Laws Of Motion Practice Questions

Given, Angle of inclination, θ = 30º Acceleration, $a=10 {\text{ms}}^{-2}$ Acceleration due to gravity, $g=10 {\text{ms}}^{-2}$ According to the question the car and bob is as shown below, Here, F'is the pseudo force acting on the bob when we considered it from car's frame and T is the tension on the string. In equilibrium, $\Σ\text{F}_{x}=0$ and $\Σ\text{F}_{y}=0$ $\Rightarrow \text{F}' \cos 30°=\text{T} \sin \alpha$ ${m a \cos 30°}{\sin \alpha }=\text{T}$ ...(i) where, m is the mass of the bob. $\text{F}' \sin 30°+m g=\text{T} \cos \alpha$ $\Rightarrow m a \sin 30°+m g=\frac{m a \cos 30°}{\sin \alpha }(\cos \alpha )$ [\because using Eq. (i)] $\Rightarrow a \sin 30°+g=\frac{a \cos 30°}{\sin \alpha }(\cos \alpha )$ $10 \times \frac{1}{2}+10=\frac{10 \times \\sqrt{3}}{2} . \alpha$ $\Rightarrow \frac{1}{2}+1=\frac{\\sqrt{3}}{2} .\alpha$ $\Rightarrow \frac{3}{2}=\frac{\\sqrt{3}}{2} . \alpha$ or$\cot \alpha =\\sqrt{3}$ $\Rightarrow \alpha =30°$

${t}=0, {u}=0$ ${a}=\frac{{F}_{0}}{{M}}-{{F}_{{o}}}/{{MT}^{2}}({t}-{T})^{2}=\frac{{d} {v}}{{dt}}$ $\int _{0} ^{{v}} {d} {v}=\int _{t=0} ^{2 {T}}({{F}_{{o}}}/{{M}}-{{F}_{{o}}}/{{MT}^{2}}({t}-{T})^{2}) {dt}$ V=[\frac{F_{o}}{M} t]_{0}^{2 T}-\frac{F_{0}}{M T^{2}}\[\frac{t^{3}}{3}-t^{2} T+T^{2} t]{0}^{2 T}$$V=\frac{4 F{0} T}{3 M}$

According to the given figure, and free body diagrammasses of two bodies $m_{1}=2 {kg} m_{2}=8 {kg}$ ,Acceleration of mass $m_{1}=a$Acceleration of mass $m_{2}=\;\frac{a}{2}$Tension $=T$Distance between $m_{2}$ and ground $=20 {cm}$ $=0.2 {m}$Initial velocity $u=0$Equation of motion of $2 {kg}$ block, $\therefore \;\; T-2 g=2 a \;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (i)$Equation of motion of $8 {kg}$ block, and $8 g-2 T=8 \;\frac{a}{2} \Rightarrow 4 g-T=2 a\;\;\;\cdot \cdot \cdot \cdot \cdot \cdot \cdot (ii)$From Eqs. (i) and (ii), $\Rightarrow \;\; T-2 g=4 g-T$ $\;\text{ Substituting the value \in Eq. (i) we get }\;$ $\;\; 3 g-2 g=2 a \Rightarrow g=2 a$ $\Rightarrow \;\; a=\;\frac{g}{2}=5 {ms}^{-2}$ $\therefore \;\; a_{1}=5 {ms}^{-2} \;\text{ and }\;_{2}=\;\frac{5}{2} {ms}^{-2}$ $\;\text{ Since, }\; s=u t+\;\frac{1}{2} a t^{2}$ $\therefore \;\; \;\frac{20}{100}=0+\;\frac{1}{2} \;\frac{5}{2} \times t^{2} \Rightarrow t^{2}=\;\frac{20 \times 4}{5 \times 100}$ $\Rightarrow \;\; t=\;\frac{2}{5}=0.4 {s}$

When both the blocks move together as a system, then acceleration of this system will be given as $a=\;\frac{F}{m+M}$...(i) Frictional force on mass, $f=m a$...(ii) From Eqs. (i) and (ii), we get $f=m(\;\frac{F}{m+M})$ For no slipping, $f \le \mu m g \;\; [ \because \mu$ being the coefficient of static friction $\Rightarrow \;\; m(\;\frac{F}{m+M}) \le \mu m g \Rightarrow F \le \mu (m+M) g$ $\therefore \;\; F_{\max }=\;\frac{3}{7}(0.5+4.5) \times 9.8 \Rightarrow F_{\max }=21 {N}$

Let's draw the free body diagram, (During the upward direction) F = 2N = (+ ve) constant (During the downward direction) F = 2N = (- ve) constan During the upward motion, the force is positive constant and duringthe downward motion the force is negative constant. Hence, the correct graph is (b).

Let ${a}_{ {CM}}$ be the acceleration of centre of mass of system and $F$ be the applied force on spring. $\therefore \;a_{C M}=\;\frac{F}{m_{A}+m_{B}}=\;\frac{m_{A} a_{A}+m_{B} a_{B}}{m_{A}+m_{B}}$ $\Rightarrow \;a_{C M}=\;\frac{F}{2 M}=\;\frac{M a_{A}+M a_{B}}{2 M} \;\; ( \because m_{A}=m_{B}=M)$ $\Rightarrow \;a_{B}=\;\frac{F-M a_{A}}{M}=\;\frac{F-M a}{M}$

Three forces $\text{F}_{1}, \text{F}_{2}$ and $\text{F}_{3}$ are acting on a body and if this body is in equilibrium, then resultant of these three forces must be zero i.e. ${\text{F}}_{ {\text{net} }}={\text{F}}_{1}+{\text{F}}_{2}+{\text{F}}_{3}=0$ $\Rightarrow {\text{F}}_{1}+{\text{F}}_{2}=-{\text{F}}_{3}$ This situation can be shown graphically by three concurrent forces at 120° with each others. or, by three forces in the same order along three sides of a triangle. Hence, both statement I and statement II are true.

Given, $\text{M}=1000 {\text{kg}}, a=20 {\text{m}} / {\text{s}}^{2}$ $v_{{\text{relative} }}=500 {\text{m}} / {\text{s}}, g=10 {\text{m}} / {\text{s}}^{2}$ The given situation is shown below $\text{F}_{{\text{thrust} }}=\frac{d}{d t}(\text{M} v_{{\text{relative} }})$ $\Rightarrow \text{F}_{{\text{thrust} }}=v_{{\text{relative} }}(\frac{d \text{M}}{d t})$ By Newton's second law of motion, $\Rightarrow \text{F}_{{\text{thrust} }}-\text{M} g=\text{M} a$ $\Rightarrow v_{{\text{relative} }}(\frac{d \text{M}}{d t})-\text{M} g=\text{M} a$ $\Rightarrow 500(\frac{d \text{M}}{d t})-1000 \times 10=1000 \times 20$ $\Rightarrow 500(\frac{d \text{M}}{d t})=1000(20+10)$ $\Rightarrow 500(\frac{d \text{M}}{d t})=1000 \times 30$ $\Rightarrow \frac{d \text{M}}{d t}=\frac{1000 \times 30}{500}=60$ $\Rightarrow \frac{d \text{M}}{d t}=60 {\text{kg}} / {\text{s}}$

Given, mass of block, $m=\sqrt{3} {kg}$ Coefficient of friction, $\mu =1 / 3 \sqrt{3}$ According to diagram,Let $F$ be the force applied on the body, $w$ be the weight $(=m g)$, $N$ be the normal reaction. Friction force $f=\mu N$ For no movement of body along $X$-axis, net force along $X$-axis should be zero. If, $F_{y}$ be the net force along $y$-axis then it will also be zero because body is not accelerating at all. $\therefore \;N=F \sin 60^{\circ}+m g$ $\Rightarrow \;N=\;\frac{\sqrt{3}}{2} F+10 \sqrt{3}$ . . . (i) $\therefore \;N=F \sin 60^{\circ}$ $\Rightarrow \;N=\;\frac{\sqrt{3}}{2} F+1$ $\;\text{ Similarly, }\; F_{x}=F \cos 60^{\circ}\;-\mu N=0$ From Eq. (i), we get $\Rightarrow \;\;\frac{F}{2}-\;\frac{1}{3 \sqrt{3}}(\;\frac{\sqrt{3}}{2} F+10 \sqrt{3})=0$ $\Rightarrow \;\;\frac{F}{2}=\;\frac{F}{6}+\;\frac{10}{3} \Rightarrow \;\frac{F}{2}-\;\frac{F}{6}=\;\frac{10}{3}$ $\Rightarrow \;F=10 {N}$ $\;\text{ Given, }\;\;F=3 x$ $\Rightarrow \;x=\;\frac{10}{3}=3.33$

The diagram and the required components of force on given block are shown belowFrom the above diagram, $N=m g-F \sin \theta$...(i) where, $N=\;\text{ normal force }\;$ and $F \cos \theta -f_{k}=m a$ $\Rightarrow \;F \cos \theta -\mu _{k} N=m a$ ...(ii) where, $f_{k}=$ kinetic friction force. From Eq. (i) and Eq. (ii), we get $F \cos \theta -\mu _{k}(m g-F \sin \theta )=m a$ $\Rightarrow \;\; a=\;\frac{F}{m} \cos \theta -\mu _{k}(g-\;\frac{F}{m} \sin \theta )$ This is the required acceleration of the block.

The free body diagram for the wooden block is shown belowUsing the condition of the equilibrium, In the $x$-direction, the summation of all the forces is to be zero. $f-T \;\; =0$ $\mu R-T \;\; =0$ $T \;\; =0.5 R$...(i) In the $y$-direction, , the summation of all the forces is to be zero. $T+R-90=0$ $\Rightarrow \;\; 0.5 R+R-90=0$ $R=60 {N}$ Hence, the normal force on the wooden block is $60 {N}$ Using the Eq. (i), $T=0.5(60)=30 {N}$ Hence, the maximum value of the tension in the rope, so that wooden block will not move is $30 {N}$.

Given, coefficient of static friction, $\mu _{s}=0.2$Various forces acting on block are shown belowFrictional force $\le m g$ $\Rightarrow \;\; N \times 0.2 \le 5$ $\Rightarrow \;\; N \le 25$ $\therefore \;\;$ Magnitude of horizontal force, $F=N=25 {N}$

Given, coefficient of static friction, µ = 0.5 Value of acceleration due to gravity, ${\text{g}}=10 {\text{ms}}^{-2}$ For complete system to move together, F = ma Here, m is total mass. F = (1 + 2) a = 3a ...(i) From free body diagram of 1 kg block, Balance forces in horizontal direction, $\text{F}_{\text{s}}=\\mu \text{R}=\text{m}' a$ ...(ii) Balance forces in vertical direction, R = m′g Put value of R in Eq. (ii), µm′g = m′a \Rightarrow 0.5 × 1 × 10 = 1 × a $\Rightarrow a=5 {\text{ms}}^{-2}$ Put the value of a in Eq. (i), we get F = 3 × 5 = 15 N Thus, the maximum horizontal force required to move block together is 15 N.

The graph for given equation is shown belowAt maximum height, the slope of tangent drawn, $\tan \theta \;=\;\frac{d y}{d x}=\;\frac{2 x}{4}=\;\frac{x}{2}$ $\Rightarrow \;\; 0.5\;=\;\frac{x}{2} \;\; [ \therefore y=\;\frac{x^{2}}{4}]$ $\Rightarrow \;\; x\;=1 {m}$ $\therefore \;\; y\;=\;\frac{x^{2}}{4}=\;\frac{1}{4}=0.25 {m}=25 {cm}$

Given, speed of projection $=v_{0}$ Damping force, $F=m a=-\alpha x^{2}$ $\Rightarrow \;\; {a}=-\alpha x^{2} / m$ Also, $\Rightarrow \;\; v d v=a d x=-\;\frac{\alpha }{m} x^{2} d x$ Integrating both sides, we get $∫_{v_{0}}^{v} v d v \;\; =∫_{0}^{x}-\;\frac{\alpha }{m} x^{2} d x$ $\Rightarrow \;\; (\;\frac{v^{2}}{2})_{v_{0}}^{0} \;\; =-\;\frac{\alpha }{m}(\;\frac{x^{3}}{3})_{0}^{x}$ $\Rightarrow \;\; 0-v_{0}^{2} / 2 \;\; =-\;\frac{\alpha }{m} \;\frac{x^{3}}{3} \Rightarrow x=(\;\frac{3 m}{2} \;\frac{v_{0}^{2}}{\alpha })^{1 / 3}$

{{d} {\vec{v}}}/{{dt}}={\vec{a}}={{\vec{F}}}/{{m}}=(8 {{i}}^{\^}+2 \hat{j}) {m} / {s}^{2} ${{d} {\vec{r}}}/{{dt}}={\vec{v}}=(8 {t} \hat{i}+2 {t} \hat{j}) {m} / {s}$ ${\vec{r}}=(8 \hat{i}+2 \hat{j}) \frac{{t}^{2}}{2} {m}$ At ${t}=10 {sec}$ {\vec{r}}=\[(8 {{i}}^{^}+2 \hat{j}) 50]$ {m}$$\Rightarrow{{r}}^{\to }=(400 \hat{i}+100 {{j}}^{^}) {m}$

Given The mass of the bullet, $m=0.1 {kg}$ The initial velocity of the bullet before hitting the wooden, $u=10 {m} / {s}$ The final velocity of the bullet after hitting the wooden, $v=0$ The distance travelled by the bullet before coming to the rest, ${s}=50 {cm}$ Using the equation of the motion, $v^{2}-u^{2}=2 a s$ $\Rightarrow \;\; 0^{2}-(10)^{2}=2 a(0.50)$ $\Rightarrow \;\; a=-100 {m} / {s}^{2}$ The uniform retardation of the bullet is $100 {m} / {s}^{2}$ The magnitude of the effective retarding force on the bullet, $F=m a$ $F=0.1(-100)$ $F=-10 {N}$ Hence, the value of $x$ to the nearest integer is 10 .

Given, mass of $man(m)=60 {kg}$ Downward acceleration of lift, $a=1.8 {ms}^{-2}$ Let $T$ be the tension in the rope connected with lift, $g$ be the acceleration due to gravity $(10 {ms}^{-2})$. As, lift is moving in downward direction $\therefore \;\; m g-T \;\; =m a$ $\Rightarrow \;\; T \;\; =m g-m a=m(g-a)$ $\;\; =60(10-1.8)=60 \times 8.2$ $\;\; =492 {N}$ Hence, the weight of the man during downward acceleration is $492 {N}$.