Third Law of Motion Questions in English

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
When a horse pulls a cart,it pushes the ground backward with its feet at an angle.
This action force exerted by the horse on the ground results in an equal and opposite reaction force exerted by the ground on the horse in the forward direction.
This reaction force allows the horse to move forward,and since the cart is attached to the horse,the cart is pulled along with it.

(N/A) The difficulty arises due to the backward reaction force of the water being ejected.
When a fireman holds a hose that ejects a large volume of water at a high velocity,the water exerts an equal and opposite reaction force on the hose in the backward direction,according to $Newton's$ third law of motion.
This backward force pushes the fireman,making it difficult for him to maintain his balance and stability while holding the hose.

(C) Mass of the rifle,$m_1 = 4\, kg$.
Mass of the bullet,$m_2 = 50\, g = 0.05\, kg$.
Initial velocity of the bullet,$v_2 = 35\, m/s$.
Let the recoil velocity of the rifle be $v_1$.
Initially,the system is at rest,so the initial momentum is $0$.
According to the law of conservation of momentum,the total momentum after firing must be equal to the total momentum before firing.
$m_1 v_1 + m_2 v_2 = 0$.
$4(v_1) + (0.05 \times 35) = 0$.
$4 v_1 + 1.75 = 0$.
$4 v_1 = -1.75$.
$v_1 = -1.75 / 4 = -0.4375\, m/s$.
The negative sign indicates that the rifle recoils in the direction opposite to the bullet,with a magnitude of $0.4375\, m/s$.

(N/A) The truck has a large mass,which results in a very high static friction between the tires and the road.
To move the truck,the applied force must exceed the maximum static frictional force.
When a person pushes the truck and it does not move,the applied force is balanced by the static frictional force acting in the opposite direction.
Therefore,the student's logic is correct in the context that the net force on the truck is zero because the applied force and the frictional force cancel each other out.

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
These two forces always act on two different bodies.
They are equal in magnitude and opposite in direction.

(C) rocket operates based on Newton's $3^{rd}$ law of motion and the principle of conservation of linear momentum.
As the fuel burns,the rocket expels hot gases at high velocity in the downward direction.
According to the law of conservation of momentum,the total momentum of the system (rocket + gases) remains constant in the absence of external forces.
Since the gases gain momentum in the downward direction,the rocket gains an equal and opposite momentum in the upward direction,which provides the necessary thrust for the rocket to lift off.

(A) According to the law of conservation of momentum,the momentum of the bullet is equal to the momentum of the rifle $(p = mv)$.
Since the force applied is the same,the impulse $(F \times t)$ imparted to both rifles is the same,resulting in the same change in momentum.
For a light rifle,the mass $(m)$ is smaller,so to achieve the same momentum,the recoil velocity $(v)$ must be significantly higher.
For a heavy rifle,the mass $(m)$ is larger,resulting in a lower recoil velocity $(v)$.
Since the kinetic energy of recoil $(KE = \frac{1}{2}mv^2)$ is proportional to the square of the velocity,the lighter rifle experiences a higher recoil velocity,which causes a greater impact force on the shoulder. Therefore,the lighter rifle will hurt the shoulder more.

(N/A) The separation between them will increase.
Initially,the momentum of both friends is zero as they are at rest.
According to the law of conservation of momentum,when the first person throws the ball,they must move in the opposite direction to the ball to conserve momentum.
When the second person catches the ball,they receive the momentum of the ball,which causes them to move backward in the direction of the ball's motion.
Consequently,both friends move away from each other,increasing the distance between them.

(N/A) The rotation of the water sprinkler is based on Newton's $Third$ Law of Motion.
When water is supplied, it rushes out through the nozzles of the sprinkler at a high velocity.
According to Newton's $Third$ Law of Motion, for every action, there is an equal and opposite reaction.
The water exerts an action force on the nozzle in the backward direction, and the nozzle exerts an equal and opposite reaction force on the water.
Due to this reaction force, the sprinkler arm experiences a torque that causes it to rotate in the opposite direction of the water flow.

(A) According to Newton's $3^{rd}$ Law of Motion,every action has an equal and opposite reaction.
In this scenario,when the man jumps forward,he exerts a backward force on the boat,which is the action.
The boat simultaneously exerts an equal and opposite forward force on the man,which is the reaction.
Therefore,the push of the man on the boat is the action,and the force exerted by the boat on the man is the reaction.

(B) According to Newton's third law of motion,for every action,there is an equal and opposite reaction.
When you kick a stone,your foot exerts a force on the stone (action).
Simultaneously,the stone exerts an equal and opposite force on your foot (reaction).
This reaction force from the hard surface of the stone causes the pain in your foot.

(C) According to Newton's Third Law of Motion,for every action,there is an equal and opposite reaction.
Since the gravitational force is a mutual interaction between two bodies,the force exerted by the Earth on the apple $(F_1)$ is equal in magnitude to the force exerted by the apple on the Earth $(F_2)$.
Therefore,$F_1 = F_2$.

(N/A) Newton's third law of motion states that to every action,there is always an equal and opposite reaction. This means that when one object exerts a force on a second object,the second object simultaneously exerts a force equal in magnitude and opposite in direction on the first object.

(N/A) Walking is a classic example. When a person walks,they push the ground backward with their feet (action). According to Newton's third law,the ground exerts an equal and opposite force on the feet (reaction). This reaction force is provided by the friction between the soles of the shoes and the surface of the ground. Without friction,the feet would slip,and it would be impossible to exert the necessary backward force to move forward.

(C) Newton's third law of motion is applied in the flight of a bird.
For flight,a bird applies force on the air in the backward direction by flapping its wings (action).
In response,the air exerts an equal and opposite force on the bird in the forward direction (reaction),which helps the bird to fly.

(B) The relation for the collision of two balls $A$ and $B$ is given by the principle of conservation of momentum:
$m_A u_A + m_B u_B = m_A v_A + m_B v_B$
Where:
$m_A, m_B$ are the masses of balls $A$ and $B$.
$u_A, u_B$ are the initial velocities of balls $A$ and $B$ before collision.
$v_A, v_B$ are the final velocities of balls $A$ and $B$ after collision.
The related law is the $\text{Law}$ $\text{of}$ $\text{Conservation}$ $\text{of}$ $\text{Momentum}$, which states that if no external unbalanced force acts on a system of two interacting objects, then the total momentum of the system remains conserved (constant).

(N/A) Newton's third law of motion states that to every action,there is an equal and opposite reaction.
Examples are:
$(i)$ When we exert a force (action) on a wall by pushing it with the palm of our hand,we experience a force (reaction) exerted by the wall on our palm.
$(ii)$ When a bullet is fired from a gun,the recoil of the gun is the reaction produced by the bullet on the gun.

(N/A) Law of conservation of momentum:
It states that in an isolated system,the total momentum of the system remains conserved (constant) if no external unbalanced force acts on it.
$(i)$ $A$ rocket taking off from ground: The chemicals inside the rocket burn and produce a high-velocity blast of hot gases. These gases are ejected through the tail nozzle in the downward direction. According to the law of conservation of momentum,the rocket gains an equal and opposite momentum in the upward direction,causing it to lift off.
$(ii)$ Flying of a jet aeroplane: In jet aeroplanes,a large volume of gases produced by the combustion of fuel is ejected through a jet in the backward direction at high velocity. The backward momentum of these gases results in an equal and opposite forward momentum for the aeroplane,which provides the necessary thrust for flight.

(N/A) $(i)$ While launching his boat into the water,the boatman pushes the river bank backward with a long pole. In accordance with Newton's third law of motion,the bank applies an equal and opposite reaction force on the pole,thereby pushing the boat forward.
$(ii)$ When passengers alight from the boat,they push the boat in the backward direction. As a result,the boat tends to move away from the shore. The boatman overcomes this by tying the boat to a rigid support on the bank.
$(iii)$ When a bullet is fired,the gun exerts a forward force on the bullet. According to Newton's third law,the bullet exerts an equal and opposite force on the gun,causing it to recoil.
$(iv)$ Rockets operate in space by ejecting burnt gases at high velocity from the rear. According to Newton's third law,the gases exert an equal and opposite reaction force on the rocket,propelling it forward.

(N/A) $(i)$ When a bullet is fired from a gun,the force exerted on the bullet is the action,and the recoil force experienced by the gun is the reaction.
$(ii)$ When a nail is hammered,the force exerted by the hammer on the nail is the action,and the force exerted by the nail on the hammer is the reaction.
$(iii)$ When a book is placed on a table,the weight of the book acting downwards on the table is the action,and the normal force exerted by the table upwards on the book is the reaction.
$(iv)$ In a moving rocket,the force exerted by the rocket on the exhaust gases is the action,and the force exerted by the gases on the rocket is the reaction.
$(v)$ When a person walks,the force exerted by the person's foot pushing the ground backwards is the action,and the frictional force exerted by the ground on the foot in the forward direction is the reaction.
$(vi)$ When a moving train collides with a stationary train,the force exerted by the moving train on the stationary train is the action,and the force exerted by the stationary train on the moving train is the reaction.

(D) The motion of the horse and cart system is governed by the following forces:
$(i)$ The weight of the horse and cart acting vertically downward.
$(ii)$ The normal reaction of the ground acting vertically upward,which balances the weight.
$(iii)$ The force exerted by the horse's feet on the ground in the backward direction (Action).
$(iv)$ The reaction force of the ground on the horse's feet in the forward direction (Reaction).
$(v)$ The force of friction acting on the cart wheels in the backward direction.
Forces $(i)$ and $(ii)$ are balanced and do not contribute to horizontal motion. The horse pushes the ground backward with force $(iii)$,and the ground exerts an equal and opposite reaction force $(iv)$ on the horse in the forward direction. The cart moves forward because the forward reaction force $(iv)$ is greater than the backward frictional force $(v)$ acting on the cart. Thus,the system moves due to the net unbalanced force in the forward direction.

(C) This phenomenon is based on $Newton's$ third law of motion,which states that for every action,there is an equal and opposite reaction.
When a person jumps from a boat towards the shore,they exert a muscular force on the boat in the backward direction (action).
In response,the boat exerts an equal and opposite force on the person,propelling them forward towards the shore.
Simultaneously,due to the action force applied by the person,the boat moves in the opposite direction,i.e.,away from the shore (reaction).

(N/A) Consider two spring balances connected together as shown in the figure.
The fixed end of balance $B$ is attached to the wall. When a force is applied through the free end of spring balance $A$,it is observed that both the spring balances show the same readings on their scales. It means that the force exerted by spring balance $A$ on spring balance $B$ is equal but opposite in direction to the force exerted by balance $B$ on balance $A$.
$(b)$ Pushkar exhibits an experimental mindset and a helpful nature.

(A) According to the law of conservation of momentum,the total momentum before the jump is equal to the total momentum after the jump.
Initially,the boat and the girl are moving together,so the initial momentum is $0$ (assuming the system starts from rest or considering the relative change).
Let $m_1 = 50\, kg$ be the mass of the girl and $v_1 = 3\, m s^{-1}$ be her velocity.
Let $m_2 = 300\, kg$ be the mass of the boat and $v_2$ be the velocity of the boat.
Using the principle of conservation of linear momentum: $m_1 v_1 + m_2 v_2 = 0$.
Substituting the values: $(50 \times 3) + (300 \times v_2) = 0$.
$150 + 300 v_2 = 0$.
$300 v_2 = -150$.
$v_2 = -\frac{150}{300} = -0.5\, m s^{-1}$.
The negative sign indicates that the boat moves in the opposite direction (backwards) with a velocity of $0.5\, m s^{-1}$.

(N/A) Momentum is defined as the total quantity of motion possessed by a body. It is equal to the product of the mass and velocity of the body. Its $SI$ unit is $kg \ m \ s^{-1}$.
$(b)$ Given:
Mass of bullet $(m) = 0.02 \ kg$
Mass of gun $(M) = 7.5 \ kg$
Velocity of bullet $(v) = 200 \ m \ s^{-1}$
Recoil velocity of gun $(V) = ?$
According to the law of conservation of momentum,the total momentum before firing is equal to the total momentum after firing:
$m \times v + M \times V = 0$
$0.02 \times 200 + 7.5 \times V = 0$
$4 + 7.5 \times V = 0$
$7.5 \times V = -4$
$V = -4 / 7.5 = -0.533 \ m \ s^{-1}$
The negative sign indicates that the gun recoils in the direction opposite to the bullet. The speed of recoil is $0.533 \ m \ s^{-1}$.

(N/A) When a bullet is fired from a gun,the gun experiences a backward jerk,which is known as the recoil of the gun.
This phenomenon is based on the Law of Conservation of Momentum.
According to this law,if no external force acts on a system,the total momentum of the system remains conserved.
Initially,both the gun and the bullet are at rest,so the initial momentum of the system is $0$.
Let $m$ be the mass of the bullet,$v$ be the velocity of the bullet,$M$ be the mass of the gun,and $V$ be the recoil velocity of the gun.
According to the law of conservation of momentum:
Initial Momentum = Final Momentum
$0 = mv + MV$
$MV = -mv$
$V = -\frac{mv}{M}$
The negative sign indicates that the direction of the recoil velocity of the gun is opposite to the direction of the bullet's velocity.

(N/A) $(a) (i)$ Spring balance $B$ will show $20 \ g \ wt$.
$(ii)$ This is due to Newton's third law of motion. When spring balance $A$ exerts a force on balance $B$,balance $B$ exerts an equal and opposite force of reaction on balance $A$. Since the force applied is $20 \ g \ wt$,the reaction force is also $20 \ g \ wt$.
$(b)$ It is observed that the balloon moves in the upward direction. Air rushes out from the balloon in the downward direction (action). As a result,there is an equal and opposite reaction in the form of upward motion of the balloon,which is in accordance with Newton's third law of motion.

(N/A) According to Newton's third law of motion,the spring exerts equal and opposite forces on both masses.
Let $F_{1}$ be the force on mass $m_{1}$ and $F_{2}$ be the force on mass $m_{2}$.
Since the forces are equal in magnitude and opposite in direction,we have $F_{1} = -F_{2}$.
Using Newton's second law,$F = ma$,we can write:
$m_{1} a_{1} = -m_{2} a_{2}$
Here,$a_{1}$ and $a_{2}$ are the accelerations of the masses $m_{1}$ and $m_{2}$ respectively.
The negative sign indicates that the accelerations are in opposite directions.
Taking the magnitude of the ratio of accelerations:
$\frac{|a_{1}|}{|a_{2}|} = \frac{m_{2}}{m_{1}}$
This proves that the ratio of their accelerations is inversely proportional to their masses.

(B) No,the boat will not move. According to Newton's $3^{rd}$ Law of Motion,when the air from the fan pushes the sail,the sail exerts an equal and opposite force on the fan. Since the fan is attached to the boat,the system (fan + boat) experiences internal forces that cancel each other out. Motion occurs only when an external unbalanced force acts on a body. Since the fan is part of the boat,no net external force is applied to the system,and therefore,the boat will not move.

(N/A) Newton's third law states that for every action,there is an equal and opposite reaction. However,these two forces act on different bodies,not the same body. Since the forces do not cancel each other out on a single object,the net force acting on a body can be non-zero,thereby allowing the body to move.

(A) According to the Law of Conservation of Momentum,the total momentum before the jump is equal to the total momentum after the jump.
Since the system (girl + boat) is initially at rest,the initial momentum is $0$.
Let $m_1 = 50 \, kg$ be the mass of the girl and $v_1 = 3 \, m s^{-1}$ be her velocity.
Let $m_2 = 300 \, kg$ be the mass of the boat and $v_2$ be the velocity of the boat.
Using the formula: $m_1 v_1 + m_2 v_2 = 0$.
Substituting the values: $(50 \times 3) + (300 \times v_2) = 0$.
$150 + 300 v_2 = 0$.
$300 v_2 = -150$.
$v_2 = -\frac{150}{300} = -0.5 \, m s^{-1}$.
The negative sign indicates that the boat moves in the opposite direction (backwards) with a velocity of $0.5 \, m s^{-1}$.

(B) According to the law of conservation of linear momentum,the total momentum before firing is equal to the total momentum after firing.
Initial momentum of the system (gun + bullet) = $0$ (since both are at rest).
Final momentum of the system = $m \times v + M \times V$,where $m = 0.01 \, kg$ (mass of bullet),$v = 400 \, m s^{-1}$ (velocity of bullet),$M = 4 \, kg$ (mass of gun),and $V$ is the recoil velocity of the gun.
Setting the initial momentum equal to the final momentum:
$0 = (0.01 \, kg \times 400 \, m s^{-1}) + (4 \, kg \times V)$
$0 = 4 + 4V$
$4V = -4$
$V = -1 \, m s^{-1}$.
The negative sign indicates that the gun recoils in the direction opposite to the bullet.

(B) The statement is False.
According to Newton's $Third$ Law of Motion, for every action, there is an equal and opposite reaction. These two forces always act on two different bodies, not on the same body. Therefore, they cannot cancel each other out.

(TRUE) The statement is True.
According to Newton's Third Law of Motion,forces always exist in pairs. Whenever one object exerts a force on a second object,the second object simultaneously exerts a force equal in magnitude and opposite in direction on the first object. Thus,an interaction between two bodies always involves a pair of forces.

(B) On a perfectly smooth ice surface, there is no friction. To move, the man must exert a force on an object and throw it in the opposite direction of the shore.
According to Newton's $third$ law of motion, for every action, there is an equal and opposite reaction.
By throwing an object away from the shore, the man experiences a reaction force in the direction of the shore, which allows him to move towards it.

(D) jet engine operates by expelling high-speed exhaust gases out of the back of the engine.
According to Newton's $third$ law of motion, for every action, there is an equal and opposite reaction.
The high-speed ejection of gases (action) creates an equal and opposite force (reaction) that pushes the jet engine forward, known as thrust.

(A) According to $Newton's$ $third$ $law$ of motion, for every action, there is an equal and opposite reaction.
These two forces always act on two different bodies.
For example, if a person pushes a wall, the person exerts a force on the wall (action), and the wall exerts an equal and opposite force on the person (reaction).
Since they act on different bodies, they do not cancel each other out.

(B) According to $Newton's$ $third$ $law$ of motion, for every action, there is an equal and opposite reaction.
These two forces always act on two different bodies.
If they acted on the same body, the net force would be zero, and no motion would be possible.
Therefore, action and reaction forces must act on different objects.

(C) According to the Law of Conservation of Momentum,the total momentum of an isolated system remains constant if no external force acts on it.
Before firing,the gun and the bullet are at rest,so the total initial momentum is $0$.
When the gun is fired,the bullet moves forward with a certain momentum. To keep the total momentum at $0$,the gun must move backward with an equal and opposite momentum.
This backward movement of the gun is known as recoil,which ensures that the total linear momentum of the system is conserved.