Brilliant Mechanics Contest - Season 1

Solution to Problem 21:

Insight Provided by Aryan Goyat

Nice solution , but i would like to add one thing to people who were getting around 19.33 as answer they were actually missing the thing that the the ball is doing pure rolling and would slip when max friction is just causing rolling and immediately after this this slipping begins .so take it analogical to sphere rolling on ground and then look for relation between f and mgsin(Q).The approach that f =mgsin(Q).is wrong.if the sphere had not been rolling then the point of contact would have already changed as the particles would have followed center of mass including the bottommost point.imagine the above case without friction i hope you got your mistake.

Solution to Problem 9:

This question is going to be solve using Conservation of Energy:

Let $\rho$ be the density of the liquid and A be the area of cross section of the column.

Change in Potential Energy in left column=$A\rho gx$=Change in Potential Energy in right column.

Therefore, total change in potential energy=$2A\rho gx$

Change in potential energy = Change in kinetic energy

$2A\rho gx$=$\frac{13\rho Axv^2}{2}$

Therefore, $v^2=\frac{4gx}{13}$ ............(1)

On differentiating both sides wrt time "t",

We get tangential acceleration $a_{t}=\frac{2g}{13}$

For Centripetal acceleration $a_{c}=\frac{6\pi}{13}$

We got this by using the formula: $a_{c}=\frac{v^2}{r}$

Therefore resultant acceleration is: $a=\frac { 2g }{ 13 } \sqrt { { 9\pi }^{ 2 }+1 }$

This question is solved by me by using a small hint from a friend.

$\text{Solution to Problem 1:}$ $x=l\cos\theta\Longrightarrow\dfrac{dx}{dt}=-l\sin\theta\times\dfrac{d\theta}{dt},similarly\\ y=l\sin\theta \Longrightarrow \dfrac{dy}{dt}=l\cos\theta\times\dfrac{d\theta}{dt},also\\ v=\dfrac{d}{dt}(2l\cos\theta) i.e\ v=-2l\sin\theta\times\dfrac{d\theta}{dt}\ using\ the\ above\ equations\ we\ get,\\ horizontal\ velocity\ of\ point\ C=\dfrac{v}{2}\\ vertical\ velocity\ of\ point\ C=\dfrac{v\cot\theta}{2}\\ \therefore\ velocity\ of\ point\ C=\dfrac{v\csc\theta}{2}$ $\text{Problem 2:}$$$ A particle executes SHM of period $7.5$ seconds.The velocity of a point $P$ $1m$ away from the centre of motion is $2ms^{-1}$.Find the amplitude of the oscillations.If the particle is passing through $P$ and moving away from the centre of oscillation,find the time that elapses before the particle next passes through $P$.$$ $\text{Answer is A=2.6m,time=2.8s}$.

$\textbf{Solution to problem 11:}$

Let velocity of $M$ be $v_{M}$ and $m$(first block) be $v_{m}$ after the first collision. By conservation of linear momentum and definition of $e$,we have:

$v_{M}= \dfrac{(M-m)v}{(M+m)}= \dfrac{(1-\phi)v}{(1+\phi)}$ $v_{m}=\dfrac{2Mv}{M+m}=\dfrac{2v}{1+\phi}$

where $\phi=\dfrac{m}{M}$

Now the for the second collision at any instant $t$ the position vectors(w.r.t ground frame) of $M$ and $m$ must be equal at that instant. If the collision is considered along the x-axis, then:

$x_{M}=v_{M}t= \dfrac{(1-\phi)vt}{(1+\phi)}$

Furthermore,The blocks connected to the spring will be executing simple harmonic motion with angular frequency $\omega=\sqrt{\frac{2k}{m}}$. Let the equation of SHM be $x=A\sin(\omega t)$. Since the CoM of the spring mass system would moving with a constant velocity of $v_{c}=\dfrac{v_{m}}{2}=\dfrac{v}{1+\phi}$, we have:

$x_{c} = \dfrac{vt}{1+\phi}$

$\omega A= \dfrac{v}{1+\phi}$

$\therefore A= \dfrac{v}{(1+\phi)\omega}$

Shifting back to ground frame, we have:

$x_{m} =x_{c}+A\sin(\omega t)= \dfrac{vt}{1+\phi}\left(1+\dfrac{\sin(\omega t)}{\omega t}\right)$

For second collision $x_{M}=x_{m}$

$\therefore M=\dfrac{-m\omega t}{\sin(\omega t)}$

Solution to Problem 13:

Image

I took minimum for h because there is a minimum height required for having motion around a complete circle.

Solution to Problem 14 :

Page 1
Page 2
Page 3
Page 4

Solution to Problem 18 :

The only forces acting on the body are it's weight $mg$ downwards and the viscous force upwards $F_{visc} = 6\pi \eta r v$ according to stoke's law.

The steady state velocity $v_{t}$ is obtained when these forces are balanced.

$6\pi \eta r v_{t} = mg \rightarrow 6\pi \eta r (0.99v_{t}) = 0.99mg$

Now, it's acceleration at any instant of time is given by,

$a = \dfrac{mg-6\pi \eta r v}{m}$

$\therefore \dfrac{dv}{dt} = \dfrac{mg-6\pi \eta r v}{m}$

$\displaystyle \int_{0}^{0.99v_{t}} \dfrac{m}{mg-6\pi \eta r v}dv = \displaystyle \int_{0}^{t} dt$

Integrating,

$\dfrac{m}{6\pi \eta r v} \left[\ln(mg - 6\pi \eta r v)\right]_{0.99v_{t}}^{0} = t$

$t = \dfrac{m}{6\pi \eta r }\cdot \ln \left(\dfrac{mg}{mg-6\pi \eta r (0.99v_{t})}\right) = \dfrac{m}{6 \pi \eta r } \cdot \ln \left( \dfrac{1}{0.01}\right) = \dfrac{m}{6 \pi \eta r } \ln(100)$

$t = \dfrac{m}{6\pi \eta r }\ln(100)$

Putting $m = \rho \dfrac{4 \pi r^{3} }{3}$

$t = \dfrac{2\rho\ln(100)r^{2}}{9\eta} = \dfrac{\rho\ln(100)d^{2}}{18\eta}$

The amplitude at any time t is given by ,

$A(t) = A_{0}e^{-\beta t}$

$\Delta A = A_{0}\left[1 - e^{\dfrac{\rho \beta d^{2} \ln(100)}{18\eta}} \right ]$

Plugging in the values we get,

$\Delta A \approx 17.4497 m$

Solution to Problem 5:

For the final position of the sleeve, net outward force = net inward force $Kx=M(l+x){ \omega }^{ 2 }$

Solving this, $x=\frac { Ml{ \omega }^{ 2 } }{ K-M{ \omega }^{ 2 } } \quad and\quad l+x=\frac { Kl }{ K-M{ \omega }^{ 2 } }$

Now total work done will be stored as the sum of elastic potential energy of the spring and the rotational kinetic energy of the sleeve, so,

$\quad \quad \quad W=\frac { 1 }{ 2 } I{ \omega }^{ 2 }+\frac { 1 }{ 2 } K{ x }^{ 2 }\\ \Rightarrow \quad W=\frac { 1 }{ 2 } M{ (l+x) }^{ 2 }{ \omega }^{ 2 }+\frac { 1 }{ 2 } K{ x }^{ 2 }$

Substituting the initially calculated values and taking some terms common,

$W=\frac { 1 }{ 2 } \frac { { l }^{ 2 }KM{ \omega }^{ 2 } }{ { (K-M{ \omega }^{ 2 }) }^{ 2 } } (K+M{ \omega }^{ 2 })$

Solution To Problem 8 :

pic

Let at any instant the length of unwinded string be $x$. Let it now rotate through an angle $d \theta$. Assume that $d \theta \rightarrow 0$. Then, we can make the approximation as shown in figure. So, from figure, $dx= R d \theta$. But now as $dx$ is very small, we can take that $\dfrac{d \theta}{dt} = \dfrac{v}{x}$.

So,

$xdx = 2Rv \, dt \\ \dfrac{x^2 }{2} = 2Rvt + c$

It is clear that from initial conditions $c=0$.

Therefore, $\boxed{l=\sqrt{2Rvt}}$.

Solution to Problem 19

Let the reaction due to collision be N, velocity of the centre of mass after collision $v$,angular velocity afterwards be $\omega$. let the time of contact during collision be $t$. Now since $t$ is very very small, the net change brought by gravity during this time interval can be neglected.

Velocity of rod just before collision(u) = $\sqrt { 2gh } =10m/s$

Also, $N \times t = m(v+u$

$N\times\frac {L}{2} \times Cos(60) = \frac { M{ L }^{ 2 }\omega }{ 12t }$

$v+\omega \frac { L }{ 2 } cos(60)\quad =\quad e\times 10$

Solving above equations, we get $v=\frac {2}{7}$ and $\omega= \frac {216}{7}$

Solution to Problem 20:

Consider the following figure before the collision. Let the mass of the target be $m$. So, the mass of the projectile is $2m$. The initial velocity of the projectile is $v$ whereas it's velocity after the collision is $v'$. Consider the angle between the two directions of the projectile as $\theta$. Also, take the velocity of the target along the initial direction, as ${ u }_{//}$ and that perpendicular to the initial direction as ${ u }_{ \bot }$.

(i) Applying principle of momentum conservation along the initial direction of motion of projectile:

$2mv = m{ u }_{//} + 2mv'\cos { \theta }$

So, by rearranging the terms:

${ u }_{//} = 2v-2v'\cos { \theta }$ $\quad \quad \quad--------- 1.$

(ii) Now, applying momentum conservation perpendicular to the initial direction of motion of projectile:

$0 = m{ u }_{ \bot } - 2mv'\sin { \theta }$

which gives us:

${u}_{ \bot } = 2v'\sin { \theta }$ $\quad \quad \quad-----------2.$

Now, using conservation of kinetic energy before and after the collision:

$\frac { 1 }{ 2 } (2m){ v }^{ 2 }=\frac { 1 }{ 2 } m{ u }^{ 2 }+\frac { 1 }{ 2 } (2m){ v' }^{ 2 }$

Now, substituting ${ u }^{ 2 }={ { u }_{ // } }^{ 2 }+{ { u }_{ \bot } }^{ 2 }$ and using the respective values from $1.$ and $2.$:

$2{ v }^{ 2 }=6{ v' }^{ 2 }+4{ v }^{ 2 }-8(v)(v')\cos { \theta }$ OR

$3{ v' }^{ 2 }-4(v)(v')\cos { \theta +{ v }^{ 2 } } =0$

For the projectile to continue on it's subsequent path, the above mentioned quadratic equation should have real roots, So:

$16{ v }^{ 2 }\cos ^{ 2 }{ \theta } -12{ v }^{ 2 }\ge 0$ OR

$\cos { \theta } \ge \frac { \sqrt { 3 } }{ 2 }$

Also, from the geometry of an n-sided polygon, we know that the angle theta is given by $\frac { 2\pi } { n }$.

So, $0\le \frac { 2\pi }{ n } \le \frac { \pi }{ 6 }$ which further implies $12\le n\le \infty$

Therefore, limiting value of $n$ = $12$, and hence, it's circumradius:

$R = \frac { L }{ 2\sin { \frac { \pi }{ n } } } =\frac { \frac { Perimeter }{ n } }{ 2\sin { \frac { \pi }{ n } } }$

Plugging in the values, we get:

$R = \frac { \frac { 144 }{ 12 } }{ 2\sin { \frac { \pi }{ 12 } } } =\frac { 6 }{ \sin { \frac { \pi }{ 12 } } } =23.1822m$

I hope this is right!! :P

Is the answer $6/sin(\pi/12)$ = $23.1822m$?

Solution to Problem 4:

Call the velocity of sphere before reaching the plank to be v .

Now i will work from the reference frame of ground .

Till pure rolling starts frictional force umg will act on the sphere in backward direction and due to third law the same force will act on the plank in forward direction .

For the plank there will be two frictional forces acting on it . the one acting in forward direction will have magnitude umg and the one acting in backward direction will have magnitude u/4 * 2mg = umg/2

let after time t pure rolling starts.

writing equations of motion for the sphere and plank .

Vs = v-ugt

Vp = ugt/2

writing torque equation about centre of mass of sphere we get

Angular acceleration = 5ug/2R

w = 5ugt/2r

For pure rolling Vs - Rw = Vp

Solving t= v/4ug .

work done by frictional force = umg/2 * displacement of plank

displacement of plank = 1/2 * ug/2 * (time)^2

Solving this we get required answer

Solution to problem 23:

(I will use $x$ for a ball of mass $2^{x}m$.)

Since collisions are elastic:

(i)coefficient of restitution = 1

(ii)velocity of approach == velocity of separation

(iii)momentum remains conserved.

Let the velocity given to the first ball be $v$.

Now using (iii), $2mv = 2mv_{1} + 4mv_{2}$ , where $v_{1} \rightarrow \text{velocity of first ball after collision}$ and $v_{2} \rightarrow \text{velocity of second ball after collision.}$

Also , using (ii), $v = v_{1} + 2v_{2}$

Solving these two equations, we get $v_{1} = -v/3 ; v_{2} = 2v/3$

[Negative sign shows that after collision, first ball will move in backward direction.]

Similarly, velocity of 3rd ball after collision with second ball will be $4v/9$.

Thus proceeding as above,velocity of ball of mass $2^{x}m$ after collision with (x-1)th ball will be $(\frac{2}{3})^{x-1} v$.

Now velocity of ball 100 after collision with ball 99 will be $\frac{2^{99}}{3^{99}} v$.

Now the ball 100 will collide with ball 101.

Using (ii) and (iii), velocity of ball 101 after collision will be $k$ where $k =\frac{2^{100}}{3^{100}} v$

and velocity of ball 100 be t , $t = \frac{-2^{99}}{3^{100}} v$.

Distance between each ball $= 1 \text{unit}$.

Time taken by ball 101 to reach ball 102 = $\dfrac{1}{k}$

Distance travelled by ball 100 in backward direction in this time = $\dfrac{t}{k}$.

Thus distance between ball 100 and ball 101 $= 1-(-\dfrac{t}{k})$ which on simplifying gives $1.5$ units.

Thus we need to find out value of $f(2a/3)$ where a is that giant number.

Note that f(q) returns q if q is a multiple of 9.

Clearly 9 divides 2a/3.

Thus f(2a/3) = 9.

Solution to problem 3 At pure rolling let speed of table cloth slipping be $v$ So $v$ will be speed of centre of mass in left direction.

Also it will be rotating with angular velocity$= \omega$ in anticlockwise.

Also $v=r \omega$

About point of contact centre of mass will have velocity=$r \omega$ due right.

So total speed of centre of mass will be

$v_{cm}=v-r \omega$

But $v=r \omega$ .Hence $v_{cm}=0$

Solution to Problem 6

$\dfrac{2m_1F}{k(m_1+m_2)}?$

$$ $\textbf{Solution to Problem 7}$ $$ Working from the Center of Mass frame, consider the system '$2 blocks + spring$'. The CM has an acceleration of $\frac{F}{m+M}$.$$ Consequently, on both blocks, a pseudo force of $Ma$ and $ma$ respectively has to be applied towards the left.$$

Thus, on the right block, there is a net external force of $F_2=F-M\frac{F}{m+M}=\frac{mF}{m+M}$ while on the left block there is a net external force of $F_1=\frac{mF}{m+m}$. $$

From the CM it will appear as though the blocks move away from each other until at the instant of maximum elongation they come to instantaneous rest. $$

Let the displacement of the left block from the CM be $x_1$ to the left and the right one, $x_2$ to the right. $$

From a modified version of the Work Energy Theorem, $$

$W_{NC}+W_{ext}=\Delta TE$

$F_1x_1+F_2x_2=\dfrac{1}{2}k(x_1+x_2)^2$

$\textbf{Solution to Problem 10:}$

Let the axis of rotation of the turntable pass through the origin and the plane of rotation be the x-y plane. At any instant let the position vector of the of the ball be $\vec{r}$. Let R be its radius.

Also the pure rolling motion is caused by the frictional force $f$. Call the angular velocity of the disc as $\vec{\omega_{t}}$ and for the ball, $\vec{\omega_{b}}$. The equations of general motion of the ball are given by:

$f =m\frac { dv }{ dt } ............(1)$(translational motion) $-{ f }\times R =I\frac { d\omega_{b} }{ dt } ........(2)$(rotational motion)

Now, define the velocity $\vec{v}$ of ball in lab frame. This is simply the vector sum of velocity of disc(at $\vec{r}$) and the velocity of ball.

$\therefore \vec{v}= \vec{\omega_{t}}\times \vec{r} + \vec{\omega_{b}}\times R.......(3)$

Differentiating Eq(3) w.r.t to time, we get: $\frac { dv }{ dt } =\omega_{t} \times \frac { dr }{ dt } +\frac { d\omega_{b} }{ dt } \times R \\ \\$

Now, work out the cross product of the above equation.Clearly the only unknown now left is $\frac { d\omega_{b} }{ dt }$. So combine Equation 1 and 2, then substitute it in Equation 3. Finally we get:$\frac { dv }{ dt } =\omega_{t} \times v-\left( \frac { m{ R }^{ 2 } }{ I } \right) \frac { dv }{ dt }$

Now rearranging and plugging the value of $I$ we get: $\vec{a}=\left( \frac { 2 }{ 7 } \omega_{t}\right)\times v$

This equation describes the trajectory of a particle moving in circular motion and the ratio of their frequencies comes out to be $2:7$

Solution to Problem 17

So, clearly, it can be seen from the diagram, that when the body reaches the point $'A'$, it will have it's weight i.e. $mg$ acting downwards. Along with this, since the body would have acquired a velocity $v$, hence it would also feel centrifugal force that would tend to press on the frame and hence add on to it's weight. All we need to do is to calculate this force.

By applying energy conservation, Loss in potential energy = Gain in kinetic energy,

So, $\frac { 1 }{ 2 } m{ v }^{ 2 }(1+\frac { 2{ r }^{ 2 } }{ 5{ r }^{ 2 } } =mgc\\ \Rightarrow v=\sqrt { \frac { 10 }{ 7 } gc }$.

Also, since the centrifugal force is equal to the centripetal force, and for that we need a radius. Now, since this curve is not exactly a circle, so we use the concept of Radius of curvature. Let that be given by $'r'$.

Formula for the radius of curvature is given by $r=\frac { { (1+{ (y' })^{ 2 }) }^{ \frac { 3 }{ 2 } } }{ y'' }$

Here, $y=\frac { { x }^{ 2 } }{ 4a }$, $y'=\frac { x }{ 2a }$, $y''=\frac { 1 }{ 2a }$ and $a$ is the focal length of the given parabola. Putting these values in the formula for $r$ and substituting $x=0$ for point $'A'$, we get $r=2a$.

Now, for calculating $a$, we use the fact that $(b,c)$ lie on the curve. Hence, ${ c }^{ 2 }=4ab\\ \Rightarrow \quad a=\frac { { c }^{ 2 } }{ 4b }$.

Hence, we get $r=\frac { { c }^{ 2 } }{ 2b }$.

Finally, centrifugal force is given by ${ F }_{ c }=\frac { m{ v }^{ 2 } }{ r } =\frac { m(\frac { 10gc }{ 7 } ) }{ \frac { { b }^{ 2 } }{ 2c } } =\frac { 20mg{ c }^{ 2 } }{ 7{ b }^{ 2 } }$.

Therefore, total force = ${ F }_{ c }+mg=\frac { 20mg{ c }^{ 2 } }{ 7{ b }^{ 2 } } +mg$

:)

Answer to problem 22 dx is small displacement.

@CH Nikhil I have solved this question.is the answer (n)(n-1)/2

Are the balls identical?

Are the balls identical ? Are they allowed to move infinitely ? Are they spaced equally ?

Ques of David morin

Solution to Problem 2:

Time period, T=7.5s

Therefore, angular velocity $\omega$=$\frac{2\pi}{T}$=$0.8377s^{-1}$

Velocity of the particle performing SHM at any position is given by: ${ v }_{ x }^{ 1}=\omega(A^2-x^2)^{\frac{1}{2}}$

where x is the distance from centre.

Therefore, amplitude, A=2.588m

Solution to Problem 12

Let the Piston move a small distance x. So, the Volume and Pressure slightly changes on both sides, causing a pressure difference, generating a force. Now, $PV^m = c \\ \text{Differentiating we get} \\ V^mdP + mV^{m-1}PdV = 0 \\ \Rightarrow \frac{dP}{P} = \frac{-mdV}{V} \\ dV = Ax , V = Al \\ \Rightarrow dP = \frac{-mxP}{l}$ Since Pressure changes both sides with opposite signs the Pressure Difference , $\Delta P$, will be $\Delta P = \frac{-2mP}{l}x = \frac{F}{A} \\ F = -\frac{2mPA}{l}x = Ma \\ a = -\frac{2mPA}{Ml}x \\ \Rightarrow \omega^2 = \frac{2mPA}{Ml} \\ \Rightarrow T = \boxed{2\pi\sqrt{\frac{Ml}{2mPA}} }$

Is it correct @Samarpit Swain ?

Solution to problem 15:

When the bullet enters the rod it will de-accelerate This means that it will undergo a force in opposite direction to its velocity with -ma According to newtons third law of motion,every action has equal and opp reaction.So ,the bullet too will exert a force on the rod In this case Torque comes into play As the Torque is given by $frsin(\theta)$ where $\theta$ is angle between force and r vector In this case none of them is zero ,So it will produce a rotational effect in the direction of force. So ,if it rotates in direction of force A will move in direction opposite to $v_0$. But if $x>l/2$ the bullet escapes the rod with no rotational effect So for $0<x<=l/2$ THE BULLET escapes the rod without any effect.(Rotational)

Solution Of Problem 15 :

Solution To Problem 16 :

Please?