Class 11 Physics (Physics Part-I) Chapter 4 Laws of Motion

Key Concept: Application to Systems

c) $a = \frac{F}{m}$

[Solution Description] According to Newton’s Second Law, the force $F$ acting on a system results in an acceleration $a$. The relationship between force, mass, and acceleration is given by the formula $F = ma$. Solving for acceleration, we have:

$a = \frac{F}{m}$

Key Concept: Motion Analysis

c) Same

[Solution Description] According to Galileo’s double inclined plane experiment, in the absence of friction, a ball released from rest will roll down and climb up the other side to the same height it started from. This is because mechanical energy is conserved, and there are no non-conservative forces doing work on the system.

Key Concept: Theoretical Application, Real-World Application

c) Gravitational force

[Solution Description] In a stable orbit, the satellite experiences gravity as a centripetal force that keeps it in its circular path. According to Galileo’s and Newton’s principles, no tangential force is required to maintain its speed; only the gravitational pull acts perpendicular to its velocity as centripetal force.

Key Concept: Basic Tension

c) 30 N

[Solution Description] Since the force is applied symmetrically and the rope is massless, the tension throughout the rope is equal to the applied force. Therefore, the tension $T = 30 \, \text{N}$.

Key Concept: Misconception Challenges, Internal vs External Forces in Systems

c) Skater A accelerates at half the rate of Skater B.

[Solution Description] When the skaters push off each other, both experience forces equal in magnitude but opposite in direction, as per Newton’s Third Law. Denote the masses of Skater A and Skater B as $m_A = 2m_B$ and $m_B$, respectively. Using Newton’s second law $F = ma$, if we denote the force exerted by Skater B on A as $F_{BA}$ and the force exerted by Skater A on B as $F_{AB}$, then:

$F_{BA} = m_A a_A$

$F_{AB} = m_B a_B$

From Newton’s third law, $F_{BA} = -F_{AB}$, hence:

$m_A a_A = m_B a_B$

Substituting $m_A = 2m_B$ gives:

$2m_B a_A = m_B a_B$

Cancelling $m_B$ results in:

$a_A = \frac{a_B}{2}$

Therefore, Skater A accelerates at half the rate of Skater B.

Key Concept: Non-Contact Forces

b) Yes, the satellite exerts a force back on Earth.

[Solution Description] Yes, according to Newton’s Third Law, the satellite exerts an equal and opposite gravitational force on Earth.

Key Concept: Newton’s Third Law

a) Equal and opposite to the force exerted on the ball

[Solution Description] By Newton’s third law, the force exerted by the ball on the bat is equal in magnitude and opposite in direction to the force exerted by the bat on the ball.

Key Concept: Proportionality Concept

b) Mass and acceleration

[Solution Description] Newton’s Second Law states that force is proportional to the product of mass and acceleration. Thus, $F = ma$ illustrates this relationship, where $F$ is the force, $m$ is the mass, and $a$ is the acceleration.

Key Concept: Real-World Application, System Analysis

b) 100 N

[Solution Description]

By Newton’s Third Law, the force that Skater A exerts on Skater B is equal and opposite to the force that Skater B exerts on Skater A.

Begin by calculating the force exerted by Skater A using the formula for force, which is $F = ma$:

For Skater A:

$F_{\text{A on B}} = m_{\text{A}} \times a_{\text{A}} = 50 \, \text{kg} \times 2 \, \text{m/s}^2$

$F_{\text{A on B}} = 100 \, \text{N}$

Therefore, the magnitude of the force exerted by Skater A on Skater B is 100 N.

Key Concept: Intermediate Calculation

c) 75 N

[Solution Description]

To find the average force, use the impulse-momentum theorem: Impulse $= \Delta p = F \times \Delta t$. The change in momentum $\Delta p$ can be calculated as:

$\Delta p = m(v_f – v_i)$

where $v_f = 0$ is the final velocity and $v_i = 25$ m/s is the initial velocity.

Substitute $$m = 0.15$ kg (converted from grams), $v_f = 0$, and $v_i = 25$ m/s:

$\Delta p = 0.15 \cdot (0 – 25) = -3.75 \text{ kg}\cdot\text{m/s}$

Now, calculate the force using $F = \frac{\Delta p}{\Delta t}$:

$F = \frac{-3.75}{0.05} = -75 \text{ N}$

The negative sign indicates the force is opposite to the direction of motion.

Key Concept: Force and Momentum Change

c) 20 Ns

[Solution Description]

Impulse is given by the product of force and time or can be calculated as the change in momentum.

First, calculate the impulse using the change in momentum:

$\text{Impulse} = \Delta p = p_{\text{final}} – p_{\text{initial}} = 28 \, \text{kg} \cdot \text{m/s} – 8 \, \text{kg} \cdot \text{m/s} = 20 \, \text{Ns}$

Therefore, the impulse imparted to the object is $20 \, \text{Ns}$.

Key Concept: Application of Newton’s First Law

d) The net external force is directed backwards.

[Solution Description]

After the engine failure, the car starts decelerating, which means there is a net external force acting on it. According to Newton’s First Law, if the velocity of an object changes (either in magnitude or direction), a net force must be acting. The frictional force now exceeds the forward thrust that was initially provided by the engine, causing deceleration.

Key Concept: Impulse

b) Force applied over a time duration

[Solution Description] Impulse is defined as the product of force and time duration, which is equal to the change in momentum. Thus, Impulse = Force $\times$ Time Duration.

Key Concept: Force Components

b) 25 N

[Solution Description]

The component of the gravitational force acting parallel to the inclined plane is given by the formula: $F_{\parallel} = mg \sin(\theta)$, where $m$ is the mass of the box, $g$ is the acceleration due to gravity, and $\theta$ is the angle of inclination. Here, the total force ($mg$) is 50 N, so we calculate:

$F_{\parallel} = 50 \times \sin(30^\circ)$

$= 50 \times 0.5$

$= 25 \, \text{N}$

Thus, the component of the gravitational force parallel to the incline is 25 N.

Key Concept: Real-World Application

b) 20 m/s to the west

[Solution Description] By the conservation of momentum, the total momentum before and after the explosion is conserved. If the rocket was initially stationary, its initial momentum was zero.

Using the momentum conservation equation:

$m_X \cdot v_X + m_Y \cdot v_Y = 0$

Given,

$2 \cdot 30 + 3 \cdot (-v_Y) = 0$

Solving for $v_Y$:

$60 – 3v_Y = 0$

$3v_Y = 60$

$v_Y = \frac{60}{3} = 20 \text{ m/s}$

So, fragment Y moves at 20 m/s to the west.

Key Concept: Basic Definition

c) The relationship between force and displacement

[Solution Description] Hooke’s Law states that the force exerted by a spring is proportional to the displacement from its equilibrium position and acts in the opposite direction. Mathematically, it is given by $F = -kx$, where $F$ is the force, $k$ is the spring constant, and $x$ is the displacement.

Key Concept: Misconception Analysis

c) It ignores the presence of friction.

[Solution Description] Aristotle claimed that a continuous external force is necessary to keep an object in motion, neglecting the role of friction, which acts as an opposing force that needs to be overcome for constant velocity. Without friction, no external force is needed to maintain constant motion.

Key Concept: Vector Nature

b) Force and acceleration have direction.

[Solution Description] Newton’s second law is a vector equation, meaning that both force and acceleration are vector quantities. This implies that they have both magnitude and direction.

Key Concept: Complex Scenario Analysis, Advanced Equilibrium

b) 12.81 N, approximately $141.34^\circ$ from the positive x-axis

[Solution Description]

For equilibrium, the sum of components in each direction must be zero.

– In the x-direction: $F_{1x} + F_{2x} + F_{3x} = 0$.

Since $F_1$ is along the x-axis and $F_2$ has no x-component,

$F_{3x} = -F_{1x} = -10 \text{ N}.$

– In the y-direction: $F_{1y} + F_{2y} + F_{3y} = 0$.

Since $F_1$ has no y-component and $F_2 = -8$$ N,

$F_{3y} = -F_{2y} = 8 \text{ N}.$

The magnitude of $F_3$ is given by:

$F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(-10)^2 + 8^2} = \sqrt{100 + 64} = \sqrt{164} \approx 12.81 \text{ N}.$

The direction $\theta$ of $F_3$ can be found using $\tan(\theta) = \frac{F_{3y}}{F_{3x}} = \frac{8}{-10}$.

$\theta = \arctan\left(\frac{-4}{5}\right) \approx -38.66^\circ.$

Therefore, $F_3$ has a magnitude of approximately 12.81 N and acts at an angle of roughly $141.34^\circ$ from the positive x-axis.

Key Concept: Basic Concept

b) It opposes impending motion

[Solution Description] Static friction acts in the opposite direction of the applied force, and its role is to oppose impending motion until the applied force exceeds it.

Key Concept: Comparative Analysis

b) Galileo believed motion could occur without external force.

[Solution Description] Aristotle believed that a continuous external force is required to keep an object in motion. In contrast, Galileo argued that once an object is in motion, it needs no further external force unless other resistive forces like friction act upon it. This marked a significant departure from Aristotelian thought.

Key Concept: Static Friction Basics

b) It opposes impending motion

[Solution Description] Static friction acts to oppose impending motion. When a body is at rest, static friction prevents it from moving by resisting any external force that attempts to cause motion.

Key Concept: Friction Reduction Techniques

b) Using ball bearings

[Solution Description] One common technique to reduce rolling friction is using ball bearings. Ball bearings help minimize the area of contact and reduce resistance, allowing smoother movement of machine parts.

Key Concept: Complex Dynamics, Real-World Application

c) 28.55 m/s

[Solution Description]

For the car to not slip, we need to consider both the gravitational force component along the plane and the frictional force. We apply the formula for maximum speed on a banked road:

$v_{max} = \sqrt{\frac{Rg(\tan \theta + \mu_s)}{1 – \mu_s \tan \theta}}$

Substituting $R = 150\,m$, $g = 9.8\,m/s^2$, $\theta = 15^\circ$, and $\mu_s = 0.25$:

$v_{max} = \sqrt{\frac{150 \times 9.8 (\tan 15^\circ + 0.25)}{1 – 0.25 \cdot \tan 15^\circ}}$

Calculate $$\tan 15^\circ \approx 0.2679$$:

$v_{max} = \sqrt{\frac{150 \times 9.8 (0.2679 + 0.25)}{1 – 0.25 \times 0.2679}}$

$v_{max} = \sqrt{\frac{150 \times 9.8 \times 0.5179}{1 – 0.066975}}$

$v_{max} = \sqrt{\frac{760.41}{0.933025}}$

$v_{max} = \sqrt{814.89} \approx 28.55\,m/s$

Key Concept: Newton’s Third Law

a) The book pushes down on the table; the table pushes up on the book.

[Solution Description]

According to Newton’s third law, every action has an equal and opposite reaction. In this scenario, the book exerts a downward force on the table due to gravity. The table, in response, exerts an equal upward force on the book. Therefore, the action-reaction pair comprises these two forces.

Key Concept: Force Direction

d) Northeast

[Solution Description]

The initial velocity vector of the car is directed towards the east, and the force applied changes the direction by adding a northward component. The resulting direction can be found using vector addition.

The car moves with an initial eastward velocity $v_0 = 20$ m/s.

When a force of 3000 N is applied northward, the acceleration $a$ is given by $\frac{F}{m}$ where $F = 3000$ N and $m = 1500$ kg.

$a = \frac{3000}{1500} = 2 \text{ m/s}^2$

This acceleration is purely in the north direction. The new velocity components after some time $t$ will be:

Eastward: $v_x = v_0 = 20$ m/s

Northward: $v_y = a \times t = 2t$

For simplicity, assume that this condition persists long enough for the velocity magnitudes to become comparable, leading to a resultant direction.

Using Pythagoras’ theorem:

$v_{resultant} = \sqrt{(v_x)^2 + (v_y)^2} = \sqrt{20^2 + (2t)^2}$

Since both velocities are present, without a specific time factor, motion remains northeast.

Thus, the correct answer is Northeast.

Key Concept: Basic Concept

c) A force that opposes relative motion between surfaces in contact

[Solution Description] Kinetic friction acts to oppose the relative motion between two surfaces that are in contact with each other. It works against the direction of movement, making it more difficult for surfaces to slide over one another.

Key Concept: Concurrent Forces Analysis

a) $F_1 = 5N$, $F_2 = 10N$, $F_3 = -15N$

[Solution Description]

To achieve equilibrium, the vector sum of forces must be zero: $F_1 + F_2 + F_3 = 0$. Let’s check each option:

a) $F_1 = 5N$, $F_2 = 10N$, $F_3 = -15N$. Sum: $5 + 10 – 15 = 0$.

b) $F_1 = 20N$, $F_2 = -10$, $F_3 = -10N$. Sum: $20 – 10 – 10 = 0$.

c) $F_1 = 8N$, $F_2 = 8N$, $F_3 = 8N$. Sum: $8 + 8 + 8 \neq 0$.

d) $F_1 = 6N$, $F_2 = -4N$, $F_3 = 2N$. Sum: $6 – 4 + 2 \neq 0$.

Options a and b satisfy the condition. However, since we need one correct answer, let’s choose a).

Key Concept: Force Analysis

d) Zero

[Solution Description] A car parked on a flat surface experiences multiple forces: gravitational force pulling it downward and normal force from the ground pushing it upward. Since these forces balance each other out and there are no horizontal forces causing acceleration, the net force on the car is zero.

Key Concept: Basic Forces

c) Gravitational force

[Solution Description] The gravitational force governs celestial motions, such as the orbits of planets around the sun. It acts at a distance and is one of the four fundamental forces in nature.

Key Concept: Real-life Application

c) Due to the inertia of motion causing them to keep moving forward

[Solution Description]

When a moving bus suddenly stops, passengers tend to lurch forward due to inertia. Their bodies were in motion along with the bus, and according to the principle of inertia, they would continue in their state of motion unless an unbalanced external force acts on them.

Key Concept: Comparative Analysis

c) Newton stated that motion can continue without external force.

[Solution Description] Aristotle believed that an external force was required to maintain motion, whereas Newton proposed that no external force is needed to maintain motion at a constant velocity; only changes in motion require an external force.

Key Concept: Momentum Change

b) 18 kg m/s

[Solution Description] The change in momentum $\Delta p$ is given by $\Delta p = m \Delta v$, where $m = 1.5 \, \text{kg}$ and $\Delta v = 12 \, \text{m/s} – 0 \, \text{m/s} = 12 \, \text{m/s}$. Thus,

$\Delta p = 1.5 \times 12 = 18 \, \text{kg m/s}$

Key Concept: Complex Scenario Analysis, System Analysis

a) Skater A experiences greater acceleration than skater B

[Solution Description] According to Newton’s Third Law, when skater A pushes on skater B, skater B pushes back on skater A with an equal and opposite force. Denote these forces by $\vec{F}_{AB}$ and $\vec{F}_{BA}$ respectively. These forces are equal in magnitude but opposite in direction: $|\vec{F}_{AB}| = |\vec{F}_{BA}|$. The accelerations of skaters A and B ($a_A$ and $a_B$) can be found using Newton’s Second Law:

$\vec{F}_{AB} = m_A \cdot a_A \quad \text{and} \quad \vec{F}_{BA} = m_B \cdot a_B$

Since $|\vec{F}_{AB}| = |\vec{F}_{BA}|$, substituting gives:

$m_A \cdot a_A = m_B \cdot a_B$

Solving for the ratio of accelerations:

$\frac{a_A}{a_B} = \frac{m_B}{m_A} = \frac{60 \ kg}{40 \ kg} = 1.5$

Thus, skater A’s acceleration is 1.5 times that of skater B.

Key Concept: Component Forces

c) $F \sin \theta$

[Solution Description] The vertical component of a force $F$ acting at an angle $\theta$ can be calculated using the sine function: $F_{\text{vertical}} = F \sin \theta$.

Key Concept: Impulse Concept

a) Equal and opposite

[Solution Description] Newton’s third law states that for every action, there is an equal and opposite reaction. Therefore, when object A exerts a force on object B, object B will exert a force that is equal in magnitude but opposite in direction on object A.

Key Concept: Basic Concept

c) 50 N

[Solution Description] The weight of the object can be calculated using the formula $W = mg$. Substituting the given values, we have $W = 5 \times 10 = 50 \, \text{N}$. Since the object is at rest and on a horizontal table, the normal force is equal in magnitude to the gravitational force. Therefore, the normal force is 50 N.

Key Concept: Momentum Calculation

b) $5000\, \text{kg m/s}$

[Solution Description] Momentum ($p$) is calculated using the formula $p = mv$, where $m$ is mass and $v$ is velocity.

$p = 1000\, \text{kg} \times 5\, \text{m/s} = 5000\, \text{kg m/s}$

Key Concept: Force Interaction

b) Gravitational and normal forces act and cancel each other.

[Solution Description]

The block is at rest, meaning that it is in equilibrium and all forces acting on it are balanced. The gravitational force acts downward, while the normal force from the surface acts upward. These two forces cancel each other out, resulting in a net external force of zero.

Key Concept: Equilibrium Condition

c) $F_1 + F_2 + F_3 = 0$

[Solution Description] For a particle in equilibrium, the vector sum of all forces acting on it must equal zero. Therefore, when three forces are considered, their sum should satisfy the equation $F_1 + F_2 + F_3 = 0$.

Key Concept: Cause and Effect

b) It changes the object’s direction.

[Solution Description]

An external force will alter the state of motion of an object with inertia of direction, typically causing it to change its current path or direction.

Key Concept: Combined Forces

c) 15178.8 N

[Solution Description]

On a banked road, we analyze the horizontal components. The equations are:

$N \cos \theta = mg + f \sin \theta$

$N \sin \theta + f \cos \theta = \frac{mv^2}{R}$

Solving these simultaneously requires determining $N$ first, using known values:

For horizontal equilibrium,

$N \sin \theta + \mu_s N \cos \theta = \frac{mv^2}{R}$

Substituting values: $m = 1000\, \text{kg}$, $v = 15\, \text{m/s}$, $R = 40\, \text{m}$, and $\theta = 10^\circ$,

$N (\sin 10^\circ + 0.2 \cos 10^\circ) = \frac{1000 \times (15)^2}{40}$

Calculate the right side:

$\frac{225000}{40} = 5625$

Solve for $N$:

$N (\sin 10^\circ + 0.2 \cos 10^\circ) = 5625$

Approximate $\sin 10^\circ \approx 0.1736$ and $\cos 10^\circ \approx 0.9848$:

$N (0.1736 + 0.19696) = 5625$

$N (0.37056) = 5625$

$N \approx \frac{5625}{0.37056} \approx 15178.8$

So the effective force due to normal and frictional components is approximately $15178.8 \, \text{N}$.

Key Concept: Equation Application

c) 4500 N

[Solution Description] The formula for centripetal force $f_c$ is given by:

$f_c = \frac{mv^2}{R}$

Substituting the values, $m = 800\, \text{kg}$, $v = 15\, \text{m/s}$, and $R = 40\, \text{m}$:

$f_c = \frac{800 \times 15^2}{40} = \frac{800 \times 225}{40} = \frac{180000}{40} = 4500\, \text{N}$

Therefore, the centripetal force is 4500 N.

Key Concept: Coefficient of Static Friction

b) 15 N

[Solution Description]

The maximum static friction force can be calculated using the formula $f_s = \mu_s N$. Here, $\mu_s = 0.3$ and $N = 50 \, \text{N}$. Therefore:

$f_s = 0.3 \times 50 = 15 \, \text{N}$

Hence, the maximum static friction force is 15 N.

Key Concept: Impulse Application

b) 10 N s

[Solution Description]

The impulse $I$ can be calculated using the change in momentum ($\Delta p$). Initially, the ball has a momentum of $p_i = m \cdot v = 1\,\text{kg} \times 5\,\text{m/s} = 5\,\text{kg}\cdot\text{m/s}$. After rebounding, its velocity is $-5\,\text{m/s}$, so the final momentum is $p_f = m \cdot (-v) = 1\,\text{kg} \times -5\,\text{m/s} = -5\,\text{kg}\cdot\text{m/s}$. Therefore, the change in momentum $\Delta p = p_f – p_i = -5\,\text{kg}\cdot\text{m/s} – 5\,\text{kg}\cdot\text{m/s} = -10\,\text{kg}\cdot\text{m/s}$. The magnitude of impulse is $|I| = |-10\,\text{N}\cdot\text{s}| = 10\,\text{N}\cdot\text{s}$.

Key Concept: Basic Identification

b) Gravitational force

[Solution Description]

The force that keeps a satellite in orbit around the Earth is the gravitational force. It acts at a distance and attracts the satellite towards the Earth’s center, balancing the inertia of the satellite to maintain its circular or elliptical motion.

Key Concept: Derivative Understanding

b) $\mathbf{F} = \frac{d\mathbf{p}}{dt}$

[Solution Description] According to Newton’s second law in terms of momentum, the force $\mathbf{F}$ can be expressed as the rate of change of momentum with respect to time. Mathematically, this is given by $\mathbf{F} = \frac{d\mathbf{p}}{dt}$.

Key Concept: Momentum Change

c) Both cars will have the same momentum.

[Solution Description] Given that Car A and Car B receive the same force $F$ for the same amount of time $t$, we can use the impulse-momentum theorem: $F \cdot t = \Delta p$. Since the initial velocities are zero, the change in momentum $\Delta p$ for each car will be equal to $F \cdot t$. Therefore, both cars will have the same momentum.

Key Concept: Force and Motion Analysis

d) It will move with a constant velocity.

[Solution Description] Once the force is removed, there are no external forces acting on the box since the surface is frictionless. According to Newton’s First Law of Motion, an object in motion will stay in motion at a constant velocity unless acted upon by an external force. Thus, the box will continue to move with a constant velocity of 10 m/s.

Key Concept: Basic Concept

c) External force

[Solution Description] To initiate the motion of a stationary object, an external force is required. This is because objects at rest will remain at rest unless acted upon by an outside influence, according to the principles outlined in Newton’s laws.

Key Concept: Complex Equilibrium, Multi-Force System

a) $F_3 = -6i – 2j + 4k$

[Solution Description]

To find $F_3$, we use the condition for equilibrium: $F_1 + F_2 + F_3 = 0$. Therefore:

$(4i – j + k) + (2i + 3j – 5k) + F_3 = 0$

Simplifying this equation gives:

$6i + 2j – 4k + F_3 = 0$

Hence, $F_3 = -6i – 2j + 4k$.

Key Concept: Intermediate Calculation

c) 5 m/s$^2$

[Solution Description]

To find the acceleration, we use Newton’s second law formula $a = \frac{F}{m}$, where $F$ is the force (thrust) and $m$ is the mass:

Thrust,$F = 75,000 \, \text{N}$ and Mass, $m = 15,000 \, \text{kg}$.

We plug these values into the formula:

$a = \frac{75,000}{15,000} = \frac{75}{15} = 5 \, \text{m/s}^2.$

Thus, the acceleration of the rocket is $5 \, \text{m/s}^2$.

Key Concept: Tension in Strings

c) 98 N

[Solution Description] The tension in the string can be calculated using the formula $T = m \cdot g$, where $m = 10$ kg is the mass of the weight and $g = 9.8$ m/s\textsuperscript{2} is the acceleration due to gravity. Thus, $T = 10 \times 9.8 = 98$ N.

Key Concept: Frictional Force Direction

a) Upwards along the slope

[Solution Description]

When a car moves slower than the optimum speed on a banked road, the tendency is to slide downwards due to insufficient centripetal force from the normal component alone. Thus, static friction acts upwards along the slope to provide the necessary centripetal force.

Key Concept: Direct Formula Application

c) -4 m/s

[Solution Description]

According to the conservation of momentum:

$p_{\text{bullet}} + p_{\text{gun}} = 0$

Where,

$p_{\text{bullet}} = m_{\text{bullet}} \times v_{\text{bullet}}$

and $p_{\text{gun}} = m_{\text{gun}} \times v_{\text{gun}}$.

Therefore,

$m_{\text{bullet}} \times v_{\text{bullet}} + m_{\text{gun}} \times v_{\text{gun}} = 0$

Rearranging for $v_{\text{gun}}$, we get:

$v_{\text{gun}} = -\left(\frac{m_{\text{bullet}} \times v_{\text{bullet}}}{m_{\text{gun}}}\right)$

Substituting the given values:

$v_{\text{gun}} = -\left(\frac{0.03 \, \text{kg} \times 400 \, \text{m/s}}{3 \, \text{kg}}\right)$

Calculating gives:

$v_{\text{gun}} = -4 \, \text{m/s}$

Hence, the recoil velocity of the gun is $-4 \, \text{m/s}$.

Key Concept: Concept Identification

b) Resistance to change in motion

[Solution Description] Inertia, as explained by Newton’s First Law, is the resistance of a body to change its state of motion or rest.

Key Concept: Static Friction

c) 27.1 m/s

[Solution Description]

To find the maximum speed $v_{max}$ that a car can have without skidding on a flat road, we use the formula $v_{max} = \sqrt{\mu_s Rg}$.

– Given:

$\mu_s = 0.5$$, $$R = 150\, \text{m}$, $g = 9.8\, \text{m/s}^2$.

– Substitute the values into the formula:

$v_{max} = \sqrt{0.5 \times 150 \times 9.8}$

– Calculate the value inside the square root:

$v_{max} = \sqrt{735}$

– Compute the square root:

$v_{max} \approx 27.1\, \text{m/s}$

Therefore, the maximum speed is approximately 27.1 m/s.

Key Concept: Real-World Application

c) Forces are balanced, resulting in no acceleration

[Solution Description] When a cyclist travels at constant speed, it means there is no acceleration. According to Newton’s First Law, this implies that the forces are balanced. The forward push by pedaling balances with the opposing forces like friction and air resistance, leading to a net force of zero.

Key Concept: Consistency with First Law

d) The acceleration is zero.

[Solution Description] When the net external force acting on a body is zero, according to Newton’s Second Law, $F = ma$. If $F = 0$, then $a = 0$. This means the acceleration of the body is zero, consistent with Newton’s First Law.

Key Concept: Friction’s Role

b) It opposes motion

[Solution Description] According to Galileo, in a frictionless environment, no external force is required to keep an object in motion. Friction opposes motion; therefore, it acts as an opposing force to maintain the motion caused by any initial force applied.

Key Concept: Static Friction Basics

b) It opposes the applied force

[Solution Description] Static friction acts to oppose the impending or initial motion of an object when an external force is applied to it.

Key Concept: Basic Forces

d) Gravitational force

[Solution Description] A non-contact force is one that acts at a distance without direct physical contact between objects. Among the given options, gravitational force does not require contact to exert its influence.

Key Concept: Force Equilibrium

a) 39.2 N

[Solution Description] The maximum static friction force ($f_s$) can be calculated using the formula $f_s = \mu_s N$, where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. Since the block is on a horizontal surface and at rest, $N = mg$. Thus,

$f_s = \mu_s \times m \times g = 0.4 \times 10 \times 9.8 = 39.2\, \text{N}$

Hence, the maximum horizontal force before the block moves is 39.2 N.

Key Concept: Force Calculation

b) 12 N

[Solution Description] To find the centripetal force, we use the formula $f_c = \frac{mv^2}{R}$.

Here, $m = 2$ kg, $v = 3$ m/s, and $R = 1.5$ m.

Calculate $v^2$

$v^2 = (3\, \text{m/s})^2 = 9\, \text{m}^2/\text{s}^2$

Now substitute the values into the formula:

$f_c = \frac{2 \times 9}{1.5} = \frac{18}{1.5} = 12\, \text{N}$

Thus, the centripetal force is 12 N.

Key Concept: Basic Concept

d) External force

[Solution Description]

An external force is needed to stop a moving body. This concept is based on the principle that an object in motion tends to stay in motion unless acted upon by an external force.

Key Concept: Momentum Calculation

d) 6 kg m/s

[Solution Description] The total momentum of the system is calculated using $p = mv$. For the 2 kg cart, its momentum is $p_A = m_A \cdot v_A = 2 \, \text{kg} \cdot 3 \, \text{m/s} = 6 \, \text{kg m/s}$. The 1 kg cart is stationary, so $p_B = 1 \, \text{kg} \cdot 0 \, \text{m/s} = 0 \, \text{kg m/s}$. Therefore, the total momentum of the system before the collision is $p_{\text{total}} = p_A + p_B = 6 \, \text{kg m/s} + 0 \, \text{kg m/s} = 6 \, \text{kg m/s}$.

Key Concept: Advanced Calculations, Problem Solving

d) 10800 N

[Solution Description]

When the elevator is decelerating while moving upwards, the net acceleration $a_{net}$ is $(g – 1)$ because it’s moving against downward gravity.

Using Newton’s second law, the net force $F_{net}$ on the elevator is given by:

$F_{net} = m \cdot (a_{net}) = 1000 \cdot (9.8 – 1) = 8800 \, \text{N}$

Now considering equilibrium, the force exerted by the cable ($T$) subtracts weight due to upward movement:

$T = m \cdot g + F_{net} = 1000 \cdot 9.8 + 8800 = 18800 \, \text{N}$

Therefore, the force exerted by the cable is 10800 N.

Key Concept: Theoretical Implications, Comparative Analysis

b) Spacecraft would require constant engine operation to sustain motion.

[Solution Description] If Aristotle’s idea that continuous force is necessary for motion were applied, current propulsion systems wouldn’t work efficiently. Spacecraft rely on initial thrust to propel and then continue moving by inertia. Under Aristotelian mechanics, engines need to constantly burn fuel which contradicts efficiency in vacuum where inertia dominates. Hence, modern physics relying on Newton’s laws would not hold true, requiring redesign of propulsion methods based on constant force application.

Key Concept: Action-Reaction Pairs

a) The gravitational force acting on the book and the normal force by the table

[Solution Description] The force exerted by the book on the table (downward due to gravity) and the force exerted by the table on the book (upward normal force) form an action-reaction pair. According to Newton’s third law, two forces that are equal in magnitude and opposite in direction constitute such a pair.

Key Concept: Component Equations

d) $F_{1x} + F_{2x} + F_{3x} = 0, \; F_{1y} + F_{2y} + F_{3y} = 0, \; F_{1z} + F_{2z} + F_{3z} = 0$

[Solution Description]

To achieve equilibrium in three dimensions, the sum of the forces in each coordinate direction must equal zero. The correct set of equations is:

$F_{1x} + F_{2x} + F_{3x} = 0, \quad F_{1y} + F_{2y} + F_{3y} = 0, \quad F_{1z} + F_{2z} + F_{3z} = 0$

This ensures that the net force in each directional component is zero.

Key Concept: Forces and Motion

d) An external force equal to friction

[Solution Description] To keep the block in uniform motion on a rough horizontal surface, an external force equal to the frictional force must be applied. Since friction opposes motion, the net external force must be zero for uniform motion, implying that the applied force should balance out the friction.

Key Concept: Momentum Change

c) 100,000 kg·m/s

[Solution Description]

The formula for momentum is $p = m \cdot v$.

Initial momentum, $p_i = 2000 \, \text{kg} \times 50 \, \text{m/s} = 100,000 \, \text{kg}\cdot\text{m/s}$.

Final momentum, $p_f = 2000 \, \text{kg} \times 100 \, \text{m/s} = 200,000 \, \text{kg}\cdot\text{m/s}$.

Change in momentum $\Delta p = p_f – p_i = 200,000 \, \text{kg}\cdot\text{m/s} – 100,000 \, \text{kg}\cdot\text{m/s} = 100,000 \, \text{kg}\cdot\text{m/s}$.