Mix Example - FORCE AND LAWS OF MOTION Questions in English

(D) Given: mass $m = 80 \ kg$,initial velocity $u = 5 \ m s^{-1}$,final velocity $v = 25 \ m s^{-1}$,and time $t = 2 \ s$.
First,calculate the acceleration $a$ using the formula:
$a = \frac{v - u}{t}$
Substituting the values:
$a = \frac{25 - 5}{2} = \frac{20}{2} = 10 \ m s^{-2}$
Now,calculate the force $F$ using Newton's second law of motion:
$F = m \times a$
Substituting the values:
$F = 80 \ kg \times 10 \ m s^{-2} = 800 \ N$
Therefore,the force required is $800 \ N$.

(A) Given:
Initial velocity, $u = 0 \, m/s$
Distance traveled, $S = 400 \, m$
Time taken, $t = 20 \, s$
Mass of the truck, $m = 7 \, \text{metric tons} = 7000 \, kg$
$(a)$ To find the acceleration $(a)$:
Using the second equation of motion, $S = ut + \frac{1}{2}at^2$
$400 = (0)(20) + \frac{1}{2} \times a \times (20)^2$
$400 = \frac{1}{2} \times a \times 400$
$a = \frac{400 \times 2}{400} = 2 \, m/s^2$
$(b)$ To find the force $(F)$:
Using Newton's second law of motion, $F = m \times a$
$F = 7000 \, kg \times 2 \, m/s^2 = 14000 \, N$

(D) Given:
Time interval, $t_1 = 1.2 \, s$
Initial speed, $u_1 = 1.8 \, m s^{-1}$
Final speed, $v_1 = 4.2 \, m s^{-1}$
Using the first equation of motion, $v = u + at$, we calculate the acceleration $(a)$:
$a = \frac{v_1 - u_1}{t_1} = \frac{4.2 - 1.8}{1.2} = \frac{2.4}{1.2} = 2 \, m s^{-2}$
Since the same force is applied to the same object, the acceleration remains constant at $a = 2 \, m s^{-2}$.
For the second interval, $t_2 = 2 \, s$:
The change in speed $(\Delta v)$ is given by $\Delta v = a \times t_2$.
$\Delta v = 2 \, m s^{-2} \times 2 \, s = 4 \, m s^{-1}$.
Thus, the speed changes by $4 \, m s^{-1}$.

(B) Given: Mass $m = 500 \, g = 0.5 \, kg$,Time $t = 10 \, s$,Force $F = 10^{-2} \, N = 0.01 \, N$,and Initial velocity $u = 0 \, m/s$.
First,calculate the acceleration $a$ using Newton's second law of motion,$F = ma$:
$a = \frac{F}{m} = \frac{0.01 \, N}{0.5 \, kg} = 0.02 \, m/s^2$.
Next,use the second equation of motion to calculate the distance $S$ travelled:
$S = ut + \frac{1}{2}at^2$
$S = 0 \times 10 + \frac{1}{2} \times 0.02 \times (10)^2$
$S = 0 + 0.01 \times 100 = 1 \, m$.
Therefore,the distance travelled by the body is $1 \, m$.

(N/A) Given: Mass $m = 5\, g = 5 \times 10^{-3}\, kg$,Initial velocity $u = 120\, m s^{-1}$,Final velocity $v = 0\, m s^{-1}$,Time $t = 0.01\, s$.
$(a)$ First,calculate the acceleration $a$ using the equation $v = u + at$:
$0 = 120 + a \times 0.01$
$a = -\frac{120}{0.01} = -12000\, m s^{-2}$ (The negative sign indicates retardation).
Now,calculate the distance of penetration $S$ using the equation $S = ut + \frac{1}{2}at^2$:
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6\, m$.
$(b)$ Calculate the average force $F$ using Newton's second law $F = ma$:
$F = (5 \times 10^{-3}\, kg) \times (12000\, m s^{-2})$
$F = 60\, N$.

(B) Given: Mass $(m) = 10 \, kg$,Initial velocity $(u) = 0 \, m s^{-1}$,Height $(S) = 0.8 \, m$,Acceleration due to gravity $(g) = 10 \, m s^{-2}$.
Using the equation of motion $v^2 - u^2 = 2gS$:
$v^2 - (0)^2 = 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4 \, m s^{-1}$.
Momentum $(p)$ is defined as the product of mass and velocity:
$p = m \times v$
$p = 10 \, kg \times 4 \, m s^{-1} = 40 \, kg \, m s^{-1}$ (or $40 \, N s$).
Thus,the dumbbell will transfer $40 \, kg \, m s^{-1}$ of momentum to the floor.

(A) Given: Initial velocity $(u) = 5 \ m s^{-1}$,Final velocity $(v) = 0$,Time $(t) = 2 \ s$.
Using the first equation of motion,$v = u + at$:
$0 = 5 + a \times 2$
$2a = -5$
$a = -2.5 \ m s^{-2}$ (retardation).
Force required to stop the car $(m = 1000 \ kg)$:
$F_{car} = m \times a = 1000 \ kg \times (-2.5 \ m s^{-2}) = -2500 \ N$.
Force required to stop the truck $(m = 10000 \ kg)$:
$F_{truck} = m \times a = 10000 \ kg \times (-2.5 \ m s^{-2}) = -25000 \ N$.
The negative sign indicates that the force is applied in the direction opposite to the motion (retarding force).

(B) Given:
Initial velocity of the ball $(u) = 0.5 \, m s^{-1}$
Final velocity of the ball $(v) = 0 \, m s^{-1}$
Time $(t) = 0.5 \, s$
Mass of the ball $(m) = 70 \, g = 0.07 \, kg$
According to Newton's second law of motion,the force $(F)$ is given by the rate of change of momentum:
$F = \frac{m(v - u)}{t}$
Substituting the values:
$F = \frac{0.07 \times (0 - 0.5)}{0.5}$
$F = \frac{0.07 \times (-0.5)}{0.5}$
$F = -0.07 \, N$
The negative sign indicates that the force applied by the player is a retarding force (opposing the motion of the ball).

(3000 N) Given:
Mass of the car $(m) = 1000 \, kg$
Initial velocity of the car $(u) = 15 \, m/s$
Final velocity of the car $(v) = 0 \, m/s$ (since it comes to rest)
Time taken $(t) = 5 \, s$
Using Newton's second law of motion,the force exerted by the tree on the car is given by:
$F = m \times a = m \times \frac{(v - u)}{t}$
Substituting the values:
$F = 1000 \times \frac{(0 - 15)}{5}$
$F = 1000 \times (-3) = -3000 \, N$
The negative sign indicates that the force exerted by the tree on the car is in the opposite direction of motion.
According to Newton's third law of motion,the force exerted by the car on the tree is equal in magnitude but opposite in direction to the force exerted by the tree on the car.
Therefore,the force exerted by the car on the tree is $3000 \, N$.

(A) Given:
Mass of the bullet $(m) = 10\, g = 0.01\, kg$
Initial velocity of the bullet $(u) = 0\, m s^{-1}$
Final velocity of the bullet $(v) = 300\, m s^{-1}$
Time taken $(t) = 0.003\, s$
Step $1$: Calculate the acceleration $(a)$ of the bullet using the formula $a = \frac{v - u}{t}$.
$a = \frac{300 - 0}{0.003} = \frac{300}{0.003} = 100,000\, m s^{-2}$.
Step $2$: Calculate the force $(F)$ exerted on the bullet using Newton's second law of motion,$F = m \times a$.
$F = 0.01\, kg \times 100,000\, m s^{-2} = 1000\, N$.
Therefore,the force exerted on the bullet by the rifle is $1000\, N$.

(N/A) Given: Mass $m = 50\, kg$,Force $F = 80\, N$.
Using Newton's second law of motion,$F = m \times a$,the acceleration $a$ is given by:
$a = F / m = 80 / 50 = 1.6\, m/s^2$.
If the mass is doubled,the new mass $m' = 2 \times 50 = 100\, kg$.
The new acceleration $a'$ is:
$a' = F / m' = 80 / 100 = 0.8\, m/s^2$.

(A) Given:
Mass of bullet $m_{B} = 50 \, g = 50 \times 10^{-3} \, kg = 0.05 \, kg$.
Velocity of bullet $v_{B} = 150 \, m s^{-1}$.
Mass of stone $m_{S} = 60 \, kg$.
Velocity of stone $v_{S} = 10 \, m s^{-1}$.
Let $n$ be the number of bullets required to stop the stone.
According to the law of conservation of momentum,the total momentum of the system must be zero to stop the stone.
$n(m_{B} v_{B}) + m_{S} v_{S} = 0$.
Taking the direction of the stone as positive,the bullets are fired in the opposite direction,so their velocity is negative relative to the stone's motion.
$n(0.05 \times 150) + 60 \times (-10) = 0$.
$n(7.5) = 600$.
$n = \frac{600}{7.5} = 80$.
Therefore,$80$ bullets must be fired to stop the stone.

(D) Given: Mass $m = 5 \ kg$,initial velocity $u = 3 \ m s^{-1}$,final velocity $v = 7 \ m s^{-1}$,time $t = 2 \ s$.
Using Newton's second law of motion,$F = m \times a = m \times \frac{(v - u)}{t}$.
Substituting the values: $F = \frac{5 \times (7 - 3)}{2} = \frac{5 \times 4}{2} = 10 \ N$.
Now,for the second part,the force $F = 10 \ N$ is applied for $t' = 5 \ s$.
The acceleration $a = \frac{F}{m} = \frac{10}{5} = 2 \ m s^{-2}$.
Using the first equation of motion,$v' = u + a \times t'$.
$v' = 3 + (2 \times 5) = 3 + 10 = 13 \ m s^{-1}$.

(N/A) Given:
Mass of the ball,$m = 20 \ g = 0.02 \ kg$.
From the velocity-time graph:
Initial velocity,$u = 20 \ m \ s^{-1}$ (at $t = 0 \ s$).
Final velocity,$v = 0 \ m \ s^{-1}$ (at $t = 10 \ s$).
Time taken,$t = 10 \ s$.
Using Newton's second law of motion,the force $F$ exerted by the ground on the ball is given by:
$F = m \times a = m \times \frac{(v - u)}{t}$
Substituting the values:
$F = 0.02 \times \frac{(0 - 20)}{10}$
$F = 0.02 \times (-2) = -0.04 \ N$.
The negative sign indicates that the force exerted by the ground (frictional force) is in the direction opposite to the motion of the ball. Thus,the magnitude of the force is $0.04 \ N$.

(N/A) Given:
Mass of engine $M_{E} = 3 \times 10^{4} \text{ kg}$
Mass of each wagon $M_{w} = 2 \times 10^{4} \text{ kg}$
Acceleration $a = 0.2 \text{ m s}^{-2}$
$(i)$ The total mass being pulled by the engine is $M_{total} = M_{E} + 2M_{w} = 3 \times 10^{4} + 4 \times 10^{4} = 7 \times 10^{4} \text{ kg}$.
The force exerted by the engine is $F = M_{total} \times a = (7 \times 10^{4} \text{ kg}) \times (0.2 \text{ m s}^{-2}) = 1.4 \times 10^{4} \text{ N}$.
$(ii)$ The force experienced by each wagon is the force required to accelerate that specific wagon. For one wagon,$F_{w} = M_{w} \times a = (2 \times 10^{4} \text{ kg}) \times (0.2 \text{ m s}^{-2}) = 0.4 \times 10^{4} \text{ N} = 4000 \text{ N}$.

(N/A) Initial momentum is calculated as $p = m \times u = 0.5 \times 25 = 12.5\, kg\, m s^{-1}$.
$(b)$ First,find the maximum height $h$ using the equation $v^2 = u^2 - 2gh$. At maximum height,$v = 0$,so $0 = (25)^2 - 2 \times 10 \times h$,which gives $h = 625 / 20 = 31.25\, m$.
The halfway mark is $H = h / 2 = 31.25 / 2 = 15.625\, m$.
Now,find the velocity $v$ at this height using $v^2 = u^2 - 2gH$:
$v^2 = (25)^2 - 2 \times 10 \times 15.625 = 625 - 312.5 = 312.5$.
$v = \sqrt{312.5} \approx 17.68\, m s^{-1}$.
The momentum at the halfway mark is $p = m \times v = 0.5 \times 17.68 = 8.84\, kg\, m s^{-1}$.

(N/A) Given: Mass $m = 5 \, g = 5 \times 10^{-3} \, kg$,initial velocity $u = 120 \, m s^{-1}$,final velocity $v = 0$,time $t = 0.01 \, s$.
$(a)$ To find the distance of penetration $(S)$,we first find the acceleration $(a)$ using the first equation of motion: $v = u + at$.
$0 = 120 + a \times 0.01$
$a = -120 / 0.01 = -12000 \, m s^{-2}$.
(The negative sign indicates retardation).
Now,using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
$S = (120 \times 0.01) + \frac{1}{2} \times (-12000) \times (0.01)^2$
$S = 1.2 - 0.6 = 0.6 \, m$.
$(b)$ The average force $(F)$ exerted on the bullet is calculated using Newton's second law: $F = ma$.
$F = (5 \times 10^{-3} \, kg) \times (12000 \, m s^{-2})$
$F = 60 \, N$ (The force is retarding in nature).

(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.

(A) Given:
Mass of man $m_{1} = 60 \ kg$
Initial velocity of man $u_{1} = 18 \ km \ h^{-1} = 18 \times (5/18) \ m \ s^{-1} = 5 \ m \ s^{-1}$
Mass of car $m_{2} = 1 \ quintal = 100 \ kg$
Initial velocity of car $u_{2} = 0 \ m \ s^{-1}$
Let $V$ be the common velocity of the car and the man after the man jumps into the car.
Applying the law of conservation of momentum:
Total momentum after jump = Total momentum before jump
$(m_{1} + m_{2}) V = m_{1} u_{1} + m_{2} u_{2}$
$(60 + 100) V = (60 \times 5) + (100 \times 0)$
$160 V = 300$
$V = 300 / 160 = 1.875 \ m \ s^{-1}$
Thus, the car will start travelling with a velocity of $1.875 \ m \ s^{-1}$.

(N/A) Given: Initial velocity $u = 0 \ m/s$,Distance $S = 800 \ m$,Time $t = 10 \ s$,Mass $m = 800 \ kg$.
Using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
Substituting the values: $800 = (0 \times 10) + \frac{1}{2} \times a \times (10)^2$.
$800 = 0 + \frac{1}{2} \times a \times 100$.
$800 = 50a$.
$a = \frac{800}{50} = 16 \ m/s^2$.
Now,calculating the force using Newton's second law: $F = ma$.
$F = 800 \ kg \times 16 \ m/s^2 = 12800 \ N$.

(N/A) Given: Force $F = -50 \ N$ (retarding force),mass $m = 200 \ kg$,initial velocity $u = 15 \ m \ s^{-1}$,final velocity $v = 0 \ m \ s^{-1}$.
$1$. Calculate acceleration $(a)$:
Using Newton's second law,$F = ma$,so $a = F / m = -50 / 200 = -0.25 \ m \ s^{-2}$.
$2$. Calculate time $(t)$:
Using the first equation of motion,$v = u + at$,we get $0 = 15 + (-0.25)t$.
$0.25t = 15$,therefore $t = 15 / 0.25 = 60 \ s$.
$3$. Calculate distance $(S)$:
Using the second equation of motion,$S = ut + 1/2 at^2$.
$S = (15 \times 60) + 1/2 \times (-0.25) \times (60)^2$.
$S = 900 - 0.125 \times 3600 = 900 - 450 = 450 \ m$.

(A) Given: Mass $m = 80 \, kg$,initial velocity $u = 30 \, m s^{-1}$,final velocity $v = 0 \, m s^{-1}$,time $t = 5 \, s$.
Using the first equation of motion: $v = u + at$
$0 = 30 + a \times 5$
$5a = -30$
$a = -6 \, m s^{-2}$.
The negative acceleration (retardation) is $6 \, m s^{-2}$.
Using Newton's second law of motion,the frictional force $F = m \times a$ (magnitude):
$F = 80 \, kg \times 6 \, m s^{-2} = 480 \, N$.

(A) The statement is $True$.
$A$ force is defined as a push or a pull acting on an object. While a force can cause an object to accelerate,change its speed,or change its direction,it does not always result in motion.
For example,if you push against a heavy wall,you are applying a force,but the wall does not move because the frictional force or the structural resistance balances your applied force. Therefore,force does not always produce motion in a body.

(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.

(B) The statement is False.
Impulse is defined as the change in momentum of a body, not the rate of change of momentum.
Mathematically, Impulse $(J) = \Delta p = F \times \Delta t$, where $\Delta p$ is the change in momentum.
The rate of change of momentum $(\frac{\Delta p}{\Delta t})$ is equal to the force $(F)$ applied to the body, according to Newton's Second Law of Motion.

(FALSE) The statement is False.
According to Newton's second law of motion,$F = m \times a$.
$1 \ N$ is defined as the force required to produce an acceleration of $1 \ m \ s^{-2}$ in a body of mass $1 \ kg$,not $1 \ g$.
Since $1 \ kg = 1000 \ g$,the force required to accelerate $1 \ g$ $(0.001 \ kg)$ at $1 \ m \ s^{-2}$ is $0.001 \ N$.

(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.

(FALSE) The statement is False.
Friction is the resistance to motion between two surfaces in contact. In general,the force of friction (specifically fluid friction or drag) is significantly lower in liquids compared to the friction between two solid surfaces. Solids have higher intermolecular forces and surface irregularities that create greater resistance to motion compared to the flow characteristics of liquids.

(TRUE) The statement is true. Newton's first law of motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external unbalanced force. This concept was originally proposed by Galileo Galilei,which is why the first law of motion is frequently referred to as the law of inertia or Galileo's law of inertia.

(D) An athlete runs before taking a long jump to gain momentum.
According to Newton's first law of motion,an object in motion tends to stay in motion unless acted upon by an external force.
By running,the athlete acquires a high velocity,which increases their inertia of motion.
This inertia of motion helps the athlete cover a greater distance during the jump,as the body continues to move forward due to the momentum gained during the run-up.

(A) When a carpet is at rest,the dust particles on it are also at rest.
According to Newton's first law of motion,an object at rest tends to remain at rest unless acted upon by an external force (Inertia of rest).
When the carpet is beaten with a stick,the carpet suddenly moves,but the dust particles tend to remain in their state of rest due to inertia.
Consequently,the carpet moves away from the dust particles,causing the dust to fall off.

(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.

(C) Inertia is the inherent property of a body to resist any change in its state of rest,uniform motion,or direction of motion.
According to Newton's First Law of Motion,an object remains in its state of rest or uniform motion in a straight line unless compelled to change that state by an external unbalanced force.
Since inertia prevents a body from changing its state of rest,uniform motion,or direction of motion on its own,all the given statements are correct.
Therefore,the correct option is $C$.

(D) force is a vector quantity. $A$ vector quantity is defined as a physical quantity that has both magnitude (size) and direction. Therefore,to completely describe a force,one must specify both how strong the push or pull is (magnitude) and the direction in which it is applied.

(A) The $SI$ unit of force is the $newton$ $(N)$.
According to $Newton's$ second law of motion, $Force = mass \times acceleration$ $(F = m \times a)$.
The unit of mass is $kilogram$ $(kg)$ and the unit of acceleration is $metre$ per $second$ squared $(m/s^2)$.
Therefore, $1 \ N = 1 \ kg \cdot m/s^2$.

(B) Inertia is the property of an object by virtue of which it resists any change in its state of rest or motion.
Inertia is directly proportional to the mass of an object.
Greater the mass,greater is the inertia.
Comparing the masses of the given objects:
$1$. $A$ pin has very little mass.
$2$. $A$ pen has very little mass.
$3$. $A$ physics book has more mass than a pin or a pen.
$4$. $A$ school bag contains books,notebooks,and other items,making its total mass significantly higher than the other options.
Since the school bag has the largest mass,it possesses the largest inertia.

(C) According to Newton's $Second$ $Law$ of $Motion$, the rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force.
Mathematically, if $p$ is momentum, $m$ is mass, and $v$ is velocity, then $p = mv$.
The rate of change of momentum is given by $\frac{dp}{dt} = \frac{d(mv)}{dt} = m \frac{dv}{dt} = ma$.
Since $F = ma$, the rate of change of momentum is equal to the force applied.

(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 second law of motion,the force $F$ acting on a body is given by the product of its mass $m$ and acceleration $a$,expressed as $F = m \times a$.
If the mass $m$ of the body remains constant,the force $F$ is directly proportional to the acceleration $a$ $(F \propto a)$.
If the acceleration is doubled $(a' = 2a)$,then the new force $F'$ will be $F' = m \times (2a) = 2 \times (m \times a) = 2F$.
Therefore,the acting force must be increased by a factor of $2$.

(C) The momentum $(p)$ of a body is defined as the product of its mass $(m)$ and its velocity $(v)$.
Mathematically,it is expressed as $p = m \times v$.
Since the mass $(m)$ of the body is given as constant,the momentum $(p)$ is directly proportional to the velocity $(v)$ or speed of the body ($p \propto v$ when $m$ is constant).
Therefore,the momentum of a body of given mass is proportional to its speed.

(D) Newton's second law of motion states that the rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force.
Mathematically,$F = ma$,where $F$ is the force,$m$ is the mass,and $a$ is the acceleration.
Thus,the second law provides a quantitative measure of force.

(A) 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 $(a)$.
This is expressed by the formula: $F = m \times a$.
Rearranging this formula to solve for acceleration,we get: $a = F / m$.
Since the mass $(m)$ of the object is constant,the acceleration $(a)$ is directly proportional to the net force $(F)$ applied to the object $(a \propto F)$.

(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.

(D) $Newton's$ second law of motion states that the rate of change of momentum of an object is directly proportional to the applied unbalanced force in the direction of the force.
Mathematically,$F = ma$,where $F$ is the force,$m$ is the mass,and $a$ is the acceleration.
This law provides a quantitative measure of force,allowing us to calculate the magnitude of force required to produce a specific acceleration in an object of a given mass.

(A) Newton's first law of motion defines force qualitatively as an external agency that changes or tends to change the state of rest or uniform motion of an object.
Newton's second law of motion provides the quantitative definition of force,which is given by the product of mass and acceleration $(F = ma)$.

(B) According to Newton's $1^{st}$ Law of Motion,if the net external force acting on a body is zero (i.e.,balanced forces),the body will maintain its state of rest or uniform motion in a straight line.
Balanced forces do not change the state of motion of an object,meaning the acceleration of the body is zero.
Therefore,the body is either at rest or moving with a constant velocity.

(C) According to Newton's $First$ Law of Motion, 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.
If the net unbalanced force acting on a body is zero, the acceleration of the body is zero ($F = ma$, so if $F = 0$, then $a = 0$).
If the acceleration is zero, the body will either remain at rest (if it was initially at rest) or continue to move with a constant velocity (if it was already in motion).
Therefore, the statement that the body moves with constant velocity is a correct description of its state of motion under zero net force.