Laws Of Motion Practice Questions

Given Force, $F=(2 \hat{i}+3 \hat{j}+5 \hat{k}) {N}$ Mass, $m=2 {kg}$ Time, $t=4 {s}$ We know that Force $=$ mass $\times$ acceleration $\Rightarrow \;\; F=m \times a \Rightarrow a=\;\frac{F}{m}$ \Rightarrow \;\; a=\;{2 \hat{i}+3 \hat{j}+5 \frac\hat{k}}{2}.....(i) From second equation of motion, $s=u t+\;\frac{1}{2} a t^{2}$...(ii) From Eqs. (i) and (ii), we get s=u t+\;\frac{1}{2} \;{(2 \hat{i}+3 \hat{j}+5 \frac\hat{k})}{2} \cdot t^{2} $=0+\;\frac{1}{4}(2 \hat{i}+3 \hat{j}+5 \hat{k}) \cdot (4)^{2} \;\; [ \because u=0$ and $t=4 {s}]$ $=0+8 \hat{i}+12 \hat{j}+20 \hat{k}=8 \hat{i}+12 \hat{j}+20 \hat{k}$ Let $s=x \hat{i}+y \hat{j}+z \hat{k}$ $\Rightarrow x \hat{i}+y \hat{j}+z \hat{k}=8 \hat{i}+12 \hat{j}+20 \hat{k}$ .....(iii) According to question,new coordinates are $(8, b, 20)$, it means $x \hat{i}+y \hat{j}+z \hat{k}=8 \hat{i}+b \hat{j}+20 \hat{k}$......(iv) Comparing Eqs. (iii) and (iv), we get $b=12$

Here, both block and wedge are moving. Consider the acceleration of the block with respect to the wedge is $\text{a}_1$ and the acceleration of the wedge is $\text{a}_2$. Given, mass of the wedge, M = 16 kg and mass of block, m = 8 kg Let's draw the free body diagram of the wedge, In the x-directions, N cos60º = $\text{Ma}_2$ $\text{N}(0.5)=16 \times \text{a}_{2}$ \Rightarrow $\text{N}=32 \text{a}_{2}$ Now, draw the free body diagram of the block with respect to thewedge. Along the perpendicular to the inclined plane, $\text{N} = 8\text{g} \cos30° - 8\text{a}2\sin30º$ $32 \text{a}_{2}=4 \\sqrt{3} \text{g}-4 \text{a}_{2}$ $36 \text{a}_{2}=4 \\sqrt{3} \text{g}$ \Rightarrow $\text{a}_{2}=\frac{\\sqrt{3}}{9} \text{g}$ Along the inclined plane, $\text{m} \text{g} \sin 30° +\text{m}\text{a}_{2} \cos 30° =\text{m} \text{a}_{1}$ $8 \text{g} \times \frac{1}{2}+8 \times \frac{\\sqrt{3}}{9} \text{g} \times \frac{\\sqrt{3}}{2}=8 \text{a}_{1}$ \Rightarrow $\text{a}_{1}=\frac{2 \text{g}}{3}$ ∴ The acceleration of the block with respect to the wedge is ${2\text{g}}/ {3}$.

${t}_{{a}}=\frac{1}{2} {t}_{{d}}$ $\\sqrt{{2 {s}}/{{a}_{{a}}}}=\frac{1}{2} \\sqrt{{2 {s}}/{{a}_{{d}}}}$ ......(i) ${a}_{{a}}={g} \sin \theta +\mu{g} \cos \theta$ $=\frac{{g}}{2}+\frac{\\sqrt{3}}{2} \mu{g}$ $a_{{d}}=g \sin \theta -\mug \cos \theta$ $=\frac{{g}}{2}-\frac{\\sqrt{3}}{2} \mu{g}$ using the above values of $a_{a}$ and $a_{d}$ and putting ineqution (i) we will gate $\mu=\frac{\\sqrt{3}}{5}$

${ \vec{F}}= {mkv}^{2}- {mg}$ ${\vec{a}}= {{\vec{F}}}/{ {m}}=-[{kv}^{2}+ {g}]$ $\Rightarrow$ v. $\frac{{d} v}{ {dh}}=-[{kv}^{2}+ {g} ]$ $\Rightarrow ∫_{ {u}} ^{0}\frac{{v}.{d} {v}}{ {kv}^{2}+ {g}}=-∫_ {0} ^{ {H}} {dh}$ \frac{1}{2 {K}} \ln \[{kv}^{2}+ {g} ]_{ {u}}^{0}=- {H}$$\Rightarrow \frac{1}{2 {K}} \ln $[\frac{{ku}^{2}+ {g}}{ {g}} ]$= {H}$

$\frac{{dm}( {t})}{ {dt}}= {bv}^{2}$ ${F}_{\text{thast }}= \frac{v}{{dm}}/{ {dt}}$ Force on statellile $=-{\vec{v} }\frac{{dm}( {t})}{ {dt}}$ ${M}( {t}) {a}=- {v}({bv}^{2} )$ $[ {a}= {a} \frac{{bv}^{3}}{ {M}( {t})}]$

${{d} v_{{x}}}/{{dt}}=\frac{{k}}{{m}} {v}_{{y}}$ ${{d} {v}_{{y}}}/{{dt}}=\frac{{k}}{{m}} {v}_{{x}}$ ${{d} {v}_{{y}}}/{{d} {v}_{{x}}}={{v}_{{x}}}/{{v}_{{y}}}$ $\Rightarrow ∫{v}_{{y}} {d} {v}_{{y}}=∫{v}_{{x}} {d} {v}_{{x}}$ ${v}_{{y}}^{2}={v}_{{x}}^{2}+{C}$ ${v}_{{y}}^{2}-{v}_{{x}}^{2}={\text{constant}}$

For equilibrium,T sin 45° = F ....(1)and T cos 45° = 10g ....(2)equation (1)/(2)we get F = 10g = 100 N

For balancing ${\text{mg}} \sin \theta ={\text{f}}$ ${\text{mg}} \sin \theta =\mu{\text{mg}} \cos \theta$ $\tan \theta =\mu$ $\tan \theta =\frac{3}{4}$ ${\text{h}}={\text{R}}-{\text{R}} \cos \theta$ $={\text{R}}-{\text{R}}(\frac{4}{5})={{\text{R}}}/{5}$ ${\text{h}}={{\text{R}}}/{5}=0.2 {\text{m}}$

Calculation: Given, Initial velocity of the ball, $u = V _{0}$ Drag force, $F_{d}=m y v^{2}$ Velocity at a maximum height $=0$ Where $m$ is the Mass of the ball, $v$ is its instantaneous velocity and $y$ is a constant We know that acceleration is the change in velocity. Force is given by, $F = ma$ Thus, the net force on the ball: $F _{ net }= mg + myv ^{2}$ $F_{\;\text{net }\;}=m(g+y v^{2})$ Acceleration is given by, $a=\;\frac{F}{m}$ $a=\;\frac{m(g+\gamma v^{2})}{m}$ $a=-(g+y v^{2})$ Thus, net acceleration is the change in the velocity, $\;\frac{ dv }{ dt }= a$ $\;\frac{d v}{d t}=-( g +\gamma v ^{2})$ $\;\frac{d v}{-( g +\gamma v ^{2})}=d t$ $\;\frac{- dv }{ g +\gamma v ^{2}}=d t$ Integrating both the sides, $∫_{ v _{0}}^{v} \;\frac{- dv }{ g +\gamma v ^{2}}=∫_{0}^{ t } dt$ Let time $t$ required to rise to its zenith $(v=0)$ so, $\Rightarrow \;\frac{1}{\gamma } ∫_{ v _{0}}^{0} \;\frac{- dv }{\frac{g}{\gamma }+ v ^{2}}=∫_{0}^{t} d t$ We know that, according to integral formula, $∫ \;\frac{1}{x^{2}+a^{2}} d x=\;\frac{1}{a} \tan ^{-1}(\;\frac{x}{a})+C$ Thus, $\Rightarrow \;\frac{1}{\gamma } ∫_{ v _{0}}^{0} \;{- dv }/{(\sqrt{\;\frac{8}{\gamma }})^{2}+(v)^{2}}=∫_{0}^{t} d t$ Here, $x=\sqrt{\;\frac{g}{\gamma }}, a=v$ $\Rightarrow t=\;\frac{1}{\gamma }\{[-\;{1}/{\sqrt{\;\frac{g}{\gamma }}} \tan ^{-1}(\;{v}/{\sqrt{\;\frac{g}{\gamma }}})]_{V_{0}}^{0}+C\}$ $[$ for $H_{\max }, v=0]$ $t=0-\;\frac{1}{\gamma }(-\sqrt{\;\frac{\gamma }{ g }} \tan ^{-1}(\;\frac{\sqrt{\gamma } V_{0}}{\sqrt{g}}))$ $\Rightarrow t =\;\frac{1}{\sqrt{\gamma g }} \tan ^{-1}(\;\frac{\sqrt{\gamma } V_{0}}{\sqrt{ g }})$ Therefore, the time taken by the ball to rise to its zenith is $\;\frac{1}{\sqrt{\gamma g }} \tan ^{-1}(\sqrt{\;\frac{\gamma }{g}} V _{0})$

at equation tan 45° = ${100}/F$ F = 100 N

2 + mg sin30 = μmg cos30° 10 = mgsin 30 + μ mg cos30° = 2μmg cos30 -2 6 = μmg cos 30 4 = mg sin 30 $3/2$ = µ × $\sqrt3$ µ = $\sqrt3/2$

$M_{A}=1 \ {kg}$ $M_{B}=3 {kg}$Pseudo force on $\ {A}$ $M_{A} a_{ { system }}=\mu m_{A} g$ $a=0.2 \\times 1 \\times 10$ $a=2 \ \times m / {sec}^{2}$Taking $(\ {A}+\ {B})$ as system$F-\mu .\ (m_{A}+m_{b}\ ) g= (m_{A}+m_{B}\ ) a_{s y s t e m}$ $F-(0.2)(1)(10)=4(2)$ $F=16 N$

$K l=\cos t$ $\ \frac{k_{1}}{k_{2}}=\ \frac{l_{2}}{l_{1}}=\ \frac{1}{n}$

$4F^{2+}9F^{2+}12F^2$ cos θ = $R^2$ $4F^{2+}36F^{2+}24F^2$ cos θ = $4R^2$ $4F^{2+}36F^{2+}24F^2$ cos θ = 4 ($13F^{2+}12F^2cos\theta$) = 52 $F^{2+}48F^2$ cos θ cos θ = $-\frac{12F^2}{24F^2}$ = - $1/2$

$V^{2}-U^{2}=2 \times \frac{F}{m} \\times s$ $\Rightarrow V=0.7 \ {m} / {s}$

mg sin 45° = 50 $\sqrt2$ µmg cos θ = 0.6 × mg × $1/\sqrt2$ = 0.6 × 50 $\sqrt2$ P = 31.28 = 32 N

$\ {CASE} :-(\ {I})$ $F_{\ {net}}=F \cos 30^{\°}-f_{K_{1}}$ $m a_{1} =20 \times \frac{\\sqrt{3}}{2}-0.2 N_{1}$ $5 a_{1} =10 \sqrt{3}-0.2 [50+20 \sin 30^{\°}\ ]$ $a_{1}=2 \\sqrt{3}-\ \frac{12}{5}=1.06 \ {m} {s}^{-2}$ $\ {CASE} :- (\ {II} )$ $F_{\ {nat}}=F \cos 30^{\°}-f_{K_{2}}$ $5 a_{2}=10 \\sqrt{3}-0.2\ [50-20 \sin 30^{\°}\ ]$ $a_{2}-a_{1}=1.86-1.06=0.8 \ {m} {s}^{-2}$

The acceleration of the object on a smooth plane is $a_1 = g \sin 45^{\circ}$, and on a rough plane, it is $a_2 = g(\sin 45^{\circ} - \mu_k \cos 45^{\circ})$.
Given the distance $s = \frac{1}{2}at^2$ is the same, we have $a_1 t_1^2 = a_2 t_2^2$. With $t_2 = n t_1$, it follows that $a_1 = n^2 a_2$, or $a_2 = a_1 / n^2$.
Substituting the acceleration expressions, we get: $g(\sin 45^{\circ} - \mu_k \cos 45^{\circ}) = \frac{g \sin 45^{\circ}}{n^2}$.
Dividing both sides by $g \cos 45^{\circ}$ (since $\sin 45^{\circ} = \cos 45^{\circ} = \frac{1}{\sqrt{2}}$), we obtain: $1 - \mu_k = \frac{1}{n^2}$.
Solving for $\mu_k$ yields $\mu_k = 1 - \frac{1}{n^2}$.

When sliding down, the equation of motion is $mg \sin \theta - f = ma$.
Substituting the given values: $2 \times 10 \times \sin 30^{\circ} - f = 2 \times 3$, which simplifies to $10 - f = 6$, yielding $f = 4 \text{ N}$.
When moving up with an external force $F$, the equation becomes $F - mg \sin \theta - f = ma$.
Rearranging to solve for $F$: $F = ma + mg \sin \theta + f$.
Substituting the values: $F = 2(3) + 2(10)(0.5) + 4 = 6 + 10 + 4 = 20 \text{ N}$.