NLM

By Newton's second law of motion, net external force applied on a system is equal to rate of change of momentum.
i.e.
$\mathbf {F} _ {\mathrm{ext}} = \frac {d \mathbf {p}}{d t}$
If $\mathbf{F}_{\mathrm{ext}} = 0$ , then $\frac{d\mathbf{p}}{dt} = 0 \Rightarrow p = \mathrm{constant}$
i.e. When $F_{ext} = 0$ , then momentum of the system remains conserved.
The internal interaction between particles of a body do not contribute to change in total momentum of body.
Hence, both Assertion and Reason are false.

For equilibrium
condition of the mass
$T\sin 45^{0} = F$ ... (i)
$T\cos 45^{0} = 100$ ... (ii)
or

$\tan 45^{\circ} = \frac{100}{F}$
$\Rightarrow F = 100N.$

$m g \sin 4 5 ^ {\circ} = \frac {1 0 0}{\sqrt {2}} = 5 0 \sqrt {2}$
$\begin{array}{r} \mu m g \cos \theta = 0.6 \times m g \times \frac {1}{\sqrt {2}} \\ = 0.6 \times 5 0 \sqrt {2} \end{array}$

$73.7 = 3 + mg sin θ$
$P = 31.28 = 32N$ (closest value)

$\frac{dp}{dt} = F = kt$
$\int_ {p} ^ {3 p} d p = \int_ {O} ^ {T} k t d t$
$2 p = \frac {K T ^ {2}}{2}$
$T = 2 \sqrt {\frac {P}{K}}.$

Momentum $p = mv$ ... (i) and for motion under gravity
$h = \frac {u ^ {2} - v ^ {2}}{2 g}\tag{... (ii}$

$h = \frac {u ^ {2} - p ^ {2} / m ^ {2}}{2 g}$

$m = 20g, u = 1m / s, v = ?$
$S = 2 0 \times 1 0 ^ {- 2} m \quad a = \frac {- 2.5 \times 1 0 ^ {- 2}}{2 0 \times 1 0 ^ {- 3}} m / s ^ {2}$
$v ^ {2} = u ^ {2} + 2 a s$
$v ^ {2} = 1 - 2 \times \frac {2.5 \times 1 0 ^ {- 2}}{2 0 \times 1 0 ^ {- 3}} \times \frac {2 0}{1 0 0}$
$v = \frac {1}{\sqrt {2}} \simeq 0. 7 m / s.$

$\begin{array}{c c}V _ {1} \leftarrow \boxed {5 0}&\boxed {2 0} \rightarrow V _ {2}\\\text { Man }&\text { Son }\end{array}$
$\Rightarrow 0 = 50V_{1} - 20V_{2}$ and $V_{1} + V_{2} = 0.7$
$\Rightarrow V _ {1} = 0.2.$

For just equilibrium
$2 = \frac {\sqrt {3} m g}{2} \mu_ {s} - \frac {m g}{2} \dots (\mathrm{i})$

In the other case
$\frac {\sqrt {3} m g}{2} \mu_ {s} + \frac {m g}{2} = 1 0 \dots (\mathrm{ii})$
equation (i)/equation (ii)
$\frac {1}{5} = \frac {\sqrt {3} \mu_ {s} - 1}{\sqrt {3} \mu_ {s} + 1}$

$\mu_ {s} = \frac {\sqrt {3}}{2}.$

$a _ {A \max} = \mu g = 2 m / s ^ {2}$
$F - 8 = 4 \times 2 = 1 6 N.$

$N _ {1} = 6 0$
$N _ {2} = 4 0$
$a _ {1} = \frac {1 0 \sqrt {3} - 0 . 2 \times 6 0}{5}$
$a _ {2} = \frac {1 0 \sqrt {3} - 0 . 2 \times 4 0}{5}$
$a _ {2} - a _ {1} = 0. 8$

According to the question, the free body diagram of the given condition will be

Since, the wedge is accelerating towards right with a, thus a pseudo force acts in the left direction in order to keep the block stationary. As, the system is in equilibrium.
$\begin{array}{c} \Sigma F _ {x} = 0 \quad \text {or} \quad \Sigma F _ {y} = 0 \\ R \sin \theta = m a \end{array}$
Similarly, $R\cos \theta = mg$
Dividing Eq. (i) by Eq (ii), we get
$\begin{array}{c} \frac {R \sin \theta}{R \cos \theta} = \frac {m a}{m g} \\ \tan \theta = \frac {a}{g} \text {or} a = g \tan \theta \end{array}$
∴ The relation between a and g for the block to remain stationary on the wedge is $a = g \tan \theta$ .

The opposing force that comes into play when one body is actually sliding over the surface of the other body is called sliding friction.
The coefficient of sliding is given as $\mu_{s}=N/F_{sliding}$ where, N is the normal reaction and $F_{sliding}$ is the sliding force. As, the dimensions of N and $F_{sliding}$ are same. Thus, $\mu_{s}$ is a dimensionless quantity. When body is rolling, then it reduces the area of contact of surfaces, hence rolling friction is smaller than sliding friction.
Hence, statement (d) is incorrect.

Let $a_{1}$ and $a_{2}$ be accelerations of a ball (upward) and rod (downward), respectively.
Clearly, from the diagram,
$2 a _ {1} = a _ {2}$
Now, for the ball, ... (i)

$2 T - \frac {9}{5} m g = \frac {9}{5} m a _ {1}\tag{... (ii}$
and for the rod, $mg - T = ma_{2}$ ... (iii)
On solving Eqs. (i), (ii) and (iii), we get
$\begin{array}{l} a _ {1} = \frac {g}{2 9} \mathrm{m} / \mathrm{s} ^ {2} \uparrow (\text { upward }) \\ a _ {2} = \frac {2 g}{2 9} \mathrm{m} / \mathrm{s} ^ {2} \downarrow (\text { downward }) \end{array}$
So, acceleration of ball w.r.t. rod = $a_{1} + a_{2} = (3g / 29)$ m/s $^{2}$
Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is 1m.
Using second equation of motion, $s = ut + \frac{1}{2} at^{2}$
$\begin{array}{l} {1 = 0 + \frac {1}{2} \times \frac {3 \times 1 0}{2 9} t ^ {2}} \\ {t = \sqrt {5 8 / 3 0} = 1.4 \mathrm{s(approx)}} \end{array}$

Angle of repose is equal to angle of limiting friction as maximum value of static friction is called the limiting friction. Hence, both Assertion and Reason are correct and Reason is the correct explanation of Assertion.

Since, $F = (M + m)a$ ... (i)

Normal, $N = (mg \cos \alpha + ma \sin \alpha)$
So, we apply pseudo force on the block by observing it from the wedge.
Now, from free body diagram of block, we get
⇒ $a = g \frac {\sin \alpha}{\cos \alpha} \Rightarrow a = g \tan \alpha\tag{... (ii}$
Now, from Eqs. (i) and (ii), we get
$F = (M + m) g \tan \alpha$

Given, $\theta = 45^{\circ}$ , $s_{1} = s_{2}$ , u = 0, $t_{1} = 2t_{2}$
On the rough incline, $a_{1}=g(\sin\theta-\mu\cos\theta)$
On the frictionless incline, $a_{2}=g\sin\theta$
$\mathrm{As}, \qquad s = u t + \frac {1}{2} a t ^ {2}$
$\begin{array}{l} \text {So,} s _ {1} = 0 + \frac {1}{2} g (\sin \theta - \mu \cos \theta) t _ {1} ^ {2} \text {and} s _ {2} = 0 + \frac {1}{2} g \sin \theta t _ {2} ^ {2} \\ \text {As,} \qquad s _ {1} = s _ {2}, \frac {1}{2} g (\sin \theta - \mu \cos \theta) t _ {1} ^ {2} = \frac {1}{2} g \sin \theta t _ {2} ^ {2} \\ \Rightarrow \qquad \frac {\sin \theta - \mu \cos \theta}{\sin \theta} = \frac {t _ {2} ^ {2}}{t _ {1} ^ {2}} \\ \Rightarrow \qquad 1 - \mu \cot \theta = \frac {t _ {2} ^ {2}}{(2 t _ {2}) ^ {2}} \qquad (\because t _ {1} = 2 t _ {2}) \\ \Rightarrow \qquad 1 - \mu \cot \theta = 1 / 4 \Rightarrow \mu \cot \theta = 1 - \frac {1}{4} = \frac {3}{4} \\ \therefore \qquad \mu = \frac {3}{4 \cot \theta} \end{array}$

Acceleration of the body down the inclined plane = g sinθ.

∴ Force applied on spring balance
$= m g \sin \theta = 5 \times 1 0 \times \sin 3 0 ^ {\circ} = 5 \times 1 0 \times \frac {1}{2} = 2 5 \mathrm{N}$

Given, $F = 10 \mathrm{~N}, v_{i} = 0, m = 0.5 \mathrm{~kg}$ and $\Delta t = 0.25 \mathrm{~s}$
As impulse, $I = p_{f} - p_{i}$
Also, $I = F\cdot \Delta t = 10\times 0.25 = 2.5\mathrm{N - s}$

As there is a load at P, so tension in AP and PB will be different. Let these be $T_{2}$ and $T_{1}$ , respectively. For vertical equilibrium at P,

$T _ {1} \cos 6 0 ^ {\circ} = M g, i. e. T _ {1} = 2 M g\tag{... (i}$
and for horizontal equilibrium at P,
$T _ {2} = T _ {1} \sin 6 0 ^ {\circ} = T _ {1} (\sqrt {3} / 2)$
Substituting the value of $T_{1}$ from Eq. (i), we get
$T _ {2} = (2 M g) \times (\sqrt {3} / 2) = \sqrt {3} M g$

$F = \frac{dp}{dt} = 2nmv\cos 45^{\circ}$ Pressure $= \frac{F}{A} = \frac{2nmv\cos 45^{\circ}}{Area}$ $= \frac{2\times 10^{23}\times 3.3\times 10^{-27}\times 10^{3}\times\left(\frac{1}{\sqrt{2}}\right)}{2\times 10^{-4}}$ $= 2.35\times 10^{3}N/m^{2}$

Coefficient of sliding friction has no dimension.

Initially system is in equilibrium with a total weight of 4mg over spring.

∴ kx = 4mg
When string is cut at the location as shown above.
Force on mass $m$ , $F_{B} = mg$
∴ Acceleration of mass m, $a_{B} = g$
For mass 3m
If $a_{A}=$ acceleration of block of mass 3m, then $F_{net}=4mg-3mg$
$\Rightarrow 3 m \cdot a _ {A} = m g \mathrm{or} a _ {A} = \frac {g}{3}$
So, accelerations of blocks A and B are $a_{A}=\frac{g}{3}$ and $a_{B}=g$

Given situation can be represented as
$F \leftarrow m _ {1} \rightarrow T _ {1} \leftarrow m _ {2} \rightarrow T _ {2} \leftarrow m _ {3} \rightarrow T _ {3} \leftarrow m _ {4}$
$\Rightarrow T _ {1} = \frac {(m _ {2} + m _ {3} + m _ {4})}{m _ {1} + m _ {2} + m _ {3} + m _ {4}} F$
Given, $m_{1} = m_{2} = m_{3} = m_{4} = m$
$\therefore T_{1} = \frac{3}{4} F$
Similarly, $T_{2} = \frac{(m_{3} + m_{4})F}{m_{1} + m_{2} + m_{3} + m_{4}}$
:
$T _ {2} = \frac {1}{2} F$
Also, $T_{3} = \frac{m_{4}F}{m_{1} + m_{2} + m_{3} + m_{4}}\Rightarrow T_{3} = \frac{1}{4} F$

According to question, the situation can be shown as

Case 1 For body projected up the plane, v = 0
Using third equation of motion, $v^{2}=u^{2}-2a_{1}s$
$0 = u ^ {2} - 2 a _ {1} s$:
$u = \sqrt {2 a _ {1} s}$
Also,
$v = u - a _ {1} t _ {1}$
$0 = u - a _ {1} t _ {1}$
$t _ {1} = u / a _ {1}$
$t _ {1} = \frac {\sqrt {2 a _ {1} s}}{a _ {1}} = \sqrt {\frac {2 s}{a _ {1}}}\tag{... (i}$
As, $ma_{1}=\mu R+mg\sin\theta$
∴Retardation, $a_{1}=\frac{\mu R+mg\sin30^{\circ}}{m}$
Also,
$R = m g \cos 3 0 ^ {\circ}$
${r l}$ $a _ {1} $ $= \frac {\mu m g \cos 3 0 ^ {\circ} + m g \sin 3 0 ^ {\circ}}{m}$ $= \mu g \frac {\sqrt {3}}{2} + \frac {g}{2} $ ....(ii)
Case 2 For body coming down the plane,
$s = u t + \frac {1}{2} a _ {2} t _ {2} ^ {2}$
As,
$t _ {2} = \sqrt {2 s / a _ {2}}$ ... (iii)
Downward acceleration,
$ a _ {2} $ $= \frac {m g \sin 3 0 ^ {\circ} - \mu R}{m}$
$ = \frac {m g \sin 3 0 ^ {\circ} - \mu m g \cos 3 0 ^ {\circ}}{m}$
$ = \frac {g}{2} - \frac {\mu g \sqrt {3}}{2}$...(iv)
Given,
$t _ {1} = t _ {2} / 2$
Substituting values of $t_{1}$ and $t_{2}$ from Eqs. (i) and (iii), we get
$\sqrt {\frac {2 s}{a _ {1}}} = \frac {1}{2} \sqrt {\frac {2 s}{a _ {2}}}$
Now, squaring both sides, we get
$a _ {1} = 4 a _ {2}\tag{... (V}$
Substituting values of $a_{1}$ and $a_{2}$ from Eqs. (ii) and (iv) in Eq. (v), we get
$\mu g \frac {\sqrt {3}}{2} + \frac {g}{2} = 4 \left(\frac {g}{2} - \frac {\mu g \sqrt {3}}{2}\right)$
Solving for $\mu$ , we get
$\mu = \frac {\sqrt {3}}{5} = \frac {1 7 3 2}{5}$
⇒
$\mu = 0.3 4 6$

The force causing the motion of A is frictional force between A and B,
$\mu M _ {B} g = M _ {A} a _ {A}$
So, acceleration of A,
$\begin{array}{c} a _ {A} = \mu \left(\frac {M _ {B}}{M _ {A}}\right) g \\ = \frac {\mu g}{2} (\text { towards right }) \end{array}$
Block B experiences friction force toward left,
$M _ {B} a _ {B} = \mu M _ {B} g$
⇒
$a _ {B} = \mu g (\text { towards left })$