NLM

Here, mass of the block, m = 200kg
coefficient of static friction, $\mu_{s} = 0.5 = \frac{1}{2}$
angle of incline plane, $\theta = 45^{\circ}$
Maximum force that each man can apply, $F = 500N$ Let $N$ number of men are required for the block to just start moving up the plane $NF = mg\sin \theta + f$ $= mg\sin \theta + \mu_s R$ $= mg\sin \theta + \mu_s mg\cos \theta = mg[\sin \theta + \mu_s\cos \theta]$ $= 200 \times 10\left[\sin 45^\circ + \frac{1}{2}\cos 45^\circ\right]$ $= 200 \times 10\left[\frac{1}{\sqrt{2}} + \frac{1}{2\sqrt{2}}\right] = \frac{200 \times 10 \times 3}{2\sqrt{2}}$ $\therefore N = \frac{200 \times 10 \times 3}{2\sqrt{2} \times 500} = 5$ .

The acceleration of the body down the smooth inclined plane is $a = g \sin \theta$ along the inclined plane, where $\theta$ is the angle of inclination.
The vertical component of acceleration a is
$a _ {(\mathrm{alongvertical})} = (g \sin \theta) \sin \theta = g \sin^ {2} \theta$
For block A
$a _ {A (\mathrm{alongvertical})} = g \sin^ {2} (6 0 ^ {\circ})$
For block B
$a _ {B (\text { along vertical })} = g \sin^ {2} (3 0 ^ {\circ})$
The relative vertical acceleration of A with respect to B is
$a _ {A B (\text { along vertical })} = a _ {A (\text { along vertical })} - a _ {B (\text { along vertical })}$
$\Rightarrow a _ {A B (\mathrm{alongvertical})} = g \sin^ {2} (6 0 ^ {\circ}) - g \sin^ {2} (3 0 ^ {\circ})$
$\Rightarrow a _ {A B (\mathrm{alongvertical})} = g \left(\left(\frac {\sqrt {3}}{2}\right) ^ {2} - \left(\frac {1}{2}\right) ^ {2}\right)$
$\Rightarrow a _ {A B (\mathrm{alongvertical})} = \frac {g}{2} = 4. 9 \mathrm{ms} ^ {- 2}$
Hence, the correct answer is (A).

When $P = mg(\sin\theta - \mu\cos\theta)$
$f = \mu m g \cos \theta$
{upwards}
when $P = mg\sin \theta ;f = 0$
and when $P = mg(\sin \theta +\mu \cos \theta)$
$f = \mu m g \cos \theta$
{downwards}
Hence, friction is first positive, then zero and then negative.
Hence, the correct answer is (A).

Here, mass of the body,
$m = 0.0 5 k g$
Acceleration $a = 9.5ms^{-2}$ , $g = 9.8ms^{-2}$
$\therefore m g - f _ {a i r} = m a \Rightarrow f _ {a i r} = m (g - a)$

$= 0.0 5 (9.8 - 9.5) = 0.0 15 N.$

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

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

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

Here, mass of bullet $m = 10g = \frac{10}{1000} kg$ .
$M = 5$
According to the conservation of linear momentum, we get
$m \times 3 0 0 + M \times 0 = m \times 0 + m v$
$\Rightarrow \frac {1 0}{1 0 0 0} \times 3 0 0 + M \times 0 = 5 v$
$\therefore v = \frac {3}{5} = 0.6 m / s = 60 c m / s.$

According to law of conservation of momentum the third piece has momentum
$= 1 \times - (3 i + 4 \hat {j}) k g m s ^ {- 1}$
$\begin{array}{l} \text { Impulse } = \text { Average force } \times \text { time } \\ \Rightarrow \text { Average force } = \frac {\text { Impulse }}{\text { time }} \xleftarrow [ 1 \times (3 \hat {i} + 4 \hat {j}) ]{F} \begin{array}{c} 1 \times 4 \hat {j} \\ \downarrow \\ W \end{array} \begin{array}{c} y \\ x \\ 1 \times 3 \hat {i} \\ \downarrow \\ W \end{array} \\ \text { Change in momentum } \end{array}$
$= \frac {\text { Change in momentum }}{\text { time }}$
$= \frac {- (3 \hat {i} + 4 \hat {j}) k g m s ^ {- 1}}{1 0 ^ {- 4} s} = - (3 \hat {i} + 4 \hat {j}) \times 1 0 ^ {4} N.$

Work done in maximum extension
$= \text { stored } P. E.$
$\Rightarrow M g \times x = \frac {1}{2} k x ^ {2}$
$\Rightarrow x = \frac {2 M g}{k}.$

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 v. 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

$m _ {A} u = (m _ {A} + m _ {B}) \nu$
$\Rightarrow v = \frac {m _ {A} u}{m _ {A} + m _ {B}} = \frac {2 \times 0 . 1 5}{2 + 3} = 0. 0 6 m s ^ {- 1}.$
According to the conservation of energy, we get
$\frac {1}{2} m _ {A} u ^ {2} = \frac {1}{2} (m _ {A} + m _ {B}) \nu^ {2} + \frac {1}{2} k x ^ {2}$
$\frac {1}{2} m _ {A} u ^ {2} - \frac {1}{2} (m _ {A} + m _ {B}) \nu^ {2} = \frac {1}{2} k x ^ {2}$
$\frac {1}{2} \times 2 \times (0. 1 5) ^ {2} - \frac {1}{2} (2 + 3) (0. 0 6) ^ {2} = \frac {1}{2} k x ^ {2}$
$0.0225 - 0.009 = \frac {1}{2} k x ^ {2} \Rightarrow 0.0135 = \frac {1}{2} k x ^ {2}$
$\Rightarrow x = \sqrt {\frac {0.0 2 7}{k}} = \sqrt {\frac {0.0 2 7}{1 0 . 8}} = 0.0 5 m.$

$2mg - T = 2ma\ldots$ (i)
$T - m g \sin 3 0 ^ {\circ} = m a $(ii)
(i) + (ii) gives,

$2 m g - \frac {m g}{2} = 3 m a \Rightarrow a = \frac {g}{2}.$

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 $8N$ .

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",
$mg \sin \theta \left( \frac{b}{2} \right) = mg \cos \theta \left( \frac{a}{2} \right)$
$\Rightarrow \tan \theta = \frac {a}{b} = \frac {1 0}{1 5} = \frac {2}{3}$
$\Rightarrow \theta = 3 3.6 9 ^ {\circ}$

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.

$\begin{array}{r l} & \text { N } \sin \theta = m g \\ & \text { N } \cos \theta = m a \\ & \Rightarrow \tan \theta = \frac {g}{a} \\ & \Rightarrow \cot \theta = \frac {a}{g} = \tan (9 0 ^ {\circ} - \theta) = \frac {d y}{d x} \\ & \Rightarrow \cot \theta = \frac {d y}{d x} \\ & \text { Since } y = k x ^ {2} \\ & \Rightarrow \frac {d y}{d x} = 2 k x \\ & \Rightarrow \frac {a}{g} = 2 k x \end{array}\tag{... (1}$
$\Rightarrow x = \frac {a}{2 \mathrm{kg}}$
Hence, the correct answer is (B).

The force is constant, i.e., the acceleration is constant. The distance is decreasing with time. Therefore it is retardation.
$1 ^ {\mathrm{st}} \text { second } u. 1 - (1 / 2) a. 1 = 5 m$....(i)
$3^{\mathrm{rd}}$ second $u.3 - (1 / 2)a.9 - (u.2 - (1 / 2)a.4) = 3m$
$\therefore u. 1 - (1 / 2) a. 5 = 3 m$....(ii)
$(i) - (i i) \text { gives } (1 / 2) a.4 = - 2 \Rightarrow a = - 1 m / s ^ {2}$
∴ Constant force = ma = 4 × 1 = 4N.
[This is a retarding force as $a$ is negative].

Force $= v\frac{dm}{dt}$
Force $= vM$
[∴ force = M v newton]

$\frac{dM}{dt} = 0.1kg / s, v_{gas} = 50m / s,$ Mass of the rocket $= 2kg, Mv =$ constant $-v\frac{dM}{dt} + M\frac{dv}{dt} = 0.\therefore \frac{dv}{dt} = \frac{1}{M} v\frac{dM}{dt}.$ $\Rightarrow$ Acceleration $= \frac{1}{2}\times 50\times 0.1 = 2.5m / s^2.$

$v_{1} = 10m / s$ ;
$m _ {1} = 1 0 k g$
$v _ {2} = 0;$
$m _ {2} = 9 k g$
$v _ {3} = v;$
$m _ {3} = 1 k g$
According to conservation of momentum
$m _ {1} v _ {1} = m _ {2} v _ {2} + m _ {3} v _ {3}$
$1 0 \times 1 0 = 9 \times 0 + 1 \times v; v = 1 0 0 m / s.$

The bullets are initially at rest. Change of momentum per second $= mvN$ . where $N$ is the number of bullets fired per second.

In terms of three significant figure Momentum $P = mv = 3.513 \times 5.00 = 17.6$ .

$m_{1} = 1000\mathrm{kg}$ . $v_{1} = 50m / s$ , $m_{2} = 1250\mathrm{kg}$ , $v_{2} = ?$ From conservation law of linear momentum $m_{1}v_{1} = m_{2}v_{2}$ $\Rightarrow 1000\times 50 = 1250\times v_{2}\Rightarrow v_{2} = 40m / s.$