NLM

For friction to be minimum,

$F\cos \theta = mg\sin \theta$
⇒
$F = m g \tan \theta = 8 \times 1 0 \times \tan 3 0 ^ {\circ} = \frac {8 0}{\sqrt {3}} \mathrm{N}$

Given, $m_{1}=10$ kg, $m_{2}=20$ kg, $F_{2}=200$ N, $a_{1}=12$ m/s $^{2}$ ,
For mass $m_{1}, F_{1} = m_{1}a_{1} = 10 \times 12 = 120\mathrm{N}$
For mass $m_{2}$ ,
⇒
$\begin{array}{c} F _ {2} - F _ {1} = m _ {2} a _ {2} \\ 2 0 0 - 1 2 0 = 2 0 \times a _ {2} \\ a _ {2} = \frac {8 0}{2 0} = 4 \mathrm{ms} ^ {- 2} \end{array}$

Suppose length of plane is L. When block descends the plane, rise in kinetic energy = fall in potential energy
=mg Lsin θ

Work done by friction
Now, work done = change in KE
$m g L \sin \theta = \mu (m g \cos \theta) \times (L / 2) \Rightarrow \tan \theta = \mu / 2$
$\mu = 2 \tan \theta$

Impulse is imparted due to change in perpendicular components of momentum of ball,
$\begin{array}{r l} & J = \Delta p = m v _ {f} - m v _ {i} \\ & \quad = m v \cos 6 0 ^ {\circ} - (- m v \cos 6 0 ^ {\circ}) \\ & \quad = 2 m v \cos 6 0 ^ {\circ} = 2 m v \times \frac {1}{2} = m v \end{array}$

$\therefore \overrightarrow{\Delta V} = \overrightarrow{V}_{\text {final}} - \overrightarrow{V}_{\text {initial}}$ $\Rightarrow \overrightarrow{\Delta V} = (-V\cos 60^0\hat{i} - V\sin 60^0\hat{j})$ $-(V\cos 60^0\hat{i} - V\sin 60^0\hat{j})$ $\Rightarrow \overrightarrow{\Delta V} = (-2V\cos 60^0\hat{i})$ Thus impulse, $I = m|\overrightarrow{\Delta V}|$ $\Rightarrow I = m(2V\cos 60^0) = mV$

By conservation of linear momentum, $0 = 3 \times 16 - 6 \times v$ $\therefore v = 8 \, m/s$ for 6 kg
∴ Kinetic energy of 6kg mass= $\frac{1}{2}$ ×6×8 $^{2}$ =192J.

$ma_{max} = \mu mg, \therefore a_{max} = \mu g$
$F _ {\max} - \mu m g = M a _ {\max}$
$\therefore F _ {\max} = (m + M) \mu g$

$\frac{v^2}{Rg} = \left(\frac{\mu_s + \tan\theta}{1 - \mu_s\tan\theta}\right)\Rightarrow v = \sqrt{Rg\left[\frac{\mu_s + \tan\theta}{1 - \mu_stan\theta}\right]}$

Force, $P = f_{m} = \mu_{s} mg$ (when body is at rest)
When the body starts moving with acceleration a, then
$P - F _ {k} = m a$
$\begin{array}{r l r} & & P - F _ {k} = m a \\ & & \mu_ {s} m g - \mu_ {k} m g = m a \\ & \Rightarrow & a = (\mu_ {s} - \mu_ {k}) g = (0. 5 - 0. 4) 1 0 \\ & & = 0. 1 \times 1 0 \mathrm{ms} ^ {- 2} = 1 \mathrm{ms} ^ {- 2} \end{array}$

For the motion of balloon,
Let upward force is F, then F - mg = ma
⇒
$F - 1 0 0 = 2 0 0 \Rightarrow F = 3 0 0 \mathrm{N}$
When 4 kg mass is removed, then weight of balloon
$= 6 \times 1 0 = 6 0 \mathrm{N}$
This is smaller than the upward force.
$F - 6 0 = 6 \times a$
where, $a =$ new upward acceleration
$\begin{array}{r l} & \Rightarrow \quad 3 0 0 - 6 0 = 6 a \Rightarrow 2 4 0 = 6 a \\ & \therefore \quad a = \frac {2 4 0}{6} = 4 0 \mathrm{ms} ^ {- 2} \end{array}$

Let weight of A is $w'$ . From the free body diagram, for equilibrium of the system,

$\begin{array}{c} {T \cos \theta = \mu N = \mu w} \\ {T \sin \theta = w ^ {\prime}} \end{array}\tag{... (i}$
... (ii)
where, T = tension in the thread lying between knot and the support.
On dividing Eq. (ii) by Eq. (i), we get
$\frac {T \sin \theta}{T \cos \theta} = \frac {w ^ {\prime}}{\mu w} \Rightarrow \tan \theta = \frac {w ^ {\prime}}{\mu w}$
$\Rightarrow \qquad w ^ {\prime} = \mu w \tan \theta$

Given, $m_{A} = 4 \, kg$ , $m_{B} = 2 \, kg$
and

So, total mass, $M = 4 + 2 + 1 = 7 \, kg$
Now,
$F = M a \Rightarrow 1 4 = 7 a \Rightarrow a = 2 \mathrm{ms} ^ {- 2}$
FBD of block A,

$F - F ^ {\prime} = 4 a \Rightarrow F ^ {\prime} = 1 4 - 4 \times 2 \Rightarrow F ^ {\prime} = 6 \mathrm{N}$
Hence, the contact force between A and B is 6N.

FBD of block A,

... (i)
FBD of block B,

$m _ {2} g - T = m _ {2} a\tag{... (ii}$
Adding Eqs. (i) and (ii), we get
$\begin{array}{r l r} & & m _ {2} g - m _ {\mathrm{f}} a = m _ {2} a + f _ {k} \\ \Rightarrow & & m _ {2} g - m _ {\mathrm{f}} a = m _ {2} a + \mu_ {k} m _ {\mathrm{l}} g \\ \Rightarrow & & a = \frac {(m _ {2} - \mu_ {k} m _ {\mathrm{l}}) g}{m _ {\mathrm{l}} + m _ {\mathrm{2}}} \end{array}$
From Eq. (ii), we get
$T = m _ {2} (g - a) = m _ {2} \left[ 1 - \frac {(m _ {2} - \mu_ {k} m _ {1})}{m _ {1} + m _ {2}} \right] g$
$T = \frac {m _ {1} m _ {2} (1 + \mu_ {k})}{(m _ {1} + m _ {2})} g$

$T\sin \theta = W_{A}$
$T \cos \theta = \mu W$
$\therefore W _ {A} = \mu W \tan \theta .$

F.B.D of blocks $A$ and $B$

for equilibrium of block A
$f _ {1} = w _ {A} = 2 0 N$
If friction is static, $f_{1} \leq \mu N_{1}$
$\text { or } 2 0 \leq . 1 \times F \Rightarrow F \geq \frac {2 0}{. 1} = 2 0 0 N$
Hence limiting value of friction between the blocks
$f _ {1} = 1 \times 2 0 0 = 2 0 N$
from F.B.D of block B
$f _ {2} = f _ {1} + w _ {B} = 2 0 + 1 0 0 = 1 2 0 N.$

$a = \frac{m_2g - \mu_km_1g}{m_1 + m_2}$
$m _ {2} g - T = (m _ {2}) (a)$

$m _ {2} g - T = (m _ {2}) \left(\frac {m _ {2} g - \mu_ {k} m _ {1} g}{m _ {1} + m _ {2}}\right)$
Solving get $T = \frac{m_1mg(1 + \mu_k)g}{m_1 + m_2}$ .

$\mu_{s} = \tan 30^{0} = \frac{1}{\sqrt{3}} = 0.57$
$\mu_ {s} = 0.5 7 = 0.6$
$S = u t + \frac {1}{2} a t ^ {2}$
$4 = \frac {1}{2} a (4) ^ {2} \Rightarrow a = \frac {1}{2} = 0.5$
$a = g \sin \theta - \mu_ {k} (g) \cos \theta \Rightarrow \mu_ {k} = \frac {0.9}{\sqrt {3}} = 0.5.$

B

When the balloon is descending down with acceleration a,
$m g - B = m \times a\tag{... (i}$
where, B = Buoyant force.
Here, we should assume that while removing some mass, the volume of balloon and hence, buoyant force will not change.
Let the new mass of the balloon is $m'$ .
So, mass removed = m - m'
When the balloon is moving up with acceleration a, then
$B - m ^ {\prime} g = m ^ {\prime} \times a\tag{... (ii}$
Adding Eqs. (i) and (ii), we get
⇒
$m g - m ^ {\prime} g = m a + m ^ {\prime} a$
→
$(m g - m a) = m ^ {\prime} (g + a)$
⇒
$m (g - a) = m ^ {\prime} (g + a)$
:
$m ^ {\prime} = \frac {m (g - a)}{(g + a)}$
So, mass removed,
$\begin{array}{c} \Delta m = m - m ^ {\prime} = m \left[ 1 - \frac {(g - a)}{(g + a)} \right] = m \left[ \frac {(g + a) - (g - a)}{(g + a)} \right] \\ = m \left[ \frac {g + a - g + a}{g + a} \right] = \frac {2 m a}{g + a} \end{array}$

Given, $u = 30\sqrt{2}$ ms $^{-1}$
Distance = s metre
Let a be the acceleration of the bullet.
According to the first condition, $(v = 0)$
From third equation of motion, $v^{2}=u^{2}-2as$
$0 = 900 \times 2-2 a s \Rightarrow a = \frac {900}{s} \mathrm{ms} ^ {- 2}$
For second condition,
$v ^ {2} = u ^ {2} - 2 a s = (30 \sqrt {2}) ^ {2} - 2 \times \frac {900}{s} \times \frac {s}{2}$
$v ^ {2} = 1800 - 900$
$v = \sqrt {900} = 30 \mathrm{ms} ^ {- 1}$

First of all consider the forces on the blocks

For the Ist block, $mg - T_{1} = m \times a$ $(\because m_{1} = m_{2} = m_{3})$ ...(i) $\Rightarrow$ Let us consider 2nd and 3rd blocks as a system.
So, $T_{1} - 2\mu mg = 2m\times a$
..(ii)
Adding Eqs. (i) and (ii), we get
$m g (1 - 2 \mu) = 3 m \times a \Rightarrow a = \frac {g}{3} (1 - 2 \mu)$

From FBD shown below, $mg \sin \theta - f = ma$

$m g \sin \theta - m a = f$
$\Rightarrow 8\times 10\sin (30^{\circ}) - 8\times 0.4 = f$
$40 - 3.2 = f \Rightarrow f = 36.8 \mathrm{N}$

If the same wedge is made rough, then time taken by it to come down becomes n times more.

For rough surface, $t = \sqrt{\frac{2s}{g\sin\theta - \mu g\cos\theta}}$
... (i)
For smooth surface, $\frac{t}{n}=\sqrt{\frac{2s}{g\sin\theta}}$ ... (ii)
Squaring Eqs. (i) and (ii) and dividing them, we get
$\mu = \left(1 - \frac {1}{n ^ {2}}\right) \tan \theta$
$\mu = \left[ 1 - \frac {1}{n ^ {2}} \right] \tan \theta = \left[ 1 - \frac {1}{2 ^ {2}} \right] \tan 4 5 ^ {\circ}$
→
$\mu = \frac {4 - 1}{4} = \frac {3}{4}$

Consider a body of mass m placed on a rough inclined surface and it is just on the point of sliding down, with coefficient of friction $\mu$ inclined at angle $\theta$ , as shown in figure.

At the equilibrium point O,
In case of limiting condition, $F = mg\sin \theta$
Normal force, $R = mg \cos \theta$
... (ii)
On dividing Eq. (i) by Eq. (ii), we get
$\underline {{F}} = \underline {{m g \sin \theta}}$
$R \quad m g \cos \theta$
$\mu = \tan \theta$
(∵ Force of friction, $F = \mu R$ )
$\Rightarrow \theta = \tan^ {- 1} (\mu)$
This is maximum value of $\theta$ for mass m to be at rest. For smaller $\theta$ , body will be at rest, i.e. in equilibrium.
So, angle of repose, i.e. $\theta = \tan^{-1}\mu$ .

$m g - B = m a\tag{... (i}$

$B - (m - m _ {0}) g = (m - m _ {0}) a\tag{... (ii}$
Equation (i) + equation (ii)
$\Rightarrow m g - m g + m _ {0} g = m a + m a - m _ {0} a$
$\Rightarrow m _ {0} = \frac {2 m a}{g + a}.$