Second Law of Motion Questions in English

(D) Mass of the first object,$m_1 = 100 \, g = 0.1 \, kg$.
Mass of the second object,$m_2 = 200 \, g = 0.2 \, kg$.
Velocity of $m_1$ before collision,$u_1 = 2 \, m/s$.
Velocity of $m_2$ before collision,$u_2 = 1 \, m/s$.
Velocity of $m_1$ after collision,$v_1 = 1.67 \, m/s$.
Let the velocity of $m_2$ after collision be $v_2$.
According to the law of conservation of momentum:
Total momentum before collision = Total momentum after collision
$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$
$(0.1 \times 2) + (0.2 \times 1) = (0.1 \times 1.67) + (0.2 \times v_2)$
$0.2 + 0.2 = 0.167 + 0.2 v_2$
$0.4 = 0.167 + 0.2 v_2$
$0.2 v_2 = 0.4 - 0.167$
$0.2 v_2 = 0.233$
$v_2 = 0.233 / 0.2 = 1.165 \, m/s$.
Thus,the velocity of the second object after the collision is $1.165 \, m/s$.

(D) Total mass of the train $(M)$ = Mass of engine + Mass of $5$ wagons.
Mass of engine = $8000 \,kg$.
Mass of $5$ wagons = $5 \times 2000 \,kg = 10000 \,kg$.
Total mass $(M)$ = $8000 + 10000 = 18000 \,kg$.
Net force $(F_{net})$ = Applied force - Frictional force.
$F_{net} = 40000 \,N - 5000 \,N = 35000 \,N$.
Using Newton's second law of motion,$F_{net} = M \times a$.
$a = \frac{F_{net}}{M} = \frac{35000}{18000} \,m/s^2$.
$a = 1.944 \,m/s^2$.

(A) Mass of the automobile vehicle,$m = 1500 \, kg$.
Acceleration of the automobile,$a = -1.7 \, m \, s^{-2}$.
From Newton's second law of motion,the force $F$ is given by the product of mass and acceleration:
$F = m \times a$
$F = 1500 \, kg \times (-1.7 \, m \, s^{-2})$
$F = -2550 \, N$.
Therefore,the force between the vehicle and the road is $-2550 \, N$,acting in the direction opposite to the motion of the vehicle.

(B) The momentum of an object is defined as the product of its mass and its velocity.
Given:
Mass of the object $= m$
Velocity of the object $= v$
Therefore, Momentum $(p) = \text{Mass} \times \text{Velocity} = mv$.

(B) Mass of the hockey ball,$m = 200 \,g = 0.2 \,kg$.
Initial velocity of the hockey ball,$u = 10 \,m \,s^{-1}$.
Initial momentum,$p_i = m \times u = 0.2 \,kg \times 10 \,m \,s^{-1} = 2 \,kg \,m \,s^{-1}$.
Since the ball returns along its original path,the final velocity is in the opposite direction,$v = -5 \,m \,s^{-1}$.
Final momentum,$p_f = m \times v = 0.2 \,kg \times (-5 \,m \,s^{-1}) = -1 \,kg \,m \,s^{-1}$.
Change in momentum,$\Delta p = p_f - p_i = -1 \,kg \,m \,s^{-1} - 2 \,kg \,m \,s^{-1} = -3 \,kg \,m \,s^{-1}$.
The magnitude of the change in momentum is $|\Delta p| = 3 \,kg \,m \,s^{-1}$.

(C) Given:
Initial velocity of the bullet,$u = 150\, m/s$
Final velocity of the bullet,$v = 0\, m/s$ (as it comes to rest)
Time taken,$t = 0.03\, s$
Mass of the bullet,$m = 10\, g = 0.01\, kg$
Using the first equation of motion,$v = u + at$:
$0 = 150 + a(0.03)$
$a = -150 / 0.03 = -5000\, m/s^2$
Using the second equation of motion to find the distance of penetration $(s)$:
$s = ut + (1/2)at^2$
$s = (150 \times 0.03) + (1/2 \times -5000 \times (0.03)^2)$
$s = 4.5 - (2500 \times 0.0009)$
$s = 4.5 - 2.25 = 2.25\, m$
Using Newton's second law of motion to find the force $(F)$:
$F = ma$
$F = 0.01\, kg \times 5000\, m/s^2 = 50\, N$
Thus,the distance of penetration is $2.25\, m$ and the force exerted is $50\, N$.

(D) Initial velocity of the object, $u = 5 \,m \,s^{-1}$.
Final velocity of the object, $v = 8 \,m \,s^{-1}$.
Mass of the object, $m = 100 \,kg$.
Time taken, $t = 6 \,s$.
Initial momentum $= m \times u = 100 \times 5 = 500 \,kg \,m \,s^{-1}$.
Final momentum $= m \times v = 100 \times 8 = 800 \,kg \,m \,s^{-1}$.
Force exerted on the object, $F = \frac{mv - mu}{t}$.
$F = \frac{800 - 500}{6} = \frac{300}{6} = 50 \,N$.
Thus, the initial momentum is $500 \,kg \,m \,s^{-1}$, the final momentum is $800 \,kg \,m \,s^{-1}$, and the force exerted is $50 \,N$.

(B) Mass of the hammer,$m = 500 \,g = 0.5 \,kg$.
Initial velocity of the hammer,$u = 50 \,m/s$.
Time taken by the nail to stop the hammer,$t = 0.01 \,s$.
Final velocity of the hammer,$v = 0 \,m/s$ (since the hammer comes to rest).
According to Newton's second law of motion,the force $F$ exerted on the hammer is given by $F = m \times a = m \times \frac{v - u}{t}$.
Substituting the values: $F = 0.5 \times \frac{0 - 50}{0.01} = 0.5 \times (-5000) = -2500 \,N$.
The negative sign indicates that the force exerted by the nail on the hammer is in the opposite direction of the hammer's motion.
According to Newton's third law of motion,the magnitude of the force exerted by the nail on the hammer is $2500 \,N$.

(A) 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, $F = \frac{dp}{dt}$, where $F$ is the force and $p$ is the momentum.
Therefore, the physical quantity that corresponds to the rate of change of momentum is $Force$.

(C) According to Newton's second law of motion,the force exerted is given by $F = \frac{\Delta p}{\Delta t}$,where $\Delta p$ is the change in momentum and $\Delta t$ is the time taken.
By pulling his hands backwards,the goalkeeper increases the time $(\Delta t)$ taken to bring the ball to rest.
Since the force is inversely proportional to time,increasing the time interval reduces the impact force exerted by the ball on the goalkeeper's hands,thereby preventing injury.

(C) According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m}$.
Initially,the acceleration of the truck is $a_1 = \frac{F}{M}$.
When the truck is loaded with an object of mass equal to the truck,the new mass becomes $M' = M + M = 2M$.
The driving force is halved,so the new force becomes $F' = \frac{F}{2}$.
The new acceleration $a_2$ is given by $a_2 = \frac{F'}{M'} = \frac{F/2}{2M} = \frac{F}{4M}$.
Comparing the new acceleration with the initial acceleration: $\frac{a_2}{a_1} = \frac{F/4M}{F/M} = \frac{1}{4}$.
Therefore,the acceleration becomes one-fourth of its original value.

(B) Derivation: According to Newton's second law,$F \propto \frac{dp}{dt}$. Since $p = mv$,$F = k \frac{d(mv)}{dt} = kma$. Taking $k=1$,we get $F = ma$.
$(i)$ Given: $m = 10 \, g = 0.01 \, kg$,$u = 10^3 \, m \, s^{-1}$,$v = 0$,$s = 5 \, cm = 0.05 \, m$.
Using $v^2 - u^2 = 2as$:
$0^2 - (10^3)^2 = 2 \cdot a \cdot 0.05$
$-10^6 = 0.1 \cdot a$
$a = -10^7 \, m \, s^{-2}$.
Resistive force $F = |ma| = 0.01 \times 10^7 = 10^5 \, N$.
$(ii)$ Using $v = u + at$:
$0 = 10^3 + (-10^7) \cdot t$
$10^7 \cdot t = 10^3$
$t = 10^{-4} \, s$.

(C) According to Newton's second law of motion,$F = ma$. The $SI$ unit of force is $kg \, m s^{-2}$,which is defined as $1 \, N$.
Given:
Force $F = 5 \, N$
Acceleration $a_1 = 8 \, m s^{-2}$ for mass $m_1$
Acceleration $a_2 = 24 \, m s^{-2}$ for mass $m_2$
Calculating individual masses:
$m_1 = \frac{F}{a_1} = \frac{5}{8} \, kg$
$m_2 = \frac{F}{a_2} = \frac{5}{24} \, kg$
When both masses are tied together,the total mass $M$ is:
$M = m_1 + m_2 = \frac{5}{8} + \frac{5}{24} = \frac{15 + 5}{24} = \frac{20}{24} = \frac{5}{6} \, kg$
Acceleration $a$ produced by the same force $F$ on the combined mass $M$ is:
$a = \frac{F}{M} = \frac{5}{5/6} = 6 \, m s^{-2}$

(N/A) Momentum is defined as the product of the mass and velocity of an object. It is a vector quantity.
Momentum $(p)$ = mass $(m)$ $\times$ velocity $(v)$
The $SI$ unit of momentum is $kg\, m\, s^{-1}$.
Force is defined as the rate of change of momentum. According to Newton's second law of motion,the force applied on an object is directly proportional to the rate of change of its momentum.
Force $(F)$ = $\frac{\Delta p}{\Delta t}$
Graphical representation:
$(a)$ Since $p = mv$,if mass $(m)$ is fixed,momentum $(p)$ is directly proportional to velocity $(v)$. The graph of momentum versus velocity is a straight line passing through the origin.
$(b)$ Since $p = mv$,if velocity $(v)$ is constant,momentum $(p)$ is directly proportional to mass $(m)$. The graph of momentum versus mass is a straight line passing through the origin.

(B) According to Newton's second law of motion,the force $(F)$ is given by the product of mass $(m)$ and acceleration $(a)$.
Formula: $F = m \times a$
Given:
Mass $(m)$ = $2 \, kg$
Acceleration $(a)$ = $5 \, m s^{-2}$
Calculation:
$F = 2 \, kg \times 5 \, m s^{-2} = 10 \, N$
Therefore,the force required is $10 \, N$.

(N/A) Momentum $(p)$ is defined as the product of mass $(m)$ and velocity $(v)$.
The mathematical formula is: $p = m \times v$.
The $SI$ unit of mass is kilogram $(kg)$ and the $SI$ unit of velocity is metre per second $(m/s)$.
Therefore,the $SI$ unit of momentum is kilogram metre per second,denoted as $kg \cdot m/s$ or $kg \cdot m \cdot s^{-1}$.

(D) According to Newton's second law of motion,the force $(F)$ is the product of mass $(m)$ and acceleration $(a)$.
Formula: $F = m \times a$
Given:
Mass $(m) = 6 \, kg$
Acceleration $(a) = 4 \, m s^{-2}$
Calculation:
$F = 6 \, kg \times 4 \, m s^{-2} = 24 \, N$
Therefore,the force required is $24 \, N$.

(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,$F = ma$,where $F$ is the force,$m$ is the mass,and $a$ is the acceleration.

(B) The $SI$ unit of force is the Newton $(N)$.
It is defined as the force required to accelerate a mass of $1 \ kg$ by $1 \ m/s^2$.

(N/A) The $CGS$ unit of force is the dyne.
$1 \text{ Newton (N)} = 10^5 \text{ dynes}$.
This relationship is derived from the fact that $1 \text{ N} = 1 \text{ kg} \cdot \text{m/s}^2$,and since $1 \text{ kg} = 10^3 \text{ g}$ and $1 \text{ m} = 10^2 \text{ cm}$,we get $1 \text{ N} = 10^3 \text{ g} \times 10^2 \text{ cm/s}^2 = 10^5 \text{ g} \cdot \text{cm/s}^2$,which is equal to $10^5 \text{ dynes}$.

(C) Glass or chinaware items are fragile and prone to breaking upon impact.
When these items are packed with straw,the straw acts as a cushion.
According to Newton's Second Law of Motion,the force exerted is inversely proportional to the time taken for the change in momentum $(F = \Delta p / \Delta t)$.
By using straw,the time taken for the change in momentum is increased during any sudden jerk or collision.
As a result,the force exerted on the glass or chinaware is significantly reduced,preventing breakage.

(A) According to Newton's second law of motion,the force exerted is given by $F = \frac{dp}{dt}$,where $dp$ is the change in momentum and $dt$ is the time taken.
When an athlete falls on a cushioned bed,the bed compresses,which increases the time $(dt)$ taken for the athlete to come to rest.
Since the force $(F)$ is inversely proportional to the time $(dt)$,an increase in time results in a significant decrease in the impulsive force exerted on the athlete.
This prevents the athlete from getting injured.

(A) We know that the force $F$ is calculated using the formula $F = m \times a$,where $m$ is mass and $a$ is acceleration.
For the first case:
$F_1 = m_1 \times a_1 = 1\, kg \times 10\, m s^{-2} = 10\, N$
For the second case:
$F_2 = m_2 \times a_2 = 2\, kg \times 4\, m s^{-2} = 8\, N$
Comparing the two forces,we find that $10\, N > 8\, N$,which means $F_1 > F_2$.
Therefore,accelerating a $1\, kg$ mass at $10\, m s^{-2}$ requires a greater force.

(B) According to Newton's second law of motion,the force exerted is given by $F = \frac{\Delta p}{\Delta t}$,where $\Delta p$ is the change in momentum and $\Delta t$ is the time interval.
When the goalkeeper stops a fast-moving ball,the momentum of the ball changes from its initial value to zero.
Since the time interval $\Delta t$ is very small,the force $F$ exerted by the ball on the goalkeeper's hands becomes very large.
This large force causes the injury to the goalkeeper's hands.

(A) The impact of a moving object depends on its momentum,which is the product of its mass $(m)$ and velocity $(v)$,given by $p = m \times v$.
$1$. $A$ table tennis ball has a very small mass,so even at high speeds,its momentum is relatively low,causing negligible impact.
$2$. $A$ cricket ball has a significantly larger mass compared to a table tennis ball. When it moves at a high velocity,its momentum $(p = mv)$ becomes very high.
$3$. According to Newton's second law,the force exerted during an impact is proportional to the rate of change of momentum. Since the cricket ball carries much higher momentum,it exerts a larger force upon impact,which can cause injury.

(B) According to Newton's second law of motion,the force $F$ is given by the product of mass $m$ and acceleration $a$,i.e.,$F = m \times a$.
For the first case:
Mass $m_1 = 2 \ kg$,Acceleration $a_1 = 5 \ m s^{-2}$
Force $F_1 = 2 \ kg \times 5 \ m s^{-2} = 10 \ N$.
For the second case:
Mass $m_2 = 5 \ kg$,Acceleration $a_2 = 3 \ m s^{-2}$
Force $F_2 = 5 \ kg \times 3 \ m s^{-2} = 15 \ N$.
Comparing the two forces,$15 \ N > 10 \ N$.
Therefore,accelerating a $5 \ kg$ mass at $3 \ m s^{-2}$ requires a greater force.

(A) The momentum $p$ of a body is defined as the product of its mass $m$ and velocity $v$,given by the formula $p = mv$.
For body $A$ with mass $m$ and velocity $v$,the momentum is $p_A = m \times v = mv$.
For body $B$ with mass $4m$ and velocity $v$,the momentum is $p_B = 4m \times v = 4mv$.
The ratio of the momentum of body $A$ to body $B$ is $\frac{p_A}{p_B} = \frac{mv}{4mv} = \frac{1}{4}$.
Therefore,the ratio is $1:4$.

(N/A) According to Newton's second law of motion,the force exerted by the ball is given by $F = m \times a$,where $a = \frac{v - u}{t}$.
By moving his hands backwards,the player increases the time $(t)$ taken to bring the ball to rest.
Since force is inversely proportional to time $(F \propto \frac{1}{t})$,increasing the time reduces the impulsive force exerted by the ball on the player's hands.
This prevents injury and makes it easier to catch the ball.

(N/A) Let a force $F$ act on a body of mass $m$ for time $t$ and change its velocity from $u$ to $v$.
Initial momentum of the body $= mu$
Final momentum of the body $= mv$
Change in momentum of the body in time $t = mv - mu = m(v - u)$
The rate of change of momentum $= \frac{\text{Change in momentum}}{\text{Time}} = \frac{m(v - u)}{t}$
Since acceleration $a = \frac{v - u}{t}$, we substitute this into the equation.
Therefore, the rate of change of momentum $= m \times a = \text{mass} \times \text{acceleration}$.
According to Newton's second law of motion, this rate of change of momentum is equal to the applied force $F$.

(N/A) $1 \text{ N} = 1 \text{ kg} \times 1 \text{ m s}^{-2}$
Since $1 \text{ kg} = 1000 \text{ g}$ and $1 \text{ m} = 100 \text{ cm}$,we substitute these values:
$1 \text{ N} = 1000 \text{ g} \times 100 \text{ cm s}^{-2}$
$1 \text{ N} = 100,000 \text{ g cm s}^{-2}$
Since $1 \text{ dyne} = 1 \text{ g cm s}^{-2}$,we get:
$1 \text{ N} = 10^{5} \text{ dynes}$.

(N/A) When a fielder stops a fast-moving ball suddenly,the velocity of the ball decreases to $0$ in a very short interval of time.
According to Newton's second law of motion,the force applied is directly proportional to the rate of change of momentum,given by $F = \frac{m(v - u)}{t}$.
Since the time interval $(t)$ is very small,the rate of change of momentum is very high.
This requires the fielder to apply a large impulsive force to stop the ball.
Consequently,the ball exerts an equal and opposite reaction force on the fielder's hands,which may cause injury to the palm.

(C) Initial momentum $p_i = mv$ and final momentum $p_f = 0$. Since $m$ and $v$ remain constant,there is no change in initial and final momentum.
$(b)$ Change of momentum $\Delta p = p_f - p_i = 0 - mv = -mv$. Since the magnitude of change depends only on initial and final states,there is no change.
$(c)$ Rate of change of momentum is given by $F = \frac{\Delta p}{t}$. If time $t$ is reduced to half $(t' = t/2)$,the new rate of change of momentum $F' = \frac{\Delta p}{t/2} = 2 \times \frac{\Delta p}{t} = 2F$. Thus,the rate of change of momentum is doubled.

(N/A) Let a body of mass $m$ be moving with velocity $v$. The initial momentum is $p_{1} = m v$.
$(i)$ When the velocity is doubled,the new velocity is $2v$. The new momentum $p_{2} = m(2v) = 2mv = 2p_{1}$. Therefore,the ratio $p_{1} : p_{2} = 1 : 2$.
$(ii)$ When the mass is halved,the new mass is $\frac{1}{2}m$. The new momentum $p_{3} = (\frac{1}{2}m)v = \frac{1}{2}mv = \frac{1}{2}p_{1}$. Therefore,the ratio $p_{1} : p_{3} = 2 : 1$.
$(iii)$ When both mass and velocity are increased by three times,the new mass is $3m$ and the new velocity is $3v$. The new momentum $p_{4} = (3m)(3v) = 9mv = 9p_{1}$. Therefore,the ratio $p_{1} : p_{4} = 1 : 9$.

(N/A) Mass of the car along with passengers,$m = 1000 \, kg$.
Initial velocity,$u = 108 \, km \, h^{-1} = \frac{108 \times 1000}{3600} \, m \, s^{-1} = 30 \, m \, s^{-1}$.
Final velocity,$v = 0 \, m \, s^{-1}$ (since the car stops).
Time taken,$t = 4 \, s$.
Acceleration,$a = \frac{v - u}{t} = \frac{0 - 30}{4} = -7.5 \, m \, s^{-2}$.
Force exerted by the brakes,$F = m \times a = 1000 \, kg \times (-7.5 \, m \, s^{-2}) = -7500 \, N$.
The negative sign indicates that the force is applied in the direction opposite to the motion of the car.
Therefore,the magnitude of the force exerted by the brakes is $7500 \, N$.

Mass of the man,$m_{1} = 60 \, kg$.
Initial speed of the man,$u_{1} = 18 \, km \, h^{-1} = \frac{18 \times 1000}{3600} \, m \, s^{-1} = 5 \, m \, s^{-1}$.
Mass of the car,$m_{2} = 100 \, kg$.
Initial speed of the car,$u_{2} = 0 \, m \, s^{-1}$.
Let the final velocity of the car with the man be $v$.
According to the law of conservation of momentum:
$m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$.
Substituting the values:
$(60 \times 5) + (100 \times 0) = (60 + 100)v$.
$300 = 160v$.
$v = \frac{300}{160} \, m \, s^{-1} = 1.875 \, m \, s^{-1}$.
Converting to $km \, h^{-1}$:
$v = 1.875 \times \frac{18}{5} \, km \, h^{-1} = 6.75 \, km \, h^{-1}$.

(N/A) The salesman's claim is scientifically impossible. If a car stops instantaneously,the time taken $(t)$ for the change in velocity is $0$. Since retardation (deceleration) is defined as $a = \frac{v - u}{t}$,if $t = 0$,the retardation becomes infinite. An infinite force would be required to produce infinite retardation,which is physically impossible.
$(b)$ Chetan demonstrates critical thinking,scientific awareness,honesty,and a sense of responsibility toward his family's well-being.

(B) $1$. Calculate the momentum of the bullet:
Mass of bullet $(m_1)$ = $10 \, g = 0.01 \, kg$
Velocity of bullet $(v_1)$ = $400 \, m s^{-1}$
Momentum $(p_1)$ = $m_1 \times v_1 = 0.01 \, kg \times 400 \, m s^{-1} = 4 \, kg \, m s^{-1}$
$2$. Calculate the momentum of the cricket ball:
Mass of cricket ball $(m_2)$ = $400 \, g = 0.4 \, kg$
Velocity of cricket ball $(v_2)$ = $90 \, km h^{-1} = 90 \times (5/18) \, m s^{-1} = 25 \, m s^{-1}$
Momentum $(p_2)$ = $m_2 \times v_2 = 0.4 \, kg \times 25 \, m s^{-1} = 10 \, kg \, m s^{-1}$
$3$. Comparison:
Since $10 \, kg \, m s^{-1} > 4 \, kg \, m s^{-1}$,the cricket ball has a higher momentum.

(N/A) Let an unbalanced force $F$ act on a body of mass $m$ moving with an initial velocity $u \, m s^{-1}$. The velocity changes to $v \, m s^{-1}$ after time $t$ seconds.
Initial momentum of the body $(p_1) = m \times u = mu$.
Final momentum of the body after $t$ seconds $(p_2) = m \times v = mv$.
Change in momentum $= p_2 - p_1 = mv - mu = m(v - u)$.
Rate of change of momentum $= \frac{m(v - u)}{t}$.
According to Newton's second law of motion, the rate of change of momentum is directly proportional to the applied force $F$:
$F \propto \frac{m(v - u)}{t}$.
Since acceleration $a = \frac{v - u}{t}$, we can write:
$F \propto ma$.
$F = kma$, where $k$ is a constant of proportionality.
By defining the unit of force such that $1 \, N$ of force produces an acceleration of $1 \, m s^{-2}$ in a body of $1 \, kg$ mass, we get $k = 1$.
Therefore, the mathematical formula for force is $F = ma$.

(N/A) Force is an external agent that produces or tends to produce acceleration in an object. Three possible effects of force are:
$(i)$ It can set a stationary object into motion.
$(ii)$ It can stop a moving object.
$(iii)$ It can change the speed or direction of motion of an object.
$(b)$ Given:
Initial velocity $u = 90 \, km \, h^{-1} = 90 \times \frac{5}{18} = 25 \, m \, s^{-1}$
Final velocity $v = 0 \, m \, s^{-1}$
Time $t = 5 \, s$
Mass $m = 200 \, kg$
Using the first equation of motion $v = u + at$:
$0 = 25 + a(5)$
$5a = -25$
$a = -5 \, m \, s^{-2}$
Using Newton's second law of motion $F = ma$:
$F = 200 \times (-5) = -1000 \, N$
The magnitude of the force exerted by the brakes is $1000 \, N$ in the direction opposite to the motion.

(A) The force required to stop a moving object is directly proportional to the rate of change of momentum $(F = \frac{dp}{dt} = ma)$.
Since both vehicles are moving with the same velocity $(v)$,the momentum $(p = mv)$ depends on the mass $(m)$ of the vehicle.
The loaded vehicle has a greater mass compared to the empty vehicle.
Therefore,the loaded vehicle possesses more momentum.
According to Newton's second law of motion,a larger force is required to change the momentum of an object with greater mass in the same amount of time.
Thus,the loaded vehicle will require a larger force to stop.

(B) Momentum $(p)$ is defined as the product of mass $(m)$ and velocity $(v)$,given by the formula: $p = m \times v$.
For body $A$:
Mass $(m_A)$ = $1 \ kg$
Speed $(v_A)$ = $10 \ m s^{-1}$
Momentum $(p_A)$ = $1 \ kg \times 10 \ m s^{-1} = 10 \ kg \ m s^{-1}$.
For body $B$:
Mass $(m_B)$ = $2 \ kg$
Speed $(v_B)$ = $9 \ m s^{-1}$
Momentum $(p_B)$ = $2 \ kg \times 9 \ m s^{-1} = 18 \ kg \ m s^{-1}$.
Comparing the two values,$18 \ kg \ m s^{-1} > 10 \ kg \ m s^{-1}$.
Therefore,body $B$ has more momentum.

(A) Given: Momentum $(p)$ = $2000 \ kg \ m \ s^{-1}$,Velocity $(v)$ = $50 \ m \ s^{-1}$.
We know that momentum is the product of mass and velocity,given by the formula: $p = m \times v$.
To find the mass $(m)$,we rearrange the formula: $m = \frac{p}{v}$.
Substituting the given values: $m = \frac{2000}{50} = 40 \ kg$.
Therefore,the mass of the missile is $40 \ kg$.

(B) Let $m$ be the mass of the body.
Given: Force $F = 40\, N$ and acceleration $a = 5\, m s^{-2}$.
According to Newton's second law of motion,the relationship is $F = m \times a$.
Substituting the given values,we get $40 = m \times 5$.
Therefore,$m = \frac{40}{5} = 8\, kg$.
The mass of the body is $8\, kg$.

(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 $800 \ N$.

(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 $4 \, m s^{-1}$.

(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).

(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:
$a' = F / m' = 80 / 100 = 0.8\, m/s^2$.

(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 $0.04 \ N$.