Q: State two advantages and disadvantages of friction.

Advantages of friction: \$1.\$ Friction enables us to walk freely.\$2.\$ It helps to support ladder against wall.\$3.\$ It becomes possible to transfer one form of energy to another.\$4.\$ Objects can be piled up without slipping.\$5.\$ Breaks of vehicles work due to friction.Disadvantages of friction:\$1.\$It always resists the motion, so extra energy is required to overcome it.\$2.\$It causes wear and tear of machines.\$3.\$It decreases the life expectancy of moving parts of vehicles.\$4.\$Since friction is very useful in some cases while harmful in some cases, friction is called a necessary evil.

Q: A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Frequency \$f=\dfrac{14}{25}Hz\$Angular velocity \$\omega =2\pi f= \dfrac{2 \times 14\pi }{25}rad/s=3.52rad/s\$Acceleration is \$a=\omega ^{2}r=3.52^{2}\times 0.8=9.9m/s^{2}\$The direction is along the radius at every point towards the centre.

Q: A body of mass \$5\$ kg is acted upon by two perpendicular forces \$8\$ N and \$6\$ N. Given the magnitude and direction of the acceleration of the body.

\$\textbf{Step 1: Drawing FBD}\$ \$\textbf{[Refer Fig. 1]}\$\$\textbf{Step 2: Finding Resultant Force}\$ \$\textbf{[Refer Fig. 2]}\$\$\large \text{Magnitude of Resultant:}\$Since \$F_1\$ and \$F_2\$ are perpendicular\$\therefore \$ Magnitude of Resultant force on body is given by \$R = \sqrt{F_1^2 + F_2^2}\$ \$=\sqrt{8^2+6^2} N\$ \$=10N\$\$\large \text{Direction of Resultant:}\$ \$\theta\ = \tan^{-1} \dfrac{F_2}{F_1}\$ \$\textbf{[Refer Fig. 2]}\$ \$=\tan^{-1} \dfrac{6}{8}\$ \$=36.9^o\$ \$\approx 37^o\$ \$\textbf{Step 3: Applying Newton's second Law for finding acceleration}\$ \$\textbf{[Refer Fig. 3]}\$ \$F_{net} = ma\$ \$\Rightarrow\$ \$10 = 5a\$ \$\Rightarrow\$ \$|\vec{a}| = 2m/s^2\$Hence magnitudes of acceleration is \$2m/s^2\$ directed at \$37^o\$ from the 8 N force.

Q: A block of mass \$2\ kg\$ rests on a rough inclined plane making an angle of \${30}^{o}\$ with the horizontal. The coefficient of static friction between the block and the plane is \$0.7\$. The frictional force on the block is

The force applied on the body that is on the inclined plane is given as, \$F = mg\sin \theta \$ \$F = 2 \times 9.8 \times \sin 30^\circ \$ \$ = 9.8\;{\rm{N}}\$ The limiting friction force between the block and the inclined plane is given as, \$f = \mu mg\cos \theta \$ \$f = 0.7 \times 2 \times 9.8\cos 30^\circ \$ \$ = 11.88\;{\rm{N}}\$ Since the limiting friction force is greater than the force that tends to slide the body. Thus, the body will be at rest and the force of friction on the block is \$9.8\;{\rm{N}}\$.

Q: A machine gun fires a bullet of mass \$40g\$ with a velocity \$1200m/s\$. The man holding it can exert a maximum force of \$144N\$ on the gun. How many bullets can be fired per second at the most?

\$\textbf{Step 1: Newton's Second Law,}\$From Newton's second law \$F = \dfrac{dP}{dt}\$ \$....(1)\$Where, \$P=mv\$ is the momentum \$\textbf{Step 2: Total bullets fired per sec}\$Let n be the total number of bullets firedThe force exerted by man must balance the change in momentum of the bullets per second. \$\therefore P = nmv\$ \$;\$ \$m\$ and \$v\$ are constant (Given) \$\Rightarrow F = \dfrac{d}{dt} (nmv)\$ \$\Rightarrow\$ \$\dfrac{dn}{dt} = \dfrac{F}{mv} = \dfrac{144 (N)}{40 \times 10^{-3}(kg) \times 1200(m/s)} = 3\$Hence total number of bullets fired per second i.e., \$\dfrac{dn}{dt} = 3\$.Hence, Option \$B\$ is correct

Q: Mention the methods used to decrease friction

\$\textbf{Cause of Friction: }\$To Identify methods of reducing friction, we first have to understand what is the cause of friction. There are two reasons behind friction:Whenever two surfaces come in contact, then the temporary bond's start forming between them, which resist the movement between the surfaces. Higher the area of contact, larger the number of bonds, higher the value of friction force.The abnormality/Roughness of the surface, more rough the surface is, higher the friction.Value of friction force is proportional to the Normal contact force. Higher the Normal force, larger is the friction\$\textbf{Methods for Reduction friction: }\$Make the surfaces little more smoother. Rough surfaces produce more friction and smooth surfaces reduce friction. Note: Making surface highly smooth will also increase the friction, as large number of bonds will form then.Lubrication is another way to make a surface smoother. A lubricant is a slippery substance designed to reduce the friction between surfaces. Make the object more streamlined. A streamline shape is one that allows air or water to flow around it easily, offering the least resistance.Reduce the Normal force acting between the surfaces in contact. The stronger the Normal force acting on the surfaces, the higher the friction.Reduce the contact between the surfaces, so that less number of bonds will be formed. Spheres are the best shape for reducing friction because very little of a spherical object is in contact with the other surface.

Question 1

State two advantages and disadvantages of friction.

Accepted Answer

Advantages of friction: $1.$ Friction enables us to walk freely.$2.$ It helps to support ladder against wall.$3.$ It becomes possible to transfer one form of energy to another.$4.$ Objects can be piled up without slipping.$5.$ Breaks of vehicles work due to friction.Disadvantages of friction:$1.$It always resists the motion, so extra energy is required to overcome it.$2.$It causes wear and tear of machines.$3.$It decreases the life expectancy of moving parts of vehicles.$4.$Since friction is very useful in some cases while harmful in some cases, friction is called a necessary evil.

Question 2

A stone tied to the end of a string 80 cm long is whirled in a horizontal circle with a constant speed. If the stone makes 14 revolutions in 25 s, what is the magnitude and direction of acceleration of the stone ?

Accepted Answer

Frequency $f=\dfrac{14}{25}Hz$Angular velocity $\omega =2\pi f= \dfrac{2 \times 14\pi }{25}rad/s=3.52rad/s$Acceleration is $a=\omega ^{2}r=3.52^{2}\times 0.8=9.9m/s^{2}$The direction is along the radius at every point towards the centre.

Question 3

A body of mass $5$ kg is acted upon by two perpendicular forces $8$ N and $6$ N. Given the magnitude and direction of the acceleration of the body.

Accepted Answer

$\textbf{Step 1: Drawing FBD}$      $\textbf{[Refer Fig. 1]}$$\textbf{Step 2: Finding Resultant Force}$      $\textbf{[Refer Fig. 2]}$$\large \text{Magnitude of Resultant:}$Since $F_1$ and $F_2$ are perpendicular$\therefore $ Magnitude of Resultant force on body is given by                  $R = \sqrt{F_1^2 + F_2^2}$                     $=\sqrt{8^2+6^2} N$ $=10N$$\large \text{Direction of Resultant:}$                $\theta\ = \tan^{-1} \dfrac{F_2}{F_1}$        $\textbf{[Refer Fig. 2]}$                                           $=\tan^{-1} \dfrac{6}{8}$ $=36.9^o$ $\approx 37^o$ $\textbf{Step 3: Applying Newton's second Law for finding acceleration}$      $\textbf{[Refer Fig. 3]}$               $F_{net} = ma$           $\Rightarrow$      $10 = 5a$           $\Rightarrow$       $|\vec{a}| = 2m/s^2$Hence magnitudes of acceleration is $2m/s^2$ directed at $37^o$ from the 8 N force.

Question 4

Three blocks with masses m, 2m and 3m are connected by strings, as shown in the figure. After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m?(g is the acceleration due to gravity)

Accepted Answer

All blocks are moving with constant velocity so net force on all blocks are zero.

Question 5

A block of mass $2\ kg$ rests on a rough inclined plane making an angle of ${30}^{o}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.7$. The frictional force on the block is

Accepted Answer

The force&#10;applied on the body that is on the inclined plane is given as,&#10;&#10;$F = mg\sin \theta $&#10;&#10;$F = 2 \times 9.8 \times \sin 30^\circ $&#10;&#10;$ = 9.8\;{\rm{N}}$&#10;&#10;The limiting&#10;friction force between the block and the inclined plane is given as,&#10;&#10;$f = \mu mg\cos \theta $&#10;&#10;$f = 0.7 \times 2 \times 9.8\cos 30^\circ $&#10;&#10;$ = 11.88\;{\rm{N}}$&#10;&#10;Since the&#10;limiting friction force is greater than the force that tends to slide the body.&#10;&#10;Thus, the body&#10;will be at rest and the force of friction on the block is $9.8\;{\rm{N}}$.

Question 6

A machine gun fires a bullet of mass $40g$ with a velocity $1200m/s$. The man holding it can exert a maximum force of $144N$ on the gun. How many bullets can be fired per second at the most?

Accepted Answer

$\textbf{Step 1: Newton's Second Law,}$From Newton's second law                    $F = \dfrac{dP}{dt}$  $....(1)$Where, $P=mv$ is the momentum           $\textbf{Step 2: Total bullets fired per sec}$Let n be the total number of bullets firedThe force exerted by man must balance the change in momentum of the bullets per second.     $\therefore P = nmv$        $;$    $m$ and $v$ are constant    (Given)           $\Rightarrow F = \dfrac{d}{dt} (nmv)$           $\Rightarrow$     $\dfrac{dn}{dt} = \dfrac{F}{mv} = \dfrac{144 (N)}{40 \times 10^{-3}(kg) \times 1200(m/s)} = 3$Hence total number of bullets fired per second i.e., $\dfrac{dn}{dt} = 3$.Hence, Option $B$ is correct

Question 7

What is banking of roads? Obtain an expression for the maximum safety speed of a vehicle moving along a curved horizontal road.

Accepted Answer

Banking of roads : To avoid risk of skidding as well as to reduce the wear and tear of the car tyres, the road surface at a bend is tilted inwards, i.e., the outer side of road is raised above its inner side. This is called 'banking of roads'.Consider a car taking a left turn along a road of radius $r$ banked at an angle $\theta$ for a designed optimum speed $V$. Let $m$ be the mass of the car. In general, the forces acting on the car are:(a) Its weight $\vec {mg}$, acting vertically down(b) The normal reaction of the road $\vec {N}$, perpendicular to the road surface(c) The frictional force $\vec {f_{s}}$ along the inclined surface of the road.Resolve $\vec {N}$ and $\vec {f_{s}}$ into two perpendicular components $N\cos \theta$ vertically up and $\vec {f_{s}} \sin \theta$ vertically down, $N\sin \theta$ and $\vec {f_{s}} \cos \theta$ horizontally towards the centre of the circular path.If $v_{max}$ is the maximum safe speed without skidding.$\dfrac {mv_{max}^{2}}{r} = N\sin \theta + f_{s}\cos \theta$               $= N\sin \theta + \mu_{s} N\cos \theta$$\dfrac {mv_{max}^{2}}{r}= N(\sin \theta + \mu_{s} \cos \theta) .... (1)$and $N\cos \theta = mg + f_{s} \sin \theta$               $= mg + \mu_{s} N\sin \theta$   $\therefore mg = N(\cos \theta - \mu_{s}\sin \theta) ... (2)$Dividing eq. $(1)$ by eq. $(2)$,$\dfrac {mv_{max}^{2}}{r.mg} = \dfrac {N(\sin \theta + \mu_{s}\cos \theta)}{N(\cos \theta - \mu_{s}\sin \theta)}$$\therefore \dfrac {v_{max}^{2}}{rg} = \dfrac {\sin \theta + \mu_{s}\cos \theta}{\cos \theta - \mu_{s}\sin \theta} = \dfrac {\tan \theta + \mu_{s}}{1 - \mu_{s}\tan \theta}$$\therefore v_{max} = \sqrt {\dfrac {rg(\tan \theta + \mu_{s})}{1 - \mu_{s}\tan \theta}} ...,. (3)$This is the expression for the maximum safe speed on a banked road.At the optimum speed, the friction between the car tyres and the road surface is not called into play. Hence, by setting $\mu_{s} = 0$ in eq. $(3)$, the optimum speed on a banked circular road is$v = \sqrt {rg \tan \theta} ... (4)$$\therefore \tan \theta = \dfrac {v^{2}}{rg}$ or $\theta = \tan^{-1}\left (\dfrac {v^{2}}{rg}\right )$From this eq. we see that $\theta$ depends upon $v, r$ and $g$. The angle of banking is independent of the mass of a vehicle negotiating the curve.

Question 8

Mention the methods used to decrease friction

Accepted Answer

$\textbf{Cause of Friction: }$To Identify methods of reducing friction, we first have to understand what is the cause of friction. There are two reasons behind friction:Whenever two surfaces come in contact, then the temporary bond's start forming between them, which resist the movement between the surfaces. Higher the area of contact, larger the number of bonds, higher the value of friction force.The abnormality/Roughness of the surface, more rough the surface is, higher the friction.Value of friction force is proportional to the Normal contact force. Higher the Normal force, larger is the friction$\textbf{Methods for Reduction friction: }$Make the surfaces little more smoother. Rough surfaces produce more friction and smooth surfaces reduce friction. Note: Making surface highly smooth will also increase the friction, as large number of bonds will form then.Lubrication is another way to make a surface smoother. A lubricant is a slippery substance designed to reduce the friction between surfaces. Make the object more streamlined. A streamline shape is one that allows air or water to flow around it easily, offering the least resistance.Reduce the Normal force acting between the surfaces in contact. The stronger the Normal force acting on the surfaces, the higher the friction.Reduce the contact between the surfaces, so that less number of bonds will be formed. Spheres are the best shape for reducing friction because very little of a spherical object is in contact with the other surface.

Question 9

State and prove the law of conservation of momentum.

Accepted Answer

Law of conservation of momentum states that total momentum of system remains conserved in the absence of external force.  Proof: Consider a body of mass m1  moving with velocity  U1,  striking against another body of mass m2  moving with velocity  U2. Let the two bodies remain in contact with each other for a small interval &#34;delta t&#34;. Let F12  be the average force exerted by mass m1 on m2, and let F21  be the force on m2 due to m1. Let v1 and v2 be the velcoities of two bodies after collision.Momentum of mass m1 before collision=m1u1Momentum of mass m2 after collision=m2u2Momentum of mass m1 after collision=m1v1By using the definition of impulse, change in momentum of mass m1 is,(refer figure).

Question 10

Two masses $8$ $kg$ and $12$ $kg$ are connected at the two ends of a light inextensible string that goes over a frictionless pulley. Find the acceleration of the masses, and the tension in the string when the masses are released.

Accepted Answer

The given system of two masses and a pulley can be represented as shown in the following figure:Smaller mass, $m_1 = 8$ kgLarger mass,$ m_2 = 12 $kgTension in the string$ = T$Mass $m_2$, owing to its weight, moves downward with acceleration, and mass $m_1$ moves upward.Applying Newton&#8217;s second law of motion to the system of each mass:For mass $m_1$:The equation of motion can be written as:$T &#8211; m_1g = ma $&#8230; (i)For mass m2:The equation of motion can be written as:$m_2g &#8211; T = m2a$ &#8230; (ii)Adding equations (i) and (ii), we get:$(m2 - m1)g = (m1 + m2)a$ $a = \dfrac{(m2 - m1) g}{(m1 + m2) }$    ....(iii)$=\dfrac{ (12 - 8)}{ (12 + 8)} &#215; 10  = \dfrac{ 4 &#215; 10}{  20}  =  2 ms-2$Therefore, the acceleration of the masses is 2 m/s2.Substituting the value of a in equation (ii), we get:$m_2g - T = m_2\dfrac{(m_2 - m_1)g }{ (m_1 + m_2)}$$T = (m2 - \dfrac{(m_2^2 - m1m2)}{(m1 + m2) }g$       $= \dfrac{2m_1m_2g }{(m_1 + m_2)}$$T= 2 &#215; 12 &#215; 8 &#215; 10 / (12 + 8)$$T= 96 N$Therefore, the tension in the string is 96 N.