F_{net} = F_a - f_k = 7\text{ N} - 1.96\text{ N} = 5.04\text{ N}$

Using Newton's Second Law ($F = ma$):

$a = \frac{F_{net}}{m} = \frac{5.04\text{ N}}{2\text{ kg}} = 2.52\text{ m/s}^2$

Step 2: Calculating Displacement ($s$)

Using the kinematic equation for displacement under constant acceleration:

$s = ut + \frac{1}{2}at^2$ [Per Kinematic Equations of Motion]

$s = (0)(10) + \frac{1}{2}(2.52)(10)^2$

$s = 0 + 1.26 \times 100 = 126\text{ m}$

Step 3: Calculating Work Done

(a) Work done by the applied force ($W_a$)

$W_a = F_a \cdot s \cdot \cos(0^\circ) = 7\text{ N} \times 126\text{ m} \times 1 = 882\text{ J}$

(b) Work done by friction ($W_f$)

$W_f = f_k \cdot s \cdot \cos(180^\circ) = 1.96\text{ N} \times 126\text{ m} \times (-1) = -246.96\text{ J}$

(c) Work done by the net force ($W_{net}$)

$W_{net} = F_{net} \cdot s = 5.04\text{ N} \times 126\text{ m} = 635.04\text{ J}$

(d) Change in Kinetic Energy ($\Delta K$)

First, find the final velocity ($v$):

$v = u + at = 0 + (2.52)(10) = 25.2\text{ m/s}$

$\Delta K = \frac{1}{2}mv^2 - \frac{1}{2}mu^2 = \frac{1}{2}(2)(25.2)^2 - 0 = 635.04\text{ J}$

Interpretation of Results

The results verify the Work-Energy Theorem, which states that the net work done on an object equals the change in its kinetic energy ($W_{net} = \Delta K$). Here, $W_a + W_f = 882 - 246.96 = 635.04\text{ J}$, which is exactly equal to the calculated change in kinetic energy.

Final Solution: (a) $882\text{ J}$, (b) $-246.96\text{ J}$, (c) $635.04\text{ J}$, (d) $635.04\text{ J}$. The results are consistent with the Work-Energy Theorem.

Step 1: Given Variables & Initial Setup

The kinetic energy ($K$) of a particle is defined by the classical relation $K = \frac{1}{2}mv^2$ [Per the work-energy theorem for non-relativistic particles].

• Kinetic Energy of electron ($K_e$): $10\text{ keV} = 10 \times 10^3 \times 1.6 \times 10^{-19}\text{ J} = 1.6 \times 10^{-15}\text{ J}$
• Kinetic Energy of proton ($K_p$): $100\text{ keV} = 100 \times 10^3 \times 1.6 \times 10^{-19}\text{ J} = 1.6 \times 10^{-14}\text{ J}$
• Mass of electron ($m_e$): $9.11 \times 10^{-31}\text{ kg}$
• Mass of proton ($m_p$): $1.67 \times 10^{-27}\text{ kg}$

Step 2: Deriving the Speed Ratio Expression

Solving the kinetic energy formula for velocity ($v$):

$v = \sqrt{\frac{2K}{m}}$

The ratio of the speed of the electron ($v_e$) to the speed of the proton ($v_p$) is given by:

$\frac{v_e}{v_p} = \frac{\sqrt{\frac{2K_e}{m_e}}}{\sqrt{\frac{2K_p}{m_p}}} = \sqrt{\frac{K_e}{K_p} \cdot \frac{m_p}{m_e}}$

Step 3: Numerical Substitution and Calculation

Substitute the given values into the ratio equation:

$\frac{v_e}{v_p} = \sqrt{\left(\frac{10\text{ keV}}{100\text{ keV}}\right) \times \left(\frac{1.67 \times 10^{-27}\text{ kg}}{9.11 \times 10^{-31}\text{ kg}}\right)}$

$\frac{v_e}{v_p} = \sqrt{0.1 \times (0.1833 \times 10^4)}$

$\frac{v_e}{v_p} = \sqrt{0.1 \times 1833} = \sqrt{183.3}$

$\frac{v_e}{v_p} \approx 13.54$

Visual Representation of Speed Comparison

Conclusion on Relative Velocity

Since $\frac{v_e}{v_p} > 1$, it is mathematically evident that the electron moves significantly faster than the proton under the specified energy conditions. [Derived from the inverse proportionality of velocity to the square root of mass for equivalent kinetic energies].

Final Solution: The electron is faster, and the ratio of the speed of the electron to that of the proton is approximately 13.54.

Step 1: Theoretical Definition of Work Done

The work done $W$ by a constant force $\vec{F}$ on a body undergoing a displacement $\vec{d}$ is defined by the scalar (dot) product:

$W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos(\theta)$

where $\theta$ is the angle between the direction of the force vector and the direction of the displacement vector [Per the definition of Scalar Product in Vector Algebra].

Step 2: Analysis of the Physical System

Consider a body sliding down an inclined plane of angle $\alpha$ with the horizontal. Let the displacement $\vec{d}$ be directed down the plane. Kinetic friction $\vec{f_k}$ acts in the direction opposing the relative motion [Per Newton’s Third Law of Motion and the nature of dissipative forces].

Step 3: Determining the Angle $\theta$

• Displacement vector ($\vec{d}$): Directed down the incline.
• Friction vector ($\vec{f_k}$): Directed up the incline (opposing sliding).

Since the force of friction is exactly opposite to the direction of displacement, the angle $\theta$ between $\vec{f_k}$ and $\vec{d}$ is exactly $180^\circ$ [By geometric construction of opposing vectors].

Step 4: Algebraic Evaluation

Substitute the value of $\theta$ into the work equation:

$W = |\vec{f_k}| |\vec{d}| \cos(180^\circ)$

Given that $\cos(180^\circ) = -1$ [Per Trigonometric Identities]:

$W = |\vec{f_k}| |\vec{d}| (-1)$

$W = -|\vec{f_k}| |\vec{d}|$

Conclusion

Because the magnitude of friction $|\vec{f_k}|$ and displacement $|\vec{d}|$ are strictly positive quantities, the product results in a negative value. This confirms that the work done by friction is negative, signifying a loss of mechanical energy from the system, typically dissipated as thermal energy [Per the Work-Energy Theorem].

Final Solution: The work done by friction on a body sliding down an inclined plane is negative.

Theoretical Foundation: Newton’s Second Law for a System of Particles

To determine the rate of change of total momentum ($\vec{P}$) for a many-particle system, we analyze the contribution of both internal and external forces acting on the system. Let the system consist of $N$ particles with individual masses $m_i$ and velocities $\vec{v}_i$.

The total momentum of the system is defined as:

$\vec{P} = \sum_{i=1}^{N} \vec{p}_i = \sum_{i=1}^{N} m_i \vec{v}_i$

Step 1: Differentiation of Total Momentum

The rate of change of the total momentum with respect to time $t$ is given by:

$\frac{d\vec{P}}{dt} = \frac{d}{dt} \sum_{i=1}^{N} \vec{p}_i = \sum_{i=1}^{N} \frac{d\vec{p}_i}{dt}$

[By the Principle of Superposition for linear momentum]

Step 2: Analysis of Forces

Per Newton’s Second Law applied to each particle $i$, the force acting on it is the sum of internal forces ($\vec{f}_{ij}$, exerted on particle $i$ by particle $j$) and external forces ($\vec{F}_i^{ext}$):

$\frac{d\vec{p}_i}{dt} = \vec{F}_i^{ext} + \sum_{j \neq i} \vec{f}_{ij}$

Step 3: Summation and Cancellation of Internal Forces

Summing over all particles $i$ from $1$ to $N$:

$\frac{d\vec{P}}{dt} = \sum_{i=1}^{N} \vec{F}_i^{ext} + \sum_{i=1}^{N} \sum_{j \neq i} \vec{f}_{ij}$

[By Newton’s Third Law, for every internal force $\vec{f}_{ij}$, there exists an equal and opposite force $\vec{f}_{ji}$ such that $\vec{f}_{ij} + \vec{f}_{ji} = 0$. Therefore, the double summation over internal forces vanishes: $\sum \sum \vec{f}_{ij} = 0$.]

Visual Representation of Force Vectors

Step 4: Conclusion of the Derivation

Since the sum of all internal forces is zero, the equation simplifies to:

$\frac{d\vec{P}}{dt} = \sum_{i=1}^{N} \vec{F}_i^{ext} = \vec{F}_{ext}^{total}$

This demonstrates that the rate of change of total momentum is strictly dependent on the net external force acting on the system, independent of internal configurations.

Final Solution: The correct alternative is the external force.

1. Analysis of the Physical System

Let $m$ be the mass of each ball bearing. The incident ball bearing (A) has an initial velocity $\vec{v}_A = V \hat{i}$, while the target ball bearings (B and C) are at rest: $\vec{v}_B = 0$ and $\vec{v}_C = 0$. Since the table is frictionless and the collision is perfectly elastic, two fundamental physical laws must be satisfied:

• Conservation of Linear Momentum: $\sum \vec{p}_{initial} = \sum \vec{p}_{final}$ [Per Newton's Second Law applied to systems]
• Conservation of Kinetic Energy: $\sum K_{initial} = \sum K_{final}$ [Per the definition of an elastic collision]

2. Mathematical Formulation

Initial state variables:

• Total Momentum $P_i = mV + m(0) + m(0) = mV$
• Total Kinetic Energy $K_i = \frac{1}{2}mV^2 + \frac{1}{2}m(0)^2 + \frac{1}{2}m(0)^2 = \frac{1}{2}mV^2$

Consider a potential final state where the incident ball (A) comes to rest, and the two target balls (B and C) move forward with velocities $v_B$ and $v_C$.

Applying Conservation of Momentum:

Applying Conservation of Kinetic Energy:

3. Logical Deduction and Solution

We test the validity of the result where balls B and C move together with a common velocity $v$. If $v_B = v_C = v$, then:

• From Eq 1: $V = v + v = 2v \implies v = \frac{V}{2}$
• Substitute into Eq 2: $V^2 = (\frac{V}{2})^2 + (\frac{V}{2})^2 = \frac{V^2}{4} + \frac{V^2}{4} = \frac{V^2}{2}$

Since $V^2 \neq \frac{V^2}{2}$, the case where both target balls move with the same velocity is physically impossible in an elastic collision.

However, if the incident ball stops, and one ball moves off while the other remains stationary or if they move with specific velocities satisfying both equations, we look for solutions to $v_B + v_C = V$ and $v_B^2 + v_C^2 = V^2$.

Squaring Eq 1: $(v_B + v_C)^2 = V^2 \implies v_B^2 + v_C^2 + 2v_Bv_C = V^2$.

Substituting Eq 2 into this expansion: $V^2 + 2v_Bv_C = V^2 \implies 2v_Bv_C = 0$.

This implies either $v_B = 0$ or $v_C = 0$.

4. Summary of Constraints

Final Solution: The only possible result that satisfies both the Law of Conservation of Momentum and the Law of Conservation of Kinetic Energy is for the incident ball A to come to rest, ball B to remain at rest, and ball C to move forward with the initial velocity $V$.

Step 1: Given Variables & Theoretical Foundation

To determine the total work done against the gravitational force, we identify the following physical constants and parameters:

• Mass of the object, $m = 10 \text{ kg}$
• Height of each lift, $h = 0.5 \text{ m}$
• Number of repetitions, $n = 1000$
• Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$ [Standard gravitational acceleration constant]

Step 2: Calculating Work for a Single Lift

According to the work-energy theorem and the definition of gravitational potential energy, the work $W_1$ required to lift a mass against gravity at a constant velocity is equal to the change in gravitational potential energy:

$W_1 = m \cdot g \cdot h$

Substituting the given values:

$W_1 = 10 \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.5 \text{ m}$

$W_1 = 49 \text{ J}$ [Per the definition of Work: $W = Fd \cos(\theta)$, where $F = mg$ and $\theta = 0^\circ$ for upward displacement]

Step 3: Calculating Total Work Done

The total work $W_{total}$ performed by the dieter over $n$ repetitions is the product of the work done per lift and the total number of repetitions:

$W_{total} = n \cdot W_1$

$W_{total} = 1000 \times 49 \text{ J}$

$W_{total} = 49,000 \text{ J}$

Step 4: Theoretical Clarification on Dissipation

The problem states that potential energy lost during the lowering phase is dissipated (i.e., converted into heat or sound through non-conservative forces). Therefore, no energy is recovered during the lowering phase. The dieter must perform the full amount of work against gravity during each of the 1000 lifts.

Final Solution: The total work done by the dieter against the gravitational force is $4.9 \times 10^4 \text{ J}$ (or $49 \text{ kJ}$).

Step 1: Establishing the Physical Model

To determine the kinetic energy of air moving through the swept area of a windmill, we define the parameters of the fluid flow. Let:

• $A$: The cross-sectional area swept by the windmill blades ($m^2$).
• $v$: The velocity of the air flow ($m/s$).
• $\rho$: The density of the air ($kg/m^3$).
• $t$: The time interval considered ($s$).

[Per the principle of continuity in fluid dynamics, the mass of air passing through the area $A$ in time $t$ forms a cylinder of volume $V = A \cdot l$, where $l$ is the distance covered by air in time $t$.]

Step 2: Calculating Mass Flow

The distance $l$ traveled by the air in time $t$ is given by $l = v \cdot t$. Therefore, the volume of air $V$ passing through the blades in time $t$ is:

$V = A \cdot v \cdot t$

The mass $m$ of this air is the product of density and volume:

$m = \rho \cdot V = \rho \cdot A \cdot v \cdot t$ [Definition of mass density $\rho = m/V$]

Step 3: Deriving Kinetic Energy

The kinetic energy ($K$) is defined by the classical mechanics formula:

$K = \frac{1}{2} m v^2$ [Per the Work-Energy Theorem for a particle of mass $m$]

Substituting the expression for mass $m$ derived in Step 2 into the kinetic energy equation:

$K = \frac{1}{2} (\rho A v t) v^2$

$K = \frac{1}{2} \rho A v^3 t$

Step 4: Visualizing the Air Cylinder

Step 5: Synthesizing the Kinetic Energy per Unit Time

Often, the kinetic energy of air is expressed as the rate of energy transfer (Power, $P$), where $P = \frac{dK}{dt}$.

Taking the derivative of $K$ with respect to $t$:

$P = \frac{d}{dt} \left( \frac{1}{2} \rho A v^3 t \right) = \frac{1}{2} \rho A v^3$

Final Solution: The kinetic energy of the air passing through the windmill in time $t$ is $K = \frac{1}{2} \rho A v^3 t$.

Step 1: Identifying Given Variables & Unit Conversions

To ensure dimensional homogeneity, we must convert all units to the SI system before calculation.

• Area of the circle swept by blades ($A$): $30 \text{ m}^2$
• Velocity of wind ($v$): $36 \text{ km/h} = 36 \times \left(\frac{5}{18}\right) \text{ m/s} = 10 \text{ m/s}$ [Conversion factor: $1 \text{ km/h} = \frac{1000 \text{ m}}{3600 \text{ s}} = \frac{5}{18} \text{ m/s}$]
• Density of air ($\rho$): $1.2 \text{ kg/m}^3$
• Efficiency of conversion ($\eta$): $25\% = 0.25$

Step 2: Determining the Mass Flow Rate of Air

The mass of air ($m$) passing through the area $A$ in time $t$ is defined by the volume of a cylinder of air moving at velocity $v$. The mass flow rate ($\dot{m}$) is given by:

$\dot{m} = \frac{m}{t} = \rho A v$

Substituting the known values:

$\dot{m} = 1.2 \text{ kg/m}^3 \times 30 \text{ m}^2 \times 10 \text{ m/s} = 360 \text{ kg/s}$

Step 3: Calculating Total Kinetic Power of the Wind

The total kinetic energy ($K$) of a mass $m$ moving with velocity $v$ is given by $K = \frac{1}{2}mv^2$ [Per the Kinetic Energy Theorem]. Consequently, the power ($P_{total}$) available in the wind is:

$P_{total} = \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt}\right) v^2 = \frac{1}{2} \rho A v \cdot v^2 = \frac{1}{2} \rho A v^3$

Calculating the total power:

$P_{total} = 0.5 \times 1.2 \text{ kg/m}^3 \times 30 \text{ m}^2 \times (10 \text{ m/s})^3$

$P_{total} = 0.5 \times 36 \times 1000 = 18,000 \text{ Watts (or 18 kW)}$

Step 4: Calculating Electrical Power Output

The actual electrical power produced ($P_{elec}$) is the product of the total power and the efficiency ($\eta$) of the conversion process:

$P_{elec} = \eta \times P_{total}$

$P_{elec} = 0.25 \times 18,000 \text{ W}$

$P_{elec} = 4,500 \text{ Watts}$

Visual Representation of Wind Power Flux

Final Solution: The electrical power produced by the windmill is 4,500 W or 4.5 kW.

Theoretical Foundation: Newton's Second Law and Impulse

To determine the conservation of linear momentum in an elastic collision, we must analyze the forces acting upon the system of two balls ($A$ and $B$) during the collision interval $\Delta t$.

According to Newton's Second Law of Motion in terms of momentum: $\vec{F}_{ext} = \frac{d\vec{p}}{dt}$ [Per Newton's Second Law of Motion].

For any system, the change in total linear momentum ($\Delta \vec{p}$) is given by the impulse delivered by external forces: $\Delta \vec{p} = \int_{t_1}^{t_2} \vec{F}_{ext} dt$ [By the Impulse-Momentum Theorem].

Analysis of the Collision Environment

• Internal Forces: During the collision, the balls exert equal and opposite impulsive forces on each other ($\vec{F}_{AB} = -\vec{F}_{BA}$) [Per Newton's Third Law of Motion]. These forces are internal to the system $(A+B)$.
• External Forces: In a short time interval $\Delta t$, the gravitational force ($\vec{F}_g = m\vec{g}$) and any normal reaction forces (e.g., from the floor) are considered negligible compared to the large, impulsive contact forces between the colliding bodies.
• Limit Condition: As $\Delta t \to 0$, the impulse due to external forces ($J_{ext} = \vec{F}_{ext} \cdot \Delta t$) approaches zero.

Visual Representation of Forces

Logical Deduction

Since the net external force acting on the system during the interval $\Delta t$ is effectively zero, the change in total linear momentum is zero ($\Delta \vec{P} = 0$). This holds true regardless of whether the collision is elastic or inelastic, provided the system is considered isolated during the interaction interval.

Final Solution: Yes, the total linear momentum is conserved. Because the impulsive contact forces during the collision are internal to the system, and external forces (like gravity or friction) are negligible over the infinitesimally small time duration of the impact, the net external force remains zero, thereby ensuring $\Delta \vec{p} = 0$ according to the Law of Conservation of Linear Momentum.

Step 1: Identification of Given Variables

• Mass of the body ($m$): $2\text{ kg}$
• Applied force ($F_{app}$): $7\text{ N}$
• Coefficient of kinetic friction ($\mu_k$): $0.1$
• Initial velocity ($u$): $0\text{ m/s}$
• Time interval ($t$): $10\text{ s}$
• Acceleration due to gravity ($g$): $9.8\text{ m/s}^2$ [Standard Earth Gravity]

Step 2: Calculation of Kinetic Frictional Force ($f_k$)

The normal reaction force ($N$) on a horizontal surface is given by $N = mg$. [Per Newton's Second Law for vertical equilibrium].

$N = 2\text{ kg} \times 9.8\text{ m/s}^2 = 19.6\text{ N}$

The frictional force ($f_k$) is calculated as:

$f_k = \mu_k \times N$

$f_k = 0.1 \times 19.6\text{ N} = 1.96\text{ N}$

Step 3: Determination of Net Acceleration ($a$)

The net force ($F_{net}$) acting on the body is the difference between the applied force and the frictional force:

$F_{net} = F_{app} - f_k$

$F_{net} = 7\text{ N} - 1.96\text{ N} = 5.04\text{ N}$

Using Newton's Second Law ($F = ma$):

$a = \frac{F_{net}}{m} = \frac{5.04\text{ N}}{2\text{ kg}} = 2.52\text{ m/s}^2$

Step 4: Calculation of Displacement ($s$) in 10 seconds

Using the kinematic equation for displacement under constant acceleration: $s = ut + \frac{1}{2}at^2$.

$s = (0 \times 10) + \frac{1}{2} \times 2.52\text{ m/s}^2 \times (10\text{ s})^2$

$s = 0.5 \times 2.52 \times 100 = 126\text{ m}$

Step 5: Calculation of Work Done by Friction ($W_f$)

Work done is defined as the scalar product of force and displacement ($W = \vec{F} \cdot \vec{s}$). Since kinetic friction acts in the direction opposite to the displacement, the angle ($\theta$) between the frictional force vector and the displacement vector is $180^\circ$.

$W_f = f_k \cdot s \cdot \cos(180^\circ)$

$W_f = 1.96\text{ N} \times 126\text{ m} \times (-1)$

$W_f = -246.96\text{ J}$

Final Solution: The work done by friction in 10 seconds is -246.96 J.

Statement Assessment

The statement "Total energy of a system is always conserved, no matter what internal and external forces on the body are present" is False.

Theoretical Justification

The principle of Conservation of Mechanical Energy states that the total mechanical energy of a system is conserved only if the internal forces are conservative and no external work is performed on the system [Per the Work-Energy Theorem].

Step 1: Analyzing the Work-Energy Theorem

The total work done ($W_{total}$) on a system is equal to the change in its kinetic energy ($\Delta K$):

$W_{total} = W_{conservative} + W_{non-conservative} + W_{external} = \Delta K$

[By the Work-Energy Theorem, the change in kinetic energy is the sum of works done by all forces.]

Step 2: Defining Total Energy vs. Mechanical Energy

If the system is isolated (no external forces, $W_{external} = 0$) and all internal forces are conservative, then the sum of kinetic energy ($K$) and potential energy ($U$) remains constant:

$E = K + U = \text{constant}$

However, if external forces act on the system, the work done by these external forces ($W_{ext}$) changes the total energy of the system:

$\Delta E = W_{ext}$

[Per the General Law of Conservation of Energy, energy can be transferred into or out of a system via external work or heat.]

Step 3: Role of Non-Conservative Forces

When non-conservative forces (e.g., friction, air resistance, or viscous drag) are present, mechanical energy is dissipated into non-mechanical forms such as thermal energy ($Q$).

$W_{non-conservative} = \Delta K + \Delta U = -\Delta Q$

The mechanical energy of the system is not conserved because it is converted into internal energy (heat), which is often excluded from the "mechanical" energy calculation of the system.

Visual Representation of Energy Flux

Summary of Factors Influencing Energy Conservation

Final Solution: The statement is False. Total energy of a system is only conserved if the system is isolated from external work and no non-conservative forces are performing work that dissipates mechanical energy into non-mechanical forms.

Step 1: Identifying Given Variables & Constants

• Volume of water ($V$): $30 \text{ m}^3$
• Time taken ($t$): $15 \text{ min} = 15 \times 60 \text{ s} = 900 \text{ s}$
• Height ($h$): $40 \text{ m}$
• Efficiency ($\eta$): $30\% = 0.30$
• Density of water ($\rho$): $1000 \text{ kg/m}^3$ [Standard physical constant]
• Acceleration due to gravity ($g$): $9.8 \text{ m/s}^2$

Step 2: Calculating Mass of Water and Output Work

The mass ($m$) of the water pumped is calculated using the density formula $m = \rho \times V$:

$m = 1000 \text{ kg/m}^3 \times 30 \text{ m}^3 = 30,000 \text{ kg}$

The useful work ($W$) done by the pump to raise the water is equal to the gravitational potential energy gained [Per the Work-Energy Theorem]:

$W = mgh$

$W = 30,000 \text{ kg} \times 9.8 \text{ m/s}^2 \times 40 \text{ m} = 11,760,000 \text{ J}$

Step 3: Determining Useful Power Output

Power is defined as the rate of doing work ($P_{\text{out}} = \frac{W}{t}$):

$P_{\text{out}} = \frac{11,760,000 \text{ J}}{900 \text{ s}} \approx 13,066.67 \text{ W}$

Step 4: Calculating Input Power Consumption

Efficiency ($\eta$) is the ratio of useful power output to actual electric power input ($P_{\text{in}}$):

$\eta = \frac{P_{\text{out}}}{P_{\text{in}}} \implies P_{\text{in}} = \frac{P_{\text{out}}}{\eta}$ [Per the definition of Efficiency]

$P_{\text{in}} = \frac{13,066.67 \text{ W}}{0.30} = 43,555.56 \text{ W}$

Visual Representation of System Energy Flow

Summary of Calculation

Final Solution: The electric power consumed by the pump is $43,555.56 \text{ W}$ or approximately $43.56 \text{ kW}$.

Step 1: Given Variables & Initial Setup

To determine the work done by the applied force, we must first analyze the dynamics of the system. The following parameters are provided:

• Mass of the body ($m$): $2\text{ kg}$
• Applied horizontal force ($F$): $7\text{ N}$
• Coefficient of kinetic friction ($\mu_k$): $0.1$
• Initial velocity ($u$): $0\text{ m/s}$
• Time duration ($t$): $10\text{ s}$
• Acceleration due to gravity ($g$): $9.8\text{ m/s}^2$

Step 2: Calculating Net Force and Acceleration

First, calculate the frictional force ($f_k$) acting in opposition to the applied force:

$f_k = \mu_k \cdot N = \mu_k \cdot mg$ [By the Law of Kinetic Friction]

$f_k = 0.1 \cdot 2\text{ kg} \cdot 9.8\text{ m/s}^2 = 1.96\text{ N}$

The net force ($F_{net}$) acting on the body is the difference between the applied force and friction:

$F_{net} = F - f_k = 7\text{ N} - 1.96\text{ N} = 5.04\text{ N}$

Applying Newton's Second Law of Motion ($F = ma$):

$a = \frac{F_{net}}{m} = \frac{5.04\text{ N}}{2\text{ kg}} = 2.52\text{ m/s}^2$

Step 3: Calculating Displacement over Time

Using the kinematic equation for displacement ($s$) under constant acceleration [Per the Equations of Kinematic Motion]:

$s = ut + \frac{1}{2}at^2$

Since $u = 0$:

$s = 0 + \frac{1}{2} \cdot 2.52\text{ m/s}^2 \cdot (10\text{ s})^2$

$s = 1.26 \cdot 100 = 126\text{ meters}$

Step 4: Calculating Work Done by Applied Force

Work done ($W$) by a constant force is defined as the scalar product of force and displacement vectors ($W = \vec{F} \cdot \vec{s} = Fs \cos \theta$). Given the force is applied horizontally in the direction of displacement ($\theta = 0^\circ, \cos 0^\circ = 1$):

$W = F \cdot s$

$W = 7\text{ N} \cdot 126\text{ m}$

$W = 882\text{ J}$

Final Solution: The work done by the applied force in 10 seconds is 882 Joules.

Step 1: Identifying the System Dynamics

To determine the rate of change of the total momentum of a many-particle system, we define the total momentum $\mathbf{P}$ of a system consisting of $N$ particles as the vector sum of individual momenta: $$\mathbf{P} = \sum_{i=1}^{N} \mathbf{p}_i = \sum_{i=1}^{N} m_i \mathbf{v}_i$$ [By the definition of linear momentum in classical mechanics].

Step 2: Differentiating with respect to Time

The rate of change of the total momentum is given by the derivative of $\mathbf{P}$ with respect to time $t$: $$\frac{d\mathbf{P}}{dt} = \sum_{i=1}^{N} \frac{d\mathbf{p}_i}{dt}$$ [By the linearity property of the derivative operator].

Step 3: Applying Newton’s Second Law

According to Newton's Second Law for each particle $i$, the force acting on it is the sum of internal forces ($\mathbf{f}_{ij}$, force exerted by particle $j$ on $i$) and external forces ($\mathbf{F}_{i, ext}$): $$\frac{d\mathbf{p}_i}{dt} = \mathbf{F}_{i, ext} + \sum_{j \neq i} \mathbf{f}_{ij}$$ Substituting this into the summation: $$\frac{d\mathbf{P}}{dt} = \sum_{i=1}^{N} \mathbf{F}_{i, ext} + \sum_{i=1}^{N} \sum_{j \neq i} \mathbf{f}_{ij}$$

Step 4: Analyzing Internal Forces

Per Newton’s Third Law of Motion, for every internal force $\mathbf{f}_{ij}$, there exists an equal and opposite reaction force $\mathbf{f}_{ji}$ such that $\mathbf{f}_{ij} = -\mathbf{f}_{ji}$. When summing over all pairs $(i, j)$ in the system, these forces cancel out exactly: $$\sum_{i=1}^{N} \sum_{j \neq i} \mathbf{f}_{ij} = 0$$ Consequently, the internal forces contribute nothing to the change in total momentum of the system. We are left with: $$\frac{d\mathbf{P}}{dt} = \sum_{i=1}^{N} \mathbf{F}_{i, ext} = \mathbf{F}_{ext}$$ where $\mathbf{F}_{ext}$ is the vector sum of all external forces acting on the system.

Visual Representation of System Forces

Conclusion

The rate of change of total momentum is equal to (and thus proportional to) the sum of external forces acting on the system. Internal forces are irrelevant to the change in the total momentum of the system.

Final Solution: The rate of change of total momentum of a many-particle system is proportional to the external force acting on the system.

Statement Analysis

The statement "In an inelastic collision, the final kinetic energy is always less than the initial kinetic energy of the system" is True.

Step 1: Theoretical Definition of Collision Types

Collisions are categorized based on the conservation of kinetic energy. [Per the Law of Conservation of Energy and Principles of Classical Mechanics]:

• Elastic Collision: Both linear momentum and total kinetic energy are conserved. The internal energy remains unchanged ($K_i = K_f$).
• Inelastic Collision: Linear momentum is conserved, but total kinetic energy is not conserved ($K_i \neq K_f$).

Step 2: Energy Dissipation Mechanism

In an inelastic collision, internal forces perform work to deform the colliding bodies or generate heat, sound, or light. This conversion of mechanical energy into other forms is irreversible [Per the First Law of Thermodynamics].

Step 3: Mathematical Derivation

For a system of two particles with masses $m_1$ and $m_2$ moving with velocities $u_1$ and $u_2$ initially, and $v_1$ and $v_2$ finally, the total initial kinetic energy is:

$K_i = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2$

The total final kinetic energy is:

$K_f = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2$

For an inelastic collision, there is a loss of kinetic energy ($\Delta K$), defined as:

$\Delta K = K_i - K_f > 0$

This implies $K_i > K_f$. [Justification: By the principle of work-energy, the lost kinetic energy is equal to the work done against internal dissipative forces $\int F_{int} \cdot dr$].

Visual Representation of Kinetic Energy Dissipation

Analytical Summary

Since the energy is dissipated into non-mechanical forms (heat, deformation, or sound), the final kinetic energy must be strictly less than the initial kinetic energy for any collision categorized as inelastic.

Final Solution: The statement is True, because kinetic energy is dissipated into other forms of energy (thermal, sound, or structural deformation) during the collision process.

Statement Analysis

Statement: "In an elastic collision of two bodies, the momentum and energy of each body is conserved."

Verdict: False.

Theoretical Justification

The definition of an elastic collision is governed by the principles of conservation of momentum and conservation of total kinetic energy for the system, not for the individual bodies involved in the collision [Per the Laws of Conservation in Classical Mechanics].

Step 1: Conservation of Momentum for the System

For a system of two bodies with masses $m_1$ and $m_2$ and velocities $\vec{u}_1, \vec{u}_2$ (initial) and $\vec{v}_1, \vec{v}_2$ (final):

$$\vec{p}_{initial} = \vec{p}_{final}$$

$$m_1\vec{u}_1 + m_2\vec{u}_2 = m_1\vec{v}_1 + m_2\vec{v}_2$$

This law applies to the entire system as a collective unit. Individual bodies experience an impulsive force during the collision, leading to a change in their individual momenta ($\Delta \vec{p}_i \neq 0$).

Step 2: Conservation of Kinetic Energy for the System

In an elastic collision, the total kinetic energy of the system is conserved:

$$\sum K_i = \sum K_f$$

$$\frac{1}{2}m_1u_1^2 + \frac{1}{2}m_2u_2^2 = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2$$

While the system’s energy remains constant, individual bodies undergo energy exchange. Body 1 may gain kinetic energy while Body 2 loses an equal amount (or vice versa), meaning $K_{1,i} \neq K_{1,f}$ and $K_{2,i} \neq K_{2,f}$.

Step 3: Logical Counter-Argument

Visual Representation of System Dynamics

Final Solution: The statement is False. In an elastic collision, linear momentum and kinetic energy are conserved only for the system as a whole, whereas individual bodies undergo changes in their respective momentum and kinetic energy due to the interaction forces during collision.

Step 1: Theoretical Foundation of Work

In classical mechanics, the work done ($W$) by a constant force ($\vec{F}$) on a body undergoing a displacement ($\vec{d}$) is defined by the dot product of the two vectors: $W = \vec{F} \cdot \vec{d} = |\vec{F}| |\vec{d}| \cos(\theta)$ where $\theta$ is the angle between the direction of the force vector and the direction of the displacement vector [Per the Work-Energy Theorem definitions].

Step 2: Analysis of the Resistive Force of Air

The resistive force of air, commonly known as air drag or air resistance, is an dissipative force that always opposes the instantaneous motion of the object through the fluid medium.

• Direction of Motion ($\vec{d}$): The pendulum bob moves along its arc of oscillation in a specific direction at any given instant.
• Direction of Force ($\vec{F}$): Because air resistance is inherently resistive, the force vector $\vec{F}$ acts in the direction exactly opposite to the velocity vector (and thus the instantaneous displacement $\vec{d}$) [By Newton’s Third Law of Motion and Fluid Dynamics Principles].

Step 3: Determination of the Angle ($\theta$)

Since the resistive force ($\vec{F}$) and the displacement ($\vec{d}$) are always anti-parallel, the angle between them is precisely $180^\circ$ (or $\pi$ radians). Substituting this into the work formula: $W = |\vec{F}| |\vec{d}| \cos(180^\circ)$ Since $\cos(180^\circ) = -1$: $W = - |\vec{F}| |\vec{d}|$ [By Trigonometric Identity for $\cos(\pi)$]

Step 4: Vector Diagram of Forces

Step 5: Conclusion on Energy Dissipation

The negative sign in the work calculation indicates that the force is removing kinetic energy from the system. As the pendulum vibrates, the work done by the resistive force is consistently negative, resulting in the conversion of mechanical energy into thermal energy within the surrounding air molecules. This continuous negative work is exactly why the pendulum's amplitude decays over time until the system is brought to rest [Per the Law of Conservation of Energy].

Final Solution: The work done by the resistive force of air on the vibrating pendulum is negative.

Initial State Configuration

We consider three identical ball bearings, each of mass $m$. Let the initial state be defined as:

• Ball 1 (moving): Velocity $\vec{u}_1 = V \hat{i}$
• Ball 2 (stationary): Velocity $\vec{u}_2 = 0$
• Ball 3 (stationary): Velocity $\vec{u}_3 = 0$

Total initial momentum $P_i = m(V) + m(0) + m(0) = mV$.

Total initial kinetic energy $K_i = \frac{1}{2}mV^2$.

Conservation Laws in Elastic Collisions

For an elastic collision, both linear momentum and kinetic energy must be conserved [Per the Law of Conservation of Momentum and the Definition of Elastic Collision].

• $\sum \vec{p}_{initial} = \sum \vec{p}_{final}$
• $\sum K_{initial} = \sum K_{final}$

Evaluation of Final State Hypotheses

Let the final velocities of the three balls be $v_1, v_2,$ and $v_3$ respectively.

Mathematical Verification of Case I

In Case I, where the first ball comes to rest and the third ball carries the total momentum $V$:

Conservation of Momentum: $mv_1 + mv_2 + mv_3 = m(0) + m(0) + m(V) = mV$. This matches the initial momentum [By Conservation of Momentum Principle].

Conservation of Kinetic Energy: $\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 + \frac{1}{2}mv_3^2 = 0 + 0 + \frac{1}{2}mV^2 = \frac{1}{2}mV^2$. This matches the initial kinetic energy [By Conservation of Energy Principle].

Vector Representation of the Elastic Collision

Conclusion of Analysis

Only the scenario where the third ball (the one on the far right) moves away with velocity $V$ while the others remain stationary satisfies both the conservation of linear momentum and the conservation of kinetic energy simultaneously. Any division of energy among multiple balls results in a loss of total kinetic energy compared to the initial state, which violates the definition of a perfectly elastic collision [Per the definition of elastic collisions in one dimension].

Final Solution: The only possible result is Case I, where the first two balls remain at rest and the third ball moves forward with velocity $V$.

Step 1: Identification of Variables and Kinematic Relations

Let the body of mass $m$ be initially at rest at $t = 0$. The body undergoes one-dimensional motion under a constant acceleration $a$. Based on the first equation of motion [Per Kinematic laws for constant acceleration]:

$v(t) = u + at$

Since the body is initially at rest, $u = 0$. Thus, the velocity at any time $t$ is:

$v(t) = at$

Step 2: Force Formulation

According to Newton's Second Law of Motion, the net force $F$ acting on the body is the product of its mass $m$ and its acceleration $a$ [Per Newton's Second Law, $F = ma$]:

$F = ma$

Since $a$ is constant, the force $F$ is also constant throughout the duration of the motion.

Step 3: Deriving Power as a Function of Time

Power $P$ is defined as the dot product of force and velocity, or for one-dimensional motion, the product of the magnitudes [Per the definition of instantaneous power $P = \vec{F} \cdot \vec{v}$]:

$P(t) = F \cdot v(t)$

Substituting the expressions derived in Step 1 and Step 2:

$P(t) = (ma) \cdot (at)$

$P(t) = ma^2t$

Step 4: Analyzing Proportionality

From the expression $P(t) = (ma^2)t$, we observe the relationship between Power ($P$) and time ($t$). Since $m$ (mass) and $a$ (acceleration) are constants, the term $(ma^2)$ is a constant factor.

$P \propto t$

Visual Representation of Power Development

The linear relationship indicates that for a body starting from rest under constant acceleration, the instantaneous power delivered increases linearly with time.

Final Solution: The power delivered to the body at time (t) is proportional to $t^1$ (i.e., directly proportional to time $t$).