Second Law of Motion Questions in English

(A) The magnitude of the force vector $\vec F = 6\hat i - 8\hat j + 10\hat k$ is given by $|\vec F| = \sqrt{6^2 + (-8)^2 + 10^2}$.
$|\vec F| = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt{2}\;N$.
According to Newton's second law,$F = ma$,so the mass $m = \frac{|\vec F|}{a}$.
Given the acceleration $a = 1\;m/s^2$,the mass $m = \frac{10\sqrt{2}}{1} = 10\sqrt{2}\;kg$.

(B) According to Newton's second law of motion,the force $F$ produced by the engine is given by $F = m \times a$,where $m$ is the mass and $a$ is the acceleration.
Since the force $F$ remains constant,we have $a \propto \frac{1}{m}$.
Initially,the mass is $m$ and the acceleration is $a_1 = 4\,m/s^2$.
When the car pulls another car of the same mass,the total mass becomes $M = m + m = 2m$.
Let the new acceleration be $a_2$. Since $F = m \times a_1 = 2m \times a_2$,we get $a_2 = \frac{a_1}{2}$.
Substituting the value,$a_2 = \frac{4}{2} = 2\,m/s^2$.

(D) Given: Initial velocity $u = 0$,Mass $m = 0.9 \,kg$,Time $t = 10 \,s$,Distance $S = 250 \,m$.
Using the second equation of motion: $S = ut + \frac{1}{2}at^2$.
Substituting the values: $250 = (0)(10) + \frac{1}{2} \times a \times (10)^2$.
$250 = 50a \Rightarrow a = \frac{250}{50} = 5 \,m/s^2$.
Now,using Newton's second law of motion: $F = ma$.
$F = 0.9 \,kg \times 5 \,m/s^2 = 4.5 \,N$.

(B) Given:
Mass of the body,$m = 5 \, kg$
Initial velocity,$u = 20 \, m/s$
Force applied,$F = 100 \, N$
Time duration,$t = 10 \, s$
According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{100}{5} = 20 \, m/s^2$.
Using the first equation of motion,$v = u + at$,we get:
$v = 20 + (20 \times 10)$
$v = 20 + 200 = 220 \, m/s$.
Therefore,the final velocity of the body is $220 \, m/s$.

(B) Given: Mass of the bullet $m = 5 \,g = 5 \times 10^{-3} \,kg$,initial velocity $u = 100 \,m/s$,final velocity $v = 0 \,m/s$,and distance $s = 6 \,cm = 0.06 \,m$.
Using the third equation of motion,$v^2 = u^2 + 2as$:
$0^2 = (100)^2 + 2 \times a \times 0.06$
$0 = 10000 + 0.12a$
$a = -\frac{10000}{0.12} = -\frac{1000000}{12} \,m/s^2$.
The negative sign indicates retardation.
Now,using Newton's second law,$F = ma$:
$F = (5 \times 10^{-3} \,kg) \times \left(\frac{1000000}{12} \,m/s^2\right)$
$F = \frac{5000}{12} \approx 416.67 \,N \approx 417 \,N$.

(B) 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,this is expressed as $\overrightarrow{F} = m\overrightarrow{a}$,where $\overrightarrow{F}$ is the force,$m$ is the mass,and $\overrightarrow{a}$ is the acceleration.
Therefore,the second law provides a quantitative measure of force.

(D) According to Newton's Second Law of Motion,the force $F$ acting on an object is given by the product of its mass $m$ and acceleration $a$,expressed as $F = m \times a$.
Let the initial force be $F_1 = m_1 \times a_1$.
If the mass is doubled $(m_2 = 2m_1)$ and the acceleration is also doubled $(a_2 = 2a_1)$,the new force $F_2$ becomes:
$F_2 = m_2 \times a_2 = (2m_1) \times (2a_1) = 4 \times (m_1 \times a_1) = 4F_1$.
Therefore,the force increases four times compared to its previous value.

(B) Given: Force $(F)$ = $5 \, N$,Weight $(W)$ = $9.8 \, N$.
We know that Weight $(W)$ = $m \times g$,where $g = 9.8 \, m/s^2$.
Therefore,mass $(m)$ = $W / g = 9.8 / 9.8 = 1 \, kg$.
According to Newton's second law of motion,$F = m \times a$.
Acceleration $(a)$ = $F / m = 5 \, N / 1 \, kg = 5 \, m/s^2$.

(B) According to Newton's second law of motion,the acceleration $a$ produced in a body is given by the formula $a = \frac{F}{m}$.
Given,force $F = 1\,N$ and mass $m = 1\,kg$.
Substituting these values into the formula:
$a = \frac{1\,N}{1\,kg} = 1\,m/s^2$.
Therefore,the body receives an acceleration of $1\,m/s^2$.

(B) Given: Mass $m = 10\, kg$,Initial velocity $u = 10\, m/s$,Final velocity $v = -2\, m/s$ (opposite direction),Time $t = 4\, s$.
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:
$F = \frac{10 \times (-2 - 10)}{4}$
$F = \frac{10 \times (-12)}{4}$
$F = \frac{-120}{4} = -30\, N$.
Thus,the force acting on the object is $-30\, N$.

(C) Given:
Acceleration due to gravity $g = 10\,m/s^2$
Force $F = 5\,kg-wt = 5 \times 10\,N = 50\,N$
Mass $m = 10\,kg$
Initial velocity $u = 0\,m/s$
Time $t = 4\,s$
Using Newton's second law,the acceleration $a$ is:
$a = \frac{F}{m} = \frac{50\,N}{10\,kg} = 5\,m/s^2$
Using the first equation of motion $v = u + at$:
$v = 0 + (5\,m/s^2 \times 4\,s) = 20\,m/s$
Therefore,the velocity of the body after $4\,s$ is $20\,m/s$.

(A) Given:
Mass of the ball,$m = 0.2 \, kg$
Initial velocity,$u = 20 \, m/s$
Final velocity,$v = 0 \, m/s$ (since it stops)
Time taken,$\Delta t = 0.1 \, s$
According to Newton's Second Law of Motion,the force $F$ is given by the rate of change of momentum:
$F = \frac{m(v - u)}{\Delta t}$
Substituting the values:
$F = \frac{0.2 \times (0 - 20)}{0.1}$
$F = \frac{0.2 \times (-20)}{0.1}$
$F = \frac{-4}{0.1} = -40 \, N$
The magnitude of the force is $40 \, N$.

(A) Given: Mass $m = 100\, kg$,Initial velocity $u = 5\, m/s$,Final velocity $v = 0\, m/s$,Time $t = \frac{1}{10}\, s = 0.1\, s$.
According to Newton's Second Law of Motion,the force required is given by $F = m \times a$,where $a$ is the acceleration (or deceleration).
The acceleration is $a = \frac{v - u}{t} = \frac{0 - 5}{0.1} = -50\, m/s^2$.
The magnitude of the force required to stop the vehicle is $F = |m \times a| = 100\, kg \times 50\, m/s^2 = 5000\, N$.

(A) The magnitude of the force vector $\vec{F} = 6\hat{i} - 8\hat{j} + 10\hat{k}$ is calculated as:
$|\vec{F}| = \sqrt{(6)^2 + (-8)^2 + (10)^2} = \sqrt{36 + 64 + 100} = \sqrt{200} \, N$.
Using Newton's second law of motion,$F = ma$,where $a = 1 \, m/s^2$:
$m = \frac{F}{a} = \frac{\sqrt{200}}{1} = \sqrt{100 \times 2} = 10\sqrt{2} \, kg$.
Therefore,the mass of the body is $10\sqrt{2} \, kg$.

(A) Given: Mass $m = 5 \,kg$,Initial velocity $u = 65 \,cm/s = 0.65 \,m/s$,Final velocity $v = 15 \,cm/s = 0.15 \,m/s$,Time $t = 0.2 \,s$.
According to Newton's second law of motion,the force $F$ is given by $F = m \cdot a = m \cdot \left( \frac{v - u}{t} \right)$.
Substituting the values: $F = 5 \cdot \left( \frac{0.15 - 0.65}{0.2} \right)$.
$F = 5 \cdot \left( \frac{-0.50}{0.2} \right) = 5 \cdot (-2.5) = -12.5 \,N$.
The negative sign indicates that the force is a resisting force (opposing motion).
Therefore,the magnitude of the average resisting force is $12.5 \,N$.

(C) Given:
Mass of the vehicle,$m = 1000\, kg$
Initial velocity,$u = 10\, m/s$
Forward force,$F_{forward} = 1000\, N$
Retarding force,$F_{friction} = 500\, N$
Time,$t = 10\, s$
Net force acting on the vehicle,$F_{net} = F_{forward} - F_{friction} = 1000 - 500 = 500\, N$
According to Newton's second law,acceleration $a = \frac{F_{net}}{m} = \frac{500}{1000} = 0.5\, m/s^2$
Using the first equation of motion,$v = u + at$
$v = 10 + (0.5 \times 10)$
$v = 10 + 5 = 15\, m/s$
Therefore,the velocity after $10\, s$ is $15\, m/s$.

(B) Given: Mass $m = 2 \, kg$,Initial velocity $u = 8 \, m/s$,Final velocity $v = 0 \, m/s$,Time $t = 4 \, s$.
According to Newton's second law of motion,the force $F$ is given by the rate of change of momentum:
$F = m \times a = m \times \frac{(v - u)}{t}$
Substituting the values:
$F = 2 \times \frac{(0 - 8)}{4}$
$F = 2 \times (-2) = -4 \, N$.
The magnitude of the force required to bring the body to rest is $4 \, N$.

(D) According to Newton's second law of motion,the net force $F$ acting on a body is equal to the rate of change of its linear momentum $p$ with respect to time $t$.
Mathematically,$F = \frac{dp}{dt}$.
Given the equation for momentum is $p = a + bt^2$.
Differentiating this equation with respect to time $t$,we get:
$F = \frac{d}{dt}(a + bt^2)$.
Since $a$ is a constant,its derivative is $0$.
$F = 0 + b(2t) = 2bt$.
Since $2$ and $b$ are constants,we can conclude that $F \propto t$.
Therefore,the net force acting on the body is proportional to $t$.

(B) Given:
Force $(F)$ = $100 \, dynes$
Mass $(m)$ = $5 \, g$
Time $(t)$ = $10 \, s$
Initial velocity $(u)$ = $0 \, cm/s$
According to Newton's second law of motion,$F = ma$,so acceleration $a = F/m$.
$a = 100 \, dynes / 5 \, g = 20 \, cm/s^2$.
Using the first equation of motion,$v = u + at$:
$v = 0 + (20 \, cm/s^2) \times (10 \, s) = 200 \, cm/s$.
Therefore,the velocity produced is $200 \, cm/s$.

(C) Given: Mass $m = 5\,kg$,initial velocity $\vec{u} = 30\hat{i} + 40\hat{j}\,m/s$,and force $\vec{F} = -1\hat{i} - 5\hat{j}\,N$.
We are interested in the $y$-component of the motion.
The initial $y$-component of velocity is $u_y = 40\,m/s$.
The $y$-component of the force is $F_y = -5\,N$.
Using Newton's second law,the acceleration in the $y$-direction is:
$a_y = \frac{F_y}{m} = \frac{-5\,N}{5\,kg} = -1\,m/s^2$.
Using the first equation of motion $v_y = u_y + a_y t$,we set the final $y$-velocity $v_y = 0$:
$0 = 40 + (-1)t$
$t = 40\,s$.
Therefore,the time taken for the $y$-component of the velocity to become zero is $40\,s$.

(B) Given:
Mass of the ball,$m = 150\; g = 0.15\; kg$
Initial velocity,$u = 20\; m/s$
Final velocity,$v = 0\; m/s$ (since the ball is caught)
Time taken,$\Delta t = 0.1\; s$
According to Newton's Second Law of Motion,the force $F$ is given by the rate of change of momentum:
$F = \frac{\Delta p}{\Delta t} = \frac{m(v - u)}{\Delta t}$
Substituting the values:
$F = \frac{0.15 \times (0 - 20)}{0.1} = \frac{0.15 \times (-20)}{0.1} = \frac{-3}{0.1} = -30\; N$
The magnitude of the force experienced by the cricketer is $30\; N$.

(C) According to Newton's second law of motion,the instantaneous force $F$ is given by the rate of change of momentum with respect to time:
$F = \frac{dp}{dt}$
This expression represents the slope of the momentum-time $(p-t)$ graph.
Therefore,the instantaneous force is maximum where the slope of the $p-t$ graph is maximum.
Looking at the provided curve,the slope is steepest at point $R$,which is the point of inflection where the curve transitions from concave up to concave down.
Thus,the force is maximum at point $R$.

(A) Given: Force $F = 6 \; N$,mass $m = 1 \; kg$,initial velocity $u = 0 \; m/s$,and final velocity $v = 30 \; m/s$.
According to Newton's second law of motion,the acceleration $a$ is given by $a = \frac{F}{m} = \frac{6 \; N}{1 \; kg} = 6 \; m/s^2$.
Using the first equation of motion,$v = u + at$,we can find the time $t$:
$30 = 0 + 6 \times t$
$t = \frac{30}{6} = 5 \; s$.
Therefore,the force acts on the body for $5 \; seconds$.

(A) Given: Mass $m = 100 \, kg$,Initial velocity $u = 5 \, m/s$,Final velocity $v = 0 \, m/s$,Time $t = 0.1 \, s$.
According to Newton's second law of motion,the force $F$ is given by the rate of change of momentum:
$F = m \cdot a = m \cdot \frac{(v - u)}{t}$
Substituting the values:
$F = 100 \cdot \frac{(0 - 5)}{0.1}$
$F = 100 \cdot \frac{-5}{0.1}$
$F = -5000 \, N$
The magnitude of the force required to stop the car is $5000 \, N$.

(C) Given force vector $\vec F = 6\hat i - 8\hat j + 10\hat k$.
The magnitude of the force is given by $|\vec F| = \sqrt{6^2 + (-8)^2 + 10^2} = \sqrt{36 + 64 + 100} = \sqrt{200} = 10\sqrt 2 \,N$.
According to Newton's second law of motion,$F = ma$,where $m$ is the mass and $a$ is the acceleration.
Given acceleration $a = 1\, m/s^2$.
Therefore,$m = \frac{F}{a} = \frac{10\sqrt 2}{1} = 10\sqrt 2 \,kg$.

(B) According to Newton's second law of motion,the force acting on a body is given by $F = ma$,where $m$ is the mass of the body and $a$ is the acceleration.
Since the mass $m$ of the body is constant and the applied force $F$ is constant,the acceleration $a = F/m$ must also be constant.
Velocity changes because acceleration is non-zero,and momentum $p = mv$ changes because velocity changes.
Therefore,the correct option is $B$.

(D) Given mass $m = 1 \, kg$ and force $\vec{F} = 2 \sin(3 \pi t) \hat{i} + 3 \cos(3 \pi t) \hat{j}$.
Acceleration $\vec{a} = \frac{\vec{F}}{m} = 2 \sin(3 \pi t) \hat{i} + 3 \cos(3 \pi t) \hat{j}$.
Velocity $\vec{v} = \int \vec{a} dt = \int (2 \sin(3 \pi t) \hat{i} + 3 \cos(3 \pi t) \hat{j}) dt = -\frac{2}{3\pi} \cos(3\pi t) \hat{i} + \frac{3}{3\pi} \sin(3\pi t) \hat{j} + \vec{C}_1$.
At $t=0$,$\vec{v}=0$,so $0 = -\frac{2}{3\pi} \hat{i} + \vec{C}_1 \implies \vec{C}_1 = \frac{2}{3\pi} \hat{i}$.
Thus,$\vec{v} = \left( \frac{2}{3\pi} - \frac{2}{3\pi} \cos(3\pi t) \right) \hat{i} + \frac{1}{\pi} \sin(3\pi t) \hat{j}$.
Position $\vec{r} = \int \vec{v} dt = \left( \frac{2t}{3\pi} - \frac{2}{9\pi^2} \sin(3\pi t) \right) \hat{i} - \frac{1}{3\pi^2} \cos(3\pi t) \hat{j} + \vec{C}_2$.
At $t=0$,$\vec{r}=0$,so $0 = 0 \hat{i} - \frac{1}{3\pi^2} \hat{j} + \vec{C}_2 \implies \vec{C}_2 = \frac{1}{3\pi^2} \hat{j}$.
The position vector is $\vec{r} = \left( \frac{2t}{3\pi} - \frac{2}{9\pi^2} \sin(3\pi t) \right) \hat{i} + \left( \frac{1}{3\pi^2} - \frac{1}{3\pi^2} \cos(3\pi t) \right) \hat{j}$.
At $t=1 \, s$,$\vec{r}(1) = \left( \frac{2}{3\pi} - 0 \right) \hat{i} + \left( \frac{1}{3\pi^2} - \frac{1}{3\pi^2} (-1) \right) \hat{j} = \frac{2}{3\pi} \hat{i} + \frac{2}{3\pi^2} \hat{j}$.
Since this result is not in the options,the correct choice is $D$.

(B) Given force $F = Be^{-Ct}$.
According to Newton's second law,$F = ma$,so $ma = Be^{-Ct}$.
Acceleration $a = \frac{dv}{dt} = \frac{B}{m} e^{-Ct}$.
Integrating with respect to time $t$ from $0$ to $t$:
$\int_{0}^{v} dv = \int_{0}^{t} \frac{B}{m} e^{-Ct} dt$.
$v = \frac{B}{m} \left[ \frac{e^{-Ct}}{-C} \right]_{0}^{t} = \frac{B}{mC} (1 - e^{-Ct})$.
Terminal velocity is the velocity as $t \to \infty$.
As $t \to \infty$,$e^{-Ct} \to 0$.
Therefore,$v_{terminal} = \frac{B}{mC} (1 - 0) = \frac{B}{mC}$.

(D) Given that the force is $F(t) = F_0e^{-bt}$.
According to Newton's second law,$F = m \frac{dv}{dt}$.
So,$m \frac{dv}{dt} = F_0e^{-bt}$.
Rearranging the terms,we get $\frac{dv}{dt} = \frac{F_0}{m} e^{-bt}$.
Integrating both sides with respect to time $t$ from $0$ to $t$,and velocity $v$ from $0$ to $v(t)$:
$\int_0^{v(t)} dv = \int_0^t \frac{F_0}{m} e^{-bt} dt$.
$v(t) = \frac{F_0}{m} \left[ \frac{e^{-bt}}{-b} \right]_0^t$.
$v(t) = \frac{F_0}{m} \left( \frac{e^{-bt}}{-b} - \frac{e^0}{-b} \right)$.
$v(t) = \frac{F_0}{mb} (1 - e^{-bt})$.
As $t \to \infty$,$v(t) \to \frac{F_0}{mb}$.
This represents an exponential growth curve that approaches a horizontal asymptote at $v = \frac{F_0}{mb}$.
Comparing this with the given options,curve $D$ represents this behavior.

(D) Given:
Mass of the ball,$m = 150 \ g = 0.15 \ kg$.
Initial velocity,$u = 20 \ m/s$.
Final velocity,$v = 0 \ m/s$ (since the ball is caught).
Time taken,$t = 0.1 \ s$.
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:
$F = \frac{0.15 \times (0 - 20)}{0.1}$
$F = \frac{0.15 \times (-20)}{0.1}$
$F = \frac{-3}{0.1} = -30 \ N$.
The magnitude of the force exerted by the ball on the hand is $30 \ N$.

(D) Given,force $F(t) = F_0e^{-bt}$.
According to Newton's second law,$F = ma = m \frac{dv}{dt}$.
So,$m \frac{dv}{dt} = F_0e^{-bt}$.
Integrating both sides with respect to time $t$ from $0$ to $t$:
$\int_0^v dv = \int_0^t \frac{F_0}{m} e^{-bt} dt$.
$v(t) = \frac{F_0}{m} \left[ \frac{e^{-bt}}{-b} \right]_0^t$.
$v(t) = \frac{F_0}{m} \left( \frac{e^{-bt} - 1}{-b} \right) = \frac{F_0}{mb} (1 - e^{-bt})$.
As $t \to 0$,$v(t) \to 0$.
As $t \to \infty$,$v(t) \to \frac{F_0}{mb}$.
The function $v(t) = \frac{F_0}{mb} (1 - e^{-bt})$ represents an increasing curve that asymptotically approaches the value $\frac{F_0}{mb}$.
This corresponds to the curve shown in option $D$.

(A) According to Newton's second law of motion,force is defined as the rate of change of momentum: $F = \frac{dp}{dt}$.
In a momentum-time $(p-t)$ graph,the force $F$ corresponds to the slope of the graph.
The negative force occurs where the slope of the graph is negative,which is in the interval $t = 4 \ s$ to $t = 8 \ s$.
The slope in this interval is calculated as:
$F = \frac{p_2 - p_1}{t_2 - t_1} = \frac{0 - 20}{8 - 4} = \frac{-20}{4} = -5 \ N$.
The magnitude of the negative force is $|-5 \ N| = 5 \ N$.

(C) Given:
Mass $m = 5\,kg$
Initial velocity $u = 5\,ms^{-1}$
Final velocity $v = 10\,ms^{-1}$
Time $t = 10\,s$
First,calculate the acceleration $a$ using the formula $a = \frac{v - u}{t}$:
$a = \frac{10\,ms^{-1} - 5\,ms^{-1}}{10\,s} = \frac{5\,ms^{-1}}{10\,s} = 0.5\,ms^{-2}$
Now,calculate the force $F$ using Newton's second law $F = ma$:
$F = (5\,kg) \times (0.5\,ms^{-2}) = 2.5\,N$
Therefore,the force acting on the body is $2.5\,N$.

(A) According to Newton's Second Law of Motion,the force $F$ applied to a body is equal to the product of its mass $m$ and acceleration $a$,given by the formula: $F = m \times a$.
To find the mass $m$,we rearrange the formula: $m = \frac{F}{a}$.
Given values are: Force $F = 10 \; N$ and acceleration $a = 1 \; m/s^2$.
Substituting these values into the formula: $m = \frac{10 \; N}{1 \; m/s^2} = 10 \; kg$.
Therefore,the mass of the body is $10 \; kg$.

(A) Given: Force $\vec{F} = 6\hat{i} - 8\hat{j} \text{ N}$ and acceleration $a = 5 \text{ m s}^{-2}$.
First,calculate the magnitude of the force vector $\vec{F}$ using the formula $F = \sqrt{F_x^2 + F_y^2}$.
$F = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \text{ N}$.
According to Newton's second law of motion,$F = ma$,therefore the mass $m = \frac{F}{a}$.
Substituting the values,$m = \frac{10 \text{ N}}{5 \text{ m s}^{-2}} = 2 \text{ kg}$.
Thus,the mass of the body is $2 \text{ kg}$.

(A) આપેલ માહિતી:
દળ $m = 5\, kg$
$t = 0\, s$ સમયે વેગ $\vec{u} = (6\hat i - 2\hat j)\, m/s$
$t = 10\, s$ સમયે વેગ $\vec{v} = 6\hat j\, m/s$
પ્રવેગ $\vec{a} = \frac{\vec{v} - \vec{u}}{t}$
$\vec{a} = \frac{6\hat j - (6\hat i - 2\hat j)}{10} = \frac{-6\hat i + 8\hat j}{10} = (-0.6\hat i + 0.8\hat j)\, m/s^2$
ન્યૂટનના ગતિના બીજા નિયમ મુજબ,બળ $\vec{F} = m\vec{a}$
$\vec{F} = 5 \times (-0.6\hat i + 0.8\hat j) = (-3\hat i + 4\hat j)\, N$
તેથી,સાચો વિકલ્પ $A$ છે.

(B) According to Newton's second law,the rate of change of momentum is equal to the applied force:
$\frac{dp}{dt} = F = kt$
Integrating both sides with respect to time from $t = 0$ to $t = T$ and momentum from $p$ to $3p$:
$\int_{p}^{3p} dp = \int_{0}^{T} kt \, dt$
$[p]_{p}^{3p} = k \left[ \frac{t^2}{2} \right]_{0}^{T}$
$3p - p = k \left( \frac{T^2}{2} - 0 \right)$
$2p = \frac{kT^2}{2}$
$4p = kT^2$
$T^2 = \frac{4p}{k}$
$T = 2\sqrt{\frac{p}{k}}$

(B) Given that the particle is at rest,the initial velocity $u = 0$.
According to Newton's second law,the acceleration $a$ is given by $a = \frac{P}{m}$.
Using the first equation of motion,$v = u + at$,we get $v = 0 + (\frac{P}{m})t = \frac{Pt}{m}$.
The kinetic energy $KE$ is given by the formula $KE = \frac{1}{2}mv^2$.
Substituting the value of $v$,we get $KE = \frac{1}{2}m(\frac{Pt}{m})^2$.
$KE = \frac{1}{2}m(\frac{P^2 t^2}{m^2}) = \frac{P^2 t^2}{2m}$.

(A) Given mass $m = 3\,kg$ and force $\vec{F} = (6t^2\hat{i} + 4t\hat{j})\,N$.
Using Newton's second law,$\vec{a} = \frac{\vec{F}}{m} = \frac{6t^2\hat{i} + 4t\hat{j}}{3} = (2t^2\hat{i} + \frac{4}{3}t\hat{j})\,m/s^2$.
Since the object starts from rest,$\vec{v}(0) = 0$. The velocity $\vec{v}$ at time $t$ is given by $\vec{v} = \int_{0}^{t} \vec{a} dt$.
$\vec{v} = \int_{0}^{3} (2t^2\hat{i} + \frac{4}{3}t\hat{j}) dt$.
$\vec{v} = [\frac{2t^3}{3}\hat{i} + \frac{4t^2}{6}\hat{j}]_{0}^{3}$.
$\vec{v} = [\frac{2(3)^3}{3}\hat{i} + \frac{4(3)^2}{6}\hat{j}] = [\frac{54}{3}\hat{i} + \frac{36}{6}\hat{j}] = 18\hat{i} + 6\hat{j}\,m/s$.

(A) The object is placed at rest in the frame $S_1$.
By definition,if an object is at rest within a specific frame of reference,its acceleration with respect to that frame is zero $(a_{rel} = 0)$.
According to Newton's Second Law of Motion,the net force acting on an object is given by $F_{net} = m \times a$.
Since the object is at rest in frame $S_1$,the acceleration of the object with respect to $S_1$ is $0 \, m/s^2$.
Therefore,the net force acting on the object with respect to the $S_1$ frame is $F_{net} = 2 \, kg \times 0 \, m/s^2 = 0 \, N$.

(B) Given: mass $m = 2\, kg$,position $x = at + bt^2 + ct^3$,$a = 3\, m/s$,$b = 4\, m/s^2$,$c = 5\, m/s^3$.
Velocity $v = \frac{dx}{dt} = a + 2bt + 3ct^2$.
Acceleration $a_c = \frac{dv}{dt} = 2b + 6ct$.
At $t = 2\, s$,acceleration $a_c = 2(4) + 6(5)(2) = 8 + 60 = 68\, m/s^2$.
Force $F = m \times a_c = 2 \times 68 = 136\, N$.

(B) Given: Thrust force $F = 10^5\, N$,Final velocity $v = 1\, km/s = 1000\, m/s$,Time $t = 10\, s$,Initial velocity $u = 0\, m/s$.
Using the first equation of motion: $v = u + at$.
$1000 = 0 + a \times 10$.
$a = \frac{1000}{10} = 100\, m/s^2$.
Using Newton's second law of motion: $F = ma$.
$m = \frac{F}{a} = \frac{10^5}{100} = 10^3\, kg$.

(C) The acceleration of the block is $0\,m/s^{2}$.
Since the problem does not specify an external force or a coefficient of friction,and assuming the block is resting on a horizontal surface without any applied force,the net force acting on the block is $0\,N$.
According to Newton's Second Law,$F_{net} = ma$,where $m = 5\,kg$.
If $F_{net} = 0$,then $a = 0/5 = 0\,m/s^{2}$.

(C) Given the linear momentum $p = 5t^2 + t + 5$ and mass $m = 5 \,kg$.
Since $p = mv$,we have $v = \frac{p}{m} = \frac{5t^2 + t + 5}{5} = t^2 + 0.2t + 1$.
The acceleration $a$ is the rate of change of velocity with respect to time: $a = \frac{dv}{dt} = \frac{d}{dt}(t^2 + 0.2t + 1) = 2t + 0.2$.
As time $t$ increases,the acceleration $a = 2t + 0.2$ also increases linearly with time.
Therefore,the body is moving with variable acceleration which is increasing with time.

(A) The position of the particle is given by the equation $y = ut + \frac{1}{2}gt^2$.
To find the velocity $v$,we differentiate the position $y$ with respect to time $t$:
$v = \frac{dy}{dt} = \frac{d}{dt}(ut + \frac{1}{2}gt^2) = u + gt$.
To find the acceleration $a$,we differentiate the velocity $v$ with respect to time $t$:
$a = \frac{dv}{dt} = \frac{d}{dt}(u + gt) = g$.
According to Newton's second law of motion,the force $F$ acting on the particle is given by $F = ma$.
Substituting the value of acceleration,we get $F = m \times g = mg$.

(A) Retarding force,$F = -50\; N$
Mass of the body,$m = 20\; kg$
Initial velocity of the body,$u = 15\; m/s$
Final velocity of the body,$v = 0\; m/s$
Using Newton's second law of motion,the acceleration $(a)$ produced in the body is given by:
$F = ma$
$-50 = 20 \times a$
$\therefore a = \frac{-50}{20} = -2.5\; m/s^2$
Using the first equation of motion,the time $(t)$ taken by the body to come to rest is:
$v = u + at$
$0 = 15 + (-2.5)t$
$2.5t = 15$
$\therefore t = \frac{15}{2.5} = 6\; s$

(A) The magnitude of the force is $0.18 \, N$,and its direction is in the direction of the motion of the body.
Given:
Mass of the body,$m = 3.0 \, kg$
Initial speed,$u = 2.0 \, m/s$
Final speed,$v = 3.5 \, m/s$
Time,$t = 25 \, s$
Using the first equation of motion,the acceleration $(a)$ produced in the body is calculated as:
$v = u + at$
$a = \frac{v - u}{t} = \frac{3.5 - 2.0}{25} = \frac{1.5}{25} = 0.06 \, m/s^2$
According to Newton's second law of motion,the force $(F)$ is given by:
$F = m \times a$
$F = 3.0 \, kg \times 0.06 \, m/s^2 = 0.18 \, N$
Since the speed of the body increases and the direction of motion remains unchanged,the force must be acting in the direction of the motion of the body.

(N/A) The magnitude of the resultant force $R$ is given by the Pythagorean theorem for two perpendicular forces:
$R = \sqrt{(8 \; N)^2 + (6 \; N)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \; N$
According to Newton's second law of motion,the magnitude of acceleration $a$ is:
$a = \frac{F}{m} = \frac{10 \; N}{5 \; kg} = 2 \; m/s^2$
The direction of the acceleration $\theta$ with respect to the $8 \; N$ force is given by:
$\tan \theta = \frac{6 \; N}{8 \; N} = 0.75$
$\theta = \tan^{-1}(0.75) \approx 36.87^{\circ}$
Thus,the acceleration is $2 \; m/s^2$ at an angle of $36.87^{\circ}$ with the $8 \; N$ force.

(D) Initial speed of the three-wheeler,$u = 36\; km/h = 36 \times \frac{5}{18} = 10\; m/s$.
Final speed of the three-wheeler,$v = 0\; m/s$.
Time taken to stop,$t = 4.0\; s$.
Total mass of the system,$M = 400\; kg + 65\; kg = 465\; kg$.
Using the first equation of motion,$v = u + at$,the acceleration $a$ is:
$a = \frac{v - u}{t} = \frac{0 - 10}{4} = -2.5\; m/s^2$.
The negative sign indicates retardation.
Using Newton's second law of motion,the retarding force $F$ is:
$F = M \times |a| = 465 \times 2.5 = 1162.5\; N$.