NLM

Here $f = mg$ and $N = m\alpha$ but $f \leq \mu N$ so $mg \leq \mu m\alpha \Rightarrow \alpha \geq \frac{g}{\mu}$ .

The various forces acting on the book are as shown in
where, $f =$ Frictional force
$W =$ Weight of a book
$N =$ Normal reaction

From figure,
The direction of force of friction between the wall and the book is upwards.

Here, $m = 20 \, \text{kg}$ , $u = 20 \, \text{m/s}$ , $v = 5 \, \text{m/s}$ , $s = 100 \, \text{m}$
Using $v^{2} = u^{2} + 2as$
$F = m a = m \times \left(\frac {v ^ {2} - u ^ {2}}{2 s}\right)$
Substituting the values, we get
$F = 2 0 \times \left[ \frac {(5) ^ {2} - (2 0) ^ {2}}{2 \times 1 0 0} \right] = - 3 7.5 N.$

According to Newton's second law :
Force = rate of change of linear momentum.

Here, mass of the body, $m = 0.05\mathrm{kg}$ Acceleration $a = 9.5ms^{-2}, g = 9.8ms^{-2}$ $\therefore mg - f_{air} = ma \Rightarrow f_{air} = m(g - a)$ $= 0.05(9.8 - 9.5) = 0.015N.$

The maximum tension = $30 \times 10N = 300N$ T - mg = ma. $300 - 10 \times 10 = 10a$ $\Rightarrow a = 20m/s^{2}$ [Maximum value]
when the mass is raised, u = 0, $a = 20m/s^{2}$ .
S = 10m, t = ? $10 = \frac{1}{2} \times 20.t^{2} \Rightarrow t = 1s.$

The acceleration of the system is

... (i)
The tension in the middle of the rope (i.e., at point $A$ ) is
$T = \left(m + \frac {m}{6}\right) a = \frac {7 m}{6} \times \frac {3 F}{4 m} \Rightarrow T = \frac {7 F}{8}.\tag{[Using (i)]}$

$\vec{p}(t)=A(\hat{i}\cos kt-\hat{j}\sin kt)$

$\vec{F}.\vec{p} = A^2 k(-\cos kt\sin kt + \sin kt\cos kt) = 0$
∴ The momentum and force are perpendicular to each other at $90^{\circ}$ .

a) Is correct $6^{\text{th}}$ coin has four coins on its top which exert a force $4\mathrm{mg}$ on it
b) Is correct. $7^{\text{th}}$ coin has three coins, placed over it.
Thus $7^{\text{th}}$ coin exerts a force $4mg$ on $6^{\text{th}}$ coin (downwards)
c) Is correct. As what is explained in b), the reaction of $6^{\text{th}}$ coin on the $7^{\text{th}}$ coin is $4mg$ (upwards)

d) Is wrong $10^{\text{th}}$ coin, which is the topmost coin, experiences a reaction force of $mg$ (upwards) from all the coins below it.

According to conservation of linear momentum $p_{3}=\sqrt{p_{1}^{2}+p_{2}^{2}}$
$\Rightarrow m \times 4 = \sqrt {(1 \times 1 2) ^ {2} + (2 \times 8) ^ {2}} = 2 0 \Rightarrow m = 5 k g.$

According to law of conservation of momentum the third piece has momentum $= 1 \times -(3i + 4\hat{j})kg ms^{-1}$ Impulse = Average force × time $\Rightarrow$ Average force = $\frac{\text{Impulse}}{\text{time}}$ $= \frac{\text{Change in momentum}}{\text{time}}$ $= \frac{-(3\hat{i} + 4\hat{j})kg ms^{-1}}{10^{-4}s} = -(3\hat{i} + 4\hat{j}) \times 10^{4} N.$

Newton's second and third laws lead to the conservation of linear momentum.

Here, mass of bullet $m = 10g = \frac{10}{1000} kg$ . Mass of ice, $M = 5kg$ According to the conservation of linear momentum, we get $m \times 300 + M \times 0 = m \times 0 + mv \Rightarrow \frac{10}{1000} \times 300 + M \times 0 = 5v \therefore v = \frac{3}{5} = 0.6m / s = 60cm / s.$

Since there in no resultant external force, linear momentum of the system remains constant.

If $1N$ and $2N$ act in the same direction and $3N$ acts in opposite direction, equilibrium is possible.

Work done in max extension
= stored P.E. $\Rightarrow Mg \times x = \frac{1}{2} kx^{2}$ $\Rightarrow x = \frac{2Mg}{k}$ .

Block A moves with velocity $0.15ms^{-1}$ compresses the spring which pushes $B$ towards right. A goes on compressing the spring till the velocity acquired by $B$ becomes equal to the velocity of $A$ . Let this velocity be $\nu$ This state occurs when the spring is in a state of maximum compression. Let $x$ be the maximum compression in this stage.
According to the law of conservation of linear momentum, we get $0.15 \, ms^{-1}$
$m _ {A} u = (m _ {A} + m _ {B}) v$

$\Rightarrow v = \frac {m _ {A} u}{m _ {A} + m _ {B}} = \frac {2 \times 0.1 5}{2 + 3} = 0.0 6 m s ^ {- 1}.$
According to the conservation of energy, we get
$\frac {1}{2} m _ {A} u ^ {2} = \frac {1}{2} (m _ {A} + m _ {B}) v ^ {2} + \frac {1}{2} k x ^ {2}$
$\frac {1}{2} m _ {A} u ^ {2} - \frac {1}{2} (m _ {A} + m _ {B}) v ^ {2} = \frac {1}{2} k x ^ {2}$
$\frac {1}{2} \times 2 \times (0.1 5) ^ {2} - \frac {1}{2} (2 + 3) (0.0 6) ^ {2} = \frac {1}{2} k x ^ {2}$
$0.0 2 2 5 - 0.0 0 9 = \frac {1}{2} k x ^ {2} \Rightarrow 0.0 1 3 5 = \frac {1}{2} k x ^ {2}$
$\Rightarrow x = \sqrt {\frac {0.0 2 7}{k}} = \sqrt {\frac {0.0 2 7}{1 0.8}} = 0.0 5 m.$

$2 m g - T = 2 m a$ ...(i)
$T - m g \sin 3 0 ^ {\circ} = m $...(ii)
$\mathrm{(i)} + \mathrm{(ii)} \text {gives},$
$2 m g - \frac {m g}{2} = 3 m a \Rightarrow a = \frac {g}{2}.$

Acceleration = $\frac{500}{25}$ = $20m/s^{2}$ in both the cases.
In fig 1, $T = 500 - 10 \times 20 = 300N$ .
In fig 2, $T' = 500 - 15 \times 20 = 200N$ .

Mass of rope, $m = 0.1 \mathrm{~kg}, \theta = 10^{\circ}$

From figure, $2T\sin \theta = mg$
$\Rightarrow T = \frac {m g}{2 \sin \theta} = \frac {0.1 \times 9.8}{2 \sin 1 0 ^ {\circ}} = 2.8 2 N.$

Gravitational field is a conservative field. Therefore work done in moving a particle from $A$ to $B$ is independent of path chosen.

The limiting force of friction is
$f _ {s} = 1 0 N$
As $F = 8N < f_{s}$ , therefore, block does not move. As static friction is a self adjusting force, therefore the frictional force on the block is $8N$ .

C

Frictional force is independent of the area of surfaces in contact, as long as normal reaction between the two bodies in contact remains the same.

(i)

(ii)
When an iron block of mass m with surface area A is placed on an inclined plane inclined at an angle $\theta$ , it will experience a frictional force f as shown in figure (i). $f = \mu R = \mu mg \cos \theta$ … (i)
When an iron block of mass $m$ with surface area $2A$ is placed on an inclined plane inclined at an angle $\theta$ , it will experience a frictional force $f'$ as shown in figure (ii).
$f ^ {\prime} = \mu R = \mu m g \cos \theta$
From (i) and (ii), we get $f' = f$ .

For rotational equilibrium about point "P",
$mg \sin \theta \left( \frac{b}{2} \right) = mg \cos \theta \left( \frac{a}{2} \right)$ $\Rightarrow \tan \theta = \frac{a}{b} = \frac{10}{15} = \frac{2}{3}$ $\Rightarrow \theta = 33.69^\circ$

i.e., toppling starts at $\theta = 33.69^{\circ}$
and angle of repose $= \tan^{-1}(\mu) = \tan^{-1}(\sqrt{3}) = 60^{\circ}$ it mean the block will remain at rest on the plane up to certain angle $\theta$ and then it will topple.