Second Law of Motion Questions in English

(N/A) Dynamics is the branch of physics that deals with the motion of objects,specifically considering the causes of that motion (forces) and the properties of the moving bodies.
It is the study of how forces affect the motion of physical systems.

(N/A) Statement: The time rate of change of momentum of a body is directly proportional to the resultant force acting on it,and this change occurs in the direction of the resultant force.
Let a resultant force $\vec{F}$ act on a body of mass $m$ for a time interval $\Delta t$. During this,let its velocity change from $\vec{v}$ to $\vec{v} + \Delta \vec{v}$.
Initial momentum: $\vec{p}_i = m\vec{v}$
Final momentum: $\vec{p}_f = m(\vec{v} + \Delta \vec{v})$
Change in momentum:
$\Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v} + \Delta \vec{v}) - m\vec{v} = m\Delta \vec{v}$
From the second law of motion:
$\vec{F} \propto \frac{\Delta \vec{p}}{\Delta t} \implies \vec{F} = k \frac{\Delta \vec{p}}{\Delta t}$
Taking the limit $\Delta t \to 0$:
$\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}$ (assuming constant mass).
Important points:
$(i)$ Newton's second law is $\vec{F} = \frac{d\vec{p}}{dt}$. If mass is constant,$\vec{F} = m\vec{a}$.
$(ii)$ If the resultant external force is zero,then $\vec{a} = 0$,meaning velocity is constant,which is consistent with Newton's first law.
$(iii)$ The law provides the magnitude of force. It is a vector quantity with components: $F_x = ma_x, F_y = ma_y, F_z = ma_z$. If force acts at an angle,only the component in the direction of velocity changes it.
$(iv)$ The equation $\vec{F} = m\vec{a}$ applies to point objects,rigid bodies,or systems of particles where $\vec{F}$ is the resultant force and $\vec{a}$ is the acceleration of the center of mass.
$(v)$ The law relates force and acceleration at a specific instant; it does not depend on the history of the object's motion.

(N/A) Newton's second law of motion states that the rate of change of linear momentum of a body is directly proportional to the external force applied to it,and this change takes place in the direction of the applied force.
Mathematically,it is expressed as $F = \frac{dp}{dt}$,where $F$ is the force and $p$ is the linear momentum.
For a body of constant mass $m$,this simplifies to $F = ma$,where $a$ is the acceleration of the body.

(B) According to Newton's second law of motion,the rate of change of linear momentum of a body is directly proportional to the external force applied to it.
Mathematically,$F = \frac{dp}{dt}$,where $dp$ is the change in linear momentum and $dt$ is the time interval.
Therefore,the ratio of the change in linear momentum to the time taken is equal to the force applied.

(B) According to Newton's second law of motion,the rate of change of linear momentum $(p)$ with respect to time $(t)$ is equal to the net force $(F)$ acting on the object.
Mathematically,this is expressed as: $F = \frac{dp}{dt}$.
Therefore,the first derivative of linear momentum with respect to time represents force.

(N/A) Newton's second law of motion states that the rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction of the force.
Mathematically,this is expressed as:
$F = \frac{dp}{dt}$
Where:
$F$ is the net force acting on the object.
$p$ is the linear momentum of the object.
$t$ is time.
For a body of constant mass $m$ moving with acceleration $a$,this simplifies to:
$F = ma$

(A) The relationship between weight $(W)$ and mass $(m)$ is given by the formula $W = m \times g$,where $g$ is the acceleration due to gravity.
Taking the standard value of $g = 9.8\,m/s^2$,we have:
$W = 1\,N$
$m = \frac{W}{g} = \frac{1}{9.8}\,kg$
$m \approx 0.102\,kg$
Therefore,the mass of the object is $0.102\,kg$.

(A) Given mass $m = 2 \, kg$ and velocity $\vec{v}(t) = 2t \hat{i} + t^2 \hat{j}$.
At $t = 2 \, s$,the velocity is:
$\vec{v}(2) = 2(2) \hat{i} + (2)^2 \hat{j} = 4 \hat{i} + 4 \hat{j} \, m/s$.
Momentum $\vec{p} = m\vec{v} = 2(4 \hat{i} + 4 \hat{j}) = (8 \hat{i} + 8 \hat{j}) \, kg \cdot m/s$.
Acceleration $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(2t \hat{i} + t^2 \hat{j}) = 2 \hat{i} + 2t \hat{j}$.
At $t = 2 \, s$,acceleration is:
$\vec{a}(2) = 2 \hat{i} + 2(2) \hat{j} = 2 \hat{i} + 4 \hat{j} \, m/s^2$.
Force $\vec{F} = m\vec{a} = 2(2 \hat{i} + 4 \hat{j}) = (4 \hat{i} + 8 \hat{j}) \, N$.

(B) Given force $\overrightarrow{F} = (20 \hat{i} + 10 \hat{j}) \, N$ and mass $m = 2 \, kg$.
Using Newton's second law,the acceleration is $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{20 \hat{i} + 10 \hat{j}}{2} = (10 \hat{i} + 5 \hat{j}) \, m/s^2$.
The initial velocity $\overrightarrow{u} = 0$. The time $t = 10 \, s$.
The displacement vector is given by $\overrightarrow{s} = \overrightarrow{u}t + \frac{1}{2} \overrightarrow{a} t^2$.
$\overrightarrow{s} = 0 + \frac{1}{2} (10 \hat{i} + 5 \hat{j}) (10)^2$.
$\overrightarrow{s} = \frac{1}{2} (10 \hat{i} + 5 \hat{j}) (100) = 50 (10 \hat{i} + 5 \hat{j}) = (500 \hat{i} + 250 \hat{j}) \, m$.
The displacement along the $x$-axis is the $i$-component of the displacement vector,which is $500 \, m$.

(B) Given: Force $\vec{F} = (40 \hat{i} + 10 \hat{j}) \, N$,mass $m = 5 \, kg$,initial velocity $\vec{u} = 0$,time $t = 10 \, s$.
Using Newton's second law,acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{40 \hat{i} + 10 \hat{j}}{5} = (8 \hat{i} + 2 \hat{j}) \, m/s^2$.
Since the body starts from rest,the position vector $\vec{r}$ at time $t$ is given by the kinematic equation $\vec{r} = \vec{u}t + \frac{1}{2} \vec{a} t^2$.
Substituting $\vec{u} = 0$,$\vec{a} = (8 \hat{i} + 2 \hat{j}) \, m/s^2$,and $t = 10 \, s$:
$\vec{r} = 0 + \frac{1}{2} (8 \hat{i} + 2 \hat{j}) (10)^2$
$\vec{r} = \frac{1}{2} (8 \hat{i} + 2 \hat{j}) (100)$
$\vec{r} = (4 \hat{i} + 1 \hat{j}) (100) = (400 \hat{i} + 100 \hat{j}) \, m$.

(D) Given that the particle is initially at rest,so at $t = 0$,$u = 0$.
According to Newton's second law,$F = M a$,so $a = \frac{F}{M}$.
Substituting the given force relation: $a = \frac{F_{0}}{M} \left(1 - \frac{(t - T)^{2}}{T^{2}}\right) = \frac{dv}{dt}$.
To find the velocity $v$ at $t = 2T$,we integrate the acceleration with respect to time from $t = 0$ to $t = 2T$:
$v = \int_{0}^{2T} \frac{F_{0}}{M} \left(1 - \frac{(t - T)^{2}}{T^{2}}\right) dt$.
Let $x = t - T$,then $dx = dt$. When $t = 0, x = -T$ and when $t = 2T, x = T$.
$v = \frac{F_{0}}{M} \int_{-T}^{T} \left(1 - \frac{x^{2}}{T^{2}}\right) dx$.
$v = \frac{F_{0}}{M} \left[ x - \frac{x^{3}}{3T^{2}} \right]_{-T}^{T}$.
$v = \frac{F_{0}}{M} \left( (T - \frac{T^{3}}{3T^{2}}) - (-T - \frac{(-T)^{3}}{3T^{2}}) \right)$.
$v = \frac{F_{0}}{M} \left( (T - \frac{T}{3}) - (-T + \frac{T}{3}) \right) = \frac{F_{0}}{M} \left( \frac{2T}{3} + \frac{2T}{3} \right) = \frac{4 F_{0} T}{3 M}$.

(A) Given: Force $\vec{F} = (10 \hat{i} + 5 \hat{j})\ N$,mass $m = 100\, g = 0.1\, kg$,initial velocity $\vec{u} = 0$,time $t = 2\ s$.
Using Newton's second law,acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{10 \hat{i} + 5 \hat{j}}{0.1} = (100 \hat{i} + 50 \hat{j})\ m/s^2$.
Using the equation of motion $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a} t^2$,since $\vec{u} = 0$:
$\vec{s} = \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (2)^2$
$\vec{s} = \frac{1}{2} (100 \hat{i} + 50 \hat{j}) (4) = 2 (100 \hat{i} + 50 \hat{j}) = (200 \hat{i} + 100 \hat{j})\ m$.
Comparing this with $(a \hat{i} + b \hat{j})\ m$,we get $a = 200$ and $b = 100$.
Therefore,$\frac{a}{b} = \frac{200}{100} = 2$.

(C) The force on the object is $F = k - \beta t^2$.
The acceleration of the particle is given by $a = \frac{F}{m} = \frac{k - \beta t^2}{m}$.
The acceleration is zero when $k - \beta t^2 = 0$,which implies $t^2 = \frac{k}{\beta} = \frac{8}{2} = 4$,so $t = 2 \, s$.
Since $a = \frac{dv}{dt}$,we have $dv = \frac{k - \beta t^2}{m} \, dt$.
Integrating from $t = -2 \, s$ to $t = 2 \, s$:
$\int_{v_i}^{v_f} dv = \int_{-2}^{2} \frac{k - \beta t^2}{m} \, dt$
$v_f - (-15) = \frac{1}{2/3} \int_{-2}^{2} (8 - 2t^2) \, dt = \frac{3}{2} \left[ 8t - \frac{2t^3}{3} \right]_{-2}^{2}$
$v_f + 15 = \frac{3}{2} \left[ (16 - 16/3) - (-16 + 16/3) \right] = \frac{3}{2} \left[ 32 - 32/3 \right] = \frac{3}{2} \left( \frac{64}{3} \right) = 32$
$v_f = 32 - 15 = 17 \, m/s$.

(A) Given the momentum of the particle as a function of time: $p = 2 + 3t^2$.
According to Newton's second law of motion,the force $F$ acting on a particle is equal to the rate of change of its momentum with respect to time:
$F = \frac{dp}{dt}$
Substituting the given expression for $p$:
$F = \frac{d}{dt}(2 + 3t^2)$
$F = 0 + 3(2t) = 6t$
Now,we need to find the force at time $t = 3 \, s$:
$F = 6 \times 3 = 18 \, N$.
Therefore,the force acting on the particle at $t = 3 \, s$ is $18 \, N$.

(C) Given: Mass $m = 2 \, kg$,initial velocity $\vec{u} = 0$,force $\vec{F} = (3t^2 \hat{i} + 4 \hat{j}) \, N$.
According to Newton's second law,$\vec{F} = m \vec{a} = m \frac{d\vec{v}}{dt}$.
Therefore,$d\vec{v} = \frac{\vec{F}}{m} dt$.
Integrating both sides from $t = 0$ to $t = 2 \, s$:
$\int_{0}^{\vec{v}} d\vec{v} = \int_{0}^{2} \frac{3t^2 \hat{i} + 4 \hat{j}}{2} dt$.
$\vec{v} = \frac{1}{2} \left[ \int_{0}^{2} 3t^2 dt \hat{i} + \int_{0}^{2} 4 dt \hat{j} \right]$.
$\vec{v} = \frac{1}{2} \left[ (t^3)_{0}^{2} \hat{i} + (4t)_{0}^{2} \hat{j} \right]$.
$\vec{v} = \frac{1}{2} [ (8 - 0) \hat{i} + (8 - 0) \hat{j} ]$.
$\vec{v} = \frac{1}{2} [ 8 \hat{i} + 8 \hat{j} ] = 4 \hat{i} + 4 \hat{j} \, m/s$.

(A) Given: Mass $m = 2\,kg$,initial velocity $u = 0$,time $t = 5\,s$,and final kinetic energy $K = 10000\,J$.
Using the formula for kinetic energy: $K = \frac{1}{2}mv^2$.
Substituting the values: $10000 = \frac{1}{2} \times 2 \times v^2$.
$v^2 = 10000$,which gives $v = 100\,m/s$.
Using the equation of motion $v = u + at$:
$100 = 0 + a \times 5$.
$a = \frac{100}{5} = 20\,m/s^2$.
Now,using Newton's second law $F = ma$:
$F = 2\,kg \times 20\,m/s^2 = 40\,N$.

(C) According to Newton's second law of motion,the force $\vec{F}$ acting on a particle is equal to the rate of change of momentum with respect to time: $\vec{F} = \frac{d\vec{p}}{dt}$.
This implies that the magnitude of the force $|\vec{F}|$ is equal to the magnitude of the slope of the $p-t$ graph: $|\vec{F}| = |\text{slope of } p-t \text{ curve}|$.
$1$. In region $a$,the slope is positive and moderate.
$2$. In region $b$,the slope is positive and small (the line is nearly horizontal).
$3$. In region $c$,the slope is negative and its magnitude is very large (the line is very steep).
Comparing the magnitudes of the slopes,the slope is maximum in region $c$ and minimum in region $b$.
Therefore,the magnitude of the force is maximum in region $c$ and minimum in region $b$.

(A) Given mass $m = 500 \, g = 0.5 \, kg$.
The velocity vector is $\vec{v} = 2t \hat{i} + 3t^2 \hat{j}$.
The acceleration $\vec{a}$ is the derivative of velocity with respect to time: $\vec{a} = \frac{d\vec{v}}{dt} = \frac{d}{dt}(2t \hat{i} + 3t^2 \hat{j}) = 2 \hat{i} + 6t \hat{j}$.
At $t = 1 \, s$,the acceleration is $\vec{a} = 2 \hat{i} + 6(1) \hat{j} = 2 \hat{i} + 6 \hat{j} \, ms^{-2}$.
Using Newton's second law,$\vec{F} = m\vec{a} = 0.5(2 \hat{i} + 6 \hat{j}) = 1 \hat{i} + 3 \hat{j} \, N$.
Comparing this with the given force $(\hat{i} + x \hat{j}) \, N$,we get $x = 3$.

(B) Given mass $m = 500\,g = 0.5\,kg$.
The velocity is given by $v = 10\sqrt{x}$.
Squaring both sides,we get $v^2 = 100x$.
Differentiating both sides with respect to $x$,we get $2v \frac{dv}{dx} = 100$.
Since acceleration $a = v \frac{dv}{dx}$,we can write $2a = 100$,which gives $a = 50\,m/s^2$.
The force acting on the body is $F = ma$.
Substituting the values,$F = 0.5\,kg \times 50\,m/s^2 = 25\,N$.

(A) The position vector is given by $\overrightarrow{r} = 10t \hat{i} + 15t^2 \hat{j} + 7 \hat{k}$.
The velocity vector $\overrightarrow{v}$ is the derivative of position with respect to time: $\overrightarrow{v} = \frac{d\overrightarrow{r}}{dt} = 10 \hat{i} + 30t \hat{j}$.
The acceleration vector $\overrightarrow{a}$ is the derivative of velocity with respect to time: $\overrightarrow{a} = \frac{d\overrightarrow{v}}{dt} = 30 \hat{j}$.
According to Newton's Second Law,$\overrightarrow{F} = m\overrightarrow{a}$. Since mass $m$ is a positive scalar,the direction of the net force $\overrightarrow{F}$ is the same as the direction of acceleration $\overrightarrow{a}$.
Since $\overrightarrow{a} = 30 \hat{j}$,the net force is directed along the positive $y$-axis.

(D) Given:
Mass of the ball,$m = 120 \,g = 0.12 \,kg$
Initial velocity,$u = 25 \,m/s$
Final velocity,$v = 0 \,m/s$ (as the ball is caught)
Time taken,$\Delta t = 0.1 \,s$
According to Newton's Second Law of Motion,the force exerted is equal to the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
Substituting the values:
$F = \frac{0.12 \times (0 - 25)}{0.1}$
$F = \frac{0.12 \times (-25)}{0.1}$
$F = \frac{-3}{0.1} = -30 \,N$
The magnitude of the force exerted by the ball on the hand is $|F| = 30 \,N$.

(C) The net force $\vec{F}_{net}$ acting on the body is the vector sum of the two forces:
$\vec{F}_{net} = \vec{F}_1 + \vec{F}_2 = (5 \hat{i} + 8 \hat{j} + 7 \hat{k}) + (3 \hat{i} - 4 \hat{j} - 3 \hat{k})$
$\vec{F}_{net} = (5+3) \hat{i} + (8-4) \hat{j} + (7-3) \hat{k} = 8 \hat{i} + 4 \hat{j} + 4 \hat{k} \ N$.
According to Newton's second law,$\vec{F} = m \vec{a}$,so the acceleration $\vec{a} = \frac{\vec{F}_{net}}{m}$.
Given mass $m = 4 \ kg$,we have:
$\vec{a} = \frac{8 \hat{i} + 4 \hat{j} + 4 \hat{k}}{4} = 2 \hat{i} + \hat{j} + \hat{k} \ m/s^2$.

(B) According to Newton's second law of motion,the force $F$ acting on a body of mass $m$ is given by $F = ma$.
Given that the force increases linearly with time $t$,we can write $F = kt$,where $k$ is a constant.
Substituting this into the equation of motion:
$kt = ma$
$a = \frac{k}{m} t$
Since $k$ and $m$ are constants,the acceleration $a$ is directly proportional to time $t$ $(a \propto t)$.
This relationship represents a straight line passing through the origin $(0,0)$ in an $a$ vs $t$ graph.
Therefore,the correct curve is the one shown in option $B$.

(D) Given mass $m = 500 \ g = 0.5 \ kg$.
The velocity is given by $v = 4 \sqrt{x}$.
Squaring both sides,we get $v^2 = 16x$.
Differentiating with respect to $x$,we get $2v \frac{dv}{dx} = 16$,which simplifies to $v \frac{dv}{dx} = 8$.
We know that acceleration $a = v \frac{dv}{dx}$. Therefore,$a = 8 \ m/s^2$.
Using Newton's second law,$F = ma = 0.5 \ kg \times 8 \ m/s^2 = 4 \ N$.

(B) Given: Mass $m = 2 \ kg$, Initial velocity $\overrightarrow{u} = 3 \hat{i} + 4 \hat{j} \ ms^{-1}$, Force $\overrightarrow{F} = 6 \hat{k} \ N$, Time $t = \frac{5}{3} \ s$.
Using Newton's second law, the acceleration is $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \ ms^{-2}$.
Using the first equation of motion, $\overrightarrow{v} = \overrightarrow{u} + \overrightarrow{a}t$.
Substituting the values: $\overrightarrow{v} = (3 \hat{i} + 4 \hat{j}) + (3 \hat{k}) \times \left(\frac{5}{3}\right)$.
$\overrightarrow{v} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \ ms^{-1}$.

(A) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (4 \hat{i} - 2 \hat{j} + 3 \hat{k}) \ N$,Initial velocity $\vec{u} = 0$,Time $t = 10 \ s$.
Using Newton's second law,acceleration $\vec{a} = \frac{\vec{F}}{m} = \frac{4 \hat{i} - 2 \hat{j} + 3 \hat{k}}{5} = (0.8 \hat{i} - 0.4 \hat{j} + 0.6 \hat{k}) \ m/s^2$.
Using the equation of motion $\vec{v} = \vec{u} + \vec{a}t$:
$\vec{v} = 0 + (0.8 \hat{i} - 0.4 \hat{j} + 0.6 \hat{k}) \times 10 = (8 \hat{i} - 4 \hat{j} + 6 \hat{k}) \ m/s$.
The magnitude of velocity is $|\vec{v}| = \sqrt{8^2 + (-4)^2 + 6^2} = \sqrt{64 + 16 + 36} = \sqrt{116} \ m/s$.
$|\vec{v}| = \sqrt{4 \times 29} = 2 \sqrt{29} \ m/s$.

(C) Given that a constant force $F$ acts on mass $m_1$,the acceleration is $A_1 = \frac{F}{m_1}$,which implies $m_1 = \frac{F}{A_1}$.
Similarly,for mass $m_2$,the acceleration is $A_2 = \frac{F}{m_2}$,which implies $m_2 = \frac{F}{A_2}$.
When the same force $F$ acts on the combined mass $(m_1 + m_2)$,the acceleration $A$ is given by $A = \frac{F}{m_1 + m_2}$.
Substituting the values of $m_1$ and $m_2$ into the equation:
$A = \frac{F}{\frac{F}{A_1} + \frac{F}{A_2}}$
$A = \frac{F}{F(\frac{1}{A_1} + \frac{1}{A_2})}$
$A = \frac{1}{\frac{A_1 + A_2}{A_1 A_2}}$
$A = \frac{A_1 A_2}{A_1 + A_2}$.

(A) Given mass $m = 5 \ kg$ and displacement $x = t^3 - 2t - 10$.
Velocity $v$ is the derivative of displacement with respect to time:
$v = \frac{dx}{dt} = \frac{d}{dt}(t^3 - 2t - 10) = 3t^2 - 2$.
Acceleration $a$ is the derivative of velocity with respect to time:
$a = \frac{dv}{dt} = \frac{d}{dt}(3t^2 - 2) = 6t$.
Force $F$ is given by Newton's second law,$F = ma$:
$F = 5 \times (6t) = 30t$.
At $t = 5 \ s$:
$F = 30 \times 5 = 150 \ N$.

(D) Given: Mass $m = 5 \ kg$,Force $\vec{F} = (-3 \hat{i} + 4 \hat{j}) \ N$,Initial velocity $\vec{u} = (6 \hat{i} - 12 \hat{j}) \ ms^{-1}$.
Using Newton's second law,$\vec{a} = \frac{\vec{F}}{m} = \left(\frac{-3}{5} \hat{i} + \frac{4}{5} \hat{j}\right) \ ms^{-2}$.
The velocity at any time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$.
$\vec{v} = (6 \hat{i} - 12 \hat{j}) + \left(-\frac{3}{5} \hat{i} + \frac{4}{5} \hat{j}\right)t$.
$\vec{v} = \left(6 - \frac{3}{5}t\right) \hat{i} + \left(-12 + \frac{4}{5}t\right) \hat{j}$.
For the velocity to be along the $y$-axis,the $x$-component of the velocity must be zero:
$6 - \frac{3}{5}t = 0$.
$\frac{3}{5}t = 6$.
$t = \frac{6 \times 5}{3} = 10 \ s$.

(A) According to Newton's second law of motion,the net force acting on the body is given by $F_{net} = ma$.
Here,the forces acting on the body are the gravitational force $(mg)$ acting downwards and the air resistance force $(F_{air})$ acting upwards.
Since the body is falling downwards with acceleration $a$,the net force equation is:
$mg - F_{air} = ma$
Rearranging the equation to solve for the air resistance force $(F_{air})$:
$F_{air} = mg - ma = m(g - a)$
Given:
Mass $(m)$ = $0.05 \ kg$
Acceleration due to gravity $(g)$ = $9.8 \ ms^{-2}$
Acceleration of the body $(a)$ = $9.5 \ ms^{-2}$
Substituting the values:
$F_{air} = 0.05 \times (9.8 - 9.5)$
$F_{air} = 0.05 \times 0.3$
$F_{air} = 0.015 \ N$

(C) Given: Mass $m = 5 \ kg$,initial velocity $\vec{u} = (30 \hat{i} + 40 \hat{j}) \ m/s$,and force $\vec{F} = -(\hat{i} + 5 \hat{j}) \ N$.
Using Newton's second law,the acceleration $\vec{a}$ is given by $\vec{a} = \frac{\vec{F}}{m} = \frac{-(\hat{i} + 5 \hat{j})}{5} = (-0.2 \hat{i} - 1 \hat{j}) \ m/s^2$.
The velocity at any time $t$ is given by $\vec{v} = \vec{u} + \vec{a}t$.
Substituting the components,the $y$-component of velocity is $v_y = u_y + a_y t$.
Here,$u_y = 40 \ m/s$ and $a_y = -1 \ m/s^2$.
We want the time $t$ when $v_y = 0$.
$0 = 40 + (-1)t$.
$t = 40 \ s$.

(B) Given: Mass $m = 1500 \,kg$,Initial velocity $u = 20 \,ms^{-1}$,Final velocity $v = 0 \,ms^{-1}$,Time $t = 5 \,s$.
Using Newton's Second Law of Motion,the force $F = ma$.
First,calculate the acceleration $a = \frac{v - u}{t} = \frac{0 - 20}{5} = -4 \,ms^{-2}$.
The negative sign indicates retardation.
Now,calculate the retarding force: $F = m \times a = 1500 \,kg \times (-4 \,ms^{-2}) = -6000 \,N$.
The magnitude of the retarding force is $6000 \,N$.

(C) According to Newton's second law of motion,the external force $F$ acting on a body is equal to the rate of change of its linear momentum,which is given by $F = ma$,where $m$ is the mass and $a$ is the acceleration produced in the body. Thus,the magnitude of the external force can be calculated directly using this law.

(D) Given: Mass $m = 20 \,g = 0.02 \,kg$,Initial velocity $u = 500 \,m/s$,Final velocity $v = 0$,Displacement $s = 1 \,cm = 0.01 \,m$.
Using the third equation of motion: $v^2 - u^2 = 2as$.
Substituting the values: $0^2 - (500)^2 = 2 \times a \times 0.01$.
$-250000 = 0.02 \times a$.
$a = -\frac{250000}{0.02} = -1.25 \times 10^7 \,m/s^2$.
The retarding force $F$ is given by Newton's second law: $F = ma$.
$F = 0.02 \,kg \times 1.25 \times 10^7 \,m/s^2 = 250 \times 10^3 \,N$.

(A) Given: Mass $m = 10 \,kg$,Initial velocity $u = 10 \,ms^{-1}$,Final velocity $v = 0 \,ms^{-1}$,Time $t = 10 \,s$.
According to Newton's second law of motion,the force $F$ is given by $F = m \times a$,where acceleration $a = \frac{v - u}{t}$.
Substituting the values: $a = \frac{0 - 10}{10} = -1 \,ms^{-2}$.
The negative sign indicates a retarding force.
Magnitude of force $F = m \times |a| = 10 \,kg \times 1 \,ms^{-2} = 10 \,N$.

(B) Given:
Mass $m = 50 \ g = 50 \times 10^{-3} \ kg = 0.05 \ kg$
Initial velocity $u = 50 \ cm \ s^{-1} = 50 \times 10^{-2} \ m \ s^{-1} = 0.5 \ m \ s^{-1}$
Final velocity $v = 0 \ m \ s^{-1}$
Time $t = 0.5 \ s$
Initial momentum $p_i = m \times u = 0.05 \ kg \times 0.5 \ m \ s^{-1} = 0.025 \ kg \ m \ s^{-1}$
Final momentum $p_f = m \times v = 0.05 \ kg \times 0 \ m \ s^{-1} = 0 \ kg \ m \ s^{-1}$
According to Newton's second law,the force $F$ is the rate of change of momentum:
$F = \frac{\Delta p}{t} = \frac{p_f - p_i}{t}$
$F = \frac{0 - 0.025 \ kg \ m \ s^{-1}}{0.5 \ s}$
$F = -0.05 \ N$
The magnitude of the force applied to stop the ball is $0.05 \ N$.

(A) Given that,the force vector is $F = (6 \hat{i} - 18 \hat{j} + 10 \hat{k}) \text{ N}$.
The magnitude of the force is calculated as:
$|F| = \sqrt{(6)^2 + (-18)^2 + (10)^2} = \sqrt{36 + 324 + 100} = \sqrt{460} \text{ N}$.
Given acceleration $a = 8 \text{ m/s}^2$.
According to Newton's second law of motion,$F = ma$,which implies $m = \frac{F}{a}$.
Substituting the values:
$m = \frac{\sqrt{460}}{8} = \frac{\sqrt{4 \times 115}}{8} = \frac{2\sqrt{115}}{8} = \frac{\sqrt{115}}{4} \text{ kg}$.

(A) According to Newton's second law of motion,the force applied to an object is equal to the rate of change of its momentum.
Mathematically,the force $F$ is given by the ratio of the change in momentum $\Delta P$ to the time interval $\Delta t$:
$F = \frac{\Delta P}{\Delta t}$
Given:
Force $F = 2 \ N$
Change in momentum $\Delta P = 0.4 \ kg \ m \ s^{-1}$
Rearranging the formula to solve for time $\Delta t$:
$\Delta t = \frac{\Delta P}{F}$
Substituting the given values:
$\Delta t = \frac{0.4}{2} = 0.2 \ s$
Therefore,the time taken is $0.2 \ s$.

(B) Given force vector $\vec{F} = (2 \hat{i} + \hat{j} - \hat{k}) \text{ N}$.
Initial velocity $\vec{u} = 0 \text{ ms}^{-1}$.
Final velocity $\vec{v} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) \text{ ms}^{-1}$.
Time $t = 20 \text{ s}$.
Using Newton's Second Law, $\vec{F} = m \vec{a} = m \left( \frac{\vec{v} - \vec{u}}{t} \right)$.
Rearranging for mass: $m = \frac{\vec{F} \cdot t}{\vec{v} - \vec{u}}$.
Since $\vec{v} - \vec{u} = (4 \hat{i} + 2 \hat{j} - 2 \hat{k}) = 2(2 \hat{i} + \hat{j} - \hat{k}) = 2 \vec{F}$.
Substituting the values: $m = \frac{\vec{F} \cdot 20}{2 \vec{F}} = \frac{20}{2} = 10 \text{ kg}$.

(A) The components of force $\vec{F_1}$ are: $F_{1x} = 4 \cos 30^{\circ} = 4 \times \frac{\sqrt{3}}{2} = 2\sqrt{3} \ N$ and $F_{1y} = 4 \sin 30^{\circ} = 4 \times \frac{1}{2} = 2 \ N$.
The components of force $\vec{F_2}$ are: $F_{2x} = 0 \ N$ and $F_{2y} = 4 \ N$.
The net force components are: $F_x = F_{1x} + F_{2x} = 2\sqrt{3} \ N$ and $F_y = F_{1y} + F_{2y} = 2 + 4 = 6 \ N$.
The magnitude of the resultant force is $F_R = \sqrt{F_x^2 + F_y^2} = \sqrt{(2\sqrt{3})^2 + 6^2} = \sqrt{12 + 36} = \sqrt{48} \approx 6.928 \ N$.
Using Newton's second law,the acceleration $a = \frac{F_R}{m} = \frac{6.928}{1} \approx 6.9 \ m \ s^{-2}$.

(B) Given: Mass $m = 1.5 \ kg$,initial velocity $\overrightarrow{v} = -8 \hat{j} \ ms^{-1}$ (towards south),force $\overrightarrow{F} = 6 \hat{i} \ N$ (towards east),time $t = 3 \ s$.
Acceleration $\overrightarrow{a} = \frac{\overrightarrow{F}}{m} = \frac{6}{1.5} \hat{i} = 4 \hat{i} \ ms^{-2}$.
Using the equation of motion for displacement: $\overrightarrow{s} = \overrightarrow{v}t + \frac{1}{2} \overrightarrow{a}t^2$.
Substituting the values: $\overrightarrow{s} = (-8 \hat{j})(3) + \frac{1}{2}(4 \hat{i})(3)^2$.
$\overrightarrow{s} = -24 \hat{j} + 2(9) \hat{i} = 18 \hat{i} - 24 \hat{j} \ m$.
The magnitude of displacement is $S = |\overrightarrow{s}| = \sqrt{(18)^2 + (-24)^2} = \sqrt{324 + 576} = \sqrt{900} = 30 \ m$.

(A) Given that:
Mass $m = 20 \ kg$
Initial velocity $u = 15 \ m s^{-1}$
Retarding force $F = 100 \ N$
Final velocity $v = 0 \ m s^{-1}$ (since the body comes to rest)
According to Newton's second law of motion,the acceleration $a$ is given by $F = ma$. Since the force is retarding,it acts in the opposite direction of motion,so $a = -F/m$.
$a = -\frac{100 \ N}{20 \ kg} = -5 \ m s^{-2}$.
Using the first equation of motion,$v = u + at$:
$0 = 15 + (-5)t$
$5t = 15$
$t = 3 \ s$.
Thus,the body will come to rest after $3 \ s$.

(B) Given: Mass of the ball $m = 0.25 \ kg$,final velocity $v = 12 \ m \ s^{-1}$,initial velocity $u = 0 \ m \ s^{-1}$,and distance $s = 0.9 \ m$.
Using the third equation of motion: $v^2 = u^2 + 2as$.
Substituting the values: $(12)^2 = (0)^2 + 2 \times a \times 0.9$.
$144 = 1.8 \times a$.
$a = 144 / 1.8 = 80 \ m \ s^{-2}$.
Now,using Newton's second law of motion: $F = ma$.
$F = 0.25 \ kg \times 80 \ m \ s^{-2} = 20 \ N$.
Therefore,the net force acting on the ball is $20 \ N$.

(A) Let the force be $F$. According to Newton's second law,$F = ma$,so $m = F/a$.
For mass $P$,$P = F/18$.
For mass $Q$,$Q = F/9$.
For mass $R$,$R = F/6$.
When the same force $F$ is applied to the combined mass $(P+Q+R)$,the acceleration $a'$ is given by $a' = F / (P+Q+R)$.
Substituting the values of $P, Q,$ and $R$:
$a' = F / (F/18 + F/9 + F/6) = F / [F(1/18 + 2/18 + 3/18)]$.
$a' = 1 / (6/18) = 1 / (1/3) = 3 \ m/s^2$.
Thus,the acceleration is $3 \ m/s^2$.

(D) Given,mass of the object,$m = 2 \ kg$.
Velocity,$v = (8t \hat{i} + 3t^2 \hat{j}) \ m/s$.
The acceleration of the object is given by:
$a = \frac{dv}{dt} = \frac{d}{dt}(8t \hat{i} + 3t^2 \hat{j}) = (8 \hat{i} + 6t \hat{j}) \ m/s^2$.
According to Newton's second law of motion,the net force on the object is:
$F = ma = 2(8 \hat{i} + 6t \hat{j}) = (16 \hat{i} + 12t \hat{j}) \ N$.
The magnitude of the force is given as $|F| = 20 \ N$.
$|F| = \sqrt{16^2 + (12t)^2} = 20$.
Squaring both sides:
$256 + 144t^2 = 400$.
$144t^2 = 144 \implies t^2 = 1 \implies t = 1 \ s$.
At $t = 1 \ s$,the force vector is:
$F = 16 \hat{i} + 12(1) \hat{j} = (16 \hat{i} + 12 \hat{j}) \ N$.
The angle $\theta$ made by the force vector with the positive $X$-axis is:
$\tan \theta = \frac{F_y}{F_x} = \frac{12}{16} = \frac{3}{4}$.
$\theta = \tan^{-1}\left(\frac{3}{4}\right)$.

(A) Given: mass $m = 4 \ kg$,initial velocity $\vec{u} = 3 \hat{i} \ m/s$,final velocity $\vec{v} = (8 \hat{i} + 10 \hat{j}) \ m/s$,and time $t = 8 \ s$.
Using the first equation of motion,$\vec{v} = \vec{u} + \vec{a}t$,we find the acceleration $\vec{a}$:
$\vec{a} = \frac{\vec{v} - \vec{u}}{t} = \frac{(8 \hat{i} + 10 \hat{j}) - 3 \hat{i}}{8} = \frac{5 \hat{i} + 10 \hat{j}}{8} \ m/s^2$.
Now,using Newton's second law,$\vec{F} = m\vec{a}$:
$\vec{F} = 4 \times \left( \frac{5 \hat{i} + 10 \hat{j}}{8} \right) = \frac{1}{2} (5 \hat{i} + 10 \hat{j}) = (2.5 \hat{i} + 5 \hat{j}) \ N$.
The magnitude of the force is $|\vec{F}| = \sqrt{(2.5)^2 + 5^2} = \sqrt{6.25 + 25} = \sqrt{31.25} = \sqrt{\frac{125}{4}} = \frac{5 \sqrt{5}}{2} \ N$.

(C) Given: Mass $m = 10 \,kg$,Initial velocity $u = 10 \,m/s$,Force $F = (3t^2 - 30) \,N$.
Using the impulse-momentum theorem,the change in momentum is equal to the integral of force over time:
$\Delta p = \int_{0}^{t} F dt = m(v - u)$
$\int_{0}^{5} (3t^2 - 30) dt = 10(v - 10)$
$[t^3 - 30t]_{0}^{5} = 10(v - 10)$
$(5^3 - 30(5)) - (0) = 10(v - 10)$
$(125 - 150) = 10(v - 10)$
$-25 = 10(v - 10)$
$-2.5 = v - 10$
$v = 10 - 2.5 = 7.5 \,m/s$.

(A) Given force $\vec{F} = a \hat{i} + b \hat{j} + c \hat{k}$.
Using Newton's second law,the acceleration $\vec{a}_{acc} = \frac{\vec{F}}{m} = \frac{a}{m} \hat{i} + \frac{b}{m} \hat{j} + \frac{c}{m} \hat{k}$.
Since the body starts from rest at the origin,the initial velocity $\vec{u} = 0$ and initial position $\vec{r}_0 = 0$.
Using the kinematic equation $\vec{s} = \vec{u}t + \frac{1}{2} \vec{a}_{acc} t^2$,we get $\vec{r} = \frac{1}{2} \vec{a}_{acc} t^2$.
Substituting the components:
$x = \frac{1}{2} (\frac{a}{m}) t^2 = \frac{at^2}{2m}$
$y = \frac{1}{2} (\frac{b}{m}) t^2 = \frac{bt^2}{2m}$
$z = \frac{1}{2} (\frac{c}{m}) t^2 = \frac{ct^2}{2m}$
Thus,the coordinates are $(\frac{at^2}{2m}, \frac{bt^2}{2m}, \frac{ct^2}{2m})$.

(C) Given,$v = \frac{k}{\sqrt{x}} = k x^{-1/2}$,where $k$ is a constant.
Acceleration $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt} = v \frac{dv}{dx}$.
First,find $\frac{dv}{dx}$:
$\frac{dv}{dx} = k \cdot (-\frac{1}{2}) x^{-3/2} = -\frac{k}{2x^{3/2}}$.
Now,substitute $v$ and $\frac{dv}{dx}$ into the acceleration formula:
$a = (k x^{-1/2}) \cdot (-\frac{k}{2x^{3/2}}) = -\frac{k^2}{2x^2}$.
According to Newton's second law,$F = ma$.
The magnitude of the force is $F = |ma| = m \cdot \frac{k^2}{2x^2}$.
Since $m$ and $k$ are constants,$F \propto \frac{1}{x^2}$.