Step 1: Identification of Given Variables

To determine the acceleration of the train, we define the parameters provided in the problem statement:

• Mass of the engine ($m_e$) = $8000\text{ kg}$
• Number of wagons ($n$) = $5$
• Mass per wagon ($m_w$) = $2000\text{ kg}$
• Applied force by engine ($F_{applied}$) = $40000\text{ N}$
• Friction force ($F_{friction}$) = $5000\text{ N}$

Step 2: Calculating Total Mass of the System

The system consists of the engine and the five wagons moving together as a single unit. Therefore, the total mass ($M_{total}$) is the sum of the mass of the engine and the total mass of all wagons.

$M_{total} = m_e + (n \times m_w)$

$M_{total} = 8000\text{ kg} + (5 \times 2000\text{ kg})$

$M_{total} = 8000\text{ kg} + 10000\text{ kg} = 18000\text{ kg}$

Step 3: Calculating Net Force

According to [Newton’s Second Law of Motion], the net force ($F_{net}$) acting on an object is the vector sum of all forces applied to it. Since friction acts in the direction opposite to the applied force:

$F_{net} = F_{applied} - F_{friction}$

$F_{net} = 40000\text{ N} - 5000\text{ N} = 35000\text{ N}$

Step 4: Calculation of Acceleration

Applying the formula derived from [Newton’s Second Law]: $F_{net} = M_{total} \times a$, we rearrange to solve for acceleration ($a$):

$a = \frac{F_{net}}{M_{total}}$

$a = \frac{35000\text{ N}}{18000\text{ kg}}$

$a = \frac{35}{18}\text{ m/s}^2 \approx 1.944\text{ m/s}^2$

Visual Representation of Force Vectors

Final Solution: The acceleration of the train is $1.944\text{ m/s}^2$.

Analysis of Newton’s Third Law of Motion

Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. Mathematically, if body $A$ exerts a force $F_{AB}$ on body $B$, then body $B$ exerts a force $F_{BA}$ on body $A$, such that: $F_{AB} = -F_{BA}$ [Per Newton's Third Law of Motion].

Evaluation of the Student's Logic

The student’s assertion that the forces "cancel each other out" is physically incorrect. The fundamental error lies in a misunderstanding of the conditions required for force cancellation. Forces can only cancel out if they act upon the same body. In this scenario:

• Force 1: The push exerted by the student on the truck ($F_{ST}$) acts on the truck.
• Force 2: The push exerted by the truck on the student ($F_{TS}$) acts on the student.

Since these forces act on two different bodies, they result in different accelerations for the respective bodies and do not sum to zero on either object. They do not cancel each other out [Per the Principle of Superposition of Forces].

Explanation: Why the Truck Does Not Move

The truck remains stationary because the external force applied by the student is insufficient to overcome the static frictional force between the truck's tires and the road surface. According to Newton’s Second Law of Motion ($F = ma$):

For an object to accelerate ($a > 0$), the net external force ($F_{net}$) must be greater than zero. The forces acting on the truck are:

• $F_{applied}$: Force applied by the student.
• $f_{static}$: Frictional force exerted by the ground on the truck.

The truck will only move if $F_{applied} > f_{static}$. As long as $F_{applied} \le f_{static}$, the net force remains zero, and the truck remains in its state of rest [Per Newton's First Law of Motion (Law of Inertia)].

Summary of Forces

Final Solution: The student's logic is flawed because action-reaction forces act on different bodies and cannot cancel each other. The truck does not move because the applied force is balanced by the static frictional force exerted by the road, resulting in a net force of zero on the truck.

Step 1: Theoretical Foundation

To determine if an object can maintain a non-zero velocity under a net zero external unbalanced force, we refer to Newton's First Law of Motion (The Law of Inertia). This law states that an object will remain at rest or continue to move at a constant velocity in a straight line unless acted upon by an external unbalanced force [Per Newton's First Law].

Step 2: Mathematical Analysis

Newton's Second Law defines the relationship between force ($F$), mass ($m$), and acceleration ($a$):

$F = m \cdot a$

Given that the net external unbalanced force $F_{net} = 0$, we substitute this into the equation:

$0 = m \cdot a$

Assuming the mass of the object $m > 0$, the acceleration must be zero ($a = 0$). Since acceleration is defined as the rate of change of velocity ($a = \frac{dv}{dt}$), $a = 0$ implies that the velocity ($v$) of the object is constant ($v = \text{constant}$).

Step 3: Verification of Condition

If $v = \text{constant}$ and that constant is non-zero ($v \neq 0$), the object is in a state of uniform motion. Therefore, it is entirely possible for an object to travel with a non-zero velocity provided the net external force remains zero.

Step 4: Required Conditions

For an object to maintain a non-zero velocity with no net external force, the following conditions must be met:

• Absence of Friction/Air Resistance: In a real-world environment, friction acts as an external force. To maintain constant velocity, the object must either be in a vacuum or on a frictionless surface.
• Directional Constancy: The motion must be in a straight line. Any change in direction constitutes acceleration, which would require an unbalanced centripetal force [Per the definition of Uniform Circular Motion].
• Constant Magnitude: The speed must not increase or decrease, as a change in speed implies an acceleration [Per Kinematic equations].

Final Solution: Yes, it is possible for the object to travel with a non-zero velocity. The conditions required are that the object must move at a constant speed in a straight line (uniform motion) and remain free from any external unbalanced forces, such as friction or air resistance, that would otherwise cause acceleration or deceleration.

Theoretical Foundation: The First Law of Motion

The phenomenon described is a classic demonstration of Newton’s First Law of Motion, often referred to as the Law of Inertia. This law states that an object at rest will remain at rest, and an object in motion will continue in motion with a constant velocity unless acted upon by an external unbalanced force [Per Newton’s First Law of Motion].

Step 1: State of the System (Static Equilibrium)

Before the interaction, both the carpet and the dust particles embedded within its fibers are at rest (velocity $v = 0$). In this state, both possess Inertia of Rest, which is the inherent property of a body to resist any change in its state of rest.

Step 2: Application of External Unbalanced Force

When the carpet is struck with a stick, an impulsive force is applied locally to the carpet. This force causes the carpet to move suddenly or accelerate rapidly [Per Newton’s Second Law, $F = ma$].

Step 3: Analytical Breakdown of Inertial Response

• The Carpet: Upon being struck, the carpet undergoes a change in its state. It moves forward due to the momentum transferred by the stick.
• The Dust Particles: Because the dust particles were not directly subjected to the same magnitude of impulsive force, they attempt to maintain their state of rest due to Inertia of Rest.

Step 4: Resultant Separation

As the carpet moves away rapidly, the dust particles fail to follow the motion of the carpet instantly. They are left behind at their initial position of rest. Gravity then acts upon the dust particles ($F_g = mg$), causing them to fall downwards. This perceived "coming out" of the dust is actually the carpet moving away from the particles that are resisting motion.

Summary of Logic

Final Solution: When the carpet is beaten, the carpet moves due to the applied force of the stick, but the dust particles remain at rest due to their inertia of rest. Consequently, the carpet moves away from the dust, causing the dust to appear to fall out of the carpet.

1. Theoretical Foundation: Newton’s First Law of Motion

The requirement to secure luggage to the roof of a moving vehicle is fundamentally grounded in Newton’s First Law of Motion, also known as the Law of Inertia. This law states that an object will remain at rest or continue to move at a constant velocity in a straight line unless acted upon by an external unbalanced force [Per Newton's First Law of Motion].

2. Analysis of Inertia in a Moving Vehicle

When luggage is placed on the roof of a bus, it possesses the same velocity as the bus. However, the luggage is not physically integrated into the structural frame of the vehicle. Its state of motion is governed by its own inertia—the tendency of an object to resist changes in its state of motion.

3. Diagrammatic Representation of Inertial Forces

4. Role of the Rope as an External Force

By tying the luggage with a rope, we introduce a tension force ($T$) that acts as the required external unbalanced force. This force effectively links the luggage to the bus frame. When the bus changes velocity or direction:

• The rope exerts a force that overcomes the luggage's inertia of motion/rest.
• The rope restricts the displacement of the luggage relative to the roof surface.
• Mathematically, the rope ensures that $F_{net}$ on the luggage remains zero relative to the vehicle, preventing displacement ($a = 0$ relative to the bus) [By Newton's Second Law: $F = ma$].

Final Solution: Luggage is tied to the roof of a bus because, according to Newton’s First Law, it possesses inertia of motion. If the bus accelerates, decelerates, or turns sharply, the luggage will attempt to maintain its state of motion, causing it to slide or fall. The rope provides the necessary external unbalanced force to tether the luggage to the bus, ensuring both move as a single system.

Given Variables & Initial Setup

To determine the momentum transferred to the floor upon impact, we must first calculate the velocity of the dumb-bell at the instant it strikes the surface. The given parameters are:

• Mass ($m$) = $10\text{ kg}$
• Height ($h$) = $80\text{ cm} = 0.8\text{ m}$ [Conversion to SI units: $1\text{ cm} = 10^{-2}\text{ m}$]
• Initial Velocity ($u$) = $0\text{ m/s}$ [As the object falls from rest]
• Acceleration ($a$) = $10\text{ m/s}^2$

Step 1: Calculating Impact Velocity ($v$)

Using the third equation of motion, which relates displacement to velocity and acceleration in the absence of time:

$v^2 - u^2 = 2as$

Substituting the known values:

$v^2 - 0^2 = 2 \times (10\text{ m/s}^2) \times (0.8\text{ m})$

$v^2 = 16\text{ m}^2/\text{s}^2$

$v = \sqrt{16} = 4\text{ m/s}$

[By the Kinematic Equation of Motion: $v^2 = u^2 + 2as$]

Step 2: Calculating Momentum ($p$)

Momentum is defined as the product of the mass and velocity of an object. The momentum transferred to the floor upon collision is equal to the momentum the object possesses immediately before impact:

$p = m \times v$

Substituting the values derived in Step 1:

$p = 10\text{ kg} \times 4\text{ m/s}$

$p = 40\text{ kg}\cdot\text{m/s}$

Visual Representation of the Motion

Summary of Data

Final Solution: The dumb-bell will transfer a momentum of 40 kg·m/s to the floor.

1. Analysis of Physical Forces acting on the Ball

To understand why a cricket ball comes to rest after rolling on level ground, we must analyze the forces acting upon it in accordance with Newton's Laws of Motion.

• Inertia of Motion: Per Newton’s First Law of Motion, an object in motion will continue to move at a constant velocity unless acted upon by an external unbalanced force. If the ball were in a vacuum without friction, it would never stop [Ref: Newton's First Law of Motion].
• External Unbalanced Force: As the ball moves across the ground, there is a contact force between the surface of the ball and the surface of the ground. This force is known as Frictional Force ($F_f$).
• Directionality: Friction acts in the direction opposite to the relative motion of the object [Ref: Coulombic Friction Model]. Since the ball is rolling forward, the force of friction is directed backward, creating negative acceleration (deceleration).

2. Visual Representation of Forces

3. Logical Deduction and Conclusion

The transition from motion to rest can be summarized through the following sequential logical steps:

Therefore, the ball stops because the ground exerts a frictional force on the ball, which opposes the motion of the ball and eventually brings it to a state of rest.

Final Solution: The ball slows to a stop because of the force of friction acting between the ball and the ground, which opposes the motion of the ball and causes it to decelerate until its velocity becomes zero.

Step 1: Given Variables and Unit Conversion

To ensure dimensional consistency, we convert all physical quantities into SI units (kg, m, s).

• Mass of bullet ($m$) = $10\text{ g} = 0.01\text{ kg}$
• Initial velocity ($u$) = $150\text{ m s}^{-1}$
• Final velocity ($v$) = $0\text{ m s}^{-1}$ (as the bullet comes to rest)
• Time taken ($t$) = $0.03\text{ s}$

Step 2: Determining Acceleration

Using the first equation of motion [Per Newton’s Equations of Linear Motion]:

$v = u + at$

$0 = 150 + a(0.03)$

$a = -\frac{150}{0.03} = -5000\text{ m s}^{-2}$

The negative sign indicates retardation caused by the wooden block.

Step 3: Calculating Distance of Penetration ($s$)

Using the second equation of motion:

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

$s = (150)(0.03) + \frac{1}{2}(-5000)(0.03)^2$

$s = 4.5 + (-2500)(0.0009)$

$s = 4.5 - 2.25 = 2.25\text{ m}$

Step 4: Calculating Force Exerted ($F$)

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

$F = (0.01\text{ kg}) \times (-5000\text{ m s}^{-2})$

$F = -50\text{ N}$

[Justification: The negative sign signifies that the force exerted by the wooden block is in the direction opposite to the motion of the bullet.]

Step 5: Summary of Results

Final Solution: The distance of penetration of the bullet is 2.25 m, and the magnitude of the force exerted by the wooden block is 50 N.

1. Theoretical Framework: Newton’s Laws of Motion

To evaluate the claims made by Akhtar, Kiran, and Rahul, we must apply Newton’s Third Law of Motion and the Law of Conservation of Momentum.

• Newton’s Third Law: For every action, there is an equal and opposite reaction. If object $A$ exerts a force $F_{AB}$ on object $B$, then $B$ exerts an equal magnitude force $F_{BA}$ on $A$, where $F_{AB} = -F_{BA}$. [Per Newton’s Third Law]
• Impulse-Momentum Theorem: The change in momentum ($\Delta p$) is equal to the product of the force applied and the duration of contact ($F \cdot \Delta t = \Delta p$).

2. Analysis of the Suggestions

3. Derivation and Logical Justification

Let $m_c$ be the mass of the car and $m_i$ be the mass of the insect. Let $v_{c1}, v_{i1}$ be initial velocities and $v_{c2}, v_{i2}$ be final velocities.
The force exerted by the car on the insect is $F_{ci} = m_i \cdot a_i = m_i \left( \frac{v_{i2} - v_{i1}}{\Delta t} \right)$.
The force exerted by the insect on the car is $F_{ic} = m_c \cdot a_c = m_c \left( \frac{v_{c2} - v_{c1}}{\Delta t} \right)$.

Since $F_{ci} = -F_{ic}$, it follows that:

$m_i(v_{i2} - v_{i1}) = -m_c(v_{c2} - v_{c1})$

This implies $\Delta p_i = -\Delta p_c$. Thus, the magnitude of change in momentum is equal for both participants.

Conclusion on the Observations

Although the forces and the change in momentum are equal in magnitude, the effect of the force is drastically different due to the disparity in masses. The insect has a negligible mass ($m_i \approx 0$), causing its acceleration ($a_i = F/m_i$) to be extremely high, resulting in its destruction. The car, having a massive inertia ($M_c \gg m_i$), experiences a negligible change in velocity, making the effect of the force imperceptible to the passengers.

Final Solution: Rahul is correct. Both the motorcar and the insect experience an equal magnitude of force and an equal magnitude of change in momentum in accordance with Newton's Third Law and the Law of Conservation of Momentum.

Given Variables & Initial Setup

To determine the change in momentum, we define the following parameters based on the problem statement:

• Mass of the hockey ball ($m$): $200\text{ g} = 0.2\text{ kg}$ [Conversion to SI units: $1\text{ kg} = 1000\text{ g}$]
• Initial velocity ($u$): $10\text{ m s}^{-1}$ (taking the initial direction as positive)
• Final velocity ($v$): $-5\text{ m s}^{-1}$ (negative sign denotes the reversal of direction along the original path)

Step 1: Formulaic Foundation

The momentum ($p$) of an object is defined as the product of its mass and velocity [Newtonian Classical Mechanics definition]:

$p = m \times v$

The change in momentum ($\Delta p$) is the difference between the final momentum ($p_f$) and the initial momentum ($p_i$):

$\Delta p = p_f - p_i = m(v - u)$

Step 2: Calculating Initial and Final Momentum

Substituting the values into the momentum formula:

• Initial momentum ($p_i$) = $m \times u = 0.2\text{ kg} \times 10\text{ m s}^{-1} = 2\text{ kg m s}^{-1}$
• Final momentum ($p_f$) = $m \times v = 0.2\text{ kg} \times (-5\text{ m s}^{-1}) = -1\text{ kg m s}^{-1}$

Step 3: Calculating Change in Momentum

Applying the change formula:

$\Delta p = (-1\text{ kg m s}^{-1}) - (2\text{ kg m s}^{-1})$

$\Delta p = -3\text{ kg m s}^{-1}$

Step 4: Magnitude Analysis

The magnitude of the change in momentum is the absolute value of the vector difference [By definition of magnitude in vector space]:

$|\Delta p| = |-3\text{ kg m s}^{-1}| = 3\text{ kg m s}^{-1}$

Visual Representation of Momentum Vector Reversal

Final Solution: The magnitude of the change in momentum occurred in the motion of the hockey ball is $3\text{ kg m s}^{-1}$.

Step 1: Identifying Given Parameters

Based on the problem statement, we define the following physical quantities:

• Mass ($m$) = $100\text{ kg}$
• Initial velocity ($u$) = $5\text{ m s}^{-1}$
• Final velocity ($v$) = $8\text{ m s}^{-1}$
• Time interval ($t$) = $6\text{ s}$

Step 2: Calculation of Initial and Final Momentum

Momentum ($p$) is defined as the product of mass and velocity [$p = m \times v$].

Initial Momentum ($p_i$):

$p_i = m \times u$

$p_i = 100\text{ kg} \times 5\text{ m s}^{-1} = 500\text{ kg m s}^{-1}$

Final Momentum ($p_f$):

$p_f = m \times v$

$p_f = 100\text{ kg} \times 8\text{ m s}^{-1} = 800\text{ kg m s}^{-1}$

Step 3: Calculating Force Exerted

According to Newton's Second Law of Motion, force is the rate of change of momentum [$F = \frac{\Delta p}{t}$] or alternatively [$F = m \times a$].

Calculating Acceleration ($a$):

Using the first equation of motion [$a = \frac{v - u}{t}$]:

$a = \frac{8\text{ m s}^{-1} - 5\text{ m s}^{-1}}{6\text{ s}}$

$a = \frac{3\text{ m s}^{-1}}{6\text{ s}} = 0.5\text{ m s}^{-2}$

Calculating Magnitude of Force ($F$):

$F = m \times a$

$F = 100\text{ kg} \times 0.5\text{ m s}^{-2}$

$F = 50\text{ N}$

Step 4: Summary Table

Final Solution: The initial momentum is $500\text{ kg m s}^{-1}$, the final momentum is $800\text{ kg m s}^{-1}$, and the magnitude of the force exerted on the object is $50\text{ N}$.

Step 1: Identifying Given Variables

According to the Principle of Conservation of Linear Momentum, the total momentum of a system remains constant if no external unbalanced force acts on it [Per Newton's Second and Third Laws of Motion].

• Mass of object 1 ($m_1$) = $1\text{ kg}$
• Velocity of object 1 ($u_1$) = $10\text{ m s}^{-1}$
• Mass of object 2 ($m_2$) = $5\text{ kg}$
• Velocity of object 2 ($u_2$) = $0\text{ m s}^{-1}$ (stationary)

Step 2: Calculating Momentum Before Impact

The total momentum before impact ($P_{initial}$) is the sum of the individual momenta of the two objects.

$P_{initial} = m_1u_1 + m_2u_2$

$P_{initial} = (1\text{ kg} \times 10\text{ m s}^{-1}) + (5\text{ kg} \times 0\text{ m s}^{-1})$

$P_{initial} = 10\text{ kg m s}^{-1} + 0\text{ kg m s}^{-1} = 10\text{ kg m s}^{-1}$

Step 3: Calculating Momentum After Impact

Since the objects stick together, the final mass is $(m_1 + m_2)$ and they move with a common final velocity ($v$). [By the Law of Conservation of Momentum]:

$P_{final} = P_{initial}$

$P_{final} = 10\text{ kg m s}^{-1}$

Step 4: Calculating Final Velocity

The total mass after collision is $M_{total} = m_1 + m_2 = 1\text{ kg} + 5\text{ kg} = 6\text{ kg}$.

Using the formula for momentum: $P_{final} = (m_1 + m_2) \times v$

$10\text{ kg m s}^{-1} = 6\text{ kg} \times v$

$v = \frac{10}{6}\text{ m s}^{-1}$

$v \approx 1.67\text{ m s}^{-1}$

Final Solution: The total momentum before and after impact is $10\text{ kg m s}^{-1}$. The velocity of the combined object is $1.67\text{ m s}^{-1}$ in the direction of the first object's initial motion.

Step 1: Identification of Given Variables

To determine the force required to decelerate the vehicle, we identify the physical quantities provided in the problem statement:

• Mass of the vehicle ($m$) = $1500\text{ kg}$
• Acceleration ($a$) = $-1.7\text{ ms}^{-2}$ (negative as it represents retardation/deceleration)

Step 2: Selection of Physical Law

According to Newton's Second Law of Motion, the force ($F$) acting on an object is equal to the product of its mass ($m$) and its acceleration ($a$). [Per Newton's Second Law: $F = m \cdot a$]

Step 3: Calculation of Force

By substituting the given values into the equation:

$F = 1500\text{ kg} \times (-1.7\text{ ms}^{-2})$

$F = -2550\text{ kg ms}^{-2}$

Since $1\text{ N} = 1\text{ kg ms}^{-2}$, the magnitude of the force is $2550\text{ N}$. The negative sign indicates that the force is applied in a direction opposite to the direction of motion (a retarding force).

Visual Representation of Forces

Step 4: Analytical Conclusion

The calculation demonstrates that the force required between the vehicle and the road to achieve the specified deceleration is $2550\text{ N}$. This force is primarily provided by the friction between the tires and the road surface.

Final Solution: The force between the vehicle and the road is $2550\text{ N}$ (acting in the direction opposite to motion).

Step 1: Theoretical Foundation

According to Newton's First Law of Motion, an object maintains its state of rest or uniform motion in a straight line unless acted upon by an unbalanced external force [Law of Inertia].

The problem states that the cabinet is being moved at a constant velocity ($v = \text{constant}$). In physics, constant velocity implies zero acceleration ($a = 0$), because acceleration is defined as the rate of change of velocity ($a = \frac{dv}{dt}$).

Step 2: Applying Newton's Second Law of Motion

Newton's Second Law states that the net force ($\sum F$) acting on an object is equal to the product of its mass ($m$) and its acceleration ($a$):

$$\sum F = m \times a$$

Since the cabinet moves at a constant velocity, the acceleration $a = 0 \text{ m/s}^2$. Substituting this into the equation:

$$\sum F = m \times 0 = 0 \text{ N}$$

This confirms that the net force acting on the cabinet must be zero for it to maintain a constant velocity [Equilibrium Condition].

Step 3: Analyzing the Force Balance

The horizontal forces acting on the cabinet are the applied force ($F_{\text{applied}}$) and the frictional force ($F_{\text{friction}}$) acting in the opposite direction. The vector sum of these forces must equal the net force:

$$F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}$$

Given $F_{\text{net}} = 0 \text{ N}$ and $F_{\text{applied}} = 200 \text{ N}$, we substitute the values:

$$0 = 200 \text{ N} - F_{\text{friction}}$$

$$F_{\text{friction}} = 200 \text{ N}$$

Vector Representation of Forces

Data Summary

Final Solution: The frictional force exerted on the cabinet is 200 N, which perfectly opposes the applied force to maintain dynamic equilibrium.

Step 1: Identifying Given Variables

Based on the kinematic parameters provided in the problem statement, we define the following variables:

• Initial velocity ($u$): $0\text{ m/s}$ [Since the truck starts from rest]
• Distance traveled ($s$): $400\text{ m}$
• Time taken ($t$): $20\text{ s}$
• Mass of the truck ($m$): $7\text{ tonnes} = 7 \times 1000\text{ kg} = 7000\text{ kg}$ [Conversion factor: $1\text{ tonne} = 1000\text{ kg}$]

Step 2: Calculating Acceleration ($a$)

To determine the constant acceleration, we utilize the second equation of motion [Per the Kinematic Equations of Uniformly Accelerated Motion]:

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

Substituting the known values into the equation:

$400 = (0 \times 20) + \frac{1}{2} \times a \times (20)^2$

$400 = 0 + \frac{1}{2} \times a \times 400$

$400 = 200a$

Solving for $a$:

$a = \frac{400}{200}$

$a = 2\text{ m/s}^2$

Step 3: Calculating Force ($F$)

To calculate the force, we apply Newton’s Second Law of Motion [Per Newton’s Second Law: $F = ma$]:

$F = m \times a$

$F = 7000\text{ kg} \times 2\text{ m/s}^2$

$F = 14000\text{ N}$

Visual Representation of Kinematic Motion

Summary Table of Calculations

Final Solution: The acceleration of the truck is $2\text{ m/s}^2$ and the force acting on it is $14000\text{ N}$.

Step 1: Identification of Given Physical Quantities

To determine the force of friction acting on the stone, we identify the following variables provided in the problem statement:

• Mass of the stone ($m$) = $1\text{ kg}$
• Initial velocity ($u$) = $20\text{ m s}^{-1}$
• Final velocity ($v$) = $0\text{ m s}^{-1}$ (as the stone comes to rest)
• Distance traveled ($s$) = $50\text{ m}$

Step 2: Calculation of Acceleration

Using the third equation of motion, which relates velocity, acceleration, and displacement independent of time [Per Kinematic Laws of Motion]:

$v^2 = u^2 + 2as$

Substituting the known values:

$0^2 = (20)^2 + 2 \cdot a \cdot 50$

$0 = 400 + 100a$

$-400 = 100a$

$a = -4\text{ m s}^{-2}$

The negative sign indicates that the acceleration is directed opposite to the motion, confirming the action of a retarding force (friction).

Step 3: Calculation of Frictional Force

According to Newton’s Second Law of Motion, the force acting on an object is the product of its mass and acceleration ($F = ma$) [Per Newton's Second Law]:

$F = 1\text{ kg} \times (-4\text{ m s}^{-2})$

$F = -4\text{ N}$

Visual Representation of Forces

Analytical Summary

Final Solution: The force of friction between the stone and the ice is 4 N in the direction opposite to the motion of the stone.

Step 1: Definition of Momentum

In classical mechanics, momentum (denoted by the symbol $p$) is a vector quantity that represents the "quantity of motion" contained within a body. It is defined as the product of the mass of an object and its velocity. [Per Newton’s Second Law of Motion foundation].

Step 2: Mathematical Formulation

Given the variables provided:

• Mass of the object = $m$ (measured in $kg$ in the SI system).
• Velocity of the object = $v$ (measured in $m/s$ in the SI system).

The mathematical relationship is defined by the equation:

$p = m \times v$

Step 3: Analytical Properties

To understand the nature of this quantity, consider the following technical characteristics:

Visual Representation of Vector Momentum

Conclusion

The momentum $p$ of an object is the product of its mass $m$ and its velocity $v$. As a vector quantity, it accounts for the total inertial movement of the body within a specific frame of reference. [Consistent with the definition of linear momentum in Newtonian dynamics].

Final Solution: The momentum of an object of mass $m$ moving with a velocity $v$ is given by the expression $p = mv$.

Step 1: Identification of Given Physical Quantities

To determine the net accelerating force, we first define the system parameters provided in the problem statement:

• Mass of the engine ($m_e$) = $8000 \text{ kg}$
• Number of wagons = $5$
• Mass of each wagon ($m_w$) = $2000 \text{ kg}$
• Applied force by the engine ($F_{applied}$) = $40000 \text{ N}$
• Frictional force acting on the entire system ($F_{friction}$) = $5000 \text{ N}$

Step 2: Theoretical Foundation

According to [Newton’s Second Law of Motion], the net force acting on an object (or a system of objects) is the vector sum of all forces applied to it. In this scenario, the motion is constrained to a horizontal track, allowing us to simplify the analysis to a one-dimensional algebraic summation:

$F_{net} = \sum F = F_{applied} - F_{friction}$

[Justification: Friction acts in the direction opposite to the applied force, thereby opposing the motion of the system.]

Step 3: Calculating the Net Accelerating Force

Substituting the given values into the equation derived in Step 2:

$F_{net} = 40000 \text{ N} - 5000 \text{ N}$

$F_{net} = 35000 \text{ N}$

Visual Representation of Force Vectors

Step 4: Analysis of the Result

The calculation reveals that even though the engine exerts $40000 \text{ N}$, a portion of this energy is dissipated by the interaction between the wheels and the tracks (friction). The resultant value, $35000 \text{ N}$, is the force effectively available to overcome the inertia of the total mass of the train ($8000 \text{ kg} + 5 \times 2000 \text{ kg} = 18000 \text{ kg}$), leading to acceleration as defined by $a = \frac{F_{net}}{m_{total}}$.

Final Solution: The net accelerating force is 35000 N.