Question 513 Marks
A ball is held at rest at position A in Fig., by two light strings. The horizontal string is cut and the ball starts swinging as a pendulum. Point B is the farthest to the right where the ball goes as it swings back and forth. What is the ratio of the tension in the supporting string in position B to its value at A before the horizontal string was cut?A ball is held at rest at position A in Fig., by two light strings. The horizontal string is cut and the ball starts swinging as a pendulum. Point B is the farthest to the right where the ball goes as it swings back and forth. What is the ratio of the tension in the supporting string in position B to its value at A before the horizontal string was cut?
AnswerIn the first case, ball is in equilibrium. So, the net force on the body in any direction should be zero. $\therefore \sum \ \vec{\text{F}}$ in the vertical direction = 0 $\therefore \text{T}_{1} \cos \theta = \text{mg}\Rightarrow \text{T}_{1} = \frac{\text{mg}}{\cos \theta}$
View full question & answer→Question 523 Marks
A block of wood of mass 3kg is resting on the surface of a rough inclined surface, inclined at an angle $\theta$ as shown in the figure:
- Name the forces $(1, 2, 3).$
- If the coefficient of static friction is 0.2, calculate the value of all the three forces. (may use $g = 10m/ s^2)$
Answer
- Force $1$ = weight = mg
Force $2$ = force of limiting friction
Force $3$ = Normal Reaction R
- $\mu=0.2,\text{ m}=3\text{kg},\theta=30^\circ$
Force 1 = weight = mg
$= 3 \times 10 = 30N$
$\text{Force }2=\text{f}_1=\text{mg}\sin\theta-\text{F}$
$\therefore\ \text{mg}\sin\theta=3\times10\times\sin30^\circ=15\text{N}$
And force of friction $\text{F}=\mu\text{R}$
$=\mu\text{mg}\cos\theta$
$=0.2\times3\times10\cos30^\circ=3\sqrt{3}\text{N}$
then force $2=\text{f}_1=15-3\sqrt{3}\approx9.8\text{N}$
Force 3 = Normal reaction R
$\therefore\ \text{R}=\text{mg}\cos\theta=3\times10\cos30^\circ$
$=15\sqrt{3}\text{N}$ View full question & answer→Question 533 Marks
For three moving objects the distances are found to be directly proportional to the times $t, t^2$ and $t^3$. What is the nature of the net force on each object?
AnswerAs F = ma $=\frac{\text{mv}}{\text{t}}=\text{m}\frac{\big(\frac{\text{x}}{\text{t}}\big)}{\text{t}}=\frac{\text{mx}}{\text{t}^2}$ When, $\text{x}\propto\text{t},\ \text{x}=\text{kt}$
$\Rightarrow\ \text{F}=\frac{\text{m}(\text{kt})}{\text{t}^2}=\frac{\text{km}}{\text{t}}$
$\text{i.e., F}\propto\frac{1}{\text{t}}$
$\therefore$ Net force on the object is inversely proportional to t. Similarly, when $\text{x}\propto\text{t}^2,$
$\therefore$ Net force on the object is independent of t.When $\text{x}\propto\text{t}^3,$
$\text{F}\propto\text{t}$
So, net force F on the object is directly proportional to the time.
View full question & answer→Question 543 Marks
State the law of conservation of momentum. Establish the same for a 'n' body system.
AnswerWhen no external force acts on a system the momentum will remain conserved. Consider a system of n bodies of masses $m_1, m_2, m_3 .... m_n$.
If $P_1, P_2, P_3 .... P_n$ are the momentum associated then the rate of change of momentum with the system, $\frac{\text{dp}}{\text{dt}}=\frac{\text{dp}_1}{\text{dt}}+\frac{\text{dp}_2}{\text{dt}}+\frac{\text{dp}_3}{\text{dt}}+......+\frac{\text{dp}_\text{n}}{\text{dt}}$ $=\frac{\text{d}}{\text{dt}}(\text{p}_1+\text{p}_2+\text{p}_3+.....\text{p}_\text{n})$ If no external force acts, $\frac{\text{dp}}{\text{dt}}=0$
$\therefore$ p = constant i.e.,$p_1 + p_2 + .....+ p_n$ = constant.
View full question & answer→Question 553 Marks
A motor car is travelling at $30m/ s$ on a circular road of radius $500m$. It is increasing its speed at the rate of $2ms^2$. What is the acceleration?
AnswerSpeed is $30m/ s$, radius = $500m$ Tangential acceleration $a_t = 2m/ s^2$ Centripetal acceleration, $\text{a}_\text{r}=\frac{\text{v}^2}{\text{r}}=\frac{900}{500}=1.8\text{ms}^{-2}$ Net acceleration $=\sqrt{\text{a}^2_\text{r}+\text{a}^2_\text{t}}$ $=\sqrt{2^2+(1.8)^2}=2.7\text{m/s}^2$
View full question & answer→Question 563 Marks
State law of conservation of momentum and prove it using third law of motion.
AnswerWhen no external force acts on the body, the momentum remains conserved. According to third law, for every action there is an equal and opposite reaction. So if $\mathrm{dp}_1$ and $\mathrm{dp}_2$ are change in momentum of two masses $\mathrm{m}_1$ and $\mathrm{m}_2$ then, $\frac{\mathrm{dp}_1}{\mathrm{dt}}=-\frac{\mathrm{dp}_2}{\mathrm{dt}}$, Since $\mathrm{F}_1=-\mathrm{F}_2 \therefore-\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{p}_1+\mathrm{p}_2\right)=0$, i.e., $\mathrm{p}_1+\mathrm{p}_2=$ constant.
View full question & answer→Question 573 Marks
For ordinary terrestrial experiments, which of the observers below are inertial and which are non-inertial:
- A child revolving in a giant wheel.
- A driver in a sports car moving with a constant high speed of $200kmh^{-1}$ on a straight road.
- The pilot of an aeroplane which is taking off.
- A cyclist negotiating a sharp turn.
- The guard of a train which is slowing down to stop at a station.
Answer
- A giant revolving wheel has accelerated (radially) motion. Therefore, a child revolving in a giant wheel is non-inertial.
- The sports car is moving with a constant speed on a straight road. Therefore, the driver in a sports car moving with a constant high speed $(200kmh^{-1})$ on a straight road is inertial.
- As the aeroplane takes off, it has accelerated motion. Therefore, the pilot of an aeroplane, which is taking off is non-inertial.
- While negotiating a sharp turn, there is change in the direction of motion of the cyclist and hence the motion is accelerated. Therefore, a cyclist negotiating a sharp turn is non-inertial.
- When a train is slowing down to stop at a station, its motion is retarding. Therefore the guard of a train which is slowing down to stop at a station acts as non-inertial observer.
View full question & answer→Question 583 Marks
Consider the bodies of mass $m _1$ and $m _2$ in contact placed on a frictionless table as shown. When force F is applied on mass $m_1$, calculate the acceleration produced, and the force of contact between the bodies. What will be the force of contact when the force $F$ is applied on mass $m_2$ ?
AnswerCase (a): Let $F$ be applied on $m_1$ and $f$ is the force of contact between the two bodies.
$\therefore F-f=m_1 a \ldots . . .(i)$
Also, $f=m _2 a \ldots$...(ii) Adding (i) and (ii),
we have $F =\left( m _1+ m _2\right)$ a or $a =\frac{ F }{\left( m _1+ m _2\right)}$
Again, $f = F - m _1 a$
$=\text{F}-\frac{\text{m}_1\text{F}}{(\text{m}_1+\text{m}_2)}=\frac{\text{m}_2\text{F}}{(\text{m}_1+\text{m}_2)}$
Case (b): Similarly, in the 2nd case.
$F - f_1 = m_2a$ and $f_1 = m_1a$
$\therefore F = (m_1 + m_2)$ aor,
$\text{a}=\frac{\text{F}}{(\text{m}_1+\text{m}_2)}$
$\therefore\ \text{f}_1=\frac{\text{m}_1\text{F}}{(\text{m}_1+\text{m}_2)}$ View full question & answer→Question 593 Marks
Show that the total linear momentum of an isolated system of interacting particles is conserve
AnswerConsider two bodies A and B, with inital momenta $\vec{\text{p}}_\text{A}$ and $\vec{\text{p}}_\text{B}$ respectively. Let the two bodies collide, get apart and have final momenta $\vec{\text{p}'}$ and $\vec{\text{p}'_\text{B}}$ respectively. By the second law of motion: Change in momentum of body $\text{A},\vec{\text{p}'_\text{A}}-\vec{\text{p}}_\text{A}=\vec{\text{F}}_\text{AB}\Delta\text{t}\dots\text{(i)}$ A where $\vec{\text{F}}_\text{AB}$ is the force acting on A due to action of B for a time $\Delta\text{t.}$ Similarly change in momentum of body, $\text{B},\vec{\text{p}}'-\text{B}-\vec{\text{p}}_\text{B}=\vec{\text{F}}_\text{BA}\Delta\text{t}\dots\text{(ii)}$ Here time $\Delta\text{t}$ the time for which two bodies A and B are in contact and interact, is same for both the forces. Moreover, from third law of motion $\vec{\text{F}}_\text{AB}=-\vec{\text{F}}_\text{BA}$ Hence, adding (i) and (ii), we obtain $\Big(\vec{\text{p}}_\text{A}-\vec{\text{p}}_\text{A}\Big)+\Big(\vec{\text{p}}_\text{B}-\vec{\text{p}}_\text{B}\Big)\text{s}$ $=\vec{\text{F}}_\text{AB=}\Delta\text{t}+\vec{\text{F}}_\text{BA}\Delta\text{t}=-\vec{\text{F}}_\text{BA}\Delta\text{t}+\vec{\text{F}}_\text{BA}\Delta\text{t}=0$ $\Rightarrow\vec{\text{p}'_\text{A}}+\vec{\text{p}'_\text{B}}=\vec{\text{p}}_\text{A}+\vec{\text{p}}_\text{B}$ Which shows that the total final momentum of the isolated system is exactly same as its initial momentum. Thus, it is proved that total momentum of an isolated system remains conserved.
View full question & answer→Question 603 Marks
The driver of a truck travelling with a velocity v suddenly notices a brick wall in front of him at a distance d. Is it better for him to apply brakes or to make a circular turn without applying brakes in order to just avoid crashing into the wall? Why?
AnswerIn applying brakes, suppose $F_B$ is the force required to stop the truck in distance (d)
$\therefore\text{F}_\text{B}\times\text{d}=\frac{1}{2}\text{mv}^2$ or $\text{F}_\text{B}=\frac{\text{mv}^2}{\text{2d}}$
In taking a turn of redius d, the force required is
$\text{F}_\text{T}=\frac{\text{mv}^2}{\text{d}}=\text{2F}_\text{B}$ or $\text{F}_\text{B}=\frac{1}{2}\text{F}_\text{T}$
Therefore, it is better to apply brakes.
View full question & answer→Question 613 Marks
Define impulse. A cricket ball of mass $150gm$ moving with speed of $12m/ s$ is hit by a bat so that the ball is turned back with a velocity of $20m/ s$. Calculate the impulse received by the ball.
AnswerThe product of force and the time on which it acts or change in momentum is called Impulse. Momentum before the hit $=150 \times 12 \times 10^{-3}=1.8 \mathrm{~kg} \mathrm{~ms}^{-1}$ Momentum after the hit $=150 \times 10^{-3} \times-20=-3 \mathrm{~kg} \mathrm{~ms}^{-1}$
$\therefore$ Impulse $=$ change in momentum $=-4.8 \mathrm{~kg} \mathrm{~ms}^{-1}$
View full question & answer→Question 623 Marks
A piece of ice slides down a $45°$ incline in twice the time it takes to slide down a frictionless $45°$ incline. What is the coefficient of friction between the ice and the incline?
AnswerHere, $\theta = 45^\circ, \text{S}_1 = \text{S}_{2}; \text{u} = 0$ On the rough incline, $\text{a}_{1} = \text{g}(\sin \theta - \mu \cos \theta )$ $t_1$= time taken On the frictionless incline, $\text{a}_{2} = \text{g} \sin \theta t_2$ = time taken From $\text{S} = \text{ut} + \frac{1}{2} \text{at} ^{2}$
$\text{S}_{1} = 0 \frac{1}{2} \text{g} (\sin \theta = \mu \cos \theta) \text{t}_{1}^{2}$ and $\text{S}_{2} = 0 +\frac{1}{2} \text{g} \ \sin \theta. \text{t}_{2}^{2}$ As $\text{S}_{1} = \text{S}_{2}$
$\therefore \frac{1}{2} \text{g} (\sin \theta - \mu \cos \theta) \text{t}_{1}^2 = \frac{1}{2} \text{g} \sin \theta. \text{t}^2_1$
$\frac{\sin \theta - \mu \cos \theta }{\sin \theta} = \frac{\text{t}^1_2}{\text{t}^2_1} = \frac{\text{t}^2_2}{(2\text{t}_2)^2} = \frac{1}{4}$
$1 - \mu \cot \theta = \frac{1}{4}$ or $\mu \cot \theta = 1 -\frac{1}{4}$
$\Rightarrow\mu \cot \theta = \frac{3}{4} \Rightarrow \mu \frac{3}{4\cot \theta}.$
View full question & answer→Question 633 Marks
A rocket with a lift-off mass $20,000kg$ is blasted upwards with an initial acceleration of $5.0ms^{-2}$. Calculate the initial thrust (force) of the blast.
AnswerGiven: Mass of the rocket, $m=20,000 \mathrm{~kg}$ Initial acceleration, $a=5 \mathrm{~m} / \mathrm{s}^2$ Acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2 U$ sing Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation: ( $F-\mathrm{mg}$ ) = ma $\mathrm{F}=$ $m(g+a)=(20000 \times(10+5))=(20000 \times 15)=3 \times 10^5 \mathrm{~N}$
View full question & answer→Question 643 Marks
Define the term 'coefficient of limiting friction' between two surfaces. A body of mass 10kg is placed on an inclined surface of angle 30°. If the coefficient of limiting friction is $\frac{1}{\sqrt{3}}$ find the force required to just push the body up the inclined surface. The force is being applied parallel to the inclined surface.
AnswerCoefficient of limiting friction between two surfaces in contact is defined as the ratio of force of limiting friction and normal reaction between them. 
$\text{m}=10\text{kg}$ $\theta=30^\circ$ $\mu=\frac{1}{\sqrt{3}}.$ $\text{R}=\text{mg}\cos\theta$ Force of friction $\text{F}=\mu\text{R}=\mu\text{ mg }\cos\theta$ $=\frac{1}{\sqrt{3}}\times10\times9.8\times\cos30^\circ$ $=\frac{1}{\sqrt{3}}\times98\times\frac{\sqrt{3}}{2}=49\text{N}$ $\text{mg }\sin\theta=10\times9.8\times\sin30^\circ$ $=49\text{N}$ Force required to push the body up inclined surface = (49 + 49) = 98N. View full question & answer→Question 653 Marks
The barrel of a gun is $1m$ long and it fires a bullet of mass $0.05kg$ with a muzzle velocity of $400ms^{-1}.$ Find:
- The acceleration,
- The force, and
- The impulse given to the bullet by the gun.
AnswerHere mass of bullet, $m =0.05 kg$, initial velocity of bullet before firing $u =0$, length of barrel of gun, moving through which the bullet is accelerated $s=1 m$, final muzzle velocity of bullet $v=400 ms^{-1}$.
- Using the relation $v^2 - u^2 =2as$, we have,
$(400)^2-(0)^2=2\times\text{a}\times1$,
$\Rightarrow\text{a}=\frac{400\times400}{2\times1}=8\times10^4\text{ms}^{-2}$,
- Force $\text{F}=\text{ma}=0.05\times8\times10^4=4000\text{N}$
- Impulse given to the bullet by the gun, J = change in momentum of bullet
$=\text{m}(\text{v}-\text{u})=0.05\times(400-0)=20\text{Ns}.$ View full question & answer→Question 663 Marks
The force required to just move a body up an inclined plane is double the force required to prevent the body from sliding down. Find the coefficient of friction.
AnswerForce required to just move the body of mass m up to an inclined plane $\text{F}_1=\text{mg}\sin\theta+\mu\text{ mg}\cos\theta$ Force required to stop the body from sliding down $\text{F}_2=\text{mg}\sin\theta-\mu\text{ mg}\cos\theta$ Given that, $F_1 = 2F_2$
$\text{mg}(\sin\theta+\mu\cos\theta)=2(\sin\theta-\mu\cos\theta)\text{mg}$
$3\mu\cos\theta=3\sin\theta\Rightarrow\ \mu=\tan\theta$
View full question & answer→Question 673 Marks
Compute the acceleration of the block and trolley system as shown. If the coefficient of kinetic friction between the trolley and the surface is $0.04$, what is the tension in the string? [Take $g = 10ms^{-2}]$
AnswerLet a be the acceleration produced in the block-trolley system.
Considering forces acting on the weight $30kg$ $30 - T = 4a ...(i)$
Kinetic friction, $F_k =$ $\mu$(mass of trolley) $\times g = 0.04 \times 30 \times 10 = 12N$
Considering forces acting on trolley of mass $30kg,$
we have $T - F_1 = 30a$ or $T - 12 = 30a ...(ii)$ Adding (i) and (ii),
we get 28 = 34aor $\text{a}=\frac{28}{34}=0.82\text{ ms}^{-2}$
Putting this value in (i), we get $T = 30 - 4 \times 0.82 = 30 - 3.28 = 26.72N$
View full question & answer→Question 683 Marks
Distinguish between static friction, limiting friction and kinetic friction. How do they vary with the applied force, explain by diagram.
AnswerStatic friction exists as long as the body stays at rest. It increases with the force applied. Limiting friction is the maximum value of static friction. Kinetic friction is the friction when the body is on the move. This will be a constant and depends on the nature of surface only. The variation of frictional force is shown below.
View full question & answer→Question 693 Marks
Why does a gun recoil on firing? What is recoil velocity? Find the expression for it.
AnswerSince no external force acts on the gun the momentum has to be conserved. So, the gun recoils. Recoil velocity: Velocity of the gun at the instant bullet is fired is known as recoil velocity. Gun recoil back opposite to direction of motion of bullet. If $m_g$ and $m_b$ are mass of gun and bullet, with velocity of bullet being $V_b$ then $m_gv_g + m_bV_b = 0\text{V}_\text{g}=-\frac{\text{m}_\text{b}\text{V}_\text{b}}{\text{m}}$
-ve sign shows that gun moves in opposite direction and $V_g$ is called recoil velocity.
View full question & answer→Question 703 Marks
A man of mass 70kg stands on a weighing scale in a lift which is moving. What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?
AnswerWhen the lift moves freely under gravity, acceleration a = g Using Newton’s second law of motion, we can write the equation of motion as: R + mg = ma R = m(g – a) = m(g - g) = 0 $\therefore$ Reading on the weighing scale $\frac{0}{\text{g}}=0\text{kg}$ The man will be in a state of weightlessness.
View full question & answer→Question 713 Marks
An electron and a proton are detected in a cosmic ray experiment, electron with kinetic energy 10 keV and proton with kinetic energy 100 keV . Which is faster: the electron or the proton? Obtain the ratio of their speeds. (Take mass of electron $=9.11 \times 101^{-31} \mathrm{~kg}$, mass of proton $\left.=1.67 \times 10^{-27} \mathrm{~kg}, \mathrm{IeV}=1.60 \times 10^{-19} \mathrm{~J}\right)$
Answer$\text{K}_\text{e}=10\text{ keV},\text{ K}_\text{p}=100\text{ keV}$$\frac{\text{K}_\text{e}}{\text{K}_\text{p}}=\frac{\text{m}_\text{e}\text{v}^2_\text{e}}{\text{m}_\text{p}\text{v}^2_\text{p}}$
$\Big(\frac{\text{v}_\text{e}}{\text{v}_\text{p}}\Big)^2=\Big(\frac{\text{m}_\text{p}}{\text{m}_\text{e}}\Big)\Big(\frac{\text{K}_\text{e}}{\text{K}_\text{p}}\Big)$
$=\frac{1.67\times10^{-27}}{9.11\times10^{-31}}\times\frac{1}{10}$
$\frac{\text{v}_\text{e}}{\text{v}_\text{p}}=\sqrt{183.31}$
$\text{v}_\text{e}=13.5\text{ v}_\text{p}$
View full question & answer→Question 723 Marks
A car is moving in a circular horizontal track of radius 10m with constant speed of 10m/s. A plumb bob is suspended from roof by a light rigid rod of length 1m. Find the angle made by the rod with the track.
AnswerThe different forces acting on the bob are shown in figure. Resolving the forces along the length and perpendicular to the rod, we have $\text{mg}\cos\theta+\frac{\text{mv}^2}{R}\sin\theta=\text{T}$ $\text{mg}\sin\theta=\frac{\text{mv}^2}{\text{R}}\cos\theta$ Now,$\tan\theta=\frac{\text{v}^2}{\text{Rg}}=\frac{(10)^2}{10\times10}=1$ $\Rightarrow\tan\theta=1\Rightarrow\theta=\tan^{-1}(1)=45^\circ.$ 
View full question & answer→Question 733 Marks
A train runs along an unbanked circular track of radius $30m$ at a speed of $54km/ h$. The mass of the train is $10^6kg$. What is the centripetal force required for this purpose? What is the angle of banking required to prevent wearing out of the rail?
AnswerGiven: $r=30 \mathrm{~m}, \mathrm{v}=54 \mathrm{~km} / \mathrm{h}=15 \mathrm{~ms}^{-1}, \mathrm{~m} 10^6 \mathrm{~kg}$ Centripetal force $=\frac{\text{mv}^2}{\text{r}}=\frac{10^6\times15^2}{30}=75\times10^5\text{N}$ Angle of banking, $\theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)=\tan^{-1}\Big(\frac{225}{30\times9.8}\Big)$ $\theta\simeq37^\circ$
View full question & answer→Question 743 Marks
Arocket with a lift off mass $20,000 \mathrm{~kg}$ is blasted upwards with an initial acceleration of $5.0 \mathrm{~ms}^{-2}$. Calculate the initial thrust (force) of the blast.
AnswerMass of the rocket, $\mathrm{m}=20,000 \mathrm{~kg}$ Acceleration to be produced in the rocket $\mathrm{a}=\mathrm{ms}^{-2}$ upward thrust required to overcome the gravitational pull, $\mathrm{F}_1=\mathrm{mg}=20,000 \times 9.8=1.96 \times 10^5 \mathrm{~N}$ Upward thrust required to impart acceleration, a $F_2=\mathrm{ma}, \mathrm{F}_2=20,000 \times 5=10^5 \mathrm{~N}$ Hence, net initial thrust on the blast $=1.96 \times 10^5 \mathrm{~N}+10^5 \mathrm{~N}=2.96 \times$ $10^5 \mathrm{~N}$
View full question & answer→Question 753 Marks
A horizontal force of $500N$ pulls two masses $10kg$ and $20kg$ (lying on a frictionless table) connected by a light string as shown. What is the tension in the string? Does the answer depend on which mass the pull is applied? 
AnswerThe acceleration produced in the body of mass $10 + 20 = 30kg$ is
given by, $\text{a}=\frac{\text{F}}{\text{m}}=\frac{500}{30}=\frac{50}{3}\text{ ms}^{-2}$
When $500N$ pull is applied on $20kg$, tension, T, produced is given by, $T_1 = m_1a$
$=10\times\frac{50}{3}=\frac{500}{3}=166.7\text{ N}$
When $500N$ pull is applied on 10kg, tension T, produced is given by, $T_2 = m_2a $
$=20\times\frac{50}{3}=\frac{1000}{3}=333.4\text{ N}$
Thus the tension depends mass-end on which the pull is applied.
View full question & answer→Question 763 Marks
A curved road of diameter $1.8km$ is banked, so that no friction is required at a speed of $30ms^{-1}$. What is the banking angle?
AnswerRadius = $0.9km = 900m$ Speed =$v = 30ms^{-1}$, $\mu=0$ We know, $\tan\theta=\frac{\text{v}^2}{\text{rg}}$
$\therefore\ \theta=\tan^{-1}\Big(\frac{\text{v}^2}{\text{rg}}\Big)$
$=\tan^{-1}(0.102)=6^\circ$
View full question & answer→Question 773 Marks
State law of conservation of linear momentum. Derive the law of conservation of momentum from Newton's third law of motion.
AnswerWhen no external force acts on a body, there is no change in linear momentum or the total momentum of an isolated system of interacting particles remain conserved. Let us consider an isolated system (which is free from the influence of any external forces) comprising of two bodies A and B with initial momentum $\vec{\text{P}}_\text{A}$ and $\vec{\text{P}}_\text{B}.$ Let them collide for a small time $\Delta\text{t}$ and separate with final momentum $\vec{\text{P}'}_\text{A}$ and $\vec{\text{P}'}_\text{B}$ respectively. During collisioin, If $\vec{\text{F}}_\text{AB}$ is force on A exerted by B, and $\vec{\text{F}}_\text{BA}$ is force on B exerted by A, then according to Newton's second law.$\vec{\text{F}}_\text{AB}\times\Delta\text{t}=$ Change in momentum of A
$=\vec{\text{P}'}_\text{A}-\vec{\text{P}}_\text{A}\dots(\text{i})$
$\vec{\text{F}}_\text{BA}\times\Delta\text{t}=$ Change in momentum of B
$=\vec{\text{P}'}_\text{B}-\vec{\text{P}}_\text{B}\dots(\text{ii})$
According to Newton's third law,$\vec{\text{F}}_\text{AB}=-\vec{\text{F}}_\text{AB}$
$\therefore$ Forms (i) and (ii),
$\vec{\text{P}'}_\text{A}-\vec{\text{P}}_\text{A}=-(\vec{\text{P}'}_\text{B}-\vec{\text{P}}_\text{B})$
Which shows that total final momentum of the isolated system is equal to its total initial momentum.$\Rightarrow\ \vec{\text{P}'}_\text{A}+\vec{\text{P}'}_\text{B}=\vec{\text{P}}_\text{A}+\vec{\text{P}}_\text{B}$
This proves the law of conservation of linear momentum.
View full question & answer→Question 783 Marks
Show that the total linear momentum of an isolated system of interacting particles is conserved.
AnswerConsider two bodies A and B, with inital momenta $\vec{\text{p}}_\text{A}$ and $\vec{\text{p}}_\text{B}$ respectively. Let the two bodies collide, get apart and have final momenta $\vec{\text{p}'}_\text{A}$ and $\vec{\text{p}'}_\text{B}$ respectively. By the second law of motion: Change in momentum of body $\text{A},\vec{\text{p}'}_\text{A}-\vec{\text{p}}_\text{A}=\vec{\text{F}}_\text{BA}\Delta\text{t}$ ...(i) Where $\vec{\text{F}}_\text{AB}$ is the force acting on A due to action of B for a time $\Delta\text{t}$. Similarly change in momentum of body $\text{B},\vec{\text{p}'}_\text{B}-\vec{\text{p}}_\text{B}=\vec{\text{F}}_\text{BA}\Delta\text{t}$ ...(ii) Here time $\Delta\text{t}$ the time for which two bodies A and B are in contact and interact, is same for both the forces. Moreover, from third law of motion $\vec{\text{F}}_\text{AB}=\vec{\text{F}}_\text{BA}$ Hence, adding (i) and (ii), we obtain $(\vec{\text{p}'}-\vec{\text{p}}_\text{A})+(\vec{\text{p}'}_\text{B}-\vec{\text{p}}_\text{B})\text{s}$ $=\vec{\text{F}}_\text{AB}\Delta\text{t}+\vec{\text{F}}_\text{BA}\Delta\text{t}$ $=-\vec{\text{F}}_\text{BA}\Delta\text{t}+\vec{\text{F}}_\text{BA}\Delta\text{t}=0$ $\Rightarrow\vec{\text{p}'}_\text{A}+\vec{\text{p}'}_\text{B}=\vec{\text{p}}_\text{A}+\vec{\text{p}}_\text{B}$ Which shows that the total final momentum of the isolated system is exactly same as its initial momentum. Thus, it is proved that total momentum of an isolated system remains conserved
View full question & answer→Question 793 Marks
A box of wood is placed on a 30° slope. If the coefficient of friction be 0.1, what is the downward acceleration of the wooden box?
Answer
Here $\theta = 30^{\circ} \text{ and } \mu = 0.1$
As shown in following figure, net accelerating force along the inclined plane is
$\text{F = m g} \sin \theta - f = \text{m g} \sin \theta - \mu \text{N}$
$= \text{m g}\sin \theta - \mu \text{ mg}\cos \theta$
But F = ma
$\Rightarrow \text{a} = \text{g}(\sin \theta - \mu \cos \theta ) $
$=9.8 \times (\sin 30^ \circ - 0.1 \times \cos 30^\circ) \simeq 4 \text{ms}^{-2}$ View full question & answer→Question 803 Marks
The displacement vector of a particle of mass m is given by $\text{r}\text{(t})=\hat{\text{i}}\text{A}\cos\omega\text{t}+\hat{\text{j}}\text{B}\sin\omega\text{t}.$Show that the trajectory is an ellipse.
AnswerThe Main concept used: To plot the graph ( r - t ) or trajectory we relate x and y coordinates. $\vec{\text{r}}\text{(t)}=\hat{\text{i}}\text{A}\cos\omega\text{t}+\hat{\text{j}}\text{B}\sin\omega\text{t}$ $\text{x}=\text{A}\cos\omega\text{t}$ and $\text{y}=\sin\omega\text{t}$ $\frac{\text{x}}{\text{A}}=\cos\omega\text{t}\ ...(\text{i})\ \frac{\text{y}}{\text{}B}=\sin\omega\text{t}\ ...(\text{ii})$ Squaring and adding (i), (ii) $\frac{\text{x}^2}{\text{A}^2}+\frac{\text{y}^2}{\text{B}^2}=\cos^2\omega\text{t}+\sin^2\omega\text{t}$ $\frac{\text{x}^2}{\text{A}^2}+\frac{\text{y}^2}{\text{B}^2}=1$ it is the equation of an ellipse. So the trajectory is an ellipse.
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A particle moves in a circle of radius $20cm$. Its linear speed at any time is given by $v = 2t$ where v is in m/s and t is in seconds. Find the radial and tangential accelerations at $t = 3$ seconds and hence calculate the total acceleration at this time.
AnswerThe linear speed at 3 seconds is $v = 2 \times 3 = 6m/s$ The radial acceleration at 3 seconds $=\frac{\text{v}^2}{\text{r}}=\frac{6\times6}{0.2=180}=180\text{m/s}^2$ The tangential acceleration is given by $\frac{\text{dv}}{\text{dt}}=2,$ since $\text{v}=2\text{t}$
$\therefore$ tangential acceleration is $2/s^2$. Total acceleraton $\sqrt{\text{a}^2_\text{r}+\text{a}^2_\text{t}}=\sqrt{180^2+2^2}$
$=\sqrt{32400+4}=\sqrt{32404}\text{ms}^{-2}.$
View full question & answer→Question 823 Marks
A particle of mass $100mg$ is moving in a circular vertical path of radius $2m$. The particle is just 'looping the loop'. What is the speed of particle and the tension in the string at the highest point of the circular path? ($g = 10ms^{-2}$)
AnswerHere, m = 100g = 0.1kg r = 2m
$\therefore$ Minimum speed of the particle at the highest point for just looping the loop $=\sqrt{\text{gr}}=\sqrt{10\times2}=\sqrt{20}=4.47\text{ms}^{-1}$ Tension in the string at highest point $=\frac{\text{mV}^2_1}{\text{r}}-\text{5mg}$ $=\text{5mg}-\text{5mg}=0$
View full question & answer→Question 833 Marks
A man weighing $60kg$ is sitting in a lift which is moving vertically with an acceleration of $2ms^{-2}$. Prove that the reaction on the base of the lift is greater when it is ascending than when it is descending. (Given $g = 9.8ms^{-2}$).
AnswerWe know, when the lift accelerates upward, the reaction $R$, on the base is given by $R_1=M g+M a=M(8+a) \Rightarrow R_1=$ $60(9.8+2)=60 \times 11.8=708 \mathrm{~N}$ When the lift accelerates downward with acceleration a, the reaction $R$, on the base given by $R_2=M g-M a=M(g-a) \Rightarrow R_2=60(9.8-2)=60 \times 7.8=468 \mathrm{~N}$ Therefore, $R_1>R_2$.
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The centre of gravity of a loaded taxi is 1.5m above the ground, and the distance between wheels is 2m. What is the maximum speed with which it can go round an unbanked curve of radius 100m without being turned upside down? What minimum value of coefficient of friction is needed at this speed?
AnswerFor rotational equilibrium, $\frac{\text{mv}^2}{\text{r}}\times\text{h}=\text{mg}(\text{x})$Given: 2x = 2m, h = 1.5m
$\therefore\ \text{v}=\sqrt{\frac{\text{gxr}}{\text{h}}}=\sqrt{\frac{9.8\times1\times100}{1.5}}$ $=25.56\text{ ms}^{-1}$ $\mu=\frac{\text{v}^2}{\text{rg}}=\frac{(25.56)^2}{100\times9.8}=0.67$
View full question & answer→Question 853 Marks
A body of mass m is suspended by two strings making angles $\alpha$ and $\beta$ with the horizonal as shown in Fig. Calculate the tensions in the two strings.
AnswerConsidering components of tensions $T_1$ and $T_2$ along the horizontal and vertical directions, We have, $-\text{T}_1\cos\alpha+\text{T}_2\cos\beta=0$
$\text{T}_1\cos\alpha=\text{T}_2\cos\beta\dots\text{(i)}$ and $\text{T}_1\sin\alpha+\text{T}_2\sin\beta=\text{mg}...\text{(ii)}$ From (i) $\text{T}_2=\frac{\text{T}_1\cos\alpha}{\cos\beta}$ and substituting it in (ii), we get $\text{T}_1\sin\alpha+\Big(\frac{\text{T}_1\cos\alpha}{\cos\beta}\Big)\sin\beta$
$=\text{mg}$ or $\text{T}_1\Big(\frac{\sin\alpha\cos\beta+\cos\alpha\sin\beta}{\cos\beta}\Big)=\text{mg}$ or $\text{T}_1\frac{\sin(\alpha+\beta)}{\cos\beta}=\text{mg}$
$\Rightarrow\text{T}_1=\frac{\text{mg}\cos\beta}{\sin(\alpha+\beta)}$ and hence $\text{T}_2=\frac{\text{T}_1\cos\alpha}{\cos\beta}$
$=\frac{\text{mg}\cos\beta}{\sin(\alpha+\beta)}.\frac{\cos\alpha}{\cos\beta}$
$=\frac{\text{mg}\cos\alpha}{\sin(\alpha+\beta)}$
View full question & answer→Question 863 Marks
Prove that Newton's second law of motion is the real law of motion.
AnswerNewton's second law is the real law of motion, as it can explain both first and third laws of motion.
- First law can be explained by using the second law.
$\because\ \text{Force, }\vec{\text{F}}=\text{m}\vec{\text{a}}$ if external force is equal to zero then $\vec{\text{a}}=0$ since $\text{m}\neq0$
$\therefore\ \vec{\text{v}}-\vec{\text{u}}=0\Rightarrow\ \vec{\text{v}}=\vec{\text{u}}$
It shows that if an object is at rest, it will remain at rest and if it is moving, then it will continue to move in uniform motion unless some external force is applied. This is Newton's first law of motion.
- Third law can also be explained by the second law. Consider an isolated system of two bodies A and B. Suppose the two bodies interact mutually with each other.
Let $\vec{\text{F}}_{\text{AB}}=\frac{\vec{\text{dp}}_\text{A}}{\text{dt}}=$ force on A exerted by B (Newton's Second law)
$\vec{\text{F}}_{\text{BA}}=\frac{\vec{\text{dp}}_\text{B}}{\text{dt}}=$ force on B exerted by A, then
$\vec{\text{F}}_\text{AB}+\vec{\text{F}}_\text{BA}=\frac{\vec{\text{dp}}_\text{A}}{\text{dt}}+\frac{\vec{\text{dp}}_\text{B}}{\text{dt}}$
$=\frac{\text{d}}{\text{dt}}(\vec{\text{p}}_\text{A}+\vec{\text{p}}_\text{B})$
Since linear momentum is conserved. therefore
$\frac{\text{d}}{\text{dt}}(\vec{\text{p}}_\text{A}+\vec{\text{p}}_\text{B})=0$ $\Big[\because\vec{\text{P}}_\text{A}+\vec{\text{P}}_\text{B}=\text{const}\Big]$
$\therefore\ \vec{\text{F}}_\text{AB}+\vec{\text{F}}_\text{BA}=0$
or $\vec{\text{F}}_\text{AB}=-\vec{\text{F}}_\text{BA}$ (Newton's Third law) View full question & answer→Question 873 Marks
A trolley of mass $20kg$ rests on a horizontal surface. A massless string tied to the trolley passes over a frictionless pulley and a load of $5kg$ is suspended from other end of string. If coefficient of kinetic friction between trolley and surface be $0.1$, find the acceleration of trolley and tension in the string. (Take $g = 10ms^{-2}$).
AnswerThe free body diagram has been shown in Fig. below, Here, $\text{M}=20\text{kg},\text{m}=5\text{kg}$ and $\mu_\text{k}=0.1$ Here net pulling force, $\text{F}=\text{mg}-\text{f}_\text{k}=\text{mg}-\mu_\text{k}.\text{N}$
$=\text{mg}-\mu_\text{k}.\text{Mg}=5\times10-0.1\times20\times10$
$=50-20=30\text{N}$
$\therefore$ Acceleration of the system $\text{a}=\frac{\text{F}}{(\text{m}+\text{M})}$
$=\frac{30\text{N}}{(5+20)\text{kg}}=1.2\text{ms}^{-2}$
$\therefore$ Tension in string $T = mg - ma = 5 \times 10 - 5 \times 10 - 5 \times 1.2 = 50 - 6 = 44N$.
View full question & answer→Question 883 Marks
When walking on ice, one should take short steps rather than long steps. Why?
Answer
Let R represent the reaction offered by the ground. The vertical component $\text{R}\cos\theta$will balance the weight of the person and the horizontal component $\text{R}\sin\theta$ will help the person to walk forward. Now, normal reaction $=\text{R}\cos\theta$ Friction force $=\text{R}\sin\theta$ Coefficient of friction, $\mu=\frac{\text{R}\sin\theta}{\text{R}\cos\theta}=\tan\theta$ In a long step, $\theta$ is more. So tan $\theta$ is more. But µ has a fixed value. So, there is danger of slipping in a long step. View full question & answer→Question 893 Marks
What is the need for banking of road? Write the expression for the maximum speed with which a vehicle can safely negotiate a curved road banked at an angle $\theta.$ The coefficient of friction between the wheels and the road is $\mu.$
AnswerNeed for banking of road: Banking of road compensate necessary centripetal force and reduce wear and tear of the tyres. When a curved road is unbanked, force of friction between the tyres and the road provides the necessary centripetal force. Friction has to be increased suitably, that cause wear and tear of the tyres.
From the force acting on the vehicle in a banked curve $(\theta).$ $\text{N}\cos\theta-\text{F}_\text{f}\sin\theta=\text{mg}$ $\text{N}\sin\theta+\text{F}_\text{f}\cos\theta=\frac{\text{mv}^2}{\text{r}}$ $\text{F}_\text{f}=\mu\text{N}$ Dividing the equations, we have $\frac{\text{V}^2}{\text{rg}}=\frac{\text{N}\sin\theta+\mu\text{N}\cos\theta}{\text{N}\cos\theta-\mu\text{N}\sin\theta}$ [Dividing each term of right side by $\text{N}\cos\theta]$$\text{V}^2=\text{rg}\Big(\frac{\tan\theta+\mu}{1-\mu\tan\theta}\Big)$
$\text{V}=\sqrt{\text{rg}\Big(\frac{\tan\theta+\mu}{1-\mu\tan\theta}\Big)}$
This is the maximum speed on a banked road. View full question & answer→Question 903 Marks
A cyclist is riding with a speed of $27km/ h$ as he approaches a circular turn (on the road) of radius $80m$, he applies brakes and reduces his speed at the constant rate $0.50m/ s^2$. Find the magnitude and direction of the net acceleration of the cyclist on circular turn?
AnswerSpeed of cyclist 'v' = 27km/ h $=27\times\frac{5}{18}\text{ms}^{-1} = 7.5ms^{-1}, r = 80m$ Centripetal acceleration ‘$a_c$' which changes the direction of linear velocity and acts along the radius towards the centre of circular path.
Also, $a_T$ = Tangential acceleration which acts along the tangential to the circular path. Resultant acceleration is given by $\text{a}=\sqrt{\text{a}^2_\text{c}+\text{a}^2_\text{T}}$ $=\sqrt{(0.7)^2+(0.5)^2}=0.86\text{ms}^{-2}$ $\tan\beta=\frac{\text{a}_\text{c}}{\text{a}_\text{T}}=\frac{0.7}{0.5}=1.4$ $\beta=\tan^{-1}(1.4)\cong54^\circ28'$ View full question & answer→Question 913 Marks
A hammer of mass $1kg$ moving with a speed of $6ms^{-1}$ strikes a wall and comes to rest in $0.1s$. Calculate:
- The impulse of force.
- The retardation of the hammer.
- The retarding force that stops the hammer.
AnswerMass of the hammer $m = 1kg$ Initial velocity, $u = 6ms^{-1}$, final velocity, $v = 0$ and $t = 0.1s$
- Impulse = Ft = m(v - u)
= 1(0 - 6) = -6Ns
- Retarding force that stops the hammer,
$\text{F}=\frac{\text{Impulse}}{\text{time}}=\frac{6}{0.1}=60\text{N}$
- Retardation of the hammer
$=\frac{\text{F}}{\text{m}}=\frac{60}{1}=60\text{ ms}^{-2}$ View full question & answer→Question 923 Marks
In the system of three blocks $A , B$ and C shown in figure,
(i) how large a force $F$ is needed to give the blocks an acceleration of $3 m / s ^2$, if the coefficient of friction between blocks and table is $0.27$
(ii) how large a force does the block $A$ exert on the block $B$ ?
AnswerIrt a be the acceleration of the system to right. All the three frictional forces $f_1 = \mu \text{m}_1\text{g}, f_2 = \mu \text{m}_2\text{g} \text{ and }f_3 = \mu \text{m}_3\text{g}$ will be directed to the left as the motion of bodies is to the right.
Hence, for the whole system
$\text{F} - \mu \text{m}_1\text{g} - \mu \text{m}_2\text{g} - \mu \text{m}_3\text{g} = (\text{m}_1 + \text{m}_2 + \text{m}_3) \text{a}$
$\text{F} = (\text{m}_1 + \text{m}_2 + \text{m}_3) (\text{a} + \mu \text{g}) $
$= (7.5 + 2 + 1)(3 + 0.2 \times 9.8) = 22.3N $
The force exerted by the 1.5kg block on the 2kg block = $\text{F} - \text{m}_1(\text{a} + \mu \text{g})$
$= 22.3 - 1.5 (3 + 0.2 \times 9.8) = 22.3 - 7.44 = 14N$ View full question & answer→Question 933 Marks
State the law of conservation of momentum. Establish the same for a 'n' body system.
AnswerWhen no external force acts on a system the momentum will remain conserved. Consider a system of a n bodies of masses $\text{m}_1,\text{m}_2,\text{m}_3....,\text{m}_\text{n}.$If $\text{p}_1,\text{p}_2,\text{p}_3,...\text{p}_\text{n}$ are the momentum associated then, the rate of change of momentum with the system, $\frac{\text{dp}}{\text{dt}}=\frac{\text{dp}_1}{\text{dt}}+\frac{\text{dp}_2}{\text{dt}}+\frac{\text{dp}_3}{\text{dt}}\\+...+\frac{\text{dp}_\text{n}}{\text{dt}}$ $\frac{\text{d}}{\text{dt}}(\text{p}_1+\text{p}_2+\text{p}_3+...+\text{p}_\text{n})$ If no external force acts, $\frac{\text{dp}}{\text{dt}}=0$ $\therefore$ p = constant, i.e., $\text{p}_1+\text{p}_2+\text{p}_3+....+\text{p}_\text{n}=$ constant.
View full question & answer→Question 943 Marks
A force of 400N acting horizontal pushes up a $20kg$ block placed on a rough inclined plane which makes an angle of $45^\circ$ with the horizontal. The acceleration experienced by the block is $0.6m/ s^2$. Find the coefficient of sliding friction between the box and incline.
Answer
The horizontally directed force $400N$ and weight $20kg$ of the block are resolved into two mutually perpendicular components,
parallel and perpendicular to the plane as shown. $\text{N}=20\text{g}\cos45^\circ+400\sin45^\circ=421.4\text{N}$
The frictional force experienced by the block F $=\mu\text{N}=\mu\times421.4=421.4\mu\text{N}$.
As the accelerated motion is taking place up the plane. $400\cos45^\circ-20\text{g}\sin45^\circ-\int=20\text{a}$
$\frac{400}{\sqrt{2}}-\frac{20\times9.8}{\sqrt{2}}-421.4\mu=20\text{a}=20$
$\mu=\Big(\frac{400}{\sqrt{2}}-\frac{196}{\sqrt{2}}-12\Big)\times\frac{1}{421.4}$
$=\frac{282.8-138.6-12}{421.4}=0.3137$
The coefficient of sliding friction between the block and the incline $= 0.3137.$ View full question & answer→Question 953 Marks
A cricket ball of mass $150g$ is moving with a velocity of $12ms^{-1}$, and is hit by a bat, so that the ball is turned back with a velocity of $20ms^{-1}$. The force of the blow acts for $0.01s$ on the ball. Find the average force exerted by the bat on the ball.
AnswerMass of the ball, $\mathrm{m}=0.15 \mathrm{~kg}$ Initial velocity, $u=12 \mathrm{~ms}^{-1}$ final velocity $v=-20 \mathrm{~ms}^{-1}, t=0.01 \mathrm{~s}$ Initial momentum of the ball $=0.15 \times 12=1.8 \mathrm{~kg} \mathrm{~ms}^{-1}$ Final momentum of the ball $=0.15 \times(-20)=-3.0 \mathrm{~kg} \mathrm{~ms}^{-1}$ Change in momentum $=$
$4.8 \mathrm{~kg} \mathrm{~ms}^{-1}$ Average force exerted by the bat on the ball $=\frac{4.8 \mathrm{~kg} \mathrm{~ms}^{-1}}{0.01 \mathrm{~s}}=480 \mathrm{~N}$
View full question & answer→Question 963 Marks
What is the acceleration of the block and trolley system shown in a Fig. 4.12(a), if the coefficient of kinetic friction between the trolley and the surface is 0.04 ? What is the tension in the string? (Take $g=$ $10 m s ^{-2}$ ). Neglect the mass of the string.
AnswerAs the string is inextensible, and the pully is smooth, the $3 kg$ block and the $20 kg$ trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. 4.12(b)),
$
30-T=3 a
$
Apply the second law to motion of the trolley (Fig. 4.12(c)),
Now
$
T-f_{ k }=20 a \text {. }
$
Here
$
\begin{aligned}
f_k & =\mu_k N, \\
\mu_k & =0.04, \\
N & =20 \times 10 \\
& =200 N .
\end{aligned}
$
Thus the equation for the motion of the trolley is
$
T-0.04 \times 200=20 a
$
Or $T-8=20 a$.
These equations give $a=\frac{22}{23} m s ^{-2}=0.96 m s ^{-2}$ and $T=27.1 N$.
View full question & answer→Question 973 Marks
See Fig. 4.11. A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
AnswerThe forces acting on a block of mass $m$ at rest on an inclined plane are (i) the weight $mg$ acting vertically downwards (ii) the normal force $N$ of the plane on the block, and (iii) the static frictional force $f_{ s }$ opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight $mg$ along the two directions shown, we have
$
m g \sin \theta=f_s, \quad m g \cos \theta=N
$
As $\theta$ increases, the self-adjusting frictional force $f_{ s }$ increases until at $\theta=\theta_{\max }, f_{ s }$ achieves its maximum value, $\left(f_{ s }\right)_{\max }=\mu_{ s } N$.
Therefore,
$
\tan \theta_{\max }=\mu_{ s } \text { or } \theta_{\max }=\tan ^{-1} \mu_{ s }
$
When $\theta$ becomes just a little more than $\theta_{\max }$, there is a small net force on the block and it begins to slide. Note that $\theta_{\max }$ depends only on $\mu_{ s }$ and is independent of the mass of the block.
For
$
\begin{aligned}
\theta_{\text {max }} & =15^{\circ} \\
\mu_s & =\tan 15^{\circ} \\
& =0.27
\end{aligned}
$
View full question & answer→Question 983 Marks
See Fig. $4.8.$ A mass of $6 \ kg$ is suspended by a rope of length $2 m$ from the ceiling. A force of $50 N$ in the horizontal direction is applied at the midpoint $P$ of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium? $($Take $g=10 m s ^{-2} )$. Neglect the mass of the rope.
AnswerFigures $4.8(b)$ and $4.8(c)$ are known as free$-$body diagrams. Figure $4.8(b)$ is the free-body diagram of $W$ and Fig. $4.8(c)$ is the free-body diagram of point $P$.
Consider the equilibrium of the weight $W$. Clearly, $T_2=6 \times 10=60 N$.
Consider the equilibrium of the point $P$ under the action of three forces - the tensions $T_1$ and $T_2$, and the horizontal force $50 N$. The horizontal and vertical components of the resultant force must vanish separately :
$T_1 \cos \theta=T_2=60 N$
$T_1 \sin \theta=50 N$
which gives that
$\tan \theta=\frac{5}{6} \text { or } \theta=\tan ^{-1}\left(\frac{5}{6}\right)=40^{\circ}$
Note the answer does not depend on the length of the rope $($assumed massless$)$ nor on the point at which the horizontal force is applied.
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