Correct option is (C)

To find the total work done by all forces applied on the particle during its displacement, we can use the work-energy theorem which states that the work done by all forces on an object is equal to the change in kinetic energy of the object. So, we first need to find the initial and final kinetic energies of the particle and then calculate the work done.

The velocity of the particle is given by $v=\alpha \sqrt{x}$,

and the kinetic energy $K$ of the particle is given by $K=\frac{1}{2}m{v}^2$. We can substitute the expression for $v$ into this formula to get the kinetic energy as a function of position $x$:

$K(x)=\frac{1}{2}m(\alpha \sqrt{x}{)}^2=\frac{1}{2}m{\alpha }^{2}x$

To find the total work done from $x=0$ to $x=d$, we need to compute the difference in kinetic energy between these two points:

$W=K(d)-K(0)$

At $x=d$,

$K(d)=\frac{1}{2}m{\alpha }^{2}d$

At $x=0$, since the particle starts from this position,

$K(0)=\frac{1}{2}m{\alpha }^2(0)=0$

So, the work done $W$ is simply the kinetic energy at $x=d$,

$W=\frac{1}{2}m{\alpha }^{2}d-0=\frac{1}{2}m{\alpha }^{2}d$

This matches with Option C:

$\frac{m{\alpha }^{2}d}{2}$.

Correct option is (D)

$\begin{array}{rl}& {\mathrm{W}}_{\mathrm{g}}+{\mathrm{W}}_{\mathrm{Fr}}+{\mathrm{W}}_{\mathrm{s}}=\mathrm{\Delta }\mathrm{KE}\\ \\ & 5\times 10\times 5-0.5\times 5\times 10\times \mathrm{x}-\frac{1}{2}{\mathrm{Kx}}^2=0-0\\ \\ & 250=25\mathrm{x}+50{\mathrm{x}}^2\\ \\ & 2{\mathrm{x}}^2+\mathrm{x}-10=0\\ \\ & \mathrm{x}=2\end{array}$

Correct option is (B)

When kinetic energy of a body becomes 36 times of its original value, the percentage increase in the momentum of the body will be :

Correct option is (C)

To find the percentage loss of kinetic energy of the bullet, we first calculate the initial kinetic energy before the bullet hits the plywood and the final kinetic energy after it emerges. The formula for kinetic energy (KE) is given by:

$KE=\frac{1}{2}m{v}^2$

where $m$ is the mass of the object and $v$ is its velocity.

Let's calculate the initial and final kinetic energies.

Initial Kinetic Energy:

$K{E}_{\text{initial}}=\frac{1}{2}\times 50\times (100{)}^2=\frac{1}{2}\times 50\times 10000=25\times 10000=250000{\text{g.m}}^2/{\text{s}}^2$

Note: To keep units consistent, we used grams and meters per second. We can also convert the mass to kilograms (by dividing by 1000) which would result in the energy being calculated in Joules, but for the purpose of finding the percentage change, the form of units does not matter as long as they are consistent, since it will be a ratio.

Final Kinetic Energy:

$K{E}_{\text{final}}=\frac{1}{2}\times 50\times (40{)}^2=\frac{1}{2}\times 50\times 1600=25\times 1600=40000{\text{g.m}}^2/{\text{s}}^2$

The loss of kinetic energy is then:

$\mathrm{\Delta }KE=K{E}_{\text{initial}}-K{E}_{\text{final}}=250000-40000=210000{\text{g.m}}^2/{\text{s}}^2$

Finally, the percentage loss of kinetic energy can be calculated using the formula:

$\text{Percentage loss of KE}=\left(\frac{\mathrm{\Delta }KE}{K{E}_{\text{initial}}}\right)\times 100\mathrm{\%}$

$\text{Percentage loss of KE}=\left(\frac{210000}{250000}\right)\times 100\mathrm{\%}=0.84\times 100\mathrm{\%}=84\mathrm{\%}$

Thus, the percentage loss of kinetic energy is 84%, which corresponds to Option C.

Correct option is (D)

The momentum $p$ of a particle is given by the product of its mass $m$ and its velocity $v$, that is, $p=m\cdot v$. For a given momentum, the relationship between mass and velocity can be understood as inversely proportional. This means that as the mass increases, the velocity decreases to maintain the same momentum, and vice versa.

The kinetic energy ($K.E.$) of a particle is given by the formula $K.E.=\frac{1}{2}m{v}^2$. This equation shows that the kinetic energy depends on both the mass of the particle and the square of its velocity.

Given that four particles $A,B,C,D$ have masses $\frac{m}{2},m,2m,4m$, respectively, and all have the same momentum, we can assume the momentum of each particle to be $p$. This common value of momentum allows us to express the velocity of each particle in terms of its mass and the common momentum $p$. The velocity $v$ of each particle will be $v=\frac{p}{m}$.

Thus, for each particle, we can determine the velocity as follows:

• For $A$: $v_A=\frac{p}{\frac{m}{2}}=\frac{2p}{m}$


• For $B$: $v_B=\frac{p}{m}$


• For $C$: $v_C=\frac{p}{2m}=\frac{p}{2m}$


• For $D$: $v_D=\frac{p}{4m}$

Now, substituting these velocities into the kinetic energy formula yields the kinetic energies for each particle:

• $K.E{.}_A=\frac{1}{2}\cdot \frac{m}{2}\cdot {\left(\frac{2p}{m}\right)}^2=\frac{1}{2}\cdot \frac{m}{2}\cdot \frac{4p^2}{m^2}=\frac{2p^2}{m}$


• $K.E{.}_B=\frac{1}{2}\cdot m\cdot {\left(\frac{p}{m}\right)}^2=\frac{1}{2}\cdot m\cdot \frac{p^2}{m^2}=\frac{p^2}{2m}$


• $K.E{.}_C=\frac{1}{2}\cdot 2m\cdot {\left(\frac{p}{2m}\right)}^2=\frac{1}{2}\cdot 2m\cdot \frac{p^2}{4m^2}=\frac{p^2}{4m}$


• $K.E{.}_D=\frac{1}{2}\cdot 4m\cdot {\left(\frac{p}{4m}\right)}^2=\frac{1}{2}\cdot 4m\cdot \frac{p^2}{16m^2}=\frac{p^2}{8m}$

Comparing these kinetic energies, we see that the particle $A$ has the maximum kinetic energy, as it is inversely related to mass in this scenario, and $A$ has the least mass but the highest velocity squared component, thus maximizing its kinetic energy. Therefore, the correct answer is:

Option D: A