Kinetic Energy Questions in English

(C) The kinetic energy $E$ of a body of mass $m$ and momentum $p$ is given by the relation $E = \frac{p^2}{2m}$.
Rearranging this formula,we get $p = \sqrt{2mE}$.
Since the kinetic energies of the two bodies are equal $(E_1 = E_2 = E)$,the momentum $p$ is directly proportional to the square root of the mass: $p \propto \sqrt{m}$.
Therefore,the ratio of their momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Thus,the ratio $p_1:p_2$ is $\sqrt{m_1}:\sqrt{m_2}$.

(A) The kinetic energy $(K.E.)$ of a body of mass $m$ and momentum $P$ is given by the formula: $K.E. = \frac{P^2}{2m}$.
Since the momenta $(P)$ of both bodies are equal,the kinetic energy is inversely proportional to the mass of the body $(K.E. \propto \frac{1}{m})$.
Therefore,the body with the smaller mass (the light body) will have a greater kinetic energy.

(B) The classical expression for kinetic energy,$K = \frac{1}{2}mv^2$,is derived from Newtonian mechanics.
According to Einstein's theory of special relativity,the total energy of a body is given by $E = \gamma mc^2$,where $\gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$.
The kinetic energy is the difference between the total energy and the rest energy: $K = E - mc^2 = mc^2(\gamma - 1)$.
Using the binomial expansion for $\gamma$ when $v \ll c$,we get $\gamma \approx 1 + \frac{1}{2}\frac{v^2}{c^2}$.
Substituting this back,$K \approx mc^2(1 + \frac{1}{2}\frac{v^2}{c^2} - 1) = \frac{1}{2}mv^2$.
Thus,the classical formula is valid only when the velocity of the body is negligible compared to the speed of light.

(D) The relationship between kinetic energy $(E)$ and momentum $(P)$ for a body of mass $(m)$ is given by the formula:
$E = \frac{P^2}{2m}$
From this equation,we can see that kinetic energy is directly proportional to the square of the momentum:
$E \propto P^2$
If the momentum $(P)$ is increased $n$ times,the new momentum becomes $P' = nP$.
Substituting this into the proportionality:
$E' \propto (nP)^2 = n^2 P^2$
Therefore,the new kinetic energy $(E')$ becomes $n^2$ times the original kinetic energy $(E)$.
Thus,the kinetic energy increases by a factor of $n^2$.

(C) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since the kinetic energies $E$ are equal for both masses,we have $P \propto \sqrt{m}$.
Therefore,the ratio of their linear momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Substituting the given values $m_1 = 1 \,g$ and $m_2 = 4 \,g$,we get $\frac{P_1}{P_2} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(A) Let the initial kinetic energy be $E_1 = E$.
Final kinetic energy $E_2 = E + 300\% \text{ of } E = E + 3E = 4E$.
We know that the relationship between momentum $P$ and kinetic energy $E$ is $P = \sqrt{2mE}$,which implies $P \propto \sqrt{E}$.
Therefore,the ratio of final momentum $P_2$ to initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{\frac{4E}{E}} = \sqrt{4} = 2$.
This means $P_2 = 2P_1$.
The percentage change in momentum is given by $\frac{P_2 - P_1}{P_1} \times 100\% = \frac{2P_1 - P_1}{P_1} \times 100\% = 100\%$.
Thus,the momentum increases by $100\%$.

(B) The relationship between kinetic energy $E$ and momentum $P$ is given by $P = \sqrt{2mE}$.
Since the kinetic energy $E$ is equal for both bodies,we have $P \propto \sqrt{m}$.
This implies that the body with the larger mass $m$ will have a greater momentum $P$.
Therefore,the heavy body will possess greater momentum.

(C) Let the initial linear momentum be $P_1 = P$.
The new linear momentum is $P_2 = P_1 + 0.5P_1 = 1.5P_1 = \frac{3}{2}P_1$.
The relationship between kinetic energy $E$ and linear momentum $P$ is given by $E = \frac{P^2}{2m}$,which implies $E \propto P^2$.
Therefore,the ratio of the new kinetic energy $E_2$ to the initial kinetic energy $E_1$ is $\frac{E_2}{E_1} = \left( \frac{P_2}{P_1} \right)^2$.
Substituting the values,$\frac{E_2}{E_1} = \left( \frac{1.5P_1}{P_1} \right)^2 = (1.5)^2 = 2.25$.
This means $E_2 = 2.25 E_1 = E_1 + 1.25 E_1$.
The percentage increase in kinetic energy is $\frac{E_2 - E_1}{E_1} \times 100\% = 1.25 \times 100\% = 125\%.$

(D) The relationship between momentum $(P)$ and kinetic energy $(E)$ is given by the formula: $P = \sqrt{2mE}$.
Since mass $(m)$ is constant,we have $P \propto \sqrt{E}$.
If the kinetic energy is doubled,the new kinetic energy $E' = 2E$.
Therefore,the new momentum $P' = \sqrt{2m(2E)} = \sqrt{2} \times \sqrt{2mE} = \sqrt{2}P$.
Thus,the momentum increases by a factor of $\sqrt{2}$.

(A) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since $m$ is constant,$P \propto \sqrt{E}$.
Taking the derivative or using the approximation for small changes: $\frac{\Delta P}{P} \approx \frac{1}{2} \frac{\Delta E}{E}$.
Given the percentage increase in kinetic energy is $\frac{\Delta E}{E} \times 100 = 0.1\%$.
The percentage increase in momentum is $\frac{\Delta P}{P} \times 100 = \frac{1}{2} \times (0.1\%) = 0.05\%$.

(C) The kinetic energy $(K.E.)$ of a body of mass $m$ moving with velocity $v$ is given by the formula: $K.E. = \frac{1}{2}mv^2$.
From this relation,it is clear that $K.E. \propto v^2$.
If the new velocity $v' = 2v$,then the new kinetic energy $K.E.'$ will be:
$K.E.' = \frac{1}{2}m(v')^2 = \frac{1}{2}m(2v)^2 = \frac{1}{2}m(4v^2) = 4 \times (\frac{1}{2}mv^2)$.
Therefore,$K.E.' = 4 \times K.E.$
Thus,the kinetic energy becomes $4$ times the previous value.

(D) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since both bodies possess the same kinetic energy $(E_A = E_B = E)$,the ratio of their linear momenta is given by:
$\frac{P_A}{P_B} = \frac{\sqrt{2m_A E}}{\sqrt{2m_B E}} = \sqrt{\frac{m_A}{m_B}}$.
Given the ratio of masses is $\frac{m_A}{m_B} = \frac{3}{1}$,we substitute this into the equation:
$\frac{P_A}{P_B} = \sqrt{\frac{3}{1}} = \sqrt{3} : 1$.
Therefore,the correct option is $D$.

(B) The kinetic energy $E$ of a body with momentum $P$ and mass $m$ is given by the formula $E = \frac{P^2}{2m}$.
Since the momentum $P$ is the same for both bodies,we have $E \propto \frac{1}{m}$.
Therefore,the ratio of their kinetic energies is $\frac{E_1}{E_2} = \frac{m_2}{m_1}$.
Given $m_1 = m$ and $m_2 = 2m$,we get $\frac{E_1}{E_2} = \frac{2m}{m} = \frac{2}{1}$.
Thus,the ratio is $2:1$.

(D) The kinetic energy $E$ of a body with mass $m$ and linear momentum $p$ is given by the relation: $E = \frac{p^2}{2m}$.
Since the linear momentum $p$ is constant for both bodies,we have $E \propto \frac{1}{m}$,which implies $m \propto \frac{1}{E}$.
Given the ratio of kinetic energies is $\frac{E_1}{E_2} = \frac{4}{1}$.
Therefore,the ratio of their masses is $\frac{m_1}{m_2} = \frac{E_2}{E_1} = \frac{1}{4}$.
Thus,the ratio of their masses is $1:4$.

(A) The relationship between kinetic energy $E$ and momentum $P$ is given by the formula $P = \sqrt{2mE}$.
Since $m$ is constant,we have $P \propto \sqrt{E}$.
Let the initial kinetic energy be $E_1$ and initial momentum be $P_1$. Then $P_1 = \sqrt{2mE_1}$.
Let the new kinetic energy be $E_2 = 4E_1$ and the new momentum be $P_2$.
Then $P_2 = \sqrt{2mE_2} = \sqrt{2m(4E_1)} = 2\sqrt{2mE_1} = 2P_1$.
Therefore,the new momentum becomes twice its initial value.

(C) The relationship between momentum $p$,mass $m$,and kinetic energy $E$ is given by $p = \sqrt{2mE}$.
For two bodies,the ratio of their momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2} \times \frac{E_1}{E_2}}$.
Given $m_1 = 2m$,$m_2 = m$,and $\frac{E_1}{E_2} = \frac{8}{1}$.
Substituting these values,we get $\frac{p_1}{p_2} = \sqrt{\frac{2m}{m} \times \frac{8}{1}} = \sqrt{2 \times 8} = \sqrt{16} = 4$.
Therefore,the ratio of their momenta is $4:1$.

(C) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Assuming mass $m$ is constant,we have $P \propto \sqrt{E}$.
Let the initial kinetic energy be $E_1$ and the final kinetic energy be $E_2 = E_1 + 0.22E_1 = 1.22E_1$.
The ratio of the final momentum $P_2$ to the initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{1.22} \approx 1.1$.
Thus,$P_2 = 1.1P_1 = (1 + 0.1)P_1 = P_1 + 0.1P_1$.
This represents an increase of $0.1 \times 100\% = 10\%$ in the momentum.

(A) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.
Since mass $m$ is constant,we have $E \propto P^2$.
Let the initial momentum be $P_1 = P$ and the final momentum be $P_2 = P + 0.20P = 1.2P$.
The ratio of kinetic energies is $\frac{E_2}{E_1} = \left( \frac{P_2}{P_1} \right)^2 = \left( \frac{1.2P}{P} \right)^2 = (1.2)^2 = 1.44$.
This implies $E_2 = 1.44 E_1$.
The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} \times 100\% = (1.44 - 1) \times 100\% = 0.44 \times 100\% = 44\%$.
Therefore,the kinetic energy increases by $44\%$.

(A) The relationship between kinetic energy $(E)$,momentum $(P)$,and mass $(m)$ is given by the formula: $E = \frac{P^2}{2m}$.
Given:
Mass $(m)$ = $2 \ kg$
Momentum $(P)$ = $2 \ Ns$
Substituting these values into the formula:
$E = \frac{(2)^2}{2 \times 2}$
$E = \frac{4}{4}$
$E = 1 \ J$.
Therefore,the kinetic energy is $1 \ J$.

(B) The relationship between kinetic energy $(E)$ and momentum $(P)$ for an object of mass $(m)$ is given by the formula:
$E = \frac{P^2}{2m}$
Given:
Mass $(m)$ = $1 \, kg$
Momentum $(P)$ = $10 \, kg \cdot m/s$
Substituting the values into the formula:
$E = \frac{(10)^2}{2 \times 1}$
$E = \frac{100}{2} = 50 \, J$
Therefore,the kinetic energy of the object is $50 \, J$.

(C) Let $M$ be the mass of the man and $m$ be the mass of the boy. Given $m = M/2$.
Let $V$ be the initial velocity of the man and $v$ be the velocity of the boy.
The kinetic energy of the man is $K_M = \frac{1}{2} M V^2$ and the kinetic energy of the boy is $K_b = \frac{1}{2} m v^2$.
According to the problem,$K_M = \frac{1}{2} K_b$,so $\frac{1}{2} M V^2 = \frac{1}{2} (\frac{1}{2} m v^2) = \frac{1}{4} m v^2$. Substituting $m = M/2$,we get $\frac{1}{2} M V^2 = \frac{1}{4} (M/2) v^2$,which simplifies to $V^2 = v^2/4$,or $v = 2V$.
When the man increases his speed by $1 \, m/s$,his new kinetic energy is $\frac{1}{2} M (V+1)^2$. This is equal to the boy's kinetic energy $K_b = \frac{1}{2} m v^2 = \frac{1}{2} (M/2) (2V)^2 = M V^2$.
Thus,$\frac{1}{2} M (V+1)^2 = M V^2$.
Dividing by $M/2$,we get $(V+1)^2 = 2V^2$.
Taking the square root,$V+1 = \sqrt{2} V$.
Rearranging gives $1 = V(\sqrt{2} - 1)$,so $V = \frac{1}{\sqrt{2} - 1} \, m/s$.

(D) The relationship between momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.
Since the kinetic energies $E$ of both substances are the same,we have $P \propto \sqrt{m}$.
Therefore,the ratio of their momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 4 \text{ g}$ and $m_2 = 9 \text{ g}$,we substitute these values:
$\frac{P_1}{P_2} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.
Thus,the ratio of their momenta is $2:3$.

(D) The kinetic energy $E$ of a body with momentum $P$ and mass $m$ is given by $E = \frac{P^2}{2m}$.
Let the initial momentum be $P_1 = P$ and the initial kinetic energy be $E_1 = \frac{P^2}{2m}$.
Since the momentum is increased by $100\%$,the new momentum $P_2 = P + 100\% \text{ of } P = P + P = 2P$.
The new kinetic energy $E_2$ is given by $E_2 = \frac{P_2^2}{2m} = \frac{(2P)^2}{2m} = 4 \left( \frac{P^2}{2m} \right) = 4E_1$.
The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} \times 100\%$.
Substituting the values,we get $\frac{4E_1 - E_1}{E_1} \times 100\% = \frac{3E_1}{E_1} \times 100\% = 300\%$.

(B) The relationship between linear momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.
Since the kinetic energy $E$ is equal for both masses,we have $P \propto \sqrt{m}$.
Therefore,the ratio of the linear momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1\,kg$ and $m_2 = 16\,kg$,we get $\frac{P_1}{P_2} = \sqrt{\frac{1}{16}} = \frac{1}{4}$.
Thus,the ratio is $1:4$.

(B) According to Newton's second law,the impulse applied to the particle is equal to the change in momentum.
Impulse $J = F \times t = \Delta P$.
Since the particle starts from rest,its initial momentum $P_i = 0$.
Therefore,the final momentum $P_f = F \times t$.
The kinetic energy $E$ of a particle is given by the formula $E = \frac{P^2}{2m}$.
Substituting the value of $P_f$,we get $E = \frac{(Ft)^2}{2m} = \frac{F^2 t^2}{2m}$.

(C) The relationship between linear momentum $(p)$,mass $(m)$,and kinetic energy $(K.E.)$ is given by the formula: $p = \sqrt{2m(K.E.)}$.
Since the kinetic energy $(K.E.)$ is equal for both bodies,we have $p \propto \sqrt{m}$.
Therefore,the ratio of their linear momentums is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = m$ and $m_2 = 4m$,we substitute these values:
$\frac{p_1}{p_2} = \sqrt{\frac{m}{4m}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.
Thus,the ratio is $1:2$.

(B) The relationship between kinetic energy $(E)$,momentum $(P)$,and mass $(m)$ is given by the formula: $E = \frac{P^2}{2m}$.
Given:
Mass $(m)$ = $3 \,kg$
Momentum $(P)$ = $2 \,N-s$
Substituting the values into the formula:
$E = \frac{(2)^2}{2 \times 3}$
$E = \frac{4}{6}$
$E = \frac{2}{3} \,J$
Therefore,the correct option is $B$.

(A) The kinetic energy $E$ of a particle with momentum $P$ and mass $m$ is given by the formula $E = \frac{P^2}{2m}$.
Since both particles have the same momentum $(P_1 = P_2 = P)$,the kinetic energy is inversely proportional to the mass: $E \propto \frac{1}{m}$.
Given that $m_1 > m_2$,it follows that the kinetic energy of the first particle must be less than that of the second particle.
Therefore,$E_1 < E_2$.

(C) The kinetic energy $(KE)$ of an object is given by the formula $KE = \frac{1}{2}mv^2$.
Since both balls are dropped from rest and fall through the same height $(h = 30\,ft)$,they will have the same final velocity $(v)$ according to the equation of motion $v^2 = u^2 + 2gh$,where $u = 0$.
Since $v$ is the same for both balls,the kinetic energy is directly proportional to the mass $(KE \propto m)$.
Therefore,the ratio of their kinetic energies is $\frac{KE_1}{KE_2} = \frac{m_1}{m_2}$.
Given $m_1 = 2\,kg$ and $m_2 = 4\,kg$,the ratio is $\frac{2}{4} = \frac{1}{2}$.

(B) The kinetic energy $E$ of a particle with momentum $P$ and mass $m$ is given by the formula $E = \frac{P^2}{2m}$.
Since the momentum $P$ is constant for all particles,we have $E \propto \frac{1}{m}$.
This implies that the particle with the smallest mass will possess the maximum kinetic energy.
Comparing the masses of the given particles: $m_{\text{electron}} < m_{\text{proton}} < m_{\text{deuteron}} < m_{\alpha\text{-particle}}$.
Therefore,the electron,being the lightest particle,has the maximum kinetic energy.

(C) Let the initial speed of the man be $v$ and his mass be $m$.
Initial kinetic energy is $E = \frac{1}{2}mv^2$ ... $(i)$
When the speed increases by $2 \, m/s$,the new speed becomes $(v + 2) \, m/s$.
The new kinetic energy is $2E = \frac{1}{2}m(v + 2)^2$ ... (ii)
Dividing equation (ii) by equation $(i)$:
$\frac{2E}{E} = \frac{\frac{1}{2}m(v + 2)^2}{\frac{1}{2}mv^2}$
$2 = \frac{(v + 2)^2}{v^2}$
Taking the square root on both sides:
$\sqrt{2} = \frac{v + 2}{v}$
$v\sqrt{2} = v + 2$
$v(\sqrt{2} - 1) = 2$
$v = \frac{2}{\sqrt{2} - 1}$
Rationalizing the denominator:
$v = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2\sqrt{2} + 2}{2 - 1} = 2 + 2\sqrt{2} \, m/s$.

(A) Energy is defined as the capacity to do work. Among the given options, $\text{Light}$ is a form of electromagnetic energy.
$\text{Pressure}$ is defined as force per unit area.
$\text{Momentum}$ is the product of mass and velocity.
$\text{Power}$ is the rate at which work is done.
Therefore, $\text{Light}$ is the correct answer.

(D) The relationship between kinetic energy $(K)$,momentum $(P)$,and mass $(m)$ is given by $K = \frac{P^2}{2m}$.
Since the kinetic energies are the same $(K_1 = K_2)$,we have $\frac{P_1^2}{2m_1} = \frac{P_2^2}{2m_2}$.
Rearranging the terms,we get $\frac{P_1^2}{P_2^2} = \frac{m_1}{m_2}$.
Taking the square root on both sides,the ratio of their momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1\, g$ and $m_2 = 9\, g$,we substitute these values:
$\frac{P_1}{P_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Therefore,the ratio of their momentum is $1:3$.

(B) The relationship between kinetic energy $E$ and momentum $P$ is given by $E = \frac{P^2}{2m}$.
Since the mass $m$ of the body is constant,we have $E \propto P^2$.
Taking the logarithmic derivative,we get $\frac{dE}{E} = 2 \frac{dP}{P}$.
For small percentage changes,the percentage increase in kinetic energy is approximately twice the percentage increase in momentum.
Given that the percentage increase in momentum is $0.01\%$,the percentage increase in kinetic energy is $2 \times 0.01\% = 0.02\%$.

(B) According to the work-energy theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
Since the object starts from rest,the initial kinetic energy is $0$.
Work done $W = F \times d$.
Therefore,the final kinetic energy $K.E. = F \times d$.
Since the force $F$ and the distance $d$ are constant,the kinetic energy acquired depends only on the product of the force and the distance.
It does not depend on the mass $m$ of the object.
Thus,the kinetic energy is independent of $m$.

(A) Let the particle be projected horizontally with velocity $u$ at $t = 0$. The horizontal component of velocity remains constant, $v_x = u$. The vertical component of velocity at time $t$ is $v_y = gt$. The total velocity $v$ is given by $v^2 = v_x^2 + v_y^2 = u^2 + (gt)^2$. The kinetic energy $E$ is given by $E = \frac{1}{2}mv^2 = \frac{1}{2}m(u^2 + g^2t^2)$. This equation represents a parabola of the form $E = At^2 + B$, where $A = \frac{1}{2}mg^2$ and $B = \frac{1}{2}mu^2$. Since $u > 0$, the initial kinetic energy at $t = 0$ is $E_0 = \frac{1}{2}mu^2 > 0$. As $t$ increases, $E$ increases parabolically. Therefore, the correct graph is shown in option $A$.

(C) The relationship between kinetic energy $E$ and momentum $p$ is given by $E = \frac{p^2}{2m}$,where $m$ is the mass of the particle.
Taking the square root on both sides,we get $\sqrt{E} = \frac{p}{\sqrt{2m}}$.
This can be rewritten as $\sqrt{E} = \frac{1}{\sqrt{2m}} \cdot p$.
However,the question asks for the graph between $\sqrt{E}$ and $\frac{1}{p}$.
From $E = \frac{p^2}{2m}$,we have $p = \sqrt{2mE}$.
Therefore,$\frac{1}{p} = \frac{1}{\sqrt{2mE}} = \frac{1}{\sqrt{2m}} \cdot \frac{1}{\sqrt{E}}$.
This implies $\sqrt{E} = \frac{1}{\sqrt{2m}} \cdot \frac{1}{(1/p)}$.
Let $y = \sqrt{E}$ and $x = \frac{1}{p}$. Then $y = \frac{k}{x}$,where $k = \frac{1}{\sqrt{2m}}$ is a constant.
This is the equation of a rectangular hyperbola. Thus,the graph between $\sqrt{E}$ and $\frac{1}{p}$ is a hyperbola.

(A) The kinetic energy $E$ of an object of mass $m$ moving with velocity $v$ is given by the formula:
$E = \frac{1}{2}mv^2$
This equation shows that $E$ is directly proportional to the square of the velocity $(E \propto v^2)$.
This relationship represents a parabola that is symmetric about the $E$-axis,where $E$ increases as the magnitude of $v$ increases in either the positive or negative direction. Therefore,the correct graph is the one showing a parabolic curve opening upwards,which corresponds to option $A$.

(C) The relationship between kinetic energy $(K)$, momentum $(p)$, and mass $(m)$ is given by the formula $K = \frac{p^2}{2m}$.
From this, we can express momentum as $p = \sqrt{2mK}$.
Given that the kinetic energies are the same $(K_1 = K_2 = K)$, the ratio of the momenta of the two particles is:
$\frac{p_1}{p_2} = \frac{\sqrt{2m_1K}}{\sqrt{2m_2K}} = \sqrt{\frac{m_1}{m_2}}$.
Given the ratio of masses is $\frac{m_1}{m_2} = \frac{1}{2}$, we substitute this into the equation:
$\frac{p_1}{p_2} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$.
Therefore, the ratio of their momenta is $1 : \sqrt{2}$.

(C) According to the law of conservation of energy,the speed of the body when it returns to the point of projection will be equal to the speed with which it was projected.
Since the body is projected upwards with a velocity $v = 2 \ m \ s^{-1}$,it will return to the ground with the same velocity $v = 2 \ m \ s^{-1}$.
The kinetic energy $K$ is given by the formula $K = \frac{1}{2}mv^2$.
Substituting the values $m = 2 \ kg$ and $v = 2 \ m \ s^{-1}$:
$K = \frac{1}{2} \times 2 \times (2)^2 = 4 \ J$.

(C) Let the initial kinetic energy be $K_1$ and the final kinetic energy be $K_2$.
Given that the kinetic energy decreases by $19\%$,so $K_2 = K_1 - 0.19 K_1 = 0.81 K_1$.
We know the relation between kinetic energy $(K)$ and momentum $(p)$ is $K = \frac{p^2}{2m}$.
Therefore,$\frac{K_2}{K_1} = \left( \frac{p_2}{p_1} \right)^2$.
Substituting the values: $0.81 = \left( \frac{p_2}{p_1} \right)^2$.
Taking the square root on both sides: $\frac{p_2}{p_1} = \sqrt{0.81} = 0.9$.
This means $p_2 = 0.9 p_1$.
The percentage decrease in momentum is given by $\frac{p_1 - p_2}{p_1} \times 100\%$.
$= \frac{p_1 - 0.9 p_1}{p_1} \times 100\% = 0.1 \times 100\% = 10\%$.

(B) According to the Work-Energy Theorem,the work done by the net force on an object is equal to the change in its kinetic energy.
$W = \Delta K.E. = K.E._{final} - K.E._{initial}$
Since the body starts from rest,the initial kinetic energy $K.E._{initial} = 0$.
Thus,$K.E. = W = F \cdot d$,where $F$ is the constant force and $d$ is the fixed distance.
Since both $F$ and $d$ are constant,the kinetic energy $K.E.$ is independent of the mass $m$ of the body.

(C) Given that the force $(F)$ is inversely proportional to the speed $(v)$,we have $F = \frac{k}{v}$,where $k$ is a constant.
Using Newton's second law,$F = ma = m \frac{dv}{dt}$.
Substituting $F$,we get $m \frac{dv}{dt} = \frac{k}{v}$.
Rearranging the terms,we get $mv \, dv = k \, dt$.
Integrating both sides,we get $\int mv \, dv = \int k \, dt$,which gives $\frac{1}{2} mv^2 = kt + C$.
Assuming the body starts from rest at $t = 0$,the constant $C = 0$.
Thus,the kinetic energy $(KE = \frac{1}{2} mv^2)$ is equal to $kt$.
Therefore,the kinetic energy is linearly related to time.

(B) The expression for kinetic energy,$K = \frac{1}{2} mv^2$,is derived from classical mechanics (Newtonian mechanics).
In classical mechanics,the mass $m$ of an object is considered a constant,independent of its velocity.
However,according to Einstein's theory of special relativity,as an object's velocity $v$ approaches the speed of light $c$,its relativistic mass increases according to the formula $m = \frac{m_0}{\sqrt{1 - v^2/c^2}}$.
Therefore,the classical formula $K = \frac{1}{2} mv^2$ is only valid when the velocity $v$ is much smaller than the speed of light $(v \ll c)$,meaning the velocity is negligible compared to the speed of light.

(C) The kinetic energy $K$ of a body with mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Rearranging this for momentum,we get $p = \sqrt{2mK}$.
Since the kinetic energy $K$ is the same for both bodies,we have $p \propto \sqrt{m}$.
Therefore,the ratio of their momenta is $\frac{p_1}{p_2} = \sqrt{\frac{m_1}{m_2}}$.
Thus,$p_1 : p_2 = \sqrt{m_1} : \sqrt{m_2}$.

(D) The kinetic energy $(KE)$ of an object is given by $KE = \frac{P^2}{2M}$,where $P$ is the momentum and $M$ is the mass.
Given that both objects have the same kinetic energy,we have $KE_1 = KE_2$.
This implies $\frac{P_1^2}{2M_1} = \frac{P_2^2}{2M_2}$.
Rearranging the terms,we get $\frac{P_1^2}{P_2^2} = \frac{M_1}{M_2}$,which simplifies to $\frac{P_1}{P_2} = \sqrt{\frac{M_1}{M_2}}$.
Since the second object is heavier,we have $M_2 > M_1$.
Therefore,$\sqrt{\frac{M_2}{M_1}} > 1$,which means $P_2 > P_1$.
Thus,the momentum of the heavy object is greater than the momentum of the light object,i.e.,$M_2 V_2 > M_1 V_1$.