Mix Examples - FORCE AND LAWS OF MOTION Questions in English

(N/A) The velocity of the ball changes $4$ times.
$1$. The first player kicks the ball: The velocity changes from $0$ to a certain value. The agent is the first player.
$2$. The second player kicks the ball towards the goal: The direction changes, hence velocity changes. The agent is the second player.
$3$. The goalkeeper collects the ball: The ball comes to rest, so velocity changes from a certain value to $0$. The agent is the goalkeeper.
$4$. The goalkeeper kicks the ball to his teammate: The velocity changes from $0$ to a certain value. The agent is the goalkeeper.

$(A)$ Initial velocity of the truck,$u = 0 \, m/s$.
Time taken,$t = 20 \, s$.
Distance covered,$s = 400 \, m$.
According to the second equation of motion,$s = ut + \frac{1}{2}at^2$.
Substituting the values: $400 = 0 \times 20 + \frac{1}{2} \times a \times (20)^2$.
$400 = 200a$.
Therefore,acceleration $a = 2 \, m/s^2$.
Mass of the truck,$m = 7 \, \text{tonnes} = 7000 \, kg$.
Force is given by $F = m \times a$.
$F = 7000 \times 2 = 14000 \, N$.

(B) Initial velocity of the stone,$u = 20\, m/s$.
Final velocity of the stone,$v = 0\, m/s$.
Distance covered by the stone,$s = 50\, m$.
According to the third equation of motion,$v^2 = u^2 + 2as$.
Substituting the values: $(0)^2 = (20)^2 + 2 \times a \times 50$.
$0 = 400 + 100a$.
$100a = -400$.
$a = -4\, m/s^2$.
The negative sign indicates that acceleration is acting against the motion of the stone.
Mass of the stone,$m = 1\, kg$.
From Newton's second law of motion,Force $F = m \times a$.
$F = 1\, kg \times (-4\, m/s^2) = -4\, N$.
Hence,the force of friction between the stone and the ice is $-4\, N$.

(D) Mass of one object,$m_1 = 1.5\, kg$.
Mass of the other object,$m_2 = 1.5\, kg$.
Velocity of $m_1$ before collision,$v_1 = 2.5\, m/s$.
Velocity of $m_2$ before collision,$v_2 = -2.5\, m/s$ (since it moves in the opposite direction).
According to the law of conservation of momentum,the total momentum before collision equals the total momentum after collision.
$m_1v_1 + m_2v_2 = (m_1 + m_2)v$
Substituting the values:
$1.5(2.5) + 1.5(-2.5) = (1.5 + 1.5)v$
$3.75 - 3.75 = 3v$
$0 = 3v$
$v = 0\, m/s$.
Thus,the velocity of the combined object after the collision is $0\, m/s$.

(D) Mass of the object,$m_1 = 1\, kg$.
Velocity of the object before collision,$v_1 = 10\, m/s$.
Mass of the stationary wooden block,$m_2 = 5\, kg$.
Velocity of the wooden block before collision,$v_2 = 0\, m/s$.
Total momentum before collision $= m_1 v_1 + m_2 v_2 = 1(10) + 5(0) = 10\, kg\, m\, s^{-1}$.
According to the law of conservation of momentum,the total momentum remains conserved.
Therefore,total momentum after collision $= 10\, kg\, m\, s^{-1}$.
Total mass of the combined system $= m_1 + m_2 = 1 + 5 = 6\, kg$.
Let the velocity of the combined object be $v$.
$10 = (m_1 + m_2)v$
$10 = 6v$
$v = 10/6 = 5/3\, m/s$.

(C) According to the law of conservation of momentum, the total momentum of the system remains conserved during the collision.
$1$. Kiran's suggestion: The insect experiences a much larger change in velocity compared to the car. Since $p = mv$, the change in momentum $(\Delta p = m \Delta v)$ for the insect is significant relative to its mass. Because the car's mass is massive, its change in velocity is negligible, meaning the insect experiences a greater change in momentum relative to its own state, but the magnitude of momentum change for both is equal.
$2$. Akhtar's suggestion: Akhtar is partially correct that the car exerts a large force, but the force exerted by the car on the insect is equal to the force exerted by the insect on the car (Newton's $3^{rd}$ Law).
$3$. Rahul's suggestion: Rahul is correct. According to Newton's $3^{rd}$ Law, the force exerted by the car on the insect is equal and opposite to the force exerted by the insect on the car. Since the time of impact is the same for both, the change in momentum $(F \times t = \Delta p)$ is equal in magnitude for both the car and the insect.

(N/A) The distance covered by the object in equal intervals of time is unequal (it follows the cube of time, $d = t^3$). This indicates non-uniform motion. Since the velocity (rate of change of distance) is increasing at an increasing rate, the acceleration is increasing.
$(b)$ According to Newton's second law of motion, $F = ma$. Since the acceleration $(a)$ of the object is increasing over time, the net force $(F)$ acting on the object must also be increasing.

(A) Mass of the motorcar $(m)$ $= 1200 \, kg$.
When two persons push the car at a uniform velocity,the net external force is zero. This means the force applied by the two persons is exactly balanced by the frictional force $(f)$.
So,$2F = f$,where $F$ is the force applied by each person.
When three persons push the car,the total applied force is $3F$. The net force $(F_{net})$ causing acceleration is the total applied force minus the frictional force:
$F_{net} = 3F - f = 3F - 2F = F$.
According to Newton's second law of motion,$F_{net} = m \times a$.
Given $a = 0.2 \, m \, s^{-2}$,we have:
$F = 1200 \, kg \times 0.2 \, m \, s^{-2} = 240 \, N$.
Therefore,each person pushes the motorcar with a force of $240 \, N$.

(C) Mass of the motor car,$m = 1200\, kg$.
Initial velocity,$u = 90\, km/h = 90 \times (5/18) = 25\, m/s$.
Final velocity,$v = 18\, km/h = 18 \times (5/18) = 5\, m/s$.
Time taken,$t = 4\, s$.
Acceleration $a = (v - u) / t = (5 - 25) / 4 = -20 / 4 = -5\, m/s^2$.
Change in momentum = $m(v - u) = 1200 \times (5 - 25) = 1200 \times (-20) = -24000\, kg\, m/s$.
Force $F = m \times a = 1200 \times (-5) = -6000\, N$.
The magnitude of the force required is $6000\, N$.

(N/A) The Law of Conservation of Momentum is applicable only to an isolated system where no external force acts on the system.
In this case,the ball is not an isolated system because it is constantly acted upon by the external gravitational force of the Earth.
This external force causes a change in the velocity and,consequently,the momentum of the ball.
Therefore,since an external force is present,the momentum of the ball alone is not conserved.

(N/A) Given:
Mass of the ball,$m = 50 \, g = 0.05 \, kg$
Initial velocity,$u = 80 \, m \, s^{-1}$ (at $t = 0 \, s$)
Final velocity,$v = 0 \, m \, s^{-1}$ (at $t = 8 \, s$)
Time taken,$t = 8 \, s$
$1$. Calculation of acceleration $(a)$:
$a = \frac{v - u}{t} = \frac{0 - 80 \, m \, s^{-1}}{8 \, s} = -10 \, m \, s^{-2}$
The negative sign indicates retardation (deceleration).
$2$. Calculation of frictional force $(F)$:
$F = m \times a$
$F = 0.05 \, kg \times (-10 \, m \, s^{-2}) = -0.5 \, N$
The negative sign indicates that the frictional force acts in the direction opposite to the motion of the ball.

(A) The $SI$ unit of force is the Newton $(N)$,which is equivalent to $kg\, m s^{-2}$ according to Newton's second law $(F = ma)$.
The $SI$ unit of momentum $(p)$ is the product of mass and velocity $(p = mv)$,which is $kg\, m s^{-1}$.
Therefore,the units for force and momentum are $kg\, m s^{-2}$ and $kg\, m s^{-1}$ respectively.

(N/A) $(i)$ Lifting a bucket; lifting a stone.
$(ii)$ Pushing open a door; pushing a luggage trolley.
$(iii)$ Pulling of a train by an engine; pulling of a cart by a horse.
$(iv)$ Stretching a rubber balloon before inflating; stretching a spring.
$(v)$ Compressing the air with a piston; a doctor compressing the plunger of a syringe while injecting.
$(vi)$ Friction resists motion; air resistance opposes the motion of falling bodies.
$(vii)$ $A$ magnet attracting a piece of iron; a stone falling towards the ground due to gravity.
$(viii)$ Squeezing out the toothpaste from a tube; squeezing wet clothes.

(B) When we stop pedalling,the external force that was maintaining the motion is removed. The bicycle begins to slow down because of the frictional force acting between the tires and the road,as well as air resistance,which opposes the motion of the bicycle.

(N/A) The statement that the force of friction always acts opposite to the direction of motion is not entirely correct. While friction generally opposes the relative motion between two surfaces,it can sometimes act in the direction of motion.
Example: Consider a person walking. When a person walks,they push the ground backward with their foot. According to Newton's third law,the ground exerts an equal and opposite force on the foot in the forward direction. This forward force is the static friction acting on the foot,which enables the person to move forward. In this case,the force of friction acts in the direction of motion.

(N/A) $(i)$ Frictional force acts in the direction opposite to the motion of the object.
$(ii)$ Gravitational force acts vertically downwards towards the center of the Earth.
$(iii)$ Centripetal force acts towards the center of the circular path along which the object moves.

(N/A) Momentum is defined as the product of the mass of a body and its velocity. It is also known as the quantity of motion contained in a body. Its $SI$ unit is $kg \cdot m \cdot s^{-1}$.
The principle of conservation of momentum states that if no external force acts on a system, the total momentum of the system remains conserved (constant).

(N/A) This is based on Newton's $\text{Second}$ Law of Motion $(F = ma)$. An empty box has less mass, and therefore less inertia, compared to a box full of books. Since force is directly proportional to mass, less force is required to accelerate the empty box.
$(b)$ This is based on Newton's $\text{Third}$ Law of Motion (Action and Reaction). When water is ejected from the hose with high velocity in the forward direction, it exerts an equal and opposite reaction force on the hose in the backward direction. This makes it difficult for the fireman to hold the hose steady.

(N/A) The force causes a change in the shape of the balloon.
$(b)$ The force causes a change in the direction and velocity of the ball.
$(c)$ The force causes a change in the shape of the spring.

(N/A) Accelerating unbalanced force: The velocity-time graph shows a constant positive slope,indicating uniform acceleration. According to Newton's second law,$F = ma$,an unbalanced force is required to produce acceleration.
$(b)$ No net force: The velocity-time graph is a horizontal line,indicating constant velocity (zero acceleration). According to Newton's first law,if the net external force is zero,an object will continue to move with a constant velocity.
$(c)$ Retarding unbalanced force: The velocity-time graph shows a constant negative slope,indicating uniform retardation (deceleration). An unbalanced force acting in the direction opposite to the motion is required to decrease the velocity.

(N/A) $(i)$ Both vehicles experience the same magnitude of force. According to Newton's third law of motion,for every action,there is an equal and opposite reaction. Thus,the force exerted by the truck on the car is equal to the force exerted by the car on the truck.
$(ii)$ The truck experiences a greater change in momentum. Momentum is defined as $p = mv$. Since the truck has a much larger mass $(M)$ compared to the car $(m)$,and both come to a halt (final velocity $= 0$),the change in momentum $\Delta p = |0 - mv| = mv$ is greater for the truck.
$(iii)$ The car experiences greater acceleration. According to Newton's second law,$F = ma$,which implies $a = F/m$. Since the force $(F)$ is the same for both,the vehicle with the smaller mass $(m)$ will experience a larger acceleration $(a)$.

(A) Mass of both cars,$m_1 = m_2 = 1000 \ kg$.
Initial velocity of the first car,$u_1 = 5 \ m s^{-1}$.
Initial velocity of the second car,$u_2 = -5 \ m s^{-1}$ (since they are moving in opposite directions).
Let $v$ be the common velocity of the combined cars after the collision.
According to the law of conservation of momentum,the total momentum before the collision is equal to the total momentum after the collision.
$m_1 u_1 + m_2 u_2 = (m_1 + m_2) v$
Substituting the values:
$1000 \times 5 + 1000 \times (-5) = (1000 + 1000) v$
$5000 - 5000 = 2000 v$
$0 = 2000 v$
$v = 0 \ m s^{-1}$.
Therefore,the combined cars will come to rest after the collision.

(N/A) $(i)$ If a person jumps out of a moving vehicle,their feet will suddenly come to rest upon touching the ground,while the rest of the body remains in motion due to inertia. This causes the person to fall forward and potentially get hurt. This danger can be minimised by running along with the moving vehicle in the same direction for some distance after jumping out,which allows the body to come to rest gradually.
$(ii)$ When the boatman pushes the water backward with the oars,the water exerts an equal and opposite force on the boat in the forward direction,according to Newton's third law of motion. This reaction force helps the boat move forward.

(N/A) $(i)$ It is difficult to walk on marshy land because when we push the ground with our feet,the ground is soft and yields,failing to provide an equal and opposite reaction force as per Newton's third law.
$(ii)$ When an air-filled balloon is punctured from below,the escaping air exerts a downward force. According to Newton's third law,the air exerts an equal and opposite upward reaction force on the balloon,causing it to rise slightly.
$(iii)$ According to Newton's third law of motion,for every action,there is an equal and opposite reaction. When a swimmer pushes the water backward (action),the water exerts an equal and opposite force on the swimmer in the forward direction (reaction),which helps the swimmer move forward.

(N/A) $(i)$ Both vehicles experience the same force of impact because action and reaction are equal and opposite (Newton's Third Law).
$(ii)$ The truck experiences a greater change in momentum because momentum change $\Delta p = m \Delta v$. Since the truck has a much larger mass $(M > m)$,the change in momentum is greater for the truck.
$(iii)$ The car experiences greater acceleration. Since $F = ma$,then $a = F/m$. For the same force $F$,the vehicle with the smaller mass (the car) experiences higher acceleration.
$(iv)$ The car suffers more damage because it experiences a much higher magnitude of deceleration (acceleration) during the impact,leading to greater structural stress.

(D) According to Newton's third law of motion, for every action, there is an equal and opposite reaction. Therefore, the force exerted by the bus on the bug is equal in magnitude to the force exerted by the bug on the bus.
Since the force $(F)$ and the time of impact $(t)$ are the same for both, the change in momentum $(\Delta p = F \times t)$ is also the same for both the bug and the bus.
However, because the mass of the bus is much larger than the mass of the bug, the change in velocity $(\Delta v = \Delta p / m)$ for the bus is negligible, whereas the change in velocity for the bug is very large.
Consequently, the bug experiences a much higher acceleration, leading to its death. Rita is correct in stating that both experience the same force and the same change in momentum.

(N/A) Momentum is defined as the product of the mass $(m)$ and velocity $(v)$ of a body. It is given by the formula $p = m \times v$.
$(b)$ At the highest point of its trajectory,the vertical velocity $(v)$ of the ball becomes $0 \ m/s$. Since momentum $p = m \times v$,substituting $v = 0$ gives $p = m \times 0 = 0$. Thus,the momentum at the highest point is zero.
$(c)$ The law of conservation of momentum states that in the absence of an external unbalanced force,the total momentum of a system of objects remains conserved (constant).

(B) If the roads go straight up,the angle of slope (i.e.,$\theta$) would be very large.
According to the laws of physics,the force required to move a vehicle up an incline is proportional to $\sin(\theta)$.
By winding the road gradually,the angle of inclination $\theta$ is significantly reduced.
This reduction in $\theta$ decreases the force required to climb the mountain,making it easier for the engine to pull the vehicle up without overheating or slipping.

(N/A) Anil explained that as the speed of a vehicle increases,the braking distance also increases. Specifically,if the speed is doubled,the braking distance becomes four times. In an emergency,if the vehicle in front were to apply brakes suddenly,Anil's father would not have been able to stop his car in the available distance,even after applying the brakes. This would have likely resulted in a life-threatening accident.
$(b)$ The values demonstrated by Anil include concern for his family's safety,responsibility,awareness of traffic safety rules,and the ability to apply scientific knowledge to real-life situations.

(A) Dynamics is the branch of mechanics that deals with the motion of objects and the forces that cause this motion.
Kinematics is the branch of mechanics that describes the motion of objects without considering the forces that cause the motion.
Statics is the branch of mechanics that deals with objects at rest or in equilibrium under the action of forces.
Therefore,the correct branch that considers the cause of motion is dynamics.

(N/A) Given: Mass $m = 0.5 \, kg$,initial velocity $u = 25 \, m s^{-1}$,acceleration due to gravity $g = -10 \, m s^{-2}$.
$(a)$ Initial momentum $p = m \times u = 0.5 \times 25 = 12.5 \, kg \, m s^{-1}$.
$(b)$ To find the momentum at half the maximum height,first calculate the maximum height $H$ using $v^2 - u^2 = 2gH$. At maximum height,final velocity $v = 0$.
$0^2 - (25)^2 = 2 \times (-10) \times H$
$-625 = -20 \times H$
$H = 625 / 20 = 31.25 \, m$.
Halfway height $h = H / 2 = 31.25 / 2 = 15.625 \, m$.
Now,find the velocity $v_h$ at height $h$ using $v_h^2 - u^2 = 2gh$:
$v_h^2 - (25)^2 = 2 \times (-10) \times 15.625$
$v_h^2 - 625 = -312.5$
$v_h^2 = 312.5$
$v_h = \sqrt{312.5} \approx 17.68 \, m s^{-1}$.
Momentum at halfway mark $p_h = m \times v_h = 0.5 \times 17.68 = 8.84 \, kg \, m s^{-1}$.

(N/A) $(i)$ The net force $F_{net}$ is the difference between the two opposing forces:
$F_{net} = F_{2} - F_{1} = 30 \, N - 20 \, N = 10 \, N$.
$(ii)$ Since $F_{2} > F_{1}$,the net force acts in the direction of the larger force,which is the direction of $F_{2}$ (towards the left).
$(iii)$ According to Newton's second law of motion,$F = ma$,where $m = 10 \, kg$ and $F = 10 \, N$:
$a = F / m = 10 \, N / 10 \, kg = 1 \, m \, s^{-2}$.

(A) $(i)$ Inertia of a body is directly proportional to its mass. Therefore,the ratio of inertia of ball $A$ to ball $B$ is $m : 2m = 1 : 2$.
$(ii)$ Momentum $p$ is defined as the product of mass and velocity $(p = mv)$. For ball $A$,$p_A = m \times 2V = 2mV$. For ball $B$,$p_B = 2m \times V = 2mV$. Thus,the ratio of their momentum is $2mV : 2mV = 1 : 1$.
$(iii)$ According to Newton's second law of motion,the force $F$ required to stop a body in a given time $t$ is $F = \frac{\Delta p}{t}$. Since both balls have the same initial momentum and are brought to rest (final momentum $= 0$) in the same time $t$,the change in momentum $\Delta p$ is the same for both. Therefore,the ratio of the force needed to stop them is $1 : 1$.

(D) The law of conservation of momentum states that in the absence of an external unbalanced force,the total linear momentum of a system remains conserved (constant).
$(b)$ Given:
Mass of the first body,$m_{1} = 2 \, kg$
Initial velocity of the first body,$u_{1} = 10 \, m s^{-1}$
Mass of the second body,$m_{2} = 5 \, kg$
Initial velocity of the second body,$u_{2} = 0 \, m s^{-1}$ (at rest)
Final velocity of the first body,$v_{1} = 1 \, m s^{-1}$
Let the final velocity of the second body be $v_{2}$.
According to the law of conservation of momentum:
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
Substituting the values:
$(2 \times 10) + (5 \times 0) = (2 \times 1) + (5 \times v_{2})$
$20 + 0 = 2 + 5v_{2}$
$20 - 2 = 5v_{2}$
$18 = 5v_{2}$
$v_{2} = \frac{18}{5} = 3.6 \, m s^{-1}$
Thus,the velocity of the second body after the collision is $3.6 \, m s^{-1}$.

(N/A) Cars are provided with seat belts to prevent injuries in case of an accident. The seat belt increases the time taken by the passenger to fall forward,thereby reducing the force exerted on them.
$(b)$ When a person jumps out of a moving bus,their body is in a state of motion due to inertia. Upon hitting the ground,the lower part of the body comes to rest suddenly due to friction,while the upper part continues to move forward,which can cause the person to fall and get injured.
$(c)$ This is because the momentum $(p = mv)$ of a vehicle is directly proportional to its velocity. At high speeds,the momentum is significantly higher,leading to a greater impulsive force during a collision,which causes more severe damage and injuries.

$(a-d)$ According to Newton's third law, the force exerted by the truck on the bus is equal in magnitude to the force exerted by the bus on the truck. Therefore, both vehicles experience the same force of impact.
$(b)$ Change in momentum $(\Delta p)$ is given by $m(v - u)$. Since the truck has a larger mass $(M > m)$, the magnitude of the change in momentum $|M(0 - V)|$ is greater for the truck than for the mini bus $|m(0 - V)|$.
$(c)$ Acceleration is given by $a = F/m$. Since both experience the same force $(F)$ but the mini bus has a smaller mass $(m)$, the mini bus experiences greater acceleration.

(N/A) This is because the wheel rotates continuously,but the mud sticking to the wheel tends to move in a straight line due to the inertia of direction. Consequently,it flies off tangentially.
$(b)$ By lowering his hands,the cricketer increases the time taken to catch the ball. This decreases the rate of change of momentum,thereby reducing the impulsive force exerted on his hands.
$(c)$ The high velocity of the pebble is reduced to zero in a very short interval of time. According to Newton's second law,this results in a very large force being exerted on the glass pane,causing it to shatter.

(N/A) 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, let an object of mass $m$ have an initial velocity $u$ and a final velocity $v$ after time $t$ under the influence of force $F$.
Initial momentum $p_1 = mu$
Final momentum $p_2 = mv$
Change in momentum = $p_2 - p_1 = m(v - u)$
Rate of change of momentum = $\frac{m(v - u)}{t}$
According to the law, $F \propto \frac{m(v - u)}{t}$
$F = k \cdot m \cdot \frac{(v - u)}{t}$
Since $\frac{(v - u)}{t} = a$ (acceleration), we have $F = kma$.
Taking $k = 1$, we get $F = ma$.
The $SI$ unit of force is the newton $(N)$, where $1 \, N = 1 \, kg \cdot m \cdot s^{-2}$.
$(b)$ Observation: The card flies away, and the coin falls into the glass.
Reason: This happens due to the inertia of rest. The coin is initially at rest and tends to remain at rest due to its inertia. When the card is flicked, it moves away, but the coin, due to its inertia, does not move with the card and falls into the glass under the influence of gravity.

(N/A) The law of conservation of momentum states that in the absence of an external unbalanced force, the total momentum of a system of bodies remains conserved.
The two factors that determine the momentum of a body are:
$(i)$ Mass of the body $(m)$
$(ii)$ Velocity of the body $(v)$
Given:
Mass $(m)$ = $0.4 \ kg$
Initial velocity $(u)$ = $10 \ m s^{-1}$
Initial momentum $(p_i)$ = $m \times u = 0.4 \ kg \times 10 \ m s^{-1} = 4 \ kg \ m s^{-1}$.
At the highest point of its reach, the final velocity $(v)$ of the ball becomes $0 \ m s^{-1}$.
Therefore, momentum at the highest point $(p_f)$ = $m \times v = 0.4 \ kg \times 0 \ m s^{-1} = 0 \ kg \ m s^{-1}$.
Change in momentum $(\Delta p)$ = $p_f - p_i = 0 - 4 = -4 \ kg \ m s^{-1}$.
Time taken $(t)$ = $1 \ s$.
Force $(F)$ = $\frac{\Delta p}{t} = \frac{-4 \ kg \ m s^{-1}}{1 \ s} = -4 \ N$.
The negative sign indicates that the force is acting in the direction opposite to the motion (retarding force).
Thus, the negatively accelerating force (retarding force) acting on the ball is $4 \ N$.

(C) Momentum is defined as the product of the mass and velocity of an object. Its $SI$ unit is $kg\, m s^{-1}$.
$(b)$ Force is given by $F = m \times a$,where $a = (v - u) / t$.
$(i)$ For interval $O$ to $A$: $u = 0\, m s^{-1}$,$v = 40\, m s^{-1}$,$t = 2\, s$. Acceleration $a = (40 - 0) / 2 = 20\, m s^{-2}$. Force $F = 5\, kg \times 20\, m s^{-2} = 100\, N$.
$(ii)$ For interval $B$ to $C$: $u = 40\, m s^{-1}$,$v = 0\, m s^{-1}$,$t = (10 - 6) = 4\, s$. Acceleration $a = (0 - 40) / 4 = -10\, m s^{-2}$. Force $F = 5\, kg \times (-10\, m s^{-2}) = -50\, N$ (The negative sign indicates a retarding force).
$(c)$ Force required for $2\, kg$ mass: $F_1 = 2\, kg \times 5\, m s^{-2} = 10\, N$.
Force required for $4\, kg$ mass: $F_2 = 4\, kg \times 2\, m s^{-2} = 8\, N$.
Since $10\, N > 8\, N$,accelerating a $2\, kg$ mass at $5\, m s^{-2}$ requires a greater force.

(N/A) $(i)$ Momentum is defined as the product of the mass and velocity of a body. Its $SI$ unit is $kg \ m/s$.
$(ii)$ Given: Mass $(m) = 50 \ kg$,Initial velocity $(u) = 4 \ m/s$,Final velocity $(v) = 8 \ m/s$,Time $(t) = 8 \ s$.
Using the first equation of motion,$v = u + at$,we find acceleration $(a)$:
$a = (v - u) / t = (8 - 4) / 8 = 4 / 8 = 0.5 \ m/s^2$.
Using Newton's second law of motion,$F = ma$:
$F = 50 \ kg \times 0.5 \ m/s^2 = 25 \ N$.
$(b)$ Newton's first law of motion states that an object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied external force.

$(100 \ kg \ m \ s^{-1})$ $(i)$ The product of mass and velocity of an object is known as its momentum. The $SI$ unit of momentum is $kg \ m \ s^{-1}$.
$(ii)$ Mass of the object, $m = 10 \ kg$.
Height, $h = 5 \ m$.
Acceleration due to gravity, $g = 10 \ m \ s^{-2}$.
Initial velocity of the object, $u = 0$.
From the third equation of motion, $v^2 - u^2 = 2gh$.
$v^2 - 0^2 = 2 \times 10 \times 5$.
$v^2 = 100$.
$v = 10 \ m \ s^{-1}$.
Momentum transferred to the floor = Change in momentum = $mv - mu$.
$= (10 \times 10) - (10 \times 0) = 100 \ kg \ m \ s^{-1}$.
$(iii)$ A karate player moves his hand at a very high speed to strike the tiles. By doing this, the large momentum of his hand is reduced to zero in a very short interval of time. According to Newton's second law, $F = \frac{\Delta p}{\Delta t}$, since the time interval $\Delta t$ is extremely small, the force $F$ exerted on the tiles becomes very large, which is sufficient to break them.

(N/A) The law of conservation of momentum states that if no external force acts on a system of particles,the total linear momentum of the system remains conserved.
Consider two particles $A$ and $B$ with masses $m_{1}$ and $m_{2}$ moving in a straight line. Let their initial velocities be $u_{1}$ and $u_{2}$ respectively,where $u_{1} > u_{2}$. They collide for a time $t$. After the collision,their final velocities become $v_{1}$ and $v_{2}$ respectively.
Initial momentum of particle $A = m_{1}u_{1}$
Initial momentum of particle $B = m_{2}u_{2}$
Final momentum of particle $A = m_{1}v_{1}$
Final momentum of particle $B = m_{2}v_{2}$
Rate of change of momentum of particle $A$ (Force exerted by $B$ on $A$,$F_{AB}$) = $\frac{m_{1}(v_{1} - u_{1})}{t}$
Rate of change of momentum of particle $B$ (Force exerted by $A$ on $B$,$F_{BA}$) = $\frac{m_{2}(v_{2} - u_{2})}{t}$
According to Newton's third law of motion,the force exerted by $A$ on $B$ is equal and opposite to the force exerted by $B$ on $A$:
$F_{AB} = -F_{BA}$
Substituting the values:
$\frac{m_{1}(v_{1} - u_{1})}{t} = -\frac{m_{2}(v_{2} - u_{2})}{t}$
Multiplying both sides by $t$:
$m_{1}v_{1} - m_{1}u_{1} = -(m_{2}v_{2} - m_{2}u_{2})$
$m_{1}v_{1} - m_{1}u_{1} = -m_{2}v_{2} + m_{2}u_{2}$
Rearranging the terms:
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
This shows that the total initial momentum of the system is equal to the total final momentum of the system,which proves the law of conservation of momentum.

(D) First Law: It states, "Everybody in the universe stays in a state of rest or of uniform motion along a straight line unless compelled by an external force to change its state."
Third Law: It states, "To every action, there is an equal (in magnitude) and opposite (in direction) reaction." In other words, action and reaction are equal and opposite.
$(b)$ Given: Mass $m = 1800 \, kg$, initial velocity $u = 10 \, m s^{-1}$, final velocity $v = 0$, and distance $S = 50 \, m$.
Using the third equation of motion: $v^2 - u^2 = 2aS$
$0^2 - (10)^2 = 2 \times a \times 50$
$-100 = 100a$
$a = -1 \, m s^{-2}$ (The negative sign indicates retardation).
Now, calculating the force using Newton's second law: $F = ma$
$F = 1800 \times (-1)$
$F = -1800 \, N$
The force acting on the car is $1800 \, N$ in the direction opposite to the motion.

(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,if $F$ is the force,$m$ is mass,$v$ is final velocity,$u$ is initial velocity,and $t$ is time:
$F \propto \frac{m(v-u)}{t}$
Since $\frac{v-u}{t} = a$ (acceleration),we get $F = kma$. By taking the constant $k = 1$,we derive the formula to measure force: $F = ma$.
$(b)$ Given:
Mass of rifle $(M)$ = $4 \ kg$
Mass of bullet $(m)$ = $50 \ g = 0.05 \ kg$
Velocity of bullet $(v)$ = $40 \ m \ s^{-1}$
Initial velocity of system = $0$
According to the law of conservation of momentum:
$MV + mv = 0$
$4 \times V + 0.05 \times 40 = 0$
$4V + 2 = 0$
$4V = -2$
$V = -0.5 \ m \ s^{-1}$
The negative sign indicates that the rifle recoils in the direction opposite to the bullet.

$(c)$ According to Newton's second law of motion, the rate of change of momentum of an object is proportional to the applied unbalanced force in the direction of the force.
Let a body of mass $m$ have an initial velocity $u$ and a final velocity $v$ after time $t$ under the influence of force $F$.
Initial momentum $p_1 = mu$.
Final momentum $p_2 = mv$.
Change in momentum = $p_2 - p_1 = m(v - u)$.
Rate of change of momentum = $\frac{m(v - u)}{t}$.
According to the law, $F \propto \frac{m(v - u)}{t}$.
Since acceleration $a = \frac{v - u}{t}$, we get $F \propto ma$, or $F = kma$. By defining the unit of force such that $k = 1$, we get $F = ma$.
$(b)$ One newton is defined as the force required to produce an acceleration of $1 \, m s^{-2}$ in an object of mass $1 \, kg$.
$(c)$ For the first case: $F_1 = m_1 \times a_1 = 0.5 \, kg \times 5 \, m s^{-2} = 2.5 \, N$.
For the second case: $F_2 = m_2 \times a_2 = 4 \, kg \times 2 \, m s^{-2} = 8 \, N$.
Since $8 \, N$ > $2.5 \, N$, the $4 \, kg$ mass requires a greater force.

(N/A) The law states that if no external force acts on a system of interacting bodies,the total momentum of the system remains constant.
$(b)$ The two factors that determine the momentum $(p)$ of a body are:
$1$. Mass $(m)$ of the body.
$2$. Velocity $(v)$ of the body.
$(c)$ Consider two objects $A$ and $B$ with masses $m_{1}$ and $m_{2}$ moving in the same direction with initial velocities $u_{1}$ and $u_{2}$ respectively,where $u_{1} > u_{2}$. They collide for a time $t$. During collision,$A$ exerts a force $F_{AB}$ on $B$,and $B$ exerts a force $F_{BA}$ on $A$.
Initial momentum of $A = m_{1}u_{1}$
Initial momentum of $B = m_{2}u_{2}$
Final momentum of $A = m_{1}v_{1}$
Final momentum of $B = m_{2}v_{2}$
According to Newton's second law,the force exerted by $A$ on $B$ is $F_{AB} = m_{2} \frac{(v_{2} - u_{2})}{t}$.
The force exerted by $B$ on $A$ is $F_{BA} = m_{1} \frac{(v_{1} - u_{1})}{t}$.
According to Newton's third law,$F_{AB} = -F_{BA}$.
$m_{2} \frac{(v_{2} - u_{2})}{t} = -m_{1} \frac{(v_{1} - u_{1})}{t}$
$m_{2}(v_{2} - u_{2}) = -m_{1}(v_{1} - u_{1})$
$m_{2}v_{2} - m_{2}u_{2} = -m_{1}v_{1} + m_{1}u_{1}$
$m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$
This proves that the total initial momentum equals the total final momentum.

(N/A) Given: Mass $m = 10 \ kg$,initial velocity $u = 0 \ m s^{-1}$,acceleration $a = g = 10 \ m s^{-2}$,and displacement $S = 0.8 \ m$.
Using the equation of motion $v^2 - u^2 = 2aS$:
$v^2 - 0^2 = 2 \times 10 \times 0.8$
$v^2 = 16$
$v = 4 \ m s^{-1}$
Now,momentum $p = m \times v = 10 \ kg \times 4 \ m s^{-1} = 40 \ kg \ m s^{-1}$.
$(b)$ According to Newton's third law of motion,for every action,there is an equal and opposite reaction. When water is ejected from the hose at a high velocity,it exerts a forward force (action). The hose,in turn,exerts an equal and opposite backward force (reaction) on the fireman. To counteract this reaction force,the fireman must hold the hose with a significant amount of force,making it difficult to control.

(C) This happens due to the inertia of direction. According to Newton's first law,an object continues to move in its original direction unless an external force acts on it. When the car turns,our body tends to maintain its original straight-line path,causing us to lean to the side.
$(b)$ Newton's first law of motion states that an object remains in a state of rest or uniform motion in a straight line unless compelled to change that state by an applied external force. Newton's third law of motion states that to every action,there is always an equal and opposite reaction.
$(c)$ Given: Force $F = 5 \ N$,acceleration $a = 10 \ m \ s^{-2}$.
Using Newton's second law,$F = m \times a$,so $m = F / a$.
$m = 5 \ N / 10 \ m \ s^{-2} = 0.5 \ kg$.
Since $1 \ kg = 1000 \ g$,the mass is $0.5 \times 1000 = 500 \ g$.

(N/A) According to Newton's second law, $F = ma$, which implies $a = F/m$. If the force $F$ remains constant and the mass $m$ is doubled $(2m)$, the new acceleration $a' = F/(2m) = a/2$. Thus, the acceleration becomes half.
$(b)$ Momentum is defined as the product of mass and velocity $(p = mv)$. Since the tennis ball has a smaller mass than the cricket ball, its momentum is lower when moving at the same speed. Therefore, it requires less force to stop it, making it easier to stop.
$(c)$ According to the law of conservation of momentum, since there is no external unbalanced force, the total momentum before the interaction equals the total momentum after the interaction.
Initial momentum = $(m_{girl} \times v_{girl}) + (m_{cart} \times v_{cart}) = (40 \times 5) + (10 \times 0) = 200 \, kg \, m s^{-1}$.
Final momentum = $(m_{girl} + m_{cart}) \times v_{final} = (40 + 10) \times v = 50v$.
Equating the two: $50v = 200$.
$v = 200 / 50 = 4 \, m s^{-1}$.