Laws of Motion

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

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