Initial State and Conservation Laws

Let the mass of the original nucleus be $M$. Since the nucleus is at rest in the laboratory frame of reference, its initial momentum $\vec{P}_{initial}$ is given by:

$\vec{P}_{initial} = M \times \vec{v} = M \times 0 = 0$

[Per the Principle of Conservation of Linear Momentum, if the net external force on a system is zero, the total linear momentum of the system remains conserved.]

Step 1: Final State Dynamics

The nucleus disintegrates into two smaller nuclei with masses $m_1$ and $m_2$ and velocities $\vec{v}_1$ and $\vec{v}_2$ respectively. The total final momentum $\vec{P}_{final}$ is the vector sum of the individual momenta of the products:

$\vec{P}_{final} = m_1\vec{v}_1 + m_2\vec{v}_2$

Step 2: Equating Momentum

By the Law of Conservation of Momentum, the initial momentum must equal the final momentum:

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

$0 = m_1\vec{v}_1 + m_2\vec{v}_2$

$m_1\vec{v}_1 = -m_2\vec{v}_2$

Step 3: Vector Interpretation

The negative sign in the relation $\vec{v}_1 = -\left(\frac{m_2}{m_1}\right)\vec{v}_2$ indicates that the velocity vector $\vec{v}_1$ points in the exact opposite direction of velocity vector $\vec{v}_2$. Since mass ($m$) is a scalar quantity and $m > 0$, the vectors are antiparallel [By the properties of vector multiplication by a scalar].

Step 4: Theoretical Conclusion

For the vector sum of two non-zero vectors to result in a null vector, the components must cancel each other out across all axes. In a one-dimensional disintegration, this necessitates that $\vec{v}_1$ and $\vec{v}_2$ possess equal and opposite directional orientation along the axis of separation.

Final Solution: Because the initial momentum of the stationary nucleus is zero, the total final momentum of the products must also be zero. This requires $m_1\vec{v}_1 = -m_2\vec{v}_2$, which mathematically necessitates that the two products move in opposite directions to satisfy the Law of Conservation of Momentum.

Theoretical Foundation: Newton’s Second Law of Motion

The motion of the cricket ball is governed by Newton's Second Law, which states that the net force ($\vec{F}$) acting on an object is equal to the rate of change of its linear momentum ($\vec{p}$):

$F = \frac{dp}{dt} = \frac{m(v_f - v_i)}{\Delta t}$

Where:

• $m$ = Mass of the cricket ball.
• $v_f$ = Final velocity of the ball (which is $0$ when caught).
• $v_i$ = Initial velocity of the ball just before impact.
• $\Delta t$ = Time duration of the collision (interaction time).

Step 1: Analyzing the Inverse Proportionality

From the derived expression $F = \frac{-m \cdot v_i}{\Delta t}$, it is observed that the impulsive force experienced by the cricketer's hands is inversely proportional to the time interval ($\Delta t$) during which the momentum of the ball is reduced to zero.

$F \propto \frac{1}{\Delta t}$ [Per the relationship between Impulse and Momentum]

Step 2: Kinematic Adjustment

When the cricketer keeps their hands stationary, the collision happens almost instantaneously. This small $\Delta t$ results in a very high impact force, which may cause injury. By moving the hands backwards, the cricketer increases the time interval ($\Delta t$) over which the ball comes to a complete rest.

Step 3: Visual Representation of Momentum Transfer

Step 4: Analytical Conclusion

By increasing the duration ($\Delta t$) of the momentum change, the magnitude of the force ($F$) exerted on the hands is significantly reduced. This application of the Impulse-Momentum Theorem allows the cricketer to absorb the energy of the ball safely.

Final Solution: The cricketer moves his hands backwards to increase the time of impact ($\Delta t$). Since force is inversely proportional to time for a constant change in momentum, increasing the time results in a smaller impact force, thereby protecting the hands from injury.

Step 1: Identification of Given Physical Parameters

• Mass of body A ($m_A$) = $5 \text{ kg}$
• Mass of body B ($m_B$) = $10 \text{ kg}$
• Applied force ($F$) = $200 \text{ N}$
• Coefficient of kinetic friction ($\mu_k$) = $0.15$
• Acceleration due to gravity ($g$) = $9.8 \text{ m/s}^2$ (standard approximation)

Step 2: Analysis of Frictional Forces

The normal force ($N$) exerted by the table on the system is equal to the total weight of the two bodies. [Per Newton's Third Law and Equilibrium condition $\sum F_y = 0$]:

$N_{total} = (m_A + m_B)g = (5 + 10) \times 9.8 = 15 \times 9.8 = 147 \text{ N}$

The total force of limiting friction ($f_{max}$) acting against the applied force is:

$f_{max} = \mu_k \times N_{total} = 0.15 \times 147 = 22.05 \text{ N}$

Step 3: Evaluating the Dynamic State of the System

The system is in contact with a rigid wall. Before calculating the reaction, we must determine if the system moves. If the applied force ($F$) is greater than the frictional force ($f_{max}$), the system would tend to slide. However, the system is blocked by a rigid wall. Thus, the system is in static equilibrium.

Total applied external force: $F = 200 \text{ N}$

Total opposing frictional force: $f_{max} = 22.05 \text{ N}$

Since $F > f_{max}$, the bodies would move if the wall were not present. Because the wall exists, it provides an additional normal reaction ($R$) to maintain static equilibrium.

Step 4: Calculation of the Wall's Reaction

Applying the condition for static equilibrium ($\sum F_x = 0$):

$F - f_{total} - R = 0$

Where $R$ is the reaction of the wall against the body B.

$R = F - f_{total} = 200 \text{ N} - 22.05 \text{ N} = 177.95 \text{ N}$

Visual Representation of Forces

Final Justification

The reaction of the wall is the force required to counteract the net imbalance between the applied horizontal force and the frictional resistance provided by the surface of the table. By the Principle of Transmissibility and Static Equilibrium, the wall exerts an equal and opposite force to the net force applied by the blocks onto the wall.

Final Solution: The reaction of the wall (partition) is 177.95 N directed opposite to the applied force.

Given Variables & Initial Setup

Let $m_1 = 10\text{ kg}$ (Body A) and $m_2 = 20\text{ kg}$ (Body B). Both bodies are connected by a light (massless) string on a smooth (frictionless) horizontal surface. An external force $F = 600\text{ N}$ is applied to body A (the 10 kg mass) in the direction of the string, pulling body B behind it.

Step 1: Calculating the Acceleration of the System

Since the bodies are tied together by a light string, they move as a single composite system with common acceleration $a$ [Per Newton's Second Law of Motion].

The total mass of the system is $M = m_1 + m_2 = 10\text{ kg} + 20\text{ kg} = 30\text{ kg}$.

Applying Newton's Second Law to the composite system:

Step 2: Free Body Diagram Analysis for Tension ($T$)

To find the tension $T$ in the string, we isolate body B ($m_2 = 20\text{ kg}$). The only horizontal force acting on body B is the tension $T$ exerted by the string [Per Newton's Third Law of Motion].

Applying Newton's Second Law to body B:

Step 3: Solving for Tension

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

Summary of Dynamics

Final Solution: The tension in the string when the force is applied to the 10 kg mass is 400 N.

Step 1: Identifying the State of the System at the Extreme Position

To determine the trajectory, we must analyze the kinematic state of the bob at the extreme position. In an oscillating pendulum, the bob reaches its maximum displacement at the extreme positions. At these points, the instantaneous velocity of the bob is zero ($v = 0 \text{ m/s}$) before it reverses direction [Per the Kinematics of Simple Harmonic Motion].

Step 2: Force Analysis upon Cutting the String

When the string is cut at the extreme position, the centripetal force ($T - mg\cos\theta$) vanishes instantaneously. The only force acting on the bob is the gravitational force ($F_g = mg$) directed vertically downwards [By Newton’s Second Law of Motion].

Step 3: Calculating Trajectory via Equations of Motion

Since the initial velocity ($u$) is $0 \text{ m/s}$ at the moment of cutting, the bob possesses no horizontal component of velocity to initiate projectile motion. Consequently, the bob will experience free-fall motion solely under the influence of gravity.

Step 4: Conclusion on Path Geometry

Because the initial velocity vector $\vec{v} = 0$, the acceleration vector $\vec{a} = \vec{g}$ (directed downwards) defines the path of the particle. According to the laws of kinematics, if a particle is released from rest in a uniform gravitational field, it follows a straight-line vertical path [Per the Equation of Motion: $s = ut + \frac{1}{2}at^2$ where $u=0$].

Final Solution: The trajectory of the bob is a straight vertical line directed downwards, representing free-fall motion.

Step 1: Identifying Given Parameters

To determine the impulse imparted to each billiard ball, we define the initial and final states of the system based on the provided data:

• Mass of each ball ($m$): $0.05\text{ kg}$
• Initial velocity of Ball 1 ($u_1$): $+6\text{ m s}^{-1}$
• Initial velocity of Ball 2 ($u_2$): $-6\text{ m s}^{-1}$
• Final velocity of Ball 1 ($v_1$): $-6\text{ m s}^{-1}$ (Rebound condition)
• Final velocity of Ball 2 ($v_2$): $+6\text{ m s}^{-1}$ (Rebound condition)

Step 2: Defining the Concept of Impulse

Impulse ($J$) is defined as the change in momentum ($\Delta p$) experienced by an object during a collision [Per the Impulse-Momentum Theorem]:

$J = \Delta p = p_{final} - p_{initial}$

$J = m(v - u)$

Step 3: Calculation for Ball 1

Substituting the values for the first ball:

$J_1 = m(v_1 - u_1)$

$J_1 = 0.05 \text{ kg} \times (-6\text{ m s}^{-1} - 6\text{ m s}^{-1})$

$J_1 = 0.05 \times (-12)$

$J_1 = -0.6\text{ kg m s}^{-1} \text{ (or N s)}$

Step 4: Calculation for Ball 2

Substituting the values for the second ball:

$J_2 = m(v_2 - u_2)$

$J_2 = 0.05 \text{ kg} \times (6\text{ m s}^{-1} - (-6\text{ m s}^{-1}))$

$J_2 = 0.05 \times (6 + 6)$

$J_2 = 0.05 \times 12$

$J_2 = +0.6\text{ kg m s}^{-1} \text{ (or N s)}$

Visual Representation of Momentum Change

Analytical Summary

The impulse imparted to each ball is equal in magnitude but opposite in direction, consistent with [Newton's Third Law of Motion], which dictates that action and reaction forces are equal and opposite.

Final Solution: The magnitude of the impulse imparted to each ball is $0.6 \text{ kg m s}^{-1}$ (or $0.6 \text{ N s}$).

Step 1: Identifying Physical Parameters

To determine the net force acting on the pebble, we list the given variables based on the parameters provided in the problem statement:

• Mass of the pebble ($m$): $0.05 \text{ kg}$
• Acceleration due to gravity ($g$): $9.8 \text{ m/s}^2$ (standard Earth gravity)
• State of motion: Moving vertically upwards.

Step 2: Theoretical Foundation

According to [Newton’s Second Law of Motion], the net force ($\vec{F}_{\text{net}}$) acting on an object is the product of its mass ($m$) and its net acceleration ($\vec{a}$):

$\vec{F}_{\text{net}} = m \vec{a}$

During the upward motion of the pebble, ignoring air resistance as is standard in introductory classical mechanics problems of this type, the only force acting on the pebble is the gravitational force ($F_g$) exerted by the Earth, directed downwards [Per the Law of Universal Gravitation].

Step 3: Calculating Magnitude and Direction

The magnitude of the gravitational force is calculated as follows:

$|F_g| = m \times g$

$|F_g| = 0.05 \text{ kg} \times 9.8 \text{ m/s}^2$

$|F_g| = 0.49 \text{ N}$

Since the only force acting on the object is gravity, the net force is equivalent to the weight of the pebble. By definition, gravity acts towards the center of the Earth [By the definition of Gravitational Field Vectors]. Therefore, the direction is vertically downwards.

Visual Representation of Forces

Step 4: Summary of Analysis

Final Solution: The net force on the pebble is $0.49 \text{ N}$ acting in the vertically downward direction.

Step 1: Identifying Given Variables

Mass of the pebble ($m$) = $0.05 \text{ kg}$

The only force acting on the object during its flight (neglecting air resistance) is the gravitational force exerted by the Earth.

Acceleration due to gravity ($g$) $\approx 9.8 \text{ m/s}^2$ [Directed vertically downwards]

Step 2: Determining the Direction of Net Force

According to Newton's Second Law of Motion ($F = ma$), the net force acting on a body is always in the direction of the acceleration [Per Newton's Second Law of Motion].

While the pebble is moving upwards, the gravitational force acts constantly towards the center of the Earth. Therefore, the direction of the net force is vertically downwards throughout the ascent.

Step 3: Calculating the Magnitude of Net Force

The magnitude of the force ($F$) is calculated using the weight formula:

$F = m \times g$

$F = 0.05 \text{ kg} \times 9.8 \text{ m/s}^2$

$F = 0.49 \text{ N}$

Step 4: Vector Representation (Visual Aid)

Summary Analysis

Despite the pebble's upward velocity ($v$), the net force remains constant because gravity is an independent field force [Per Newton's Law of Universal Gravitation]. The upward motion is simply a result of the initial momentum imparted to the pebble, which is gradually depleted by the constant downward force.

Final Solution: The magnitude of the net force is $0.49 \text{ N}$ and its direction is vertically downwards.

Step 1: Calculating the Net Force on the Man

Given parameters for the man:

• Mass ($m$) = $65\ kg$
• Acceleration of the conveyor belt ($a$) = $1\ m/s^2$

As the man is stationary with respect to the belt, he moves with the same acceleration as the belt. According to Newton's Second Law of Motion [$\vec{F}_{net} = m\vec{a}$]:

$F_{net} = 65\ kg \times 1\ m/s^2 = 65\ N$

The net force acting on the man is provided by the static frictional force between the shoes and the belt.

Step 2: Determining Maximum Acceleration for Stationary State

The man remains stationary relative to the belt as long as the required frictional force does not exceed the maximum limiting static friction ($f_{s,max}$).

• Coefficient of static friction ($\mu_s$) = $0.2$
• Acceleration due to gravity ($g$) = $9.8\ m/s^2$

The normal force ($N$) exerted by the belt on the man is equal to his weight [Per Newton's Third Law]:

$N = mg = 65\ kg \times 9.8\ m/s^2 = 637\ N$

The limiting static friction force is calculated as:

$f_{s,max} = \mu_s N = 0.2 \times 637\ N = 127.4\ N$

Step 3: Calculating Critical Acceleration ($a_{max}$ )

To remain stationary, the force of friction must satisfy the condition $F_{applied} \leq f_{s,max}$. Thus, the maximum acceleration ($a_{max}$) is constrained by:

$ma_{max} = \mu_s mg$

$a_{max} = \mu_s g$

$a_{max} = 0.2 \times 9.8\ m/s^2 = 1.96\ m/s^2$

Visual Representation of Forces

Final Solution: The net force on the man is $65\ N$. The man can continue to remain stationary relative to the belt up to a maximum acceleration of $1.96\ m/s^2$.

Step 1: Given Variables & Initial Setup

The system comprises a body of mass $m = 5 \text{ kg}$. The forces applied are $F_1 = 8 \text{ N}$ and $F_2 = 6 \text{ N}$, acting perpendicular to each other. [Per Newton's Second Law of Motion: $\vec{F} = m\vec{a}$].

Step 2: Calculating the Net Force

Since the forces are perpendicular, the magnitude of the resultant force $F_{net}$ is calculated using the Pythagorean theorem, treating the forces as vectors in a 2D Cartesian plane [By Vector Addition Rule]:

$F_{net} = \sqrt{F_1^2 + F_2^2}$

$F_{net} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ N}$

Step 3: Calculating Acceleration Magnitude

Using Newton's Second Law, $a = \frac{F_{net}}{m}$:

$a = \frac{10 \text{ N}}{5 \text{ kg}} = 2 \text{ m/s}^2$

Step 4: Calculating Direction of Acceleration

The direction of acceleration $\theta$ coincides with the direction of the net force. Defining $\theta$ as the angle with respect to the $8 \text{ N}$ force:

$\tan(\theta) = \frac{F_2}{F_1} = \frac{6}{8} = 0.75$

$\theta = \arctan(0.75) \approx 36.87^\circ$

Analytical Data Table

Final Solution: The magnitude of the acceleration is $2 \text{ m/s}^2$, and it is directed at an angle of $36.87^\circ$ from the direction of the $8 \text{ N}$ force.

Given Variables & Initial Setup

To determine the magnitude and direction of the force, we identify the following physical parameters provided in the problem statement:

• Mass of the body ($m$): $3.0\text{ kg}$
• Initial velocity ($u$): $2.0\text{ m s}^{-1}$
• Final velocity ($v$): $3.5\text{ m s}^{-1}$
• Time interval ($\Delta t$): $25\text{ s}$

Step 1: Calculating Acceleration

Acceleration ($a$) is defined as the rate of change of velocity over time. [Per the definition of average acceleration in kinematics]:

$a = \frac{v - u}{\Delta t}$

Substituting the given values:

$a = \frac{3.5\text{ m s}^{-1} - 2.0\text{ m s}^{-1}}{25\text{ s}}$

$a = \frac{1.5\text{ m s}^{-1}}{25\text{ s}}$

$a = 0.06\text{ m s}^{-2}$

Step 2: Calculating Magnitude of Force

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 acceleration ($a$):

$F = m \times a$ [Per Newton's Second Law: $F = \frac{dp}{dt} = ma$]

$F = 3.0\text{ kg} \times 0.06\text{ m s}^{-2}$

$F = 0.18\text{ N}$

Visual Representation of Force Vector

Step 3: Determining Direction

Since the acceleration ($a = 0.06\text{ m s}^{-2}$) is positive and the initial velocity ($u = 2.0\text{ m s}^{-1}$) is also positive, the velocity vector is increasing in magnitude while maintaining its original orientation. [Per the relationship $\vec{F} = m\vec{a}$]:

• The force acts in the same direction as the acceleration.
• Since the body's speed increased, the force must be acting in the direction of the body's motion.

Summary of Calculation

Final Solution: The magnitude of the force is 0.18 N, and it acts in the direction of the motion of the body.

Given Variables & Initial Setup

To determine the net force acting on the cork, we identify the following physical parameters:

• Mass of the cork ($m$): $10\text{ g} = 0.01\text{ kg}$
• State of motion: Floating (Equilibrium state)
• Gravitational acceleration ($g$): $9.8\text{ m/s}^2$ (acting downwards)

Step 1: Identifying Active Forces

Per Newton’s First Law of Motion, an object at rest or in uniform linear motion remains in that state unless acted upon by an external net force. Since the cork is floating at rest on the water surface, it is in a state of static equilibrium [Per the First Law of Motion].

The forces acting on the cork are:

1. Weight ($W$): The gravitational force acting vertically downward ($W = mg$).
2. Buoyant Force ($F_b$): The upward force exerted by the displaced water (Archimedes' Principle).

Step 2: Vector Analysis & Force Equilibrium

In equilibrium, the vector sum of all forces must be zero. Let $\vec{F}_{net}$ be the net force:

$\vec{F}_{net} = \vec{W} + \vec{F}_b = 0$

Substituting the magnitudes:

$\vec{F}_{net} = (-mg)\hat{j} + (F_b)\hat{j} = 0$

Since the cork is floating, by the Law of Flotation, the weight of the displaced water equals the weight of the object. Thus, $F_b = mg$.

Step 3: Calculating Net Force

Calculating the magnitude of the gravitational force:

$|W| = 0.01\text{ kg} \times 9.8\text{ m/s}^2 = 0.098\text{ N}$

Since the forces are equal in magnitude and opposite in direction ($F_b = 0.098\text{ N}$ upwards, $W = 0.098\text{ N}$ downwards), the net force is:

$F_{net} = 0.098\text{ N} - 0.098\text{ N} = 0\text{ N}$

Visual Representation of Force Equilibrium

Summary of Deductions

Final Solution: The magnitude of the net force acting on the cork is 0 N. Consequently, there is no direction for the net force.

Step 1: Analyzing Forces at the Lowest Point

At the lowest point of the vertical circle, the stone is subjected to two primary forces acting along the vertical axis:

• Tension ($T_1$) acting vertically upwards towards the center of the circle.
• Weight ($mg$) acting vertically downwards.

The net force provides the necessary centripetal force directed towards the center. Therefore, the equation of motion is:

$T_1 - mg = \frac{mv_1^2}{R}$ [Per Newton's Second Law of Motion]

The question asks for the net force directed downwards. Since the centripetal force is directed upwards (towards the center at the bottom), the downward force is the negative of the centripetal force:

$F_{\text{net, down}} = -(T_1 - mg) = mg - T_1$

Step 2: Analyzing Forces at the Highest Point

At the highest point of the vertical circle, both forces act in the same direction:

• Tension ($T_2$) acting vertically downwards towards the center.
• Weight ($mg$) acting vertically downwards.

The sum of these forces provides the centripetal force required for circular motion:

$T_2 + mg = \frac{mv_2^2}{R}$ [Per Newton's Second Law of Motion]

Since both forces are acting vertically downwards, the net force directed downwards is simply the sum:

$F_{\text{net, down}} = T_2 + mg$

Step 3: Visual Representation of Forces

Step 4: Comparative Summary

Final Solution: The net force directed vertically downwards at the lowest point is $(mg - T_1)$ and at the highest point is $(T_2 + mg)$.

Given Variables & Initial Setup

• Mass of the particle ($m$) = $4 \text{ kg}$
• The motion is one-dimensional.
• The graph provided is a position-time ($x-t$) graph.
• Impulse ($J$) is defined as the change in momentum: $J = \Delta p = m(v_f - v_i)$.
• Velocity ($v$) is defined as the slope of the $x-t$ graph: $v = \frac{dx}{dt}$.

Step 1: Analysis of the Position-Time Graph

Based on the standard physics problem description for this figure (where $x=0$ at $t < 0$, $x=3t$ for $0 < t < 4$, and $x=12$ for $t > 4$):

Step 2: Calculation of Impulse at $t = 0 \text{ s}$

At $t = 0$, the particle transitions from rest to a velocity of $3 \text{ m/s}$.

• Initial velocity ($v_i$) = $0 \text{ m/s}$
• Final velocity ($v_f$) = $3 \text{ m/s}$
• Impulse ($J_{t=0}$) = $m(v_f - v_i)$ [Per the Impulse-Momentum Theorem]
• $J_{t=0} = 4 \text{ kg} \times (3 \text{ m/s} - 0 \text{ m/s})$
• $J_{t=0} = 12 \text{ kg}\cdot\text{m/s}$

Step 3: Calculation of Impulse at $t = 4 \text{ s}$

At $t = 4$, the particle transitions from a velocity of $3 \text{ m/s}$ to rest ($0 \text{ m/s}$).

• Initial velocity ($v_i$) = $3 \text{ m/s}$
• Final velocity ($v_f$) = $0 \text{ m/s}$
• Impulse ($J_{t=4}$) = $m(v_f - v_i)$
• $J_{t=4} = 4 \text{ kg} \times (0 \text{ m/s} - 3 \text{ m/s})$
• $J_{t=4} = -12 \text{ kg}\cdot\text{m/s}$

Visual Representation of $x-t$ Motion

Final Solution: The impulse at $t = 0 \text{ s}$ is $12 \text{ kg}\cdot\text{m/s}$, and the impulse at $t = 4 \text{ s}$ is $-12 \text{ kg}\cdot\text{m/s}$.

Step 1: Identification of Given Variables

First, we convert all given quantities into SI units for consistency [Per International System of Units standards].

• Initial velocity ($u$): $36 \text{ km/h} = 36 \times \frac{5}{18} \text{ m/s} = 10 \text{ m/s}$
• Final velocity ($v$): $0 \text{ m/s}$ (since the vehicle comes to rest)
• Time taken ($t$): $4.0 \text{ s}$
• Mass of three-wheeler ($m_1$): $400 \text{ kg}$
• Mass of driver ($m_2$): $65 \text{ kg}$
• Total mass ($M$): $m_1 + m_2 = 400 + 65 = 465 \text{ kg}$

Step 2: Kinematic Analysis

To find the average retarding force, we must first determine the deceleration ($a$) of the vehicle using the first equation of motion [By Kinematic Equation of Motion]:

$v = u + at$

$0 = 10 + a(4.0)$

$4a = -10$

$a = -2.5 \text{ m/s}^2$

The negative sign indicates a retardation (deceleration) of $2.5 \text{ m/s}^2$.

Step 3: Force Calculation

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

$F = M \times a$

$F = 465 \text{ kg} \times (-2.5 \text{ m/s}^2)$

$F = -1162.5 \text{ N}$

Step 4: Visualizing the Dynamics

Step 5: Analysis of Results

The retarding force represents the resistive force applied by the brakes against the direction of motion. Since the magnitude is requested, we express the force as a positive retarding value.

Final Solution: The average retarding force acting on the vehicle is 1162.5 N.

1. Given Variables & Initial Conditions

• Mass of the man ($m$): $70 \text{ kg}$
• Radius of the cylinder ($r$): $3 \text{ m}$
• Coefficient of static friction ($\mu_s$): $0.15$
• Acceleration due to gravity ($g$): $9.8 \text{ m/s}^2$
• Condition: The man must not slide downwards.

2. Force Analysis & Free Body Diagram

When the floor is removed, the man is held against the wall by the normal reaction ($N$) provided by the wall. The forces acting on the man are:

• Weight ($mg$) acting vertically downwards.
• Frictional force ($f$) acting vertically upwards [Per Newton’s Third Law].
• Normal reaction ($N$) acting horizontally towards the center, providing the necessary centripetal force.

3. Mathematical Derivations

For the man to remain in equilibrium vertically, the static friction must balance the weight:

$f = mg$ [Condition for equilibrium]

We know the maximum limiting friction is $f \leq \mu_s N$. Therefore, to prevent sliding:

$\mu_s N \geq mg$

The normal force provides the centripetal acceleration ($a_c = \omega^2 r$):

$N = m\omega^2 r$ [By Newton’s Second Law for circular motion]

Substituting $N$ into the friction inequality:

$\mu_s (m\omega^2 r) \geq mg$

$\omega^2 \geq \frac{g}{\mu_s r}$

4. Calculation of Minimum Angular Velocity ($\omega$)

$\omega_{min} = \sqrt{\frac{g}{\mu_s r}}$

$\omega_{min} = \sqrt{\frac{9.8}{0.15 \times 3}}$

$\omega_{min} = \sqrt{\frac{9.8}{0.45}} \approx \sqrt{21.777} \approx 4.667 \text{ rad/s}$

5. Conversion to Rotational Speed ($n$) in rev/min

Relationship: $\omega = 2\pi n$ where $n$ is in rev/s.

For rev/min ($N_{rpm}$): $N_{rpm} = \frac{\omega \times 60}{2\pi}$

$N_{rpm} = \frac{4.667 \times 60}{2 \times 3.14159} \approx \frac{280.02}{6.283} \approx 44.57 \text{ rev/min}$

Final Solution: The minimum rotational speed required for the man to remain stuck to the wall is approximately 44.57 rev/min.

Step 1: Identifying the Physical Context

The system under consideration is a high-speed electron located in deep space. The environment is explicitly defined as void of material objects, electric fields ($\vec{E} = 0$), and magnetic fields ($\vec{B} = 0$).

Step 2: Applying Newton’s First Law of Motion

Newton’s First Law of Motion, also known as the Law of Inertia, states that an object will remain at rest or continue to move at a constant velocity unless acted upon by a net external force [Per Newton’s First Law of Motion].

Step 3: Evaluating the Net Force Equation

The net force $\vec{F}_{net}$ acting on a charged particle in the presence of electromagnetic fields is defined by the Lorentz Force Law:

$\vec{F}_{net} = q(\vec{E} + \vec{v} \times \vec{B})$

Given the constraints:

• Charge of the electron: $q = -e$ (where $e \approx 1.602 \times 10^{-19} \, \text{C}$)
• Electric field: $\vec{E} = 0$
• Magnetic field: $\vec{B} = 0$

Substituting these values into the Lorentz equation:

$\vec{F}_{net} = -e(0 + \vec{v} \times 0)$

$\vec{F}_{net} = 0$

Step 4: Vector and Magnitude Analysis

Since the resultant vector of the net force is the null vector ($\vec{0}$), the magnitude of the force is zero. Because the magnitude is zero, the direction of the net force is undefined or irrelevant, as there is no force vector to orient.

Step 5: Theoretical Justification

In the absence of any external agents capable of doing work or exerting a pressure on the electron, the momentum of the electron remains invariant. Since $\vec{F} = \frac{d\vec{p}}{dt}$ [Per Newton's Second Law], and the rate of change of momentum is zero for an isolated particle moving at constant velocity, the net force must be zero.

Final Solution: The magnitude of the net force is $0 \, \text{N}$. The direction is undefined, as no force is acting upon the electron.

Step 1: Identifying Known Variables

To determine the reading on the weighing scale, we must identify the forces acting on the man of mass $m$ within the non-inertial reference frame of the accelerating lift.

• Mass of the man ($m$): $70\text{ kg}$
• Acceleration due to gravity ($g$): $9.8\text{ m/s}^2$ (standard approximation)
• Acceleration of the lift ($a$): $5\text{ m/s}^2$ (directed downwards)
• Reading on the scale ($R$): Equal to the normal force ($N$) exerted by the floor of the lift on the man.

Step 2: Free Body Diagram & Force Analysis

The forces acting on the man are:

1. The gravitational force ($mg$) acting downwards.
2. The normal reaction force ($R$) acting upwards.

[Per Newton’s Second Law of Motion: $F_{net} = ma$]

Step 3: Formulation of the Equation of Motion

Since the lift is accelerating downwards, the net force is directed downwards. Applying Newton's Second Law along the vertical axis:

$mg - R = ma$

[Rearranging for $R$]:

$R = mg - ma$

$R = m(g - a)$

Step 4: Calculation

Substitute the given numerical values into the derived expression:

$R = 70 \times (9.8 - 5)$

$R = 70 \times (4.8)$

$R = 336\text{ N}$

Step 5: Conversion to Equivalent Scale Reading

Weighing scales calibrated for mass typically output the value in kg. To find the effective mass ($m_{eff}$) shown on the scale, we divide the normal reaction by the acceleration due to gravity ($g$):

$m_{eff} = \frac{R}{g} = \frac{336}{9.8} \approx 34.28\text{ kg}$

Final Solution: The reading on the scale would be $336\text{ N}$ (or an equivalent mass reading of approximately $34.28\text{ kg}$).

Step 1: Theoretical Foundation of Newtonian Mechanics

To analyze the motion of a horse-cart system, we must apply Newton’s Third Law of Motion: "To every action, there is always an equal and opposite reaction." [Per Newton’s Third Law]. In the context of locomotion, the horse exerts a backward horizontal force ($F_{h}$) on the ground. Simultaneously, the ground exerts an equal and opposite forward reaction force ($F_{g}$) on the horse's hooves. This reaction force is the primary external force responsible for the horse's acceleration.

Step 2: Analysis of the "Empty Space" Condition

In "empty space" (vacuum, devoid of a terrestrial surface), the following conditions prevail:

• Absence of Interaction Surface: There is no ground or physical surface for the horse to push against.
• Force Vanishing: Since $F_{h} = 0$, the corresponding reaction force $F_{g}$ from the ground must also be $0$ [Per Newton’s Third Law].
• Acceleration Dynamics: According to Newton’s Second Law, $F_{net} = ma$. If $F_{net} = 0$, then the acceleration ($a$) of the horse-cart system must be $0$.

Step 3: Vector Representation of Force Interaction

The following diagram illustrates the necessary interaction between the horse and the ground in a standard environment, which is impossible in empty space.

Step 4: Logical Deduction

In empty space, the horse lacks the requisite external force to initiate motion. Even if the horse exerts internal effort (muscle contraction), these forces are internal to the horse-cart system. According to the principle of conservation of momentum, internal forces cannot change the total momentum of a system [Per Newton’s Third Law and the Law of Conservation of Momentum].

Final Solution: A horse cannot pull a cart in empty space because locomotion requires an external reaction force from the ground. Since empty space lacks a surface to push against, the reaction force ($F_{g}$) is non-existent, resulting in a net force of zero and, consequently, zero acceleration of the horse-cart system.

Step 1: Identifying Physical State at the Mean Position

Given parameters at the mean position ($x = 0$):

• Mass of the bob ($m$) = $0.1 \, \text{kg}$
• Length of the string ($l$) = $2 \, \text{m}$
• Velocity at mean position ($v$) = $1 \, \text{m/s}$
• Position: The bob is at the lowest point of its swing.

At the mean position, the velocity vector $\vec{v}$ is directed horizontally (tangent to the arc of the swing). Thus, $\vec{v} = 1 \, \hat{i} \, \text{m/s}$ (assuming the bob is moving in the $x-y$ plane).

Step 2: Kinematic Analysis post-cut

The moment the string is cut, the tension force $T$ becomes zero. The bob is subjected only to the gravitational force $\vec{F}_g = m\vec{g} = -mg \, \hat{j}$.

According to [Newton's First Law of Motion], the object will maintain its current velocity vector unless acted upon by an external force. Per [Newton's Second Law], the acceleration acting on the bob is solely the acceleration due to gravity, $g \approx 9.8 \, \text{m/s}^2$ downward.

Step 3: Defining the Trajectory Equation

The bob acts as a projectile launched horizontally from a height $h = 2 \, \text{m}$ (length of the string) with initial velocity $u_x = 1 \, \text{m/s}$ and $u_y = 0 \, \text{m/s}$.

The position coordinates at time $t$ are given by:

$x(t) = u_x t = 1 \cdot t$

$y(t) = h - \frac{1}{2}gt^2 = 2 - 4.9t^2$

Eliminating $t$ by substituting $t = x$ into the $y$ equation yields the parabolic trajectory:

$y = 2 - 4.9x^2$

Step 4: Geometric Representation

Step 5: Theoretical Conclusion

Since the bob possesses a horizontal velocity component at the moment of detachment and is subject to a constant downward acceleration, the bob follows a parabolic trajectory. This is defined by the kinematics of horizontal projectile motion under constant gravity [per the Equivalence Principle and Newtonian Dynamics].

Final Solution: The trajectory of the bob is a parabola, as it undergoes horizontal motion with constant velocity while simultaneously accelerating vertically downward under gravity.