Work Power & Energy

At point x and a of circular path, the points are at same height but less than h. So the velocity is more than point y.

So required centripetal force = $\frac{mv^{2}}{r}$ is more.

Given, a particle at rest is subjected to force. Motion of particle is along x-axis.

Change in kinetic energy is given as

$K = {{dK} \over {dt}} = \gamma t$ ..... (1)

where $\gamma$ is positive constant.

We know that kinetic energy is given by

$K = {1 \over 2}m{v^2}$...... (2)

Now, differentiate Eq. (2), we get

${{dK} \over {dt}} = {2 \over 2}mv{{dv} \over {dt}} = mv{{dv} \over {dt}}$

${{dK} \over {dt}} = mv{{dv} \over {dt}}$ ....... (3)

From Eqs. (1) and (3), we get

$mv{{dv} \over {dt}} = \gamma t$

$ \Rightarrow mv\,dv = \gamma t\,dt$ ...... (4)

Integrating Eq. (4), we get

$\int {mv\,dv = \int {\gamma t\,dt \Rightarrow m\int {v\,dv = \gamma \int {t\,dt} } } } $

${{m{v^2}} \over 2} = {{\gamma {t^2}} \over 2} \Rightarrow m{v^2} = \gamma {t^2}$

$ \Rightarrow {v^2} = {\gamma \over m}{t^2} \Rightarrow v = \sqrt {{\gamma \over m}} t$

$ \Rightarrow v \propto t$

Since, $\gamma$ and m are constant so the speed of particle is proportional to time therefore option (B) is true.

Also, $a = {{dv} \over {dt}} \Rightarrow a = \sqrt {{\gamma \over m}} $ = constant

F = ma = constant

So, option (A) is true.

Also, $v = {{dr} \over {dt}}$

$ \Rightarrow {{dr} \over {dt}} = \sqrt {{\gamma \over m}} t$

$ \Rightarrow dr = \sqrt {{\gamma \over m}} t\,dt$ ...... (5)

Integrating Eq. (5), we get

$\int {dr = \int {\sqrt {{\gamma \over m}} t\,dt \Rightarrow r = \sqrt {{\gamma \over m}} {{{t^2}} \over 2}} } $

$ \Rightarrow r \propto {t^2}$

So, the distance of the particle from the origin increases at t² so option (C) is false.

A force is conservative if work done by the force in any closed loop is zero (i.e., work done is independent of the path). If force is constant (both magnitude and direction) then work done by the force in a closed loop is given by

$W = \oint {\overrightarrow F \,.\,d\overrightarrow r = \overrightarrow F \,.\,\oint {d\overrightarrow r = \overrightarrow F \,.\,\overrightarrow 0 } = 0} $.

Note that we are able to take $\overrightarrow F $ out of integral sign because $\overrightarrow F $ is constant and $\oint {d\overrightarrow r } $ is zero because integration is carried out over a closed loop. The argument 'since force is constant, hence it is conservative' should be understood properly. The kinetic friction force and viscous drag force (when body moves with a terminal speed) are both constant (in magnitude) but are non-conservative. These forces cannot start the particle motion from the rest. The situation given in this problem is similar to a particle dropped from rest under the gravitational force. This force is constant, can initiate the motion given in the problem and is conservative.

The linear momentum $\overrightarrow p $ of a particle is conserved if force on it is zero i.e., $\overrightarrow F = m\overrightarrow a = \overrightarrow 0 $. If $\overrightarrow p $ is conserved then kinetic energy $K = {p^2}/2m$ is also conserved. The angular momentum $\overrightarrow L $ of a particle is conserved if torque on it is zero i.e., $\overrightarrow \tau = \overrightarrow r \times \overrightarrow F = \overrightarrow 0 $. The total energy E is conserved for conservative forces.

For P : $\overrightarrow r (t) = \alpha t\widehat i + \beta t\widehat j$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta \widehat j$ = constant

$ \Rightarrow \overrightarrow p $ is constant (conserved)

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}} $ = constant

Therefore, $K = {1 \over 2}m{v^2}$ is constant (conserved)

Now, $\overrightarrow a = {{d\overrightarrow v } \over {dt}} = 0 \Rightarrow \overrightarrow F = m\overrightarrow a = 0$

Thus, $F = \nabla U \Rightarrow U$ is constant (conserved)

Now, $E = U + K \Rightarrow E$ is constant, since U and K are constant

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = 0 \Rightarrow \overrightarrow L $ is constant (conserved)

Thus, the correct mapping is $P \to 1,2,3,4,5$.

For Q : $\overrightarrow r (t) = \alpha \cos \omega t\,\widehat i + \beta \sin \omega t\,\widehat j$

$ \Rightarrow v = {{d\overrightarrow r } \over {dt}} = - \alpha \omega \sin \omega t\,\widehat i + \beta \omega \cos \omega t\,\widehat j$

So v is not a constant $ \Rightarrow \overrightarrow p $ is not conserved for the path.

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}{{\sin }^2}\omega t + {\beta ^2}{\omega ^2}\sin \omega t} $ which is not constant. Therefore,

$K = {1 \over 2}m{v^2}$ is not constant (not conserved)

Now, $\overrightarrow a = {{d\overrightarrow v } \over {dt}} = - \alpha {\omega ^2}\cos \omega t\,\widehat i + \beta {\omega ^2}\sin \omega t\,\widehat j$

$ \Rightarrow \overrightarrow a $ is not constant

$ \Rightarrow \overrightarrow F $ is not constant

$ \Rightarrow \overrightarrow U $ is not constant (not conserved)

E = U + K is constant (conserved)

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega \alpha \beta \widehat k$

$ \Rightarrow \overrightarrow L $ is constant (conserved)

Thus, the correct mapping is $Q \to 2,5$.

For R : $\overrightarrow r (t) = \alpha (\cos \omega t\,\widehat i + \sin \omega t\,\widehat j)$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \omega ( - \sin \omega t\,\widehat i + \cos \omega t\,\widehat j)$

$ \Rightarrow \overrightarrow v $ is not a constant

$ \Rightarrow \overrightarrow p $ is not a constant (not conserved)

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2}{\omega ^2}({{\sin }^2}\omega t + {{\cos }^2}\omega t)} = \alpha \omega $

$ \Rightarrow \left| {\overrightarrow v } \right|$ is constant

$ \Rightarrow K = {1 \over 2}m{v^2}$ is a constant (conserved)

Now, $\overrightarrow a = {{dv} \over {dt}} = \alpha {\omega ^2}( - \sin \omega t\,\widehat i - \cos \omega t\,\widehat j)$

$ \Rightarrow \overrightarrow a $ is not constant

Since $ E = U + K$ is constant (conserved)

$\Rightarrow U$ is constant (conserved)

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = m\omega {\alpha ^2}\widehat k$

$ \Rightarrow \overrightarrow L $ is constant (conserved)

Thus, the correct mapping is $R \to 2,3,4,5.$

For S : $\overrightarrow r (t) = \alpha t\,\widehat i + {\beta \over 2}{t^2}\,\widehat j$

$\overrightarrow v = {{d\overrightarrow r } \over {dt}} = \alpha \widehat i + \beta t\,\widehat j$

$ \Rightarrow \overrightarrow v $ is not conserved

$ \Rightarrow \overrightarrow p $ is not conserved

Now, $\left| {\overrightarrow v } \right| = \sqrt {{\alpha ^2} + {\beta ^2}\,{t^2}} $

$ \Rightarrow \left| {\overrightarrow v } \right|$ is not conserved

$ \Rightarrow K = {1 \over 2}m{v^2}$ is not conserved

$\overrightarrow a = {{d\overrightarrow v } \over {dt}} = \beta \widehat j$

$ \Rightarrow E = U + K$ is conserved

But, U is not conserved.

Now, $\overrightarrow L = m(\overrightarrow r \times \overrightarrow v ) = {1 \over 2}\alpha \beta {t^2}\widehat k$

$ \Rightarrow \overrightarrow L $ is not conserved.

Thus, the correct mapping is $S \to 5$.

Therefore, option (A) is correct.

It is given that the average speed of gas molecules is u; the speed of the plate is v; the force on the plate if F.

$\bullet$ Just before the collision : The gas molecules are approaching the leading and trailing faces of the plate with speed u.

$\bullet$ Just after the collision : The gas molecules bounce back with speed u₁ and u₂.

$\bullet$ At trailing face

The speed before collision is u $-$ v.

The speed after collision is u₂ + v.

From law of conservation of linear momentum, we have

u₂ + v = u $-$ v

That is,

u₂ = u $-$ 2v ....... (1)

and

$\Delta$u₂ = 2u $-$ 2v ....... (2)

$\bullet$ At leading face

The speed before collision is u + v.

The speed after collision is u₁ $-$ v.

From law of conservation of linear momentum, we have

u₁ $-$ v = u + v

That is,

u₁ = u + 2v ..... (3)

and

$\Delta$u₁ = 2u + 2v ...... (4)

Now, from Newton's second law of motion, we have

$F = {{dp} \over {dt}}$

$\bullet$ For leading face : ${F_1} = {{d{p_1}} \over {dt}}$.

The volume swept by the plate is A(u + v). Therefore,

${F_1} = {{d{p_1}} \over {dt}} = \rho A(u + v)\Delta {u_1}$

${F_1} = \rho A(u + v)(2u + 2v)$ ....... (5)

$\bullet$ For trailing face : ${F_2} = {{d{p_2}} \over {dt}}$

The volume swept by the plate is A(u $-$ v). Therefore,

${F_2} = {{d{p_2}} \over {dt}} = \rho A(u - v)\Delta {u_2}$

${F_2} = \rho A(u - v)(2u - 2v)$ ....... (6)

The net force on the plate is

${F_{net}} = {F_1} - {F_2}$

$ = \rho A(u + v)(2u + 2v) - \rho A(u - v)(2u - 2v)$

$ = 2\rho A(u + v)(u + v) - 2\rho A(u - v)(u - v)$

$ = 2\rho A{(u + v)^2} - 2\rho A{(u - v)^2}$

$ = 2\rho A[{(u + v)^2} - {(u - v)^2}]$

$ = 2\rho A[{u^2} + {v^2} + 2uv - ({u^2} + {v^2} - 2uv)]$

$ = 2\rho A({u^2} + {v^2} + 2uv - {u^2} - {v^2} + 2uv)$

$ = 2\rho A(4uv)$

$ = 8\rho Auv$

Therefore, the pressure difference is

${{{F_{net}}} \over A} = {{8\rho Auv} \over A} = 8\rho uv$

Thus, the pressure difference between the leading and trailing faces of the plate is proportional to uv. Hence, option (A) is correct.

Now,

${F_{net}} = 8\rho Auv$

The net resistive force is

$F - (8\rho Au)v = {{mdv} \over {dt}}$

where F is the constant force applied by the gas molecules on the plate.

The resistive force experienced by the plate is proportional to v.

Hence, option (B) is also correct.

At a later time, the velocity v becomes sufficient and the external force $F( = 8\rho Auv)$ balances the resistive force.

Hence, option (D) is also correct.

As tennis ball is dropped, so its initial velocity, u = 0

Suppose its velocity at any time t = v

Acceleration due to gravity = g

$\therefore$ v = u + gt

$\Rightarrow$ v = gt ..... (i)

We know, kinetic energy,

$K = {1 \over 2}m{v^2} = {1 \over 2}m({g^2}{t^2})$ [Using eqn. (i)]

$ \Rightarrow K = {1 \over 2}m{g^2}{t^2}$

$ \Rightarrow K \propto {t^2}$ (As m and g are constant)

This shows that relation between kinetic energy of tennis ball K and time t is parabolic.

During the collision, velocity of the ball falls sharply to zero as it is compressed and regains maximum velocity in the same short time interval.

This relation is best illustrated by the option (b).

At point Q, the forces acting on the particle are normal reaction $N_Q$, gravitational force $m g$ and frictional force $f$. Resolve these in the directions parallel and perpendicular to the path. The net force towards the centre $(P)$ provides the centripetal acceleration. Apply Newton's second law in a direction normal to the path to get

${N_Q} - mg\sin 30^\circ = {{m{v^2}} \over R}$

$ \Rightarrow $ ${N_Q} = mg\sin 30^\circ +{{m{v^2}} \over R}$

$ = 1 \times 10 \times {1 \over 2} + {{100} \over {40}}$

$ = 5 + {5 \over 2} = {{15} \over 2}$

$ \Rightarrow {N_Q} = 7.5$ N

Hence, option (A) is correct.

Here we will use the concept of work-energy theorem. By work-energy theorem,

Total work done = Change in KE

$\Rightarrow$ Workdone by gravity + workdone by friction = $\Delta KE$

$ \Rightarrow mgR\sin 30^\circ - 150 = {1 \over 2}m{v^2} - 0$

$ \Rightarrow 1 \times 10 \times 40 \times {1 \over 2} - 150 = {1 \over 2}{v^2}$

$ \Rightarrow 200 - 150 = {1 \over 2}{v^2}$

$ \Rightarrow {v^2} = 100$

$ \Rightarrow v = 10$ m/s

Hence, option (B) is correct.

The work done on a particle of mass m by a force

$ \vec{F}=K\left[\frac{x}{\left(x^2+y^2\right)^{3/2}} \hat{i}+\frac{y}{\left(x^2+y^2\right)^{3/2}} \hat{j}\right] $

(with K being a constant with appropriate dimensions), when the particle is taken from the point (a,0) to the point (0,a) along a circular path of radius a about the origin in the x-y plane is:

Let $ \vec{r}=x \hat{i}+y \hat{j} $

Therefore, $ d \vec{r}=d x \hat{i}+d y \hat{j} $ and $ r=\sqrt{x^2+y^2} $

Work done by the force in moving a particle from A to B is

$\begin{aligned} W & =\int_{r_A}^{r_B} \vec{F} \cdot d \vec{r} \\\\ & =\int_{r_A}^{r_B} K \left[\frac{x}{(x^2+y^2)^{3/2}} \hat{i}+\frac{y}{(x^2+y^2)^{3/2}} \hat{j}\right] \cdot (d x \hat{i}+d y \hat{j}) \\\\ & =K \int_{r_A}^{r_B} \frac{x d x}{(x^2+y^2)^{3/2}} + \frac{y d y}{(x^2+y^2)^{3/2}} \\\\ & =K \int_{r_A}^{r_B} \frac{1}{(x^2+y^2)^{3/2}} \left[d\left(\frac{x^2}{2}\right) + d\left(\frac{y^2}{2}\right)\right] \\\\ & =K \int_{r_A}^{r_B} \frac{1}{2(x^2 + y^2)^{3/2}} d(x^2 + y^2) \\\\ & =K \int_{r_A}^{r_B} \frac{1}{2 r^3} d(r^2) = K \int_{r_A}^{r_B} \frac{2 r d r}{2 r^3} = K \int_{r_A}^{r_B} \frac{d r}{r^2} \quad (\because r=\sqrt{x^2 + y^2}) \\\\ & =K\left[-\frac{1}{r}\right]_{r_A}^{r_B} = K\left[\frac{1}{r_A} - \frac{1}{r_B}\right] \end{aligned}$

Here, $ r_A = a $ and $ r_B = a $

Therefore, $ W = 0 $

Let M be the mass of the ring and m that of the ball and let V and $\upsilon $ be their velocity before collision. The initial momentum of the system (ring and ball) in the horizontal direction is

${\overrightarrow p _i} = M\overrightarrow V + m\overrightarrow \upsilon $

$ = 2 \times 1 + 0.1 \times ( - 20)$

$ = 2 - 2 = 0$

From conservation of momentum, the final momentum of the system ${\overrightarrow p _f} = 0$ in the horizontal direction. Hence ${V_{cm}} = 0$ for the ring, i.e. the ring has pure rotation about its centre of mass. So choice (a) is correct.

The total initial angular momentum of the system about the point of collision is

${L_i} = m\upsilon r - I\omega $

$ = m\upsilon r - M{R^2}{V \over R}$

$ = m\upsilon r - MRV$

$ = 0.1 \times 20 \times 0.75 - 2 \times 0.5 \times 1$

$ = 1.5 - 1 = 0.5$ kg m² s^$-$1

From the conservation of angular momentum, the final angular velocity must be anticlockwise. Hence the friction between the ring and the ground is to the left. So the correct choices are (a) and (c).