Kinetic Energy Questions in English

(A) The kinetic energy $K$ of a particle with mass $m$ and linear momentum $p$ is given by the formula $K = \frac{p^2}{2m}$.
Since the kinetic energies of the two particles are equal,we have $K_1 = K_2$.
Substituting the formula,we get $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.
Given $m_1 = 4\, g$ and $m_2 = 16\, g$,we have $\frac{p_1^2}{2 \times 4} = \frac{p_2^2}{2 \times 16}$.
Simplifying this,$\frac{p_1^2}{4} = \frac{p_2^2}{16}$,which implies $\frac{p_1^2}{p_2^2} = \frac{4}{16} = \frac{1}{4}$.
Taking the square root of both sides,$\frac{p_1}{p_2} = \frac{1}{2}$.
We are given that the ratio of the magnitudes of their linear momentum is $n : 2$,so $\frac{p_1}{p_2} = \frac{n}{2}$.
Comparing $\frac{n}{2} = \frac{1}{2}$,we find $n = 1$.

(A) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = 4K_1$.
Since kinetic energy $K = \frac{P^2}{2m}$,where $P$ is momentum and $m$ is mass,we have $P = \sqrt{2mK}$.
Therefore,the ratio of final momentum $P_2$ to initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{\frac{4K_1}{K_1}} = \sqrt{4} = 2$.
This implies $P_2 = 2P_1$.
The percentage change in momentum is given by $\frac{P_2 - P_1}{P_1} \times 100$.
Substituting the values,we get $\frac{2P_1 - P_1}{P_1} \times 100 = \frac{P_1}{P_1} \times 100 = 100\%$.

(B) The kinetic energy $K$ and momentum $P$ are related by the formula $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.
Given that the kinetic energies are equal $(K_1 = K_2 = K)$,the ratio of their momenta is:
$\frac{P_1}{P_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the given masses $m_1 = 8\,kg$ and $m_2 = 2\,kg$:
$\frac{P_1}{P_2} = \sqrt{\frac{8}{2}} = \sqrt{4} = 2$.
Thus,the ratio is $2:1$.

(A) Given: Mass $m = 200\,g = 0.2\,kg$,Initial kinetic energy $K_i = 90\,J$,Final kinetic energy $K_f = 40\,J$ at time $t = 1\,s$.
Using the relation $K = \frac{1}{2}mv^2$,we find the velocities:
Initial velocity $u = \sqrt{\frac{2K_i}{m}} = \sqrt{\frac{2 \times 90}{0.2}} = \sqrt{900} = 30\,m/s$.
Final velocity $v = \sqrt{\frac{2K_f}{m}} = \sqrt{\frac{2 \times 40}{0.2}} = \sqrt{400} = 20\,m/s$.
Assuming constant retardation,the acceleration $a$ is:
$a = \frac{v - u}{t} = \frac{20 - 30}{1} = -10\,m/s^2$.
To find the distance $s$ required to come to rest $(v_{final} = 0)$:
Using $v_{final}^2 = u^2 + 2as$,we get $0 = (30)^2 + 2(-10)s$.
$20s = 900 \implies s = 45\,m$.

(C) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$.
Let the initial momentum be $P$ and the initial kinetic energy be $K = \frac{P^2}{2m}$.
The new momentum $P'$ is increased by $20\%$,so $P' = P + 0.20P = 1.2P$.
The new kinetic energy $K'$ is $K' = \frac{(P')^2}{2m} = \frac{(1.2P)^2}{2m} = 1.44 \times \frac{P^2}{2m} = 1.44K$.
The percentage increase in kinetic energy is given by $\frac{K' - K}{K} \times 100$.
Substituting the values,we get $\frac{1.44K - K}{K} \times 100 = 0.44 \times 100 = 44\%$.

(B) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$,which implies $P = \sqrt{2mK}$.
Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2 = K_1 + 0.44 K_1 = 1.44 K_1$.
Let the initial momentum be $P_1$ and the final momentum be $P_2$.
Since $P \propto \sqrt{K}$,we have $\frac{P_2}{P_1} = \sqrt{\frac{K_2}{K_1}} = \sqrt{1.44} = 1.2$.
Therefore,$P_2 = 1.2 P_1$.
The percentage increase in momentum is $\frac{P_2 - P_1}{P_1} \times 100 = (1.2 - 1) \times 100 = 20 \%$.

(C) The kinetic energy $K$ of a body of mass $m$ and momentum $P$ is given by the relation $K = \frac{P^2}{2m}$.
Since both bodies have the same kinetic energy,we can write:
$K_1 = K_2$
$\frac{P_1^2}{2m_1} = \frac{P_2^2}{2m_2}$
Rearranging the terms to find the ratio of momenta $\frac{P_1}{P_2}$:
$\frac{P_1^2}{P_2^2} = \frac{m_1}{m_2}$
Taking the square root on both sides:
$\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$
Therefore,the correct option is $C$.

(B) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$.
Let the initial momentum be $p_i$. The initial kinetic energy is $KE_i = \frac{p_i^2}{2m}$.
The momentum is increased by $50 \%$,so the final momentum $p_f = p_i + 0.50 p_i = 1.5 p_i$.
The final kinetic energy is $KE_f = \frac{p_f^2}{2m} = \frac{(1.5 p_i)^2}{2m} = \frac{2.25 p_i^2}{2m} = 2.25 KE_i$.
The percentage increase in kinetic energy is given by $\frac{KE_f - KE_i}{KE_i} \times 100$.
Substituting the values: $\frac{2.25 KE_i - KE_i}{KE_i} \times 100 = 1.25 \times 100 = 125 \%$.

(C) The kinetic energy $K$ of a body is related to its linear momentum $P$ and mass $m$ by the formula $K = \frac{P^2}{2m}$.
Given that both bodies have equal kinetic energies, we have $K_1 = K_2$.
Therefore, $\frac{P_1^2}{2m_1} = \frac{P_2^2}{2m_2}$.
Rearranging the terms to find the ratio of their momenta, we get $\frac{P_1^2}{P_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides, $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the given masses $m_1 = 4 \,g$ and $m_2 = 25 \,g$, we get $\frac{P_1}{P_2} = \sqrt{\frac{4}{25}} = \frac{2}{5}$.
Thus, the ratio of their linear momentum is $2: 5$.

(D) Initial kinetic energy $K_{i} = \frac{1}{2} m v_{i}^2 = \frac{1}{2} m (100)^2 = 5000 m \ J$.
Final kinetic energy $K_{f} = \frac{1}{2} m v_{f}^2 = \frac{1}{2} m (40)^2 = 800 m \ J$.
Percentage loss of kinetic energy $= \frac{K_{i} - K_{f}}{K_{i}} \times 100$.
$= \frac{5000 m - 800 m}{5000 m} \times 100$.
$= \frac{4200}{5000} \times 100 = 84 \%$.

(C) The kinetic energy $KE$ of a particle is related to its momentum $p$ and mass $m$ by the formula: $KE = \frac{p^2}{2m}$.
Since all four particles have the same momentum $p$,the kinetic energy is inversely proportional to the mass $(KE \propto \frac{1}{m})$.
Therefore,the particle with the smallest mass will have the maximum kinetic energy.
Comparing the masses: $\frac{m}{2} < m < 2m < 4m$.
The particle $A$ has the smallest mass,$\frac{m}{2}$.
Thus,particle $A$ has the maximum kinetic energy.

(A) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$,where $m$ is the mass of the body.
From this,we can write $P = \sqrt{2mK}$.
Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f = 36 K_i$.
The initial momentum is $P_i = \sqrt{2mK_i}$ and the final momentum is $P_f = \sqrt{2mK_f} = \sqrt{2m(36K_i)} = 6\sqrt{2mK_i} = 6P_i$.
The percentage increase in momentum is given by $\frac{P_f - P_i}{P_i} \times 100 \%$.
Substituting the values,we get $\frac{6P_i - P_i}{P_i} \times 100 \% = \frac{5P_i}{P_i} \times 100 \% = 500 \%$.

(A) The kinetic energy $KE$ of a body is related to its linear momentum $P$ and mass $m$ by the formula $KE = \frac{P^2}{2m}$.
Since the kinetic energies are equal, we have $P^2 = 2m(KE)$, which implies $P = \sqrt{2m(KE)}$.
Since $2$ and $KE$ are constants, the linear momentum is directly proportional to the square root of the mass: $P \propto \sqrt{m}$.
Given masses are $m_A = 400 \,g = 0.4 \,kg$, $m_B = 1.2 \,kg$, and $m_C = 1.6 \,kg$.
The ratio of their momenta is $P_A : P_B : P_C = \sqrt{m_A} : \sqrt{m_B} : \sqrt{m_C}$.
Substituting the values: $P_A : P_B : P_C = \sqrt{0.4} : \sqrt{1.2} : \sqrt{1.6}$.
Multiplying by $\sqrt{10}$ to simplify: $\sqrt{4} : \sqrt{12} : \sqrt{16} = 2 : 2\sqrt{3} : 4$.
Dividing by $2$, we get $1 : \sqrt{3} : 2$.

(B) Let the maximum height reached by the ball be $H$. At the maximum height,the final velocity is $0$. Using the equation of motion $v_f^2 = v_i^2 - 2gH$,we get $0 = v^2 - 2gH$,which implies $H = \frac{v^2}{2g}$.
At half the height,$h = \frac{H}{2} = \frac{v^2}{4g}$.
Using the equation $v'^2 = v^2 - 2gh$,the velocity $v'$ at height $h$ is $v'^2 = v^2 - 2g(\frac{v^2}{4g}) = v^2 - \frac{v^2}{2} = \frac{v^2}{2}$.
Thus,$v' = \frac{v}{\sqrt{2}}$.
The kinetic energy $E'$ at this point is $E' = \frac{1}{2}mv'^2 = \frac{1}{2}m(\frac{v^2}{2}) = \frac{1}{2}(\frac{1}{2}mv^2) = \frac{E}{2}$.

(C) The relationship between kinetic energy $(K)$ and momentum $(P)$ is given by $K = \frac{P^2}{2m}$.
$1$. For the graph between $K$ and $P$,the equation is $K = \frac{1}{2m} P^2$. This is of the form $y = ax^2$,which represents a parabola opening upwards. Thus,graph $(A)$ is correct.
$2$. For the graph between $K$ and $P^2$,let $X = P^2$. Then the equation becomes $K = \frac{1}{2m} X$. This is of the form $y = mx$,which represents a straight line passing through the origin. Thus,graph $(C)$ is correct.
Therefore,graphs $(A)$ and $(C)$ are correct.

(A) The kinetic energy $(K.E.)$ of a body is related to its momentum $(p)$ by the formula: $K.E. = \frac{p^2}{2m}$.
When the momentum increases by $20 \%$,the new momentum $p'$ becomes:
$p' = p + 0.20p = 1.2p$.
The new kinetic energy $(K.E.')$ is:
$K.E.' = \frac{(p')^2}{2m} = \frac{(1.2p)^2}{2m} = \frac{1.44p^2}{2m}$.
Since $K.E. = \frac{p^2}{2m}$,we have $K.E.' = 1.44 \times K.E$.
The percentage increase in kinetic energy is given by:
$\Delta K.E. \% = \frac{K.E.' - K.E.}{K.E.} \times 100 \%$
$= \frac{1.44 K.E. - K.E.}{K.E.} \times 100 \%$
$= 0.44 \times 100 \% = 44 \%$.

(D) The kinetic energy $(KE)$ of a body is given by $KE = \frac{1}{2} mv^2 = \frac{p^2}{2m}$,where $p = mv$ is the momentum.
Since both bodies have the same kinetic energy,$KE_1 = KE_2$.
Therefore,$\frac{p_1^2}{2M_1} = \frac{p_2^2}{2M_2}$.
This implies $\frac{p_1^2}{p_2^2} = \frac{M_1}{M_2}$,or $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}}$.
Since the body with mass $M_2$ is heavier,$M_2 > M_1$,which means $\frac{M_1}{M_2} < 1$.
Consequently,$\frac{p_1}{p_2} < 1$,which implies $p_1 < p_2$.
Since $p = mv$,we have $M_1 V_1 < M_2 V_2$ or $M_2 V_2 > M_1 V_1$.

(B) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Rearranging this formula,we get $p = \sqrt{2mK}$.
Since the kinetic energy $K$ is the same for both bodies,the momentum $p$ is directly proportional to the square root of the mass,i.e.,$p \propto \sqrt{m}$.
Therefore,the body with the larger mass will have a greater momentum.
Thus,the heavy body has greater momentum.

(B) The kinetic energy $K$ is related to momentum $p$ and mass $m$ by the formula $K = \frac{p^2}{2m}$.
Rearranging this formula for momentum,we get $p^2 = 2mK$,which implies $p = \sqrt{2mK}$.
Since all three bodies have equal kinetic energy ($K$ is constant),the momentum is directly proportional to the square root of the mass: $p \propto \sqrt{m}$.
Comparing the masses: $m_P = m$,$m_Q = 2m$,and $m_R = 3m$.
Since $m_R > m_Q > m_P$,it follows that $p_R > p_Q > p_P$.
Therefore,the body with the greatest mass,which is $R$,will have the greatest momentum.

(D) Given that the body is initially at rest,the initial velocity $u = 0$.
According to Newton's second law,the acceleration $a = \frac{F}{m}$.
Using the first equation of motion,the velocity $v$ at time $t$ is $v = u + at = 0 + \left(\frac{F}{m}\right)t = \frac{Ft}{m}$.
The kinetic energy $K$ is given by the formula $K = \frac{1}{2}mv^{2}$.
Substituting the value of $v$,we get $K = \frac{1}{2}m\left(\frac{Ft}{m}\right)^{2}$.
$K = \frac{1}{2}m \frac{F^{2}t^{2}}{m^{2}} = \frac{F^{2}t^{2}}{2m}$.

(B) The correct option is $B$.
Concept: According to the work-energy theorem, the total work done by all forces acting on a body is equal to the change in its kinetic energy $(K.E.)$.
Given: Mass $m = 5 \,kg$, initial velocity $u = 0 \,m/s$, final velocity $v = 20 \,m/s$.
Work done $W = \Delta K.E. = K.E._{final} - K.E._{initial}$.
$W = \frac{1}{2} mv^2 - \frac{1}{2} mu^2$.
$W = \frac{1}{2} \times 5 \times (20)^2 - 0$.
$W = \frac{1}{2} \times 5 \times 400 = 5 \times 200 = 1000 \,J = 10^3 \,J$.

(A) The kinetic energy $K$ and linear momentum $p$ of a body of mass $m$ are related by the formula: $K = \frac{p^2}{2m}$.
From this,we can express momentum as: $p = \sqrt{2mK}$.
Given that both masses have equal kinetic energy $(K_1 = K_2 = K)$,the ratio of their momenta is:
$\frac{p_1}{p_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the given masses $m_1 = 4 \,kg$ and $m_2 = 1 \,kg$:
$\frac{p_1}{p_2} = \sqrt{\frac{4}{1}} = \frac{2}{1}$.
Thus,the ratio of the magnitude of their linear momenta is $2: 1$.

(A) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Given that the kinetic energies of the two masses are equal,we have $K_1 = K_2$.
Substituting the expression for kinetic energy,we get $\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.
Rearranging the terms to find the ratio of momenta,we get $\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides,the ratio of momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1 \ g$ and $m_2 = 4 \ g$,we substitute these values:
$\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Therefore,the ratio of their momenta is $1: 2$.

(C) Given,the kinetic energy $(K)$ of two particles of masses $m_{1}$ and $m_{2}$ is equal.
We know that kinetic energy is related to momentum $(p)$ by the formula $K = \frac{p^{2}}{2m}$.
Since $K_{1} = K_{2}$,we have $\frac{p_{1}^{2}}{2m_{1}} = \frac{p_{2}^{2}}{2m_{2}}$.
Rearranging the terms to find the ratio of momenta $\frac{p_{1}}{p_{2}}$,we get $\frac{p_{1}^{2}}{p_{2}^{2}} = \frac{m_{1}}{m_{2}}$.
Taking the square root on both sides,we obtain $\frac{p_{1}}{p_{2}} = \sqrt{\frac{m_{1}}{m_{2}}} = \frac{\sqrt{m_{1}}}{\sqrt{m_{2}}}$.
Therefore,the ratio of their momenta is $\sqrt{m_{1}}: \sqrt{m_{2}}$.

(B) The kinetic energy $K$ of a body is related to its linear momentum $p$ by the formula $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Since the mass $m$ remains constant,we have $K \propto p^2$.
Let the initial momentum be $p_1 = p$ and the final momentum be $p_2 = p + 0.50p = 1.5p$.
The ratio of the final kinetic energy $K_2$ to the initial kinetic energy $K_1$ is given by:
$\frac{K_2}{K_1} = \left(\frac{p_2}{p_1}\right)^2 = \left(\frac{1.5p}{p}\right)^2 = (1.5)^2 = 2.25$.
To find the percentage increase in kinetic energy,we use the formula:
$\text{Percentage increase} = \left(\frac{K_2}{K_1} - 1\right) \times 100 \%$.
Substituting the value: $(2.25 - 1) \times 100 \% = 1.25 \times 100 \% = 125 \%$.

(C) Given: mass $m = 4 \,kg$, momentum $p = 6 \,Ns$.
The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.
Substituting the given values into the formula:
$K = \frac{6^2}{2 \times 4} = \frac{36}{8} = 4.5 \,J$.
Therefore, the kinetic energy of the body is $4.5 \,J$.

(B) The kinetic energy $K$ of a body with momentum $P$ and mass $m$ is given by the relation $K = \frac{P^2}{2m}$.
Given that both bodies have the same kinetic energy,we have $K_H = K_L$.
Substituting the formula,we get $\frac{P_H^2}{2 m_H} = \frac{P_L^2}{2 m_L}$.
Rearranging the terms,we get $\frac{P_H^2}{P_L^2} = \frac{m_H}{m_L}$,which implies $\frac{P_H}{P_L} = \sqrt{\frac{m_H}{m_L}}$.
Since the body is heavy,$m_H > m_L$,which means $\frac{m_H}{m_L} > 1$.
Therefore,$\frac{P_H}{P_L} > 1$,which leads to $P_H > P_L$.

(B) Let the mass of the body be $m$ and the velocity of the body be $v$,such that it has momentum $p$ and kinetic energy $E$.
Momentum is given by $p = mv$.
Squaring both sides,we get $p^2 = m^2 v^2$.
Kinetic energy is given by $E = \frac{1}{2} mv^2$.
We can rewrite this as $E = \frac{m^2 v^2}{2m} = \frac{p^2}{2m}$.
Since $m$ is constant for a given body,we have $E \propto p^2$.
This equation is of the form $y = kx^2$,which represents a parabola.
Therefore,the nature of the graph between the momentum and kinetic energy of a body is a parabola.

(A) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$,which implies $p = \sqrt{2mK}$.
Let the initial kinetic energy be $K_1 = K$ and the final kinetic energy be $K_2 = 4K$.
The initial momentum is $p_1 = \sqrt{2mK_1} = \sqrt{2mK}$.
The final momentum is $p_2 = \sqrt{2mK_2} = \sqrt{2m(4K)} = 2\sqrt{2mK}$.
Comparing the two,we get $p_2 = 2p_1$.
Therefore,the momentum increases by $2$ times.

(C) The kinetic energy $K$ of an object of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Given that the kinetic energies are equal,we have $K_1 = K_2$.
Therefore,$\frac{p_1^2}{2m_1} = \frac{p_2^2}{2m_2}$.
Rearranging the terms to find the ratio of momenta,we get $\frac{p_1^2}{p_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides,we obtain $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}} = \frac{\sqrt{m_1}}{\sqrt{m_2}}$.
Thus,the ratio $p_1: p_2$ is $\sqrt{m_1}: \sqrt{m_2}$.

(D) Let the initial velocity be $v$.
Initial kinetic energy $E = \frac{1}{2}mv^2$.
When the velocity is increased by $1 \,m/s$, the new velocity becomes $v' = v + 1$.
The new kinetic energy $E' = \frac{1}{2}m(v+1)^2$.
According to the problem, the kinetic energy is doubled, so $E' = 2E$.
Substituting the expressions: $\frac{1}{2}m(v+1)^2 = 2 \times (\frac{1}{2}mv^2)$.
Simplifying the equation: $(v+1)^2 = 2v^2$.
Expanding the left side: $v^2 + 2v + 1 = 2v^2$.
Rearranging the terms to form a quadratic equation: $v^2 - 2v - 1 = 0$.
Using the quadratic formula $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:
$v = \frac{2 \pm \sqrt{(-2)^2 - 4(1)(-1)}}{2(1)} = \frac{2 \pm \sqrt{4 + 4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{2 \pm 2\sqrt{2}}{2} = 1 \pm \sqrt{2}$.
Since velocity must be positive, we take $v = (1 + \sqrt{2}) \,m/s$.

(C) Given mass, $m = 100 \,g = 0.1 \,kg$.
Velocity vector, $\vec{v} = (\hat{i} + 2\hat{j} - 3\hat{k}) \,m/s$.
The magnitude of the velocity squared is given by $v^2 = |\vec{v}|^2 = (1)^2 + (2)^2 + (-3)^2 = 1 + 4 + 9 = 14 \,m^2/s^2$.
Kinetic energy $K$ is calculated using the formula $K = \frac{1}{2}mv^2$.
Substituting the values, $K = \frac{1}{2} \times 0.1 \,kg \times 14 \,m^2/s^2 = 0.05 \times 14 = 0.7 \,J$.

(C) The kinetic energy $(K.E.)$ of an object is defined by the formula $K.E. = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the speed of the body.
Since the mass $m$ is constant,the relationship between kinetic energy and speed is $K.E. \propto v^2$.
This equation is of the form $y = kx^2$,which represents a parabola.
Therefore,the graph plotted between speed $(v)$ on the $x$-axis and kinetic energy $(K.E.)$ on the $y$-axis is a parabola.

(C) The kinetic energy $(K.E.)$ of a body is given by the formula $K.E. = \frac{1}{2}mv^2$.
Since the bodies are moving with the same linear velocity $(v_1 = v_2 = v)$,the kinetic energy is directly proportional to the mass $(K.E. \propto m)$.
Therefore,the ratio of their kinetic energies is equal to the ratio of their masses:
$\frac{K.E._1}{K.E._2} = \frac{m_1}{m_2}$.
Given that $\frac{K.E._1}{K.E._2} = \frac{4}{1}$,it follows that $\frac{m_1}{m_2} = \frac{4}{1}$.
Thus,the ratio of their masses is $4: 1$.

(C) The kinetic energy $(KE)$ of a body is given by the formula: $KE = \frac{1}{2} m v^2$,where $m$ is the mass and $v$ is the speed.
Let the initial mass be $m$ and the initial speed be $v$. Then the initial kinetic energy is $KE_i = \frac{1}{2} m v^2$.
According to the problem,the mass is doubled $(m' = 2m)$ and the speed is doubled $(v' = 2v)$.
The final kinetic energy $KE_f$ is given by:
$KE_f = \frac{1}{2} (2m) (2v)^2$
$KE_f = \frac{1}{2} (2m) (4v^2)$
$KE_f = 8 \times (\frac{1}{2} m v^2)$
$KE_f = 8 KE_i$
Therefore,the kinetic energy becomes eight times the initial kinetic energy.

(B) Let the mass of the man be $m$. Then the mass of the boy is $m_b = \frac{m}{2}$.
Let the initial speed of the man be $v_m$ and the speed of the boy be $v_b$.
According to the problem,the kinetic energy of the man is half the kinetic energy of the boy:
$\frac{1}{2} m v_m^2 = \frac{1}{2} (\frac{1}{2} m_b v_b^2) = \frac{1}{2} (\frac{1}{2} \cdot \frac{m}{2} v_b^2) = \frac{1}{8} m v_b^2$.
Simplifying,$v_m^2 = \frac{1}{4} v_b^2$,which gives $v_m = \frac{v_b}{2}$.
When the man speeds up by $1 \,ms^{-1}$,his new speed is $v_m' = v_m + 1 = \frac{v_b}{2} + 1$.
Now,his kinetic energy equals the boy's kinetic energy:
$\frac{1}{2} m (\frac{v_b}{2} + 1)^2 = \frac{1}{2} (\frac{m}{2}) v_b^2$.
Dividing both sides by $\frac{m}{2}$,we get $(\frac{v_b}{2} + 1)^2 = \frac{v_b^2}{2}$.
Taking the square root of both sides: $\frac{v_b}{2} + 1 = \frac{v_b}{\sqrt{2}}$.
$1 = v_b (\frac{1}{\sqrt{2}} - \frac{1}{2}) = v_b (\frac{2 - \sqrt{2}}{2\sqrt{2}})$.
$v_b = \frac{2\sqrt{2}}{2 - \sqrt{2}} = \frac{2\sqrt{2}(2 + \sqrt{2})}{4 - 2} = \frac{4\sqrt{2} + 4}{2} = 2\sqrt{2} + 2 = 2(\sqrt{2} + 1) \,ms^{-1}$.

(B) Let the particle fall from height $H$ and reach a height $h$ from the ground.
At this height $h$,the potential energy is $PE = mgh$.
The distance fallen by the particle is $x = H - h$.
The kinetic energy at this point is $KE = mgx = mg(H - h)$.
According to the problem,$KE = 2(PE)$.
Substituting the expressions,we get $mg(H - h) = 2(mgh)$.
$H - h = 2h \Rightarrow H = 3h \Rightarrow h = \frac{H}{3}$.
Now,to find the speed $v$ at height $h = \frac{H}{3}$,we use the equation of motion $v^2 = u^2 + 2ax$,where $u = 0$ and $x = H - h = H - \frac{H}{3} = \frac{2H}{3}$.
$v^2 = 2g(\frac{2H}{3}) = \frac{4gH}{3}$.
$v = \sqrt{\frac{4gH}{3}} = 2\sqrt{\frac{gH}{3}}$.
Thus,the height is $\frac{H}{3}$ and the speed is $2\sqrt{\frac{gH}{3}}$.

(D) The kinetic energy $(K)$ of a body is given by the formula $K = \frac{1}{2}mv^2$, where $m$ is the mass and $v$ is the magnitude of the velocity.
First, calculate the magnitude of the velocity vector $\vec{v} = (2 \hat{i} + 3 \hat{j} - 4 \hat{k}) \text{ m/s}$.
$v = |\vec{v}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \text{ m/s}$.
Thus, $v^2 = 29 \text{ m}^2/\text{s}^2$.
Given $K = 87 \text{ J}$, substitute the values into the kinetic energy formula:
$87 = \frac{1}{2} \times m \times 29$.
$87 = 14.5 \times m$.
$m = \frac{87}{14.5} = 6 \text{ kg}$.
Therefore, the mass of the body is $6 \text{ kg}$.

(C) Given: Mass $m = 4 \ kg$. Velocity vector $\vec{v} = (2 \hat{i} - 4 \hat{j} - \hat{k}) \ ms^{-1}$.
First,calculate the magnitude of the velocity $v = |\vec{v}| = \sqrt{(2)^2 + (-4)^2 + (-1)^2}$.
$v = \sqrt{4 + 16 + 1} = \sqrt{21} \ ms^{-1}$.
The kinetic energy $K.E.$ is given by the formula $K.E. = \frac{1}{2} mv^2$.
Substituting the values: $K.E. = \frac{1}{2} \times 4 \times (\sqrt{21})^2$.
$K.E. = 2 \times 21 = 42 \ J$.

(C) Given: Masses $m_1 = 1 \,g$ and $m_2 = 4 \,g$. Kinetic energies are equal, $K_1 = K_2$.
We know the relation between kinetic energy $K$ and linear momentum $p$ is $K = \frac{p^2}{2m}$, which implies $p = \sqrt{2mK}$.
Since $K_1 = K_2$, the ratio of momenta is $\frac{p_1}{p_2} = \frac{\sqrt{2m_1K_1}}{\sqrt{2m_2K_2}} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the values: $\frac{p_1}{p_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus, the ratio of their linear momenta is $1:2$.

(A) The relationship between kinetic energy $(K)$ and linear momentum $(p)$ is given by $K = \frac{p^2}{2m}$,which implies $p = \sqrt{2mK}$.
Given masses are $m_1 = m$ and $m_2 = 4m$.
Given kinetic energies are $K_1$ and $K_2$ such that $\frac{K_1}{K_2} = \frac{2}{1}$.
The ratio of their linear momenta is $\frac{p_1}{p_2} = \sqrt{\frac{2m_1 K_1}{2m_2 K_2}} = \sqrt{\frac{m_1}{m_2} \times \frac{K_1}{K_2}}$.
Substituting the values: $\frac{p_1}{p_2} = \sqrt{\frac{m}{4m} \times \frac{2}{1}} = \sqrt{\frac{1}{4} \times 2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Thus,the ratio of their linear momenta is $\frac{1}{\sqrt{2}}$.

(D) The relationship between kinetic energy $K$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.
Let the initial kinetic energy be $K_i$ and the final kinetic energy be $K_f$. Given that the kinetic energy changes by $20 \%$,we have $K_f = 1.20 K_i$.
Since $K \propto p^2$,we can write $\frac{K_f}{K_i} = \left( \frac{p_f}{p_i} \right)^2$.
Substituting the values,$1.20 = \left( \frac{p_f}{p_i} \right)^2$.
Taking the square root on both sides,$\frac{p_f}{p_i} = \sqrt{1.20} \approx 1.0954$.
The percentage change in momentum is given by $\frac{p_f - p_i}{p_i} \times 100 = (1.0954 - 1) \times 100 = 9.54 \%$.
Since $9.54 \%$ is not among the given options,the correct answer is $D$.

(B) The relationship between kinetic energy $K$ and momentum $P$ is given by $K = \frac{P^2}{2m}$.
Let the initial momentum be $P$ and the initial kinetic energy be $K = \frac{P^2}{2m}$.
If the momentum is increased by $20 \%$,the new momentum $P'$ becomes $P' = P + 0.20P = 1.2P$.
The new kinetic energy $K'$ is given by $K' = \frac{(P')^2}{2m} = \frac{(1.2P)^2}{2m} = \frac{1.44P^2}{2m}$.
Substituting $K = \frac{P^2}{2m}$,we get $K' = 1.44K$.
The percentage increase in kinetic energy is given by $\frac{K' - K}{K} \times 100 \% = \frac{1.44K - K}{K} \times 100 \% = 0.44 \times 100 \% = 44 \%$.