Laws of Motion

$ a=\mu g=0.12 \times 10=1.2 \mathrm{~m} / \mathrm{s}^2 $

$ \begin{aligned} & F_{\text {net }}=\sqrt{6^2+8^2} \\ & a=\frac{F_{\text {net }}}{m}=\frac{10}{5}=2 \mathrm{~m} / \mathrm{s}^2 \\ & \tan \theta=\frac{6}{8} \\ & \theta=\tan ^{-1}\left(\frac{3}{4}\right) \text { from } 8 \mathrm{~N} \end{aligned} $

$ \begin{aligned} & \tan \theta=\frac{v^2}{r g}=\frac{10 \times 10}{50 \times 10}=\frac{1}{5} \\ & \theta=\tan ^{-1}\left(\frac{1}{5}\right) \end{aligned} $

The time taken to slide down the rough surface is twice that of the smooth surface: $t_{\text{rough}} = 2t_{\text{smooth}}$.

The acceleration on the smooth surface is given by:

$ a_{\text{smooth}} = g \sin \theta $

The time to slide down an inclined plane is inversely proportional to the square root of the acceleration:

$ t \propto \frac{1}{\sqrt{a}} \Rightarrow t_{\text{smooth}} \propto \frac{1}{\sqrt{g \sin \theta}} $

The acceleration on the rough surface is:

$ a_{\text{rough}} = g \sin \theta - \mu_k g \cos \theta $

Relating the times on both surfaces, we have:

$ \frac{t_{\text{rough}}}{t_{\text{smooth}}} = \frac{\sqrt{\sin \theta}}{\sqrt{\sin \theta - \mu_k \cos \theta}} = 2 $

Squaring both sides results in:

$ \frac{\sin \theta}{\sin \theta - \mu_k \cos \theta} = 4 $

Substituting $\theta = 45^{\circ}$ where $\sin \theta = \cos \theta = \frac{1}{\sqrt{2}}$, we find:

$ \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}} - \mu_k \times \frac{1}{\sqrt{2}}} = 4 $

Simplifying further:

$ 1 - \mu_k = \frac{1}{4} $

Solving for $\mu_k$:

$ \mu_k = \frac{3}{4} = 0.75 $

Thus, the coefficient of kinetic friction is $0.75$.

$T-M g \sin 30=M a$

$\begin{aligned} & 30-5 \times 10 \times \frac{1}{2}=5 a \\ & a=1 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

$ \begin{aligned} & F=\left(M_1+M_2\right) a \\ & a=\frac{10}{2+3}=2 \mathrm{~ms}^{-2} \\ & F^{\prime}=M_2(2)=3 \times 2 \mathrm{~N}=6 \mathrm{~N} \end{aligned} $

When a particle moves along a circular path with a uniform speed, it is important to understand the motion characteristics in terms of both velocity and acceleration. Velocity is a vector quantity which means it has both magnitude and direction, while acceleration is the rate of change of velocity with respect to time.

First, let's analyze the velocity:

Although the speed (magnitude of the velocity vector) remains constant in uniform circular motion, the direction of the velocity vector continuously changes as the particle progresses along the circle. Since velocity includes both the magnitude and the direction, any change in either results in a change in velocity. Consequently, in uniform circular motion, the velocity of the particle is not constant but varies due to the continuous change in direction.

Next, consider the acceleration:

In circular motion, there is always an acceleration directed towards the center of the circle, known as centripetal acceleration. This acceleration is responsible for changing the direction of the velocity vector, thereby keeping the particle moving in a circle, despite the speed being constant. The formula for centripetal acceleration is:

$ a_c = \frac{v^2}{r} $

where $ v $ is the speed of the particle and $ r $ is the radius of the circle. This acceleration is always directed towards the center of the circle and varies with the square of the speed and inversely with the radius of the circle.

Therefore, given that both the velocity and acceleration change, the correct choice is:

Option D - Varying velocity and varying acceleration

It states that in uniform circular motion, both the velocity and acceleration of the particle vary – velocity due to continuous changes in direction and acceleration due to its consistent inward (centripetal) direction towards the center of the circle.

$\mu=\frac{1}{\sqrt3}$

$\tan30^\circ=\frac{1}{\sqrt3}$

as $\mu=\tan\theta$

the block is at rest and net force on it must be zero.

$\vec{V}_{\mathrm{i}}=(\mathrm{V})$ Southward

$\overrightarrow{\mathrm{V}}_{\mathrm{F}}=(\mathrm{V})$ Eastward

$\overrightarrow{\Delta \mathrm{V}}=\overrightarrow{\mathrm{V}}_{\mathrm{F}}-\overrightarrow{\mathrm{V}}_{\mathrm{i}}$

= Along North - East

To find the maximum acceleration ($a_{\max}$) of the car that allows a body to stay stationary relative to the car, we use the concept of static friction. Static friction ($F_s$) is what keeps the body from sliding on the car's floor. It acts in the opposite direction of the potential movement of the body.

The maximum static friction force is given by $F_{s_{\max}} = \mu_s \times N$ where $\mu_s$ is the coefficient of static friction and $N$ is the normal force. In this scenario, the normal force is equal to the gravitational force on the body ($mg$), where $m$ is the mass of the body and $g$ is the acceleration due to gravity.

Since the maximum force of static friction equals the product of the mass and the maximum acceleration ($ma_{\max}$), we have:

$\mu_s mg = ma_{\max}$

By canceling out the mass $m$ on both sides, we get:

$a_{\max} = \mu_s g$

Substituting the given values ($\mu_s = 0.15$ and $g = 10 \,\mathrm{m/s}^2$):

$a_{\max} = 0.15 \times 10 = 1.5 \,\mathrm{m/s}^2$

Therefore, the maximum acceleration of the car to ensure the body remains stationary with respect to the car's floor is $1.5 \,\mathrm{m/s}^2$.

$B y v^{2}=u^{2}+2 a s$

$\left(\frac{u}{3}\right)^{2}=u^{2}-2 a x$

$2 a x=u^{2}-\frac{u^{2}}{9}$

$2 \mathrm{ax}=\frac{8 \mathrm{u}^{2}}{9}$ ..... (1)

Similarly from starting

$v^{2}=u^{2}+2 a x$

$0=u^{2}-2 a x_{2}$

$2 \mathrm{ax}_{2}=\mathrm{u}^{2}$ ..... (2)

$\mathrm{By}~(1) /(2)$

$\frac{x}{x_{2}}=\frac{8}{9}$

$\frac{24}{x_{2}}=\frac{8}{9}$

$\mathrm{x}_{2}=27 \mathrm{~cm}$

Given, $\overrightarrow F = 2\widehat i + \widehat j - \widehat k$ and $\overrightarrow r = 3\widehat i + 2\widehat j - 2\widehat k$

$\overrightarrow F \,.\,\overrightarrow r = 2(3) + 2(1) + ( - 2)( - 1)$

$ = 6 + 2 + 2 = 10$

$\overrightarrow F \times \overrightarrow r = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 3 & 2 & { - 2} \cr } } \right|$

$ = \widehat i( - 2 + 2) - \widehat j( - 4 + 3) + \widehat k(4 - 3)$

$ \Rightarrow \overrightarrow F \times \overrightarrow r = \widehat j + \widehat k$

$|\overrightarrow F \times \overrightarrow r | = \sqrt {{1^2} + {1^2}} = \sqrt 2 $

Assuming all three blocks as system

Hence, acceleration of system

$a = {{60 - 30 - 18} \over 6} = {{12} \over 6} = 2$ m/s²

$a = 2$ m/s², upward along the incline.

Now, for 1 kg block

$ \Rightarrow {N_{12}} - 18 - 5 = 1 \times 2$

$ \Rightarrow {N_{12}} - 23 = 2$

$ \Rightarrow {N_{12}} = 25$ N

Given, Force applied by gun,

$F=\left(100-0.5 \times 10^5 t\right) \mathrm{N}$

Speed of bullet, $v=400 \mathrm{~m} / \mathrm{s}$

when $\quad F=0$

$\Rightarrow \quad 100-0.5 \times 10^5 t=0 \Rightarrow t=2 \times 10^{-3} \mathrm{~s}$

$\therefore \text { Impulse, } I=\int F d t$

$\begin{aligned} & =\int_\limits0^{2 \times 10^{-3}}\left(100-0.5 \times 10^5 t\right) d t=\left[100 t-0.5 \times 10^5 \frac{t^2}{2}\right]_0^{2 \times 10^{-3}} \\ & =100 \times 2 \times 10^{-3}-\frac{0.5 \times 10^5}{2} \times 4 \times 10^{-6}-0 \\ & =0.2-0.1=0.1 \mathrm{~Ns} \end{aligned}$

Both the assertion and the reason are true, and the reason is the correct explanation of the assertion.

When a glass ball is dropped on a concrete floor, it comes to rest in a much shorter time compared to when it is dropped on a wooden floor. The hard, rigid nature of concrete means that it does not deform much upon impact, causing a rapid deceleration of the glass ball. This rapid deceleration results in a large force being exerted on the glass ball due to Newton's Second Law of Motion :

$ F = ma $

where :

$ F $ is the force,

$ m $ is the mass of the glass ball,

$ a $ is the acceleration (or deceleration) of the glass ball.

Since the time to stop ($ t $) is shorter, the deceleration ($ a $) is higher. According to :

$ F = m \frac{\Delta v}{\Delta t} $

where $ \Delta v $ is the change in velocity, and $ \Delta t $ is the change in time, a smaller $ \Delta t $ results in a larger force $ F $.

On the other hand, a wooden floor may deform slightly upon impact, increasing the time over which the glass ball comes to rest. This longer time reduces the deceleration and, consequently, the force exerted on the glass ball, making it less likely to break.

Thus, the correct option is :

Option A

If both assertion and reason are true and reason is the correct explanation of assertion.

$\text { Since, } F=(M+m) a$ .... (i)

So, apply pseudo force on the block by observing, it from the wedge.

Now, as in free body diagram of block, we get

$\begin{aligned} & m a \cos \alpha=m g \sin \alpha \\ \Rightarrow \quad & a=g \frac{\sin \alpha}{\cos \alpha} \Rightarrow a=g \tan \alpha \quad \text{.... (ii)} \end{aligned}$

Now, from Eqs. (i) and (ii), we get

$F=(M+m) g \tan \alpha$

Given, $\theta=45^{\circ}, s_1=s_2, u=0$

On the rough incline, $a_1=g(\sin \theta-\mu \cos \theta)$

$t_1=$ time taken

On the frictionless incline, $a_2=g \sin \theta$

$\begin{aligned} & t_2=\text { time taken, } t_1=2 t_2 \\ & s=u t+\frac{1}{2} a t^2 \\ & s_1=0+\frac{1}{2} g(\sin \theta-\mu \cos \theta) t_1^2 \\ & s_2=0+\frac{1}{2} g \sin \theta t_2^2 \end{aligned}$

$\text { As } s_1=s_2 \text {, }$

$\begin{aligned} & \frac{1}{2} g(\sin \theta-\mu \cos \theta) t_1^2=\frac{1}{2} g \sin \theta t_2^2 \\ & \frac{\sin \theta-\mu \cos \theta}{\sin \theta}=\frac{t_2^2}{t_1^2} \\ & \Rightarrow \quad 1-\mu \cot \theta=\frac{t_2^2}{\left(2 t_2\right)^2} \\ & \Rightarrow \quad 1-\mu \cot \theta=\frac{1}{4} \\ & \Rightarrow \quad \mu \cot \theta=1-\frac{1}{4}=\frac{3}{4} \\ & \therefore \quad \mu=\frac{3}{4 \cot \theta} \\ & \end{aligned}$

Acceleration of the body down the rough inclined plane $=g \sin \theta$

$\therefore$ Force applied on spring balance

$\begin{aligned} & =m g \sin \theta=5 \times 10 \times \sin 30^{\circ} \\\\ & =5 \times 10 \times \frac{1}{2}=25 \mathrm{~N} \end{aligned}$

When the system is in equilibrium, then the spring force is 3 mg. When the string is cut, then net force on block A

$=3mg-2mg=mg$

Hence, acceleration of block A at that instant

$a=\frac{\text { force on block } A}{\text { mass of block } A}=\frac{m g}{2 m}=\frac{g}{2}$

When string is cut, then block $B$ falls freely with an acceleration equal to $g$.

Let $a_1$ and $a_2$ be accelerations of a ball (upward) and rod (downward), respectively.

Clearly, from the diagram

$2 a_1=a_2$ ..... (i)

Now, for the ball

$2 T-\frac{9}{5} m g=\frac{9}{5} m a$ ..... (ii)

and for the rod,

$m g-T=m a_2$ ...... (iii)

On solving Eqs. (i), (ii) and (iii), we get

$\begin{aligned} & a_1=\frac{g}{29} \mathrm{~m} / \mathrm{s}^2 \uparrow \text { (upward) } \\\\ & a_2=\frac{2 g}{29} \mathrm{~m} / \mathrm{s}^2 \downarrow(\text { downward ) } \end{aligned}$

So, acceleration of ball w.r.t. rod $=a_1+a_2=\frac{3 g}{29}$

Now, displacement of ball w.r.t. rod when it reaches the upper end of rod is $1 \mathrm{~m}$.

Using equation of motion,

$\begin{aligned} & s=u t+\frac{1}{2} a t^2 \\\\ & s=0+\frac{1}{2} \times \frac{3 \times 10}{29} t^2 \\\\ & t=\sqrt{\frac{58}{30}}=1.4 \mathrm{~s} \text { (approx) } \end{aligned}$

Angle or repose is equal to angle of limiting friction and maximum value of static friction is called the limiting friction.

If man slides down with same acceleration then, its apparent weight decreases. As the critical condition given is that, rope can bear only $2 / 3$ of his weight. If $a$ is minimum acceleration,

Breaking strength $=m(g-a)$

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

Given situation can be represented as,

$\begin{aligned} F & \leftarrow m_1 \rightarrow T_1 \leftarrow m_2 \rightarrow T_2 \leftarrow m_3 \rightarrow T_3 \leftarrow m_4 \\ T_1 & =\frac{\left(m_2+m_3+m_4\right)}{m_1+m_2+m_3+m_4} F \end{aligned}$

Given, $m_1=m_2=m_3=m_4=m$

$\begin{array}{ll} \therefore & T_1=\frac{3}{4} F \\ \text { Similarly, }, & T_2=\frac{\left(m_3+m_4\right) F}{m_1+m_2+m_3+m_4} \\ \therefore & T_2=\frac{1}{2} F \\ \text { Also, } & T_3=\frac{m_4 F}{m_1+m_2+m_3+m_4} \\ \Rightarrow \quad & T_3=\frac{1}{4} F \end{array}$

According to question, the situation can be shown

Case 1: For body projected up the plane

$\begin{aligned} & v=0 \\ & v^2=u^2-2 a_1 s \\ & \text { i.e. } \quad 0=u^2-2 a_1 s \\ & \therefore \quad u=\sqrt{2 a_1 s} \\ & \text { Also, } \quad v=u-a_1 t \\ & \text { i.e. } \quad 0=u-a_1 t_1 \\ & t_1=u / a_1 \\ & \therefore \quad t_1=\frac{\sqrt{2 a_1 s}}{a_1}=\sqrt{\frac{2 s}{a_1}} \quad \text{...... (i)}\\ \end{aligned}$

Retardation

$a_1=\frac{\mu R+m g \sin 30^{\circ}}{m}$

$\begin{aligned} \text{As} \quad R & =m g \cos 30^{\circ} \\ a_1 & =\frac{\mu m g \cos 30^{\circ}+m g \sin 30^{\circ}}{m} \\ & =\mu g \frac{\sqrt{3}}{2}+\frac{g}{2} \quad \text{..... (ii)} \end{aligned}$

Case 2: For body coming down the plane

$\begin{aligned} & s=u t+\frac{1}{2} a_2 t_2^2 \\ & u=0 \\ & t_2=\sqrt{\frac{2 s}{a_2}} \quad \text{..... (iii)} \end{aligned}$

Downward acceleration

$a_2=\frac{m g \sin 30^{\circ}-\mu R}{m}$

$\begin{aligned} & =\frac{m g \sin 30^{\circ}-\mu m g \cos 30^{\circ}}{m} \\ & =\frac{g}{2}-\frac{\mu g \sqrt{3}}{2} \end{aligned}$

Given, $t_1=\frac{t_2}{2}$

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

$a_1=4 a_2$ ..... (iv)

Substituting values of $a_1$ and $a_2$ from Eqs. (ii) and (iv), 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

$\begin{aligned} \mu & =\frac{\sqrt{3}}{5}=\frac{1.732}{5} \\ & =0.346 \end{aligned}$