NLM

Impulse = Area under F - t curve
A) $\frac{1}{2} \times 1 \times 0.5 = \frac{1}{4} N.s$
B) $0.5 \times 2 = 1N.s$ (maximum)
C) $\frac{1}{2} \times 1 \times 0.75 = \frac{3}{8} N.s$
D) $\frac{1}{2} \times 2 \times 0.5 = \frac{1}{2} N.s$

$\frac{1}{2} mv_A^2 + 0 = \frac{1}{2} mv_B^2 + mgL$
$\frac {1}{2} m (2 v) ^ {2} = \frac {1}{2} m v ^ {2} + 2 m g L$

$\frac {1}{2} m v ^ {2} \times 3 = 2 m g L$
$v = \sqrt {\frac {4 g L}{3}} = 2 \sqrt {\frac {g L}{3}}.$
$\frac {1}{2} m v _ {C} ^ {2} + m g L = \frac {1}{2} m \frac {4 g L}{3} \times 4$
$\frac {1}{2} m v _ {C} ^ {2} + m g L = \frac {8 m g L}{3}$
$\Rightarrow v _ {C} ^ {2} = \frac {5 g L \times 2}{3} \Rightarrow v _ {C} = \sqrt {\frac {1 0 g L}{3}}.$

Kinetic energy,
$K = \frac {1}{2} m v ^ {2} = \frac {1}{2} m \frac {4 g L}{3} = \frac {2 m g L}{3}.$

$N = Mg - F\sin 30^{\circ}$
$= m g - \frac {F}{2} = 1 0 0 - \frac {F}{2} = \frac {2 0 0 - F}{2}$
$F \cos 3 0 ^ {\circ} = \mu N$
$\sqrt {3} \frac {F}{2} = 0.2 5 \times \left(\frac {2 0 0 - F}{2}\right)$
$4 \sqrt {3} F = 2 0 0 - F$
$F = \frac {2 0 0}{4 \sqrt {3} + 1} = 2 5.2 2.$

Friction always opposes relative motion.

Acceleration of mass at distance x
$a = g (\sin \theta - \mu_ {0} x \cos \theta)$
Speed is maximum,
$\text { when } a = 0$
$g (\sin \theta - \mu_ {0} x \cos \theta) = 0$
$x = \frac {\tan \theta}{\mu_ {0}}$

$a = v\frac{dv}{dx} = (g\sin \theta -\mu_0xg\cos \theta)$
$\frac {v ^ {2}}{2} = (g \sin \theta x - \frac {\mu_ {0} x ^ {2}}{2} g \cos \theta)$
$v = 0 \Rightarrow x = \frac {2 \tan \theta}{\mu_ {0}}.$

at $x = \frac{\tan \theta}{\mu_0}v = \sqrt {\frac {g \sin \theta \tan \theta}{\mu_ {0}}}.$

$m = 2 k g $<br /> $ \vec {v} = 2 \hat {i}$
$ \vec {F} = 3 \hat {j} $

$\begin{array}{r l} & \Rightarrow \vec {a} = \frac {\vec {F}}{m} = \frac {3}{2} \hat {j} \\ & \quad t = 2 s \\ & \text {Since} \vec {s} = \vec {v} t + \frac {1}{2} \vec {a} t ^ {2} \\ & \Rightarrow \vec {s} = (2 \hat {i}) 2 + \frac {1}{2} \left(\frac {3}{2}\right) (4) \hat {j} \\ & \Rightarrow \vec {s} = 4 \hat {i} + 3 \hat {j} \\ & \Rightarrow | \vec {s} | = \sqrt {4 ^ {2} + 3 ^ {2}} = 5 m \end{array}$
Hence, the correct answer is (B).

Drawing free body diagram of block with respect to plane.

Acceleration of the block up the plane is
$a = \frac {m g \cos (3 0 ^ {\circ}) - m g \sin (3 0 ^ {\circ})}{m}$
$\Rightarrow \quad a = \left(\frac {\sqrt {3} - 1}{2}\right) g = 3.6 6 \mathrm{ms} ^ {- 2}$
Since, $s=\frac{1}{2}at^{2}$
$\Rightarrow t = \sqrt {\frac {2 s}{a}} = \sqrt {\frac {2 \times 1}{3.66}} = 0.7 4 \mathrm{s}$
Hence, the correct answer is (B).

Since the string is light, so at the point of application of force F, and when the threads make an angle $\theta$ with the horizontal, we have
$F - 2 T \sin \theta = 0$
$\Rightarrow F = 2 T \sin \theta$

For mass m to move, say with acceleration A, we have
$\begin{array}{r l} & T \cos \theta = m A \\ \Rightarrow & \left(\frac {F}{2 \sin \theta}\right) \cos \theta = m A \\ \Rightarrow & A = \left(\frac {F}{2 m}\right) \cot \theta \\ \text {where} \tan \theta = \frac {\sqrt {a ^ {2} - x ^ {2}}}{x} \\ \Rightarrow & A = \frac {F}{2 m} \left(\frac {x}{\sqrt {a ^ {2} - x ^ {2}}}\right) \end{array}$
Hence, the correct answer is (C).

Let $f$ be the force exerted by the rope on the block $M$ . Then
$f - M g = M a$...(i)
We must note that once P acts on free end of the rope, then tension due to pull P at the end of rope attached to M is zero. So, M will move up only due to the interaction force f.

Now, for mass m, we have
$P - f - m g = m a\tag{... (2}$
Adding (1) and (2), we get
$\begin{array}{r l} & P - (M + m) g = (M + m) a \\ \Rightarrow & P = (M + m) (g + a) \\ \Rightarrow & a = \frac {P}{M + m} - g \\ & f = M (g + a) \\ \Rightarrow & f = \frac {M P}{M + m} \end{array}$
Hence, the correct answer is (D).

$N + F \sin \theta = W$ and
$F \cos \theta = \mu N$
$\Rightarrow \mu = \frac {F \cos \theta}{N}$

$\Rightarrow \quad \mu = \frac {F \cos \theta}{W - F \sin \theta}$
Since $\sin\theta=\frac{h}{L}$ and $\cos\theta=\frac{\sqrt{L^{2}-h^{2}}}{L}$
$\Rightarrow \mu = \frac {F \sqrt {L ^ {2} - h ^ {2}}}{W L - F h}$
Hence, the correct answer is (A).

Let us draw the free body diagrams of C and A.

$W =$ Weight of each sphere
$N_{2}=$ Normal reaction between A and inclined plane
$N_{1}=$ Normal reaction between A and C
$N_{1}=$ Normal reaction between B and C
Free body diagram of $C$
Resolving vertically $2N_{1}\cos30^{\circ}=W$
$\Rightarrow N _ {1} = \frac {W}{\sqrt {3}}\tag{... (1}$
When the arrangement is on the point of collapsing, the reaction between A and B is zero.
Free body diagram of $A$
Resolving horizontally and vertically
$N _ {2} \sin \theta = N _ {1} \sin 3 0 ^ {\circ}$
$\Rightarrow \quad N _ {2} \sin \theta = \frac {W}{2 \sqrt {3}}\tag{\( \ldots(2) \}$
$N _ {2} \cos \theta = W + N _ {1} \cos 3 0 ^ {\circ} = \frac {3 W}{2}\tag{... (3}$
Dividing (2) by (3), we get
$\tan \theta = \frac {1}{3 \sqrt {3}}$
$\Rightarrow \theta = \tan^ {- 1} \left(\frac {1}{3 \sqrt {3}}\right)$
Hence, the correct answer is (C).

As soon as the inclination of slanted side reaches the angle of friction, sand begins to slide down.

For maximum height, we have
$\mu = \tan \alpha = \frac {h}{R}$
$\Rightarrow \quad h = \mu R$
Hence, the correct answer is (C).

Free body diagram of the two bodies are as follows


Maximum friction between the two blocks is
$f _ {\mathrm{max}} = \mu m g \text {where} m = 2 \mathrm{kg}$
$\Rightarrow f _ {\mathrm{max}} = (0.5) (2) (1 0) = 1 0 \mathrm{N}$
Let acceleration of both the blocks be a towards left. Then
$\begin{array}{r l} & a = \frac {f - 2}{2} = \frac {2 0 - f}{4} \\ \Rightarrow & 2 f - 4 = 2 0 - f \\ \Rightarrow & f = 8 \mathrm{N} \end{array}$
Now since $f < f_{max}$ , i.e., static region of 2 kg on 4 kg, hence friction force between the two blocks is 8 N.
Hence, the correct answer is (C).

$m g \sin \alpha - m a _ {0} \cos \alpha = m a$
$\Rightarrow \quad a = g \sin \alpha - a _ {0} \cos \alpha$
Hence, the correct answer is (D).

$mg\sin\alpha-ma_0\cos\alpha=ma$
$\Rightarrow a=g\sin\alpha-a_0\cos\alpha$
Hence, the correct answer is (D).

$\sin \theta = \frac {3}{5}$
$\Rightarrow \cos \theta = \frac {4}{5}$

$mg\sin\theta=(10)(10)\left(\frac{3}{5}\right)=60\ \text{N}$
$\mu mg\cos\theta=\left(\frac{3}{4}\right)(10)(10)\left(\frac{4}{5}\right)=60\ \text{N}$
i.e., when a force of $30\ \text{N}$ is applied in upward direction, friction force of $30\ \text{N}$ acts upwards. Normal reaction is perpendicular to plane. Therefore, resultant force will be along $OB$.
Hence, the correct answer is (B).

$mg \sin \theta = 60 \, N$ , $F = 30 \, N$ and $f = 30 \, N$ . So, from the free body diagram, we observe that the net force on the block is zero and hence the acceleration is zero.

Hence, the correct answer is (D).

For upward motion
$- F h = - \frac {1}{2} m v ^ {2} + m g h$
For downward motion
$- F h = \frac {1}{2} m v ^ {\prime 2} - m g h$
$\Rightarrow v ^ {\prime} = v \sqrt {\frac {m g - F}{m g + F}}$
Hence, the correct answer is (C).

Net horizontal force on block B is zero.
Hence, the given figure (a) can be replaced by figure (b)

Maximum value of friction
$f _ {\mathrm{max}} = \mu m _ {A} g = (0. 5) (2) (1 0) = 1 0 \mathrm{N}$
Block B moves due to friction
Therefore, maximum common acceleration of the two blocks can be
$\begin{array}{r l} & a _ {\max} = \frac {f _ {\max}}{m _ {B}} = \frac {1 0}{2} = 5 \mathrm{ms} ^ {- 2} \\ \Rightarrow & F _ {\max} = (m _ {A} + m _ {B}) a _ {\max} \\ \Rightarrow & F _ {\max} = (2 + 2) (5) = 2 0 \mathrm{N} \end{array}$
Hence, the correct answer is (D).
$1 4.\quad 2 T \cos \theta = \sqrt {2} M g\tag{... (1}$
$T \sin \theta$ and $T \sin \theta$ cancel out as both are equal in magnitude and are directed opposite to each other. Further,
$\begin{array}{r l} & T = M g \\ \Rightarrow & 2 M g \cos \theta = \sqrt {2} M g \\ \Rightarrow & \cos \theta = \frac {1}{\sqrt {2}} \\ \Rightarrow & \theta = 4 5 ^ {\circ} \end{array}$....(2)
Hence, the correct answer is (C).

C

Mass $m_{1}$ moves with constant velocity if tension in the lower string
$T _ {1} = m _ {1} g = (1) (1 0) = 1 0 \mathrm{N}$...(i)
So, tension in the upper string is
$T _ {2} = 2 T _ {1} = 2 0 \mathrm{N}$....(2)
Acceleration of block M is therefore,
$a = \frac {T _ {2}}{M} = \frac {2 0}{M}$.. (3)
This is also the acceleration of pulley 2
Absolute acceleration of mass $m_{1}$ is zero. Thus, acceleration of $m_{1}$ relative to pulley 2 is a upwards or acceleration of $m_{2}$ with respect to pulley 2 is a downwards. Drawing free body diagram of $m_{2}$ with respect to pulley 2.

Equation of motion gives
$2 0 - \frac {4 0}{M} - 1 0 = 2 a = \frac {4 0}{M}$
Solving this we get M = 8 kg
Hence, the correct answer is (C).