NLM

Given, $m_{1} = 4 \, kg$ , $m_{2} = 6 \, kg$ and a = ?
The free body diagram of given system is shown below

The balancing equations for given system are
$T - m _ {1} g = m _ {1} a$ and ... (i)
$m _ {2} g - T = m _ {2} a\tag{... (ii}$
Adding Eqs. (i) and (ii), we get
$m _ {2} g - m _ {1} g = m _ {1} a + m _ {2} a$
$a = \left(\frac {m _ {2} - m _ {1}}{m _ {1} + m _ {2}}\right) \cdot g = \left(\frac {6 - 4}{4 + 6}\right) \times g = \frac {g}{5}$

The given situation is shown in the following diagram.

If a be the acceleration of the system,
then equation of motion of 10 kg trolley,
$\begin{array}{r l r} & & T - \mu R = 1 0 a \\ & \Rightarrow & T - 0.0 5 \times 1 0 g = 1 0 a \quad (\text {Given,} \mu = 0.05, R = 1 0 g) \\ & \Rightarrow & T - 5 = 1 0 a \qquad \dots \end{array}\tag{... (i}$
Equation of motion of 2kg block,
$\begin{array}{c} {2 g - T = 2 a} \\ {2 0 - T = 2 a} \end{array}\tag{... (ii}$
Adding Eqs. (i) and (ii), we get
$15 = 12 a \Rightarrow a = \frac {1 5}{1 2} = \frac {5}{4} = 1.25 \mathrm{ms} ^ {- 2}$

As the truck move to the right, so the bob will move to the left due to inertia of rest with acceleration a.
Thus, the given situation can be drawn as

:
Now, from equilibrium of forces in above diagram (b), we get
$ma \cos \theta = mg \sin \theta$
⇒
$\frac {\sin \theta}{\cos \theta} = \frac {m a}{m g}$
⇒
$\tan \theta = \frac {a}{g} \Rightarrow \theta = \tan^ {- 1} \left(\frac {a}{g}\right)$

The situation can be drawn as

The frictional force, $f = \mu N = \mu mg$ ( $\because N = mg$ )
From free body diagram, the resultant force is
$\begin{array}{r} {| \textbf {F} | = \sqrt {N ^ {2} + f ^ {2}}} \\ {= \sqrt {(m g) ^ {2} + (\mu m g) ^ {2}} = m g \sqrt {1 + \mu^ {2}}} \end{array}$
This is the minimum force required to move the object. But as the body is not moving.
$| \textbf {F} | \leq m g \sqrt {1 + \mu^ {2}}$

As the three forces are represented by three sides of a triangle taken in order, then they will be in equilibrium.

$\begin{array}{l} \mathrm {F_ {net}} = \mathrm {F_ {PQ}} + \mathrm {F_ {QR}} + \mathrm {F_ {RP}} = 0 \\ \mathrm {F_ {net}} = m \times \mathbf {a} = m \frac {d \mathbf {v}}{d t} = 0 \end{array}$
$\frac {d \mathbf {v}}{d t} = 0 \quad \mathrm{or} \quad \mathbf {v} = \mathrm{constant}$
So, the velocity of particle remain constant.

Force exerted by concrete floor is more as compared to wooden floor due to greater change in momentum.
Since on concrete floor, glass ball will take less time to come to rest, so a glass ball dropped on concrete floor can easily get broken compared, if it is dropped on wooden floor.
Hence, both Assertion and Reason are true and Reason is the correct explanation of Assertion.

Force applied by gun, $F = (100 - 0.5 \times 10^{5} t)$ N
Speed of bullet, v = 400 m/s
When $F = 0$ , then $100 - 0.5 \times 10^{5} t = 0$
$\Rightarrow \quad t = 2 \times 1 0 ^ {- 3} \mathrm{s}$
Impulse, $I = \int F dt = \int_{0}^{2 \times 10^{-3}} (100 - 0.5 \times 10^{5} t) dt$
$\begin{array}{l} = \left[ 1 0 0 t - 0.5 \times 1 0 ^ {5} \cdot \frac {t ^ {2}}{2} \right] _ {0} ^ {2 \times 1 0 ^ {- 3}} \\ = 1 0 0 \times 2 \times 1 0 ^ {- 3} - \frac {0 .5 \times 1 0 ^ {5}}{2} \times (4 \times 1 0 ^ {- 6} - 0) \\ = 0.2 - 0.1 = 0.1 \mathrm{N-s} \end{array}$

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.

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
$T _ {2} = (2 M g) \times (\sqrt {3} / 2) = \sqrt {3} M g$

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,
$M _ {B} a _ {B} = \mu M _ {B} g$
⇒
$a _ {B} = \mu g (\text { towards left })$

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,
$\begin{array}{r l} & J = \Delta p = m v _ {f} - m v _ {i} \\ & \quad = m v \cos 6 0 ^ {\circ} - (- m v \cos 6 0 ^ {\circ}) \\ & \quad = 2 m v \cos 6 0 ^ {\circ} = 2 m v \times \frac {1}{2} = m v \end{array}$