Work Power & Energy

Apply energy conservation,

${1 \over 2}mv_0^2 = mgh + {1 \over 2}m{v^2}$

${1 \over 2}m{v^2} = {1 \over 2}mv_0^2 - mgh$ ..... (i)

At top point y, the centripetal force provided by the component of mg,

mg sin$\theta$ = centripetal force = ${{m{v^2}} \over R}$

${{m{v^2}} \over R} = mg\sin \theta $

Putting this value in Eq. (i), we get

${{mgR\sin \theta } \over 2} = {1 \over 2}mv_0^2 - mgh$

${{mgR} \over 2}{1 \over 2} = {1 \over 2}mv_0^2 - mgh$

${{gR} \over 2} = v_0^2 - 2gh$

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.

Change in speed = $(2v + {v_0} - {v_0}) = 2v$ In every collision it acquires 2v,



$f = {v \over {2x}}$

${{dv} \over {dt}} = f \times 2v$

$dv = {v \over {2x}}2vdt$

$dv = {v \over {2x}}2( - dx)$

Integration on both sides limits v₀ to v, we get

$\int\limits_{{v_0}}^v {{{dv} \over v}} = \int\limits_l^x {{{ - dx} \over x}} $

$ \Rightarrow \ln {v \over {{v_0}}} = - \ln {x \over l}$

$ \Rightarrow v = {{{v_0}l} \over x}$

where, $x = {l \over 2},v = 2{v_0}$

so, $f = {{2{v_0}} \over {2{l \over 2}}} = {{2{v_0}} \over l}$

$ \therefore $ ${{{k_f}} \over {{k_i}}} = 4$

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.

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.

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).

The work done by the friction force is zero in rolling without slipping. The surface BC is frictionless. Thus mechanical energy of the balls is conserved.

i.e., it is same at point $\mathrm{A}, \mathrm{B}$ and C .

If $K_A, K_B$ and $K_C$ represent the total oenergies (transtational plus rotational)

Then, $\mathrm{K}_{\mathrm{A}}+\mathrm{mgh}_{\mathrm{A}}=\mathrm{K}_{\mathrm{B}}=\mathrm{K}_{\mathrm{C}}+\mathrm{mgh}_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

The equation (i) gives $K_B>K_C$ and $K_B>K_A$ irrespective of the relation between $\mathrm{h}_{\mathrm{A}}$ and $\mathrm{h}_{\mathrm{C}}$.

Let us analyses the problem a little more.

Let $K_B{ }^{\text {tans }}$ and $K_C{ }^{\text {trans }}$ are translational kinetics energies and $K_B{ }^{\text {rot }}$ and $K_C{ }^{\text {rot }}$ rotational kinetic energies of balls.

The equation (i) can be written as

$ \begin{aligned} \mathrm{K}_{\mathrm{A}}+m g h_A & =\left(\mathrm{K}_{\mathrm{B}}\right)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{B}}\right)_{\mathrm{rot}} \\ & =\left(\mathrm{K}_{\mathrm{C}}\right)_{\operatorname{tran}}+\left(\mathrm{K}_{\mathrm{C}}\right)_{\mathrm{rot}}+m g h_{\mathrm{C}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii) \end{aligned} $

Track $B C$ is frictionless and there is no torque on the ball about its centre of mass. Hence, its angular acceleration is zero, i.e., it moves with a constant angular velocity, which implies.

$ \left(\mathrm{K}_{\mathrm{B}}\right)_{\mathrm{rot}}=\left(\mathrm{K}_{\mathrm{C}}\right)_{\mathrm{rot}} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

The equation (iii) and (iv) gives $\left(\mathrm{K}_{\mathrm{B}}\right)_{\text {trans }}>\left(\mathrm{K}_{\mathrm{C}}\right)_{\text {trans }}$.