Laws of Motion

As the body is moving with constant velocity so forces acting on the body must be balanced.

$\Rightarrow$ Contact force from incline should balance weight of the body.

$\Rightarrow$ | F_contact | = Mg

${\theta _1} = 30^\circ ,\,{\theta _2} = 45^\circ $

${a_1} = g\sin {\theta _1} = 5$ m/s², ${a_2} = g\sin {\theta _2} = 5\sqrt 2 $ m/s²

${{{t_1}} \over {{t_2}}} = {{\sqrt {{{2l} \over {{a_1}}}} } \over {\sqrt {{{2l} \over {{a_2}}}} }} = \sqrt {{{{a_2}} \over {{a_1}}}} $

${{{t_1}} \over {{t_2}}} = {(2)^{1/4}}$

${t_2} = {(2)^{3/4}}$

$ \approx 1.68$ s

${a_1} = {{{M_2} - {M_1}} \over {{M_2} + {M_1}}} \times g = {{2{M_1} - {M_1}} \over {3{M_1}}} \times g$

$ = {g \over 3}$

And, ${a_2} = {{3{M_1} - {M_1}} \over {4{M_1}}} \times g = {g \over 2}$

$\therefore$ ${{{a_1}} \over {{a_2}}} = {{g/3} \over {g/2}} = {2 \over 3}$

Here we will use the Newton's laws of motion.

Let acceleration of 100 kg block w.r.t. ground is a.

Now, Let's draw FBD of 100 kg block w.r.t. ground,

By NLM,

$F-T-N_1=100a\quad \text{... (1)}$

FBD of 20 kg block w.r.t. 100 kg block,

By NLM,

${N_1} = {m_2}a = 20a \Rightarrow {N_1} = 20a\quad$ ... (2)

$T - {m_2}g = {m_2}(2)$

$ \Rightarrow T - 20g = 20(2)$

$ \Rightarrow T = 40 + 200 = 240$

$ \Rightarrow T = 240N\quad$ .... (3)

Now, FBD of 10 kg block w.r.t. 100 kg block,

By NLM,

${m_1}a - T = {m_1}(2)$

$ \Rightarrow 10a - T = 20$

$ \Rightarrow 10a - 240 = 20\quad$ [from (3)]

$ \Rightarrow 10a = 260$

$ \Rightarrow a = 26\,m{s^{ - 2}}$

So, from (2), ${N_1} = 20(26) = 520N$

Now, from (1), $F - 240 - 520 = 100(26)$

$ \Rightarrow F = 2600 + 240 + 520$

$ \Rightarrow F = 3360\,N$

T_down = 50 $\times$ (10 $-$ 4)

= 50 $\times$ 6

= 300 N

T_up = 50 $\times$ (10 + 5)

= 50 $\times$ 15

= 750 N

$\Rightarrow$ Rope will break while climbing up.

Resultant of already applied forces $ = - \widehat i - \widehat j$

$\Rightarrow$ Force required to balance $ = \widehat i + \widehat j$

$\Rightarrow$ Force required $ = \sqrt 2 $ N in magnitude at angle 45$^\circ$ with +ve x-axis

Here F_max force is trying to move the block upward so friction force will be applied towards downward.

Along horizontal direction,

N = 50 N

Along vertical diretion,

F_max = mg + f_max

= 2 $\times$ 10 + $\mu$N

= 20 + 0.5 $\times$ 50

= 20 + 25

= 45 N

${f_{{r_{\max }}}} = \mu N$

$ = 0.02 \times 400$

= 8 N

Let the acceleration is a as shown then.

$40 - T = 4a$

$T - 8 = 40a$

$ \Rightarrow a = {{32} \over {44}} = {8 \over {11}}$ m/s²

Using Newton's second law

$mg - {{mg} \over 4} = ma$

$ \Rightarrow a = {{3g} \over 4}$

$F = - 12x$

$mv{{dv} \over {dx}} = - 12x$

$\int_4^v {vdv = - 6\int_{0.5}^{1.5} {xdx} } $ ($m = 2$ kg)

${{{v^2} - 16} \over 2} = - 6\left[ {{{{{1.5}^2} - {{0.5}^2}} \over 2}} \right]$

${{{v^2} - 16} \over 2} = - 6$

$v = 2$ m/sec

$\therefore$ a_max = $\mu$g

= 0.5 $\times$ 9.8 = 4.9 m/s²

$\therefore$ F_max = (8 + 2) $\times$ 4.9 = 49 N

From virtual work done method,

4T $\times$ a₁ + 2T $\times$ a₂ + T $\times$ a₃ + T $\times$ a₄ = 0

$\Rightarrow$ 4a₁ + 2a₂ + a₃ + a₄ = 0

N₁ = mg

N₂ = mg + ma

N₃ = mg $-$ ma

N₄ = mg

N₅ = mg

Let time taken to ascent is t₁ and that to descent is t₂. Height will be same so

$H = {1 \over 2} \times 12t_1^2 = {1 \over 2}\times8t_2^2$

$ \Rightarrow {{{t_1}} \over {{t_1}}} = {{\sqrt 2 } \over {\sqrt 3 }}$

$S = {{{u^2}} \over {2a}} = {{{u^2}} \over {2(\mu g)}}$

$ = {{{{(9.8)}^2}} \over {2 \times 0.5 \times (9.8)}}$

$ = {{9.8} \over 1}$

$ = 9.8$ m

The given situation is shown below

Body loose contact with surface when reaction is zero, thus

$ \begin{aligned} & F \sin 45^{\circ}=m g \\ \Rightarrow \quad & F=\frac{m g}{\sin 45^{\circ}}=\frac{1 \times 10}{\frac{1}{\sqrt{2}}} \text { or } F=10 \sqrt{2} \end{aligned} $

But, $F=\alpha t$

Hence, $\alpha t=10 \sqrt{2}$ or $t=\frac{10 \sqrt{2}}{\alpha}$

Now, acceleration of body, $a=\frac{F}{m} \cos 45^{\circ}$

$ \begin{array}{ll} \Rightarrow & \frac{d v}{d t}=\frac{\alpha t}{m} \cos 45^{\circ} \\ \text { or } & d v=\frac{\alpha}{\sqrt{2} \times 1} \times t \times d t \\ \Rightarrow & d v=t \cdot \frac{\alpha}{\sqrt{2}} \cdot d t \\ \Rightarrow & \int_0^v d v=\frac{\alpha}{\sqrt{2}} \int_0^{10 \sqrt{2}} t \cdot d t \\ \Rightarrow & v-0=\frac{\alpha}{\sqrt{2}}\left(\frac{t^2}{2}\right)_0^{\frac{10 \sqrt{2}}{\alpha}} \\ \Rightarrow & v=\frac{\alpha}{\sqrt{2}} \times \frac{100 \times 2}{2 \times \alpha^2}=\frac{50 \sqrt{2}}{\alpha} \mathrm{~m} / \mathrm{s} \end{array} $

Force applied on block $A, F=10 \mathrm{~N}$

Acceleration produced in block $A$.

$ a=20 \mathrm{~m} / \mathrm{s}^2 $

According to Newton's second equation of motion,

$ \begin{aligned} F & =m_A a \\ m_A & =\frac{F}{a}=\frac{10}{20}=0.5 \mathrm{~kg} \end{aligned} $

Again, block $A$ is kept in contact with block $B$, where $m_B=1.5 \mathrm{~kg}$.

If $a_1$ be the common acceleration in the block system and $R$ be reaction force on block $A$ and $B$, then from free body diagram of block $A$ and $B$,

Equation of motion of block $A$ is

$ \begin{aligned} &\begin{aligned} & & F^{\prime}-R & =m_A \cdot a_1 \\ \Rightarrow & & 20-R & =0.5 a_1 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned}\\ &\text { For block } B, R=m_B a_1 \end{aligned} $

$ R=1.5 a_1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii) $

From Eqs. (i) and (ii), we have

$ \begin{aligned} & & 20-1.5 a_1 & =0.5 a_1 \\ \Rightarrow & & 20 & =2 a_1 \\ \Rightarrow & & a_1 & =10 \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

∴ From Eqs. (ii),

$ R=1.5 a_1=1.5 \times 10=15 $

Hence, force on block $B$ will be 15 N .

Initial velocity of motor car,

$ u=7 \mathrm{~m} / \mathrm{s} $

Stopping distance, $s=10 \mathrm{~m}$

Final velocity, $v=0 \mathrm{~m} / \mathrm{s}$

If retardation of car is $a \mathrm{~m} / \mathrm{s}^2$, due to applying brakes, then

$ \begin{aligned} a & =\frac{v^2-u^2}{2 s} \\ & =\frac{0^2-7^2}{2 \times 10}=\frac{-4.9}{2} \mathrm{~m} / \mathrm{s}^2 \end{aligned} $

Negative sign indicates that retardation is applied in opposite direction of motion.

$ \text { } \begin{aligned}∴ Resistance,\,\, R & =m a=\frac{w}{9.8} \times \frac{-4.9}{2} \\ & =-\left(\frac{w}{4}\right) \end{aligned} $

The given situation is shown below.

Let's assume the block is stationary only till point $P$. Slope of point $P=\tan \theta$ (where, $\theta=$ angle of repose)

also : $\frac{d y}{d x}=\frac{d}{d x}\left(\frac{x^2}{20}\right)=\frac{x}{10}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

We know that, the block will remain stationary only up till angle of repose.

We know, angle of repose $\theta=\tan ^{-1}\left(\mu_s\right)$

$ \begin{array}{ll} \Rightarrow & \theta=\tan ^{-1}(0.5) \\ \Rightarrow & \tan \theta=0.5 \end{array} $

Now, equating $(d y / d x)$ at $P$ with $\tan \theta$ from Eqs. (i) and (ii), we get

$ \begin{gathered} \frac{d y}{d x}=\tan \theta \\ \Rightarrow \quad \frac{x}{10}=0.5 \Rightarrow x=5 \,\,...(ii)\\ \therefore \quad y=\frac{x^2}{20}=\frac{25}{20}=1.25 \mathrm{~m} \end{gathered} $

According to the question,

$ \begin{aligned} f_1 & =\text { Friction on } 1 \mathrm{~kg}=\mu, N_1=0.4 \times \frac{10}{\sqrt{2}} \mathrm{~N} \\ & =\frac{4}{\sqrt{2}} \mathrm{~N} \\ f_2 & =\text { Friction on } 2 \mathrm{~kg} \\ & =\mu, N_2=0.4 \times \frac{20}{\sqrt{2}}=\frac{8}{\sqrt{2}} \mathrm{~N} \end{aligned} $

Applying Newton's second law along $X$-axis. Since, both are connected, so assuming common acceleration $=a$

$ \begin{aligned} &\begin{aligned} \therefore \quad & \left(\frac{10}{\sqrt{2}}+\frac{20}{\sqrt{2}}\right)-\left(f_1+f_2\right)=1 a+2 a \\ & \frac{30}{\sqrt{2}}-\left(\frac{4}{\sqrt{2}}+\frac{8}{\sqrt{2}}\right)=3 a \\ & \frac{30}{\sqrt{2}}-\frac{12}{\sqrt{2}}=3 a \\ & \frac{18}{\sqrt{2}}=3 a \\ & a=\frac{6}{\sqrt{2}} \end{aligned}\\ &\text { Since, given in question, } a=\alpha \sqrt{2}\\ &\begin{aligned} \therefore & & \alpha \sqrt{2} & =\frac{6}{\sqrt{2}} \\ \Rightarrow & & \alpha & =\frac{6}{2}=3 \end{aligned} \end{aligned} $

Given,

$ \begin{aligned} & m=100 \mathrm{~g}=0.1 \mathrm{~kg} \\ & \Delta t=2 \mathrm{~ms}=2 \times 10^{-3} \mathrm{~s} \end{aligned} $

Initial speed, $u=10 \mathrm{~m} / \mathrm{s}$

Let final speed be $v$.

We know that, area under force-time graph gives impulse or change in momentum.

∴ Area of the triangle $=|\Delta \mathbf{p}|$ (where, $\mathbf{p}=$ momentum)

$ \begin{aligned} & \Rightarrow \quad \frac{1}{2} \times \text { Base × Height }=\left|\mathbf{p}_f-\mathbf{p}_i\right| \\ & \left(\text { where, } \mathbf{p}_f=\right.\text { final momentum and } \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \left.\mathbf{p}_i=\text { initial momentum }\right) \\ & \Rightarrow \frac{1}{2} \times\left(2 \times 10^{-3}\right) \times(20 \times 1000)=0.1 v-0.1 u \\ &\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, (\because 20 \mathrm{kN}=20000 \mathrm{~N}) \end{aligned} $

$ \begin{array}{lc} \Rightarrow & 2 \times 10=0.1(v-10) \quad(\because u=10 \mathrm{~m} / \mathrm{s}) \\ \Rightarrow & v-10=200 \\ \Rightarrow & v=210 \mathrm{~m} / \mathrm{s} \end{array} $

Initial speed of body on the horizontal rough surfaces, $u=10 \mathrm{~ms}^{-1}$

Final velocity, $v=4 \mathrm{~ms}^{-1}$ and $t=2 \mathrm{~s}$.

If $a$ be the retardation due to friction, then by using equation,

$\begin{aligned} & v=u-a t \\ & \Rightarrow \quad 4=10-a \times 2 \\ \end{aligned}$

$\Rightarrow \quad 2 a=6 \Rightarrow a=3 \mathrm{~ms}^{-2}$

According to given situation,

$\begin{array}{ll} \text { friction force }=m a \\ \Rightarrow \mu_k m g=m a \\ \Rightarrow \mu_k=\frac{a}{g}=\frac{3}{10}=0.3 \end{array}$

Mass of cricket ball, $m=50 \mathrm{~g}=0.05 \mathrm{~kg}$

Initial velocity of ball, $u=50 \mathrm{cms}^{-1}=0.5 \mathrm{~ms}^{-1}$

time, $t=0.5 \mathrm{~s}$

If $a$ be the retardation applied on the cricket ball, then by using

$\begin{aligned} v & =u-a t \\ \Rightarrow \quad 0 & =0.5-a \times 0.5 \Rightarrow a=\frac{0.5}{0.5} \\ & =1 \mathrm{~m} / \mathrm{s}^2 \end{aligned}$

Force applied,

$\begin{aligned} F & =\text { mass } \times \text { acceleration } \\ & =0.05 \times 1=0.05 \mathrm{~N} \end{aligned}$

FBD of the situation is given below.

From Newton's law, $\Sigma \mathbf{F}=$ ma

Force on $M_1$,

in $x$-direction, $T=M_1 a$ ... (i)

in $y$-direction, $N_1-w_1=0$ .... (ii)

Similarly, forces on $M_2$,

in $x$-direction, $M_2 g \sin \theta-T=M_2 a$ .... (iii)

in $y$-direction, $N_2-m_2 g \cos \theta=0$ .... (iv)

Adding to eq. (i) with eq. (ii), we get

$\begin{aligned} M_2 g \sin \theta & =M_1 a+M_2 a \\ \Rightarrow \quad a & =\frac{M_2 g \sin \theta}{M_1+M_2} \end{aligned}$

According to question, when $M_1$ is doubled and $M_2$ is halved

$a^{\prime}=\frac{\frac{M_2}{2} g \sin \theta}{2 M_1+\frac{M_2}{2}}=\frac{M_2 g \sin \theta}{4 M_1+M_2}$

Now, $\frac{a^{\prime}}{a}=\frac{\frac{M_2 g \sin \theta}{4 M_1+M_2}}{\frac{M_2 g \sin \theta}{M_1+M_2}} \Rightarrow a^{\prime}=\left(\frac{M_1+M_2}{4 M_1+M_2}\right) a$

The given situation is shown below

Let $a$ be the common acceleration in the system of combination of blocks, then equation of motion of block $m_1$ is

$\begin{aligned} & T=m_1 a \\ & T=40 a \quad \text{.... (i)} \end{aligned}$

and equation of motion of block $m_2$ is

$\begin{aligned} F-T & =m_2 a \\ \Rightarrow 1000-T & =m_2 a \\ \Rightarrow 1000-T & =60 \times\left(\frac{T}{40}\right) \quad \text{[from Eq. (i)]}\\ \Rightarrow 1000-T & =\frac{3}{2} T \Rightarrow 2000-2 T=3 T \\ \Rightarrow \quad & T=\frac{2000}{5}=400 \mathrm{~N} \end{aligned}$