Kinetic Energy Questions in English

(D) According to the work-energy theorem,the work done $(W)$ on a particle is equal to the change in its kinetic energy $(K)$.
Since the particle starts from rest,the initial kinetic energy is $0$.
Therefore,$W = K = \frac{1}{2}mv^2$.
Here,$m$ is the mass of the particle,which is constant.
Thus,$W \propto v^2$.
This relationship represents a parabola opening upwards along the $W$-axis.
Comparing this with the given options,the graph in option $D$ represents a parabolic curve where $W$ increases with the square of $v$.

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

(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.
Work done $(W)$ = Force $(F)$ $\times$ Displacement $(s)$.
Given: $F = 10 \ N$,$s = 1 \ m$.
$W = 10 \ N \times 1 \ m = 10 \ J$.
Since the object starts from rest (implied),the change in kinetic energy is equal to the work done.
Therefore,the kinetic energy gained = $10 \ J$.

(B) Given:
Mass,$m = 3 \ kg$
Momentum,$p = 2 \ Ns$
The relationship between kinetic energy $(K)$ and momentum $(p)$ is given by the formula:
$K = \frac{p^2}{2m}$
Substituting the given values:
$K = \frac{(2)^2}{2 \times 3}$
$K = \frac{4}{6}$
$K = \frac{2}{3} \ J$
Therefore,the kinetic energy of the object is $\frac{2}{3} \ J$.

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

(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 object.
From this,we can write $p = \sqrt{2mK}$.
Let the initial kinetic energy be $K_1$ and the initial momentum be $p_1 = \sqrt{2mK_1}$.
Let the new kinetic energy be $K_2 = 4K_1$.
The new momentum $p_2$ is given by $p_2 = \sqrt{2mK_2} = \sqrt{2m(4K_1)}$.
$p_2 = \sqrt{4} \times \sqrt{2mK_1} = 2 \times p_1$.
Therefore,the new momentum is double its initial value.

(D) The average velocity is given by $\bar{v} = \frac{d}{t}$.
Given $d = 100 \ m$ and $t = 10 \ s$,we have $\bar{v} = \frac{100}{10} = 10 \ m/s$.
Assuming the mass of an athlete $(m)$ ranges from $50 \ kg$ to $100 \ kg$,we calculate the kinetic energy $(K = \frac{1}{2}mv^2)$:
For $m_1 = 50 \ kg$: $K_1 = \frac{1}{2} \times 50 \times (10)^2 = 25 \times 100 = 2,500 \ J$.
For $m_2 = 100 \ kg$: $K_2 = \frac{1}{2} \times 100 \times (10)^2 = 50 \times 100 = 5,000 \ J$.
Thus,the estimated range of kinetic energy is $2,000 \ J - 5,000 \ J$.

(C) Let the initial speed be $u$ and the mass of the man be $m$.
The initial kinetic energy is $E_1 = \frac{1}{2}mu^2$.
The final speed is $(u + 2) \ m/s$.
The final kinetic energy is $E_2 = \frac{1}{2}m(u + 2)^2$.
Given that the kinetic energy doubles, $E_2 = 2E_1$.
Substituting the expressions: $\frac{1}{2}m(u + 2)^2 = 2 \times \frac{1}{2}mu^2$.
Simplifying: $(u + 2)^2 = 2u^2$.
Taking the square root on both sides: $u + 2 = \sqrt{2}u$.
Rearranging for $u$: $\sqrt{2}u - u = 2 \implies u(\sqrt{2} - 1) = 2$.
Solving for $u$: $u = \frac{2}{\sqrt{2} - 1}$.
Rationalizing the denominator: $u = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2(\sqrt{2} + 1)}{2 - 1} = 2\sqrt{2} + 2 \ m/s$.

(C) Let the original speed of the vehicle be $v$.
The kinetic energy $K_1 = \frac{1}{2}mv^2$.
When the speed increases by $2 \ m/s$,the new speed is $(v + 2) \ m/s$.
The new kinetic energy $K_2 = \frac{1}{2}m(v + 2)^2$.
According to the problem,$K_2 = 2K_1$.
Therefore,$\frac{1}{2}m(v + 2)^2 = 2 \times (\frac{1}{2}mv^2)$.
$(v + 2)^2 = 2v^2$.
Taking the square root on both sides: $v + 2 = \sqrt{2}v$.
Rearranging the terms: $2 = \sqrt{2}v - v = v(\sqrt{2} - 1)$.
$v = \frac{2}{\sqrt{2} - 1}$.
Rationalizing the denominator: $v = \frac{2(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{2(\sqrt{2} + 1)}{2 - 1} = 2(\sqrt{2} + 1) \ m/s$.

(A) The relation between kinetic energy $(KE)$ and momentum $(p)$ is given by $KE = \frac{p^2}{2m}$,which implies $p = \sqrt{2m \cdot KE}$.
Let the initial kinetic energy be $KE_1 = KE$ and the initial momentum be $p_1 = \sqrt{2m \cdot KE}$.
The kinetic energy increases by $300\%$,so the new kinetic energy is $KE_2 = KE + 300\% \text{ of } KE = KE + 3KE = 4KE$.
The new momentum is $p_2 = \sqrt{2m \cdot KE_2} = \sqrt{2m \cdot 4KE} = 2 \sqrt{2m \cdot KE} = 2p_1$.
The percentage increase in momentum is given by $\frac{p_2 - p_1}{p_1} \times 100\%$.
Substituting the values: $\frac{2p_1 - p_1}{p_1} \times 100\% = \frac{p_1}{p_1} \times 100\% = 100\%$.

(C) The kinetic energy $E$ is related to momentum $p$ by the formula $E = \frac{p^2}{2m}$.
Let the initial momentum be $p_1 = p$ and the initial kinetic energy be $E_1 = \frac{p^2}{2m}$.
Given that the momentum increases by $100\ \%$,the new momentum $p_2 = p + 100\% \text{ of } p = p + p = 2p$.
The new kinetic energy $E_2 = \frac{p_2^2}{2m} = \frac{(2p)^2}{2m} = 4 \left( \frac{p^2}{2m} \right) = 4E_1$.
The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} \times 100\ \%$.
Substituting the values: $\frac{4E_1 - E_1}{E_1} \times 100\ \% = \frac{3E_1}{E_1} \times 100\ \% = 300\ \%$.

(A) The kinetic energy $E$ of a particle with momentum $p$ and mass $m$ is given by the relation $E = \frac{p^2}{2m}$.
Since the momentum $p$ is constant for both particles,we have $E \propto \frac{1}{m}$.
Given that $m_1 > m_2$,it follows that the kinetic energy is inversely proportional to the mass.
Therefore,$E_1 < E_2$.

(B) The relation between kinetic energy $(E)$ and momentum $(p)$ is given by $E = \frac{p^2}{2m}$.
Since mass $(m)$ is constant,$E \propto p^2$.
Let the initial momentum be $p_1 = 100$ and initial kinetic energy be $E_1$.
If the momentum increases by $50\%$,the new momentum is $p_2 = 100 + 50 = 150$.
The new kinetic energy is $E_2 = \frac{p_2^2}{2m}$.
The percentage increase in kinetic energy is given by $\frac{E_2 - E_1}{E_1} \times 100$.
Substituting the values: $\frac{(150)^2 - (100)^2}{(100)^2} \times 100$.
$= \frac{22500 - 10000}{10000} \times 100 = \frac{12500}{10000} \times 100 = 125\%$.
Thus,the kinetic energy increases by $125\%$.

(A) The relationship between linear momentum $p$ and kinetic energy $E$ is given by $p = \sqrt{2mE}$.
Since mass $m$ is constant,$p \propto \sqrt{E}$.
Let the initial kinetic energy be $E_1 = E$.
The final kinetic energy is $E_2 = E + 300\% \text{ of } E = E + 3E = 4E$.
The initial momentum is $p_1 = \sqrt{2mE}$.
The final momentum is $p_2 = \sqrt{2m(4E)} = 2\sqrt{2mE} = 2p_1$.
The percentage increase in momentum is given by $\frac{p_2 - p_1}{p_1} \times 100\%$.
Substituting the values: $\frac{2p_1 - p_1}{p_1} \times 100\% = \frac{p_1}{p_1} \times 100\% = 100\%$.

(D) The formula for centripetal force is $F_c = \frac{mv^2}{r}$.
Given $F_c = 10 \, N$ and $r = 20 \, cm = 0.2 \, m$.
Substituting the values: $\frac{mv^2}{0.2} = 10$.
This implies $mv^2 = 10 \times 0.2 = 2 \, J$.
The kinetic energy $K$ is given by $K = \frac{1}{2}mv^2$.
Substituting $mv^2 = 2$,we get $K = \frac{1}{2} \times 2 = 1 \, J$.

(C) Let the mass of the man be $M$ and his initial speed be $V$. Let the mass of the boy be $m = M/2$ and his speed be $v$.
Given,kinetic energy of the man is half that of the boy:
$\frac{1}{2} M V^2 = \frac{1}{2} \left( \frac{1}{2} m v^2 \right)$
Substituting $m = M/2$:
$\frac{1}{2} M V^2 = \frac{1}{2} \left( \frac{1}{2} \cdot \frac{M}{2} \cdot v^2 \right) \implies V^2 = \frac{v^2}{4} \implies v = 2V$ ... $(i)$
When the man increases his speed by $1 \ m/s$,his kinetic energy becomes equal to the boy's kinetic energy:
$\frac{1}{2} M (V + 1)^2 = \frac{1}{2} m v^2$
$\frac{1}{2} M (V + 1)^2 = \frac{1}{2} \left( \frac{M}{2} \right) v^2$
$(V + 1)^2 = \frac{v^2}{2}$
Taking the square root: $V + 1 = \frac{v}{\sqrt{2}}$ ... (ii)
Substitute $v = 2V$ from $(i)$ into (ii):
$V + 1 = \frac{2V}{\sqrt{2}} = \sqrt{2} V$
$1 = V(\sqrt{2} - 1)$
$V = \frac{1}{\sqrt{2} - 1} \ m/s$.

(B) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.
Since $m$ is constant,$K \propto p^2$.
Taking the logarithmic derivative,we get $\frac{\Delta K}{K} = 2 \frac{\Delta p}{p}$.
Given the percentage change in momentum $\frac{\Delta p}{p} \times 100 = 0.01\%$.
Therefore,the percentage change in kinetic energy is $\frac{\Delta K}{K} \times 100 = 2 \times (0.01\%) = 0.02\%$.

(C) The kinetic energy $E$ is related to momentum $P$ by the formula $E = \frac{P^2}{2m}$.
Let the initial momentum be $P_1 = P$ and the initial kinetic energy be $E_1 = \frac{P^2}{2m}$.
Since the momentum increases 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 $\frac{4E_1 - E_1}{E_1} \times 100 \% = \frac{3E_1}{E_1} \times 100 \% = 300 \%$.

(C) The relationship between momentum $P$,mass $m$,and kinetic energy $E$ is given by the formula $P = \sqrt{2mE}$.
Since the kinetic energy $E$ is the same for both bodies,we have $P \propto \sqrt{m}$.
Therefore,the ratio of their momenta is $\frac{P_1}{P_2} = \sqrt{\frac{m_1}{m_2}}$.
Given $m_1 = 1 \ g$ and $m_2 = 9 \ g$,we substitute these values:
$\frac{P_1}{P_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}$.
Thus,the ratio of their momenta is $1 : 3$.

(C) Given: Mass $m = 300 \ g = 0.3 \ kg$.
Velocity vector $\vec{v} = (3\hat{i} + 4\hat{j}) \ m/s$.
The magnitude of velocity is $v = |\vec{v}| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \ m/s$.
The formula for kinetic energy is $K = \frac{1}{2}mv^2$.
Substituting the values: $K = \frac{1}{2} \times 0.3 \times (5)^2$.
$K = 0.15 \times 25 = 3.75 \ J$.

(C) The relationship between momentum $P$,mass $m$,and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since the kinetic energy $E$ 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,the correct option is $C$.

(C) The kinetic energy $K$ of a body of mass $m$ moving with speed $v$ is given by the formula $K = \frac{1}{2}mv^2$.
Since the mass $m$ is constant,we have $K \propto v^2$.
This equation is of the form $y = ax^2$,which represents a parabola.
Therefore,the graph plotted between speed $v$ (on the x-axis) and kinetic energy $K$ (on the y-axis) is a parabola.

(B) The relationship between momentum $P$ and kinetic energy $E$ is given by $P = \sqrt{2mE}$.
Since $m$ is constant,$P \propto \sqrt{E}$.
If the kinetic energy decreases by $36\%$,the new kinetic energy $E_2$ is $100\% - 36\% = 64\%$ of the initial kinetic energy $E_1$.
So,$E_2 = 0.64 E_1$.
The ratio of the new momentum $P_2$ to the initial momentum $P_1$ is $\frac{P_2}{P_1} = \sqrt{\frac{E_2}{E_1}} = \sqrt{0.64} = 0.8$.
This means $P_2 = 0.8 P_1$,which is $80\%$ of the initial momentum.
The decrease in momentum is $100\% - 80\% = 20\%$.

(A) The relation between kinetic energy $(K)$ and momentum $(p)$ is given by $K = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Taking the natural logarithm on both sides:
$\ln K = \ln \left( \frac{p^2}{2m} \right)$
$\ln K = \ln(p^2) - \ln(2m)$
$\ln K = 2 \ln p - \ln(2m)$
Rearranging the terms to express $\ln p$ in terms of $\ln K$:
$2 \ln p = \ln K + \ln(2m)$
$\ln p = \frac{1}{2} \ln K + \frac{1}{2} \ln(2m)$
This equation is of the form $y = mx + c$,where $y = \ln p$,$x = \ln K$,the slope $m = \frac{1}{2}$,and the intercept $c = \frac{1}{2} \ln(2m)$.
Since the slope is positive and constant,the graph of $\ln p$ versus $\ln K$ is a straight line.

(C) The kinetic energy $(KE)$ of a body is related to its momentum $(p)$ by the formula: $KE = \frac{p^2}{2m}$,where $m$ is the mass of the body.
Let the initial momentum be $p_1 = p$. Then the initial kinetic energy is $KE_1 = \frac{p^2}{2m}$.
When the momentum increases by $100\%$,the new momentum $p_2$ becomes:
$p_2 = p + 100\% \text{ of } p = p + p = 2p$.
The new kinetic energy $KE_2$ is:
$KE_2 = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4 \times KE_1$.
The percentage increase in kinetic energy is given by:
$\text{Percentage Increase} = \frac{KE_2 - KE_1}{KE_1} \times 100\%$
$= \frac{4 KE_1 - KE_1}{KE_1} \times 100\%$
$= \frac{3 KE_1}{KE_1} \times 100\% = 300\%$.

(A) The average speed of the athlete is $v = \frac{100 \ m}{10 \ s} = 10 \ m/s$.
The kinetic energy $(K.E.)$ is given by the formula $K.E. = \frac{1}{2}mv^2$.
Assuming the mass $(m)$ of an athlete typically ranges between $40 \ kg$ and $100 \ kg$:
For $m = 40 \ kg$: $K.E. = \frac{1}{2} \times 40 \times (10)^2 = 20 \times 100 = 2000 \ J$.
For $m = 100 \ kg$: $K.E. = \frac{1}{2} \times 100 \times (10)^2 = 50 \times 100 = 5000 \ J$.
Thus,the range of kinetic energy is $2000 \ J - 5000 \ J$.

(B) The relationship between kinetic energy $K$ and momentum $P$ is given by the formula:
$K = \frac{P^2}{2m}$
From the given graph,we can identify a point on the curve where the kinetic energy $K = 4$ and the momentum $P = 4$.
Substituting these values into the formula:
$4 = \frac{4^2}{2m}$
$4 = \frac{16}{2m}$
$4 = \frac{8}{m}$
Solving for $m$:
$m = \frac{8}{4} = 2 \ kg$
Therefore,the mass of the moving particle is $2 \ kg$.

(D) The relationship between kinetic energy $E$ and momentum $p$ for an object of mass $m$ is given by $E = \frac{p^2}{2m}$.
When the momentum becomes $p' = 2p$,the new kinetic energy $E'$ is given by:
$E' = \frac{(p')^2}{2m} = \frac{(2p)^2}{2m} = \frac{4p^2}{2m}$.
Substituting $E = \frac{p^2}{2m}$ into the equation for $E'$,we get:
$E' = 4 \times \left( \frac{p^2}{2m} \right) = 4E$.
Therefore,the new kinetic energy will be $4E$.

(D) From the graph,the kinetic energy $(K.E.)$ is directly proportional to time $(t)$:
$K.E. = \frac{1}{2} m v^{2} = k t$ (where $k$ is a constant)
$v^{2} \propto t \Rightarrow v \propto \sqrt{t}$
Let $v = \alpha \sqrt{t}$,where $\alpha$ is a constant.
The acceleration $a$ is the rate of change of velocity:
$a = \frac{d v}{d t} = \frac{d}{d t} (\alpha \sqrt{t}) = \frac{\alpha}{2 \sqrt{t}}$
The force $F$ acting on the body is given by Newton's second law:
$F = m a = m \left( \frac{\alpha}{2 \sqrt{t}} \right)$
Since $v = \alpha \sqrt{t}$,we have $\sqrt{t} = \frac{v}{\alpha}$.
Substituting this into the expression for force:
$F = m \left( \frac{\alpha}{2 (v / \alpha)} \right) = \frac{m \alpha^{2}}{2 v}$
Thus,$F \propto \frac{1}{v}$.
Therefore,the force is inversely proportional to the velocity.

(D) The kinetic energy $K$ of a body of mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Since the kinetic energies of the two bodies are equal,we have $K_1 = K_2$.
Therefore,$\frac{p_1^2}{2M_1} = \frac{p_2^2}{2M_2}$.
Rearranging the terms,we get $\frac{p_1^2}{p_2^2} = \frac{M_1}{M_2}$.
Taking the square root on both sides,we obtain $\frac{p_1}{p_2} = \sqrt{\frac{M_1}{M_2}} = \sqrt{M_1} : \sqrt{M_2}$.

(C) The kinetic energy $K$ is related to momentum $p$ by the formula $K = \frac{p^2}{2m}$.
Since mass $m$ is constant,we have $K \propto p^2$.
Let the initial momentum be $p_1 = p$ and the final momentum be $p_2 = p + 100\% \text{ of } p = 2p$.
The ratio of kinetic energies is $\frac{K_2}{K_1} = \left(\frac{p_2}{p_1}\right)^2 = \left(\frac{2p}{p}\right)^2 = 4$.
Thus,$K_2 = 4K_1$.
The percentage increase in kinetic energy is given by $\frac{K_2 - K_1}{K_1} \times 100\% = \frac{4K_1 - K_1}{K_1} \times 100\% = 300\%$.

(D) The kinetic energy $(K.E.)$ of an object is given by the formula: $K.E. = \frac{1}{2} mv^2$.
First,convert the mass from grams to kilograms: $m = 120\, g = 0.12\, kg$.
Next,calculate the square of the velocity magnitude $(v^2)$ using the dot product of the velocity vector with itself: $v^2 = \vec{v} \cdot \vec{v} = (2\hat{i} + 5\hat{j}) \cdot (2\hat{i} + 5\hat{j}) = (2^2 + 5^2) = 4 + 25 = 29\, m^2/s^2$.
Now,substitute these values into the kinetic energy formula:
$K.E. = \frac{1}{2} \times 0.12\, kg \times 29\, m^2/s^2$
$K.E. = 0.06 \times 29 = 1.74\, J$.
Therefore,the kinetic energy is $1.74\, J$.

(A) The kinetic energy $(K.E.)$ of a particle with mass $m$ and momentum $p$ is given by the formula: $K.E. = \frac{p^2}{2m}$.
Given that both particles have the same momentum $(p_1 = p_2 = p)$,the ratio of their kinetic energies is:
$\frac{K.E._1}{K.E._2} = \frac{p^2 / (2m_1)}{p^2 / (2m_2)} = \frac{m_2}{m_1}$.
Substituting the given masses $m_1 = 1\,kg$ and $m_2 = 5\,kg$:
$\frac{K.E._1}{K.E._2} = \frac{5}{1} = 5:1$.

(C) According to the principle of conservation of mechanical energy,the loss in gravitational potential energy is equal to the gain in kinetic energy.
Given that the body starts from rest,the initial kinetic energy is $0$.
Let $m$ be the mass of the body.
The loss in potential energy is given as $U$.
The gain in kinetic energy is $\frac{1}{2}mv^2 - 0 = \frac{1}{2}mv^2$.
Equating the two: $U = \frac{1}{2}mv^2$.
Solving for $m$: $m = \frac{2U}{v^2}$.

(C) The relationship between kinetic energy $E$ and linear momentum $p$ is given by $E = \frac{p^2}{2m}$.
For small percentage changes,we can use the differential method: $\frac{\Delta E}{E} = 2 \left( \frac{\Delta p}{p} \right)$.
Given $\frac{\Delta p}{p} = 5\% = 0.05$,the change in kinetic energy is $\frac{\Delta E}{E} = 2 \times 5\% = 10\%$.
Alternatively,using the exact formula: $E' = \frac{(1.05p)^2}{2m} = 1.1025 E$.
The percentage increase is $(1.1025 - 1) \times 100 = 10.25\%$. Since $10.25\%$ is the precise value,we select the closest standard approximation or the exact value if provided.

(C) The initial kinetic energy $(K_i)$ of the bullet is given by $K_i = \frac{1}{2} m v_i^2$.
Given $m = 50.0 \; g = 0.050 \; kg$ and $v_i = 200 \; m s^{-1}$.
$K_i = \frac{1}{2} \times 0.050 \times (200)^2 = 0.025 \times 40000 = 1000 \; J$.
The bullet emerges with $10 \%$ of its initial kinetic energy,so the final kinetic energy $(K_f)$ is:
$K_f = 0.10 \times 1000 \; J = 100 \; J$.
Let $v_f$ be the emergent speed of the bullet.
Using the formula $K_f = \frac{1}{2} m v_f^2$:
$100 = \frac{1}{2} \times 0.050 \times v_f^2$.
$100 = 0.025 \times v_f^2$.
$v_f^2 = \frac{100}{0.025} = 4000$.
$v_f = \sqrt{4000} \approx 63.25 \; m s^{-1}$.
Thus,the emergent speed of the bullet is approximately $63.2 \; m s^{-1}$.

(N/A) The most important factor to control the speed of a vehicle in areas like schools or hospitals is the application of $braking$ $force$ to reduce $kinetic$ $energy$. By applying the brakes,a negative acceleration (deceleration) is produced,which opposes the motion of the vehicle. According to the equation of motion $v^2 = u^2 + 2as$,where $v$ is the final velocity,$u$ is the initial velocity,$a$ is acceleration,and $s$ is displacement,reducing the velocity $v$ requires a negative value of $a$. This ensures safety by allowing the driver to stop the vehicle within a shorter distance in case of an emergency.

(A) Yes,$I$ agree. The kinetic energy $(K)$ of an object is given by the formula $K = \frac{p^2}{2m}$,where $p$ is the momentum and $m$ is the mass of the object.
If $K = 0$,then $\frac{p^2}{2m} = 0$,which implies $p^2 = 0$,and therefore $p = 0$.
Since momentum $p = mv$,if $p = 0$ and $m \neq 0$,then the velocity $v$ must be $0$.
Thus,an object with no kinetic energy must be at rest,and consequently,it has no momentum.

(A) The relationship between kinetic energy $(K)$ and linear momentum $(p)$ of a body of mass $(m)$ is given by the formula: $K = \frac{p^2}{2m}$.
From this equation,we can express momentum as: $p = \sqrt{2mK}$.
Since mass $(m)$ is constant,the momentum $(p)$ is directly proportional to the square root of the kinetic energy $(p \propto \sqrt{K})$.
Therefore,if the kinetic energy $(K)$ of a body increases,its momentum $(p)$ must also increase.

(N/A) The energy of a body is the ability (or capacity) of the body to do work.
Half of the product of the mass of a body and the square of its velocity is defined as kinetic energy $(K)$ of the body.
$\therefore K = \frac{1}{2} mv^2$,where $m$ is the mass of the body and $v$ is the velocity or speed of the body.
The unit of kinetic energy is the same as the unit of work. Its $SI$ unit is Joule $(J)$ and its $CGS$ unit is erg. The dimensional formula of kinetic energy is $[M^1 L^2 T^{-2}]$.
Kinetic energy is a scalar quantity. It indicates the magnitude of work that a body can perform due to its motion.
Examples of work done using kinetic energy:
$1$. The kinetic energy of a fast-flowing stream has been used to grind grains.
$2$. Sailing ships employ the kinetic energy of the wind.
$3$. Electricity is generated by wind mills using the kinetic energy of air.

(A) Kinetic energy is defined as the energy possessed by an object due to its motion.
It is given by the formula $K = \frac{1}{2}mv^2$,where $m$ is the mass and $v$ is the speed of the object.
Since mass is a scalar and the square of speed $(v^2)$ is also a scalar,the product $\frac{1}{2}mv^2$ results in a scalar quantity.
Kinetic energy does not have a direction associated with it; it only has magnitude.
Therefore,kinetic energy is a scalar quantity.

(B) The kinetic energy $K$ of a body with mass $m$ and momentum $p$ is given by the relation $K = \frac{p^2}{2m}$.
Since the momentum $p$ is the same for both bodies,we have $K \propto \frac{1}{m}$.
This implies that the kinetic energy is inversely proportional to the mass of the body.
Therefore,the body with the smaller mass (the lighter body) will have more kinetic energy.

(C) The kinetic energy $K$ of an object with mass $m$ and momentum $p$ is given by $K = \frac{p^2}{2m}$.
If the momentum is doubled,the new momentum $p' = 2p$.
The new kinetic energy $K'$ is given by $K' = \frac{(p')^2}{2m} = \frac{(2p)^2}{2m} = \frac{4p^2}{2m} = 4K$.
The increase in kinetic energy is $\Delta K = K' - K = 4K - K = 3K$.
The percentage increase in kinetic energy is $\frac{\Delta K}{K} \times 100 = \frac{3K}{K} \times 100 = 300\%$.

(N/A) The kinetic energy $K$ of an object of mass $m$ moving with velocity $v$ is given by $K = \frac{1}{2}mv^2$.
Multiply and divide the expression by $m$:
$K = \frac{1}{2} \frac{m^2v^2}{m} = \frac{(mv)^2}{2m}$.
Since momentum $p = mv$,we can substitute $p$ into the equation:
$K = \frac{p^2}{2m}$.
Rearranging for $p$:
$p^2 = 2mK$.
Therefore,the momentum is $p = \sqrt{2mK}$.

(A) The formula for kinetic energy $(K)$ is given by $K = \frac{1}{2} m v^2$.
Given:
Mass $(m)$ = $2 \, g = 2 \times 10^{-3} \, kg$.
Velocity $(v)$ = $500 \, m/s$.
Substituting the values into the formula:
$K = \frac{1}{2} \times (2 \times 10^{-3} \, kg) \times (500 \, m/s)^2$.
$K = 10^{-3} \times 250,000$.
$K = 250 \, J$.

(B) The total mechanical energy $E$ of a system is given by the sum of kinetic energy $K$ and potential energy $V$,i.e.,$E = K + V$.
Rearranging this,we get $K = E - V$.
Since kinetic energy $K$ must always be greater than or equal to zero $(K \ge 0)$,the expression $E - V$ cannot be negative.
Therefore,the condition $E - V < 0$ is not possible.

(B) The kinetic energy $K$ of an object is given by $K = \frac{1}{2} mv^2$.
We know that linear momentum $p = mv$,which implies $m = \frac{p}{v}$.
Substituting the value of $m$ into the kinetic energy formula:
$K = \frac{1}{2} (\frac{p}{v}) v^2$
$K = \frac{1}{2} pv$.

(B) Given: Mass $m = 0.5\, kg$,Initial speed $u = 20\, ms^{-1}$.
Initial kinetic energy $K_i = \frac{1}{2} mu^2 = \frac{1}{2} \times 0.5 \times (20)^2 = 100\, J$.
After deflection,the final kinetic energy $K_f$ is $5\%$ of $K_i$.
$K_f = \frac{5}{100} \times 100 = 5\, J$.
Let the final speed be $v$. Then $K_f = \frac{1}{2} mv^2$.
$5 = \frac{1}{2} \times 0.5 \times v^2$.
$5 = 0.25 \times v^2$.
$v^2 = \frac{5}{0.25} = 20$.
$v = \sqrt{20} \approx 4.47\, ms^{-1}$.

(B) The kinetic energy $K$ of a body with mass $m$ and linear momentum $P$ is given by the formula $K = \frac{P^2}{2m}$.
Since both solids $A$ and $B$ have equal linear momentum,we have $P_A = P_B = P$.
Therefore,the kinetic energy is inversely proportional to the mass: $K \propto \frac{1}{m}$.
Taking the ratio of the kinetic energies of $A$ and $B$,we get $\frac{(K.E.)_A}{(K.E.)_B} = \frac{m_B}{m_A}$.
Given $m_A = 1\, kg$ and $m_B = 2\, kg$,we substitute these values: $\frac{(K.E.)_A}{(K.E.)_B} = \frac{2}{1}$.
Comparing this with the given ratio $\frac{A}{1}$,we find that $A = 2$.