Important Questions for CBSE Class 11 Physics Chapter 5 - Law of Motion

1. What is the unit of coefficient of friction?

Ans: Since, $Coefficient pf friction = Frictional Force × [Normal Force]^{-1}$ ……(1)

Also, we know that

\[\text{Force}\left( \text{F} \right)\text{ = }\text{Mass}\text{ }\!\!\times\!\!\text{ }\text{acceleration }\!\!~\!\!\text{  = }\text{Mass}\text{ }\!\!\times\!\!\text{ }\text{Velocity}\text{ }\!\!\times\!\!\text{ }{{\text{ }\!\![\!\!\text{ Time }\!\!]\!\!\text{ }}^{\text{ - 1}}}\]

And

${\text{Velocity}}\,{\text{ = }}\,{\text{Displacement}}\,\, \times \,\,{[{\text{ Time }}]^{ - 1}}$

$\therefore $ The dimensions of Force $ = [{\text{M}}] \times \left[ {{\text{L}}{{\text{T}}^{ - 1}}} \right] \times {[{\text{T}}]^{ - 1}} = \left[ {{{\text{M}}^1}\;{{\text{L}}^1}\;{{\text{T}}^{ - 2}}} \right]$             ……(2)

On putting the equation $(2)$ in equation $(1)$ we get,

$\mu  = \left[ {{{\text{M}}^1}\;{{\text{L}}^1}\;{{\text{T}}^{ - 2}}} \right] \times {\left[ {{{\text{M}}^1}\;{{\text{L}}^1}\;{{\text{T}}^{ - 2}}} \right]^{ - 1}} = \left[ {{{\text{M}}^0}\;{{\text{L}}^0}\;{{\text{T}}^0}} \right]$.

Since mass, velocity and time all having zero dimensions. Hence, the coefficient of friction has no units.

2. Name the factor on which coefficient of friction depends?

Ans: The coefficient of friction will mainly depends upon two factor, they are as following:

i. The materials of the surfaces in contact.

ii. The characteristics of the surfaces.

3. What provides the centripetal force to a car taking a turn on a level road?

Ans: The frictional contact between the tyres and the road provides centripetal force.

4. Why is it desired to hold a gun tight to one's shoulder when it is being fired?

Ans: As the gun recoils after shooting, it must be held softly on the shoulder. Here the gun and the shoulder are one mass system, due to this the back kick will be reduced. A gunman must keep his weapon securely against his shoulder when shooting.

5. Why does a swimmer push the water backwards?

Ans: From the Newton's 3rd law of motion, we know that "when one body exerts a force on the other body, the first body experiences a force equivalent in magnitude in the opposite direction of the force exerted". As a result, in order to swim ahead, the swimmer pushes water backward with his hands.

6. Friction is a self-adjusting force. Justify.

Ans: Friction is a self-adjusting force that changes in magnitude from zero to maximum to limit friction.

7. A thief jumps from the roof of a house with a box of weight ${\text{W}}$ on his head. What will be the weight of the box as experienced by the thief during jump?

Ans: The thief is in free fall during the jump. Both he/she and the box will be weightless during that time. So, the weight of the box experience by the thief during the jump will be zero.

So, mathematically it can be written as:

Weight of the box, W = m(g - a) = m(g - g) = 0.

8. Which of the following is scalar quantity? Inertia, force and linear momentum.

Ans: The resistance of a body to its own acceleration is measured by its inertia. As a result, mass becomes a qualitative indicator of inertia. Because mass is a scalar quantity, linear inertia is also a scalar quantity. Hence, inertia will be the scalar quantity among them.

9. Action and reaction forces do not balance each other. Why?

Ans: Because a force of action and response always operates on two separate bodies, action and reaction do not balance each other.

10. If force is acting on a moving body perpendicular to the direction of motion, then what will be its effect on the speed and direction of the body?

Ans: When a force acts in a perpendicular direction on a moving body, the work done by the force is zero.

Since 

$W = F \cdot S\cos \theta$, where $S = 90^\circ$ and $\cos 90^\circ  = 0$, therefore $W = 0$.

As a result, the magnitude of the body's velocity (or speed) will remain unchanged. The direction of motion of the body, however, will be altered.

11. The two ends of spring - balance are pulled each by a force of 10kg.wt. What will be the reading of the balance?

Ans: As the spring balancing is based on the tension in the spring, it gauges weight. Now, if both ends are pulled by a 10kg weight, the tension is 10kg, and the reading will be 10kg.

12. A lift is accelerated upward. Will the apparent weight of a person inside the lift increase, decrease or remain the same relative to its real weight? If the lift is going with uniform speed, then?

Ans: There will be an increase in perceived weight. The apparent weight will stay the same as the true weight if the lift moves at a constant pace.

13. One end of a string of length $l$ is connected to a particle of mass $m$ and the other to a small peg on a smooth horizontal table. If the particle moves in a circle with speed $v$ the net force on the particle (directed towards the centre) is:

i.

ii. $T - \dfrac{{m{v^2}}}{l}$

iii. $T + \dfrac{{m{v^2}}}{l}$

iv. 0

$T$ is the tension in the string. (Choose the correct alternative).

Ans: (i) The tension created in the string provides the centripetal force when a particle attached to a string spins in a circular motion around a centre. As a result, in the given situation, the particle's net force is tension. $T$, i.e.

$F = T = \dfrac{{m{v^2}}}{l}$

Where, $F$ is the net force acting on the particle.

14. If, in Exercise 5.21, the speed of the stone is increased beyond the maximum permissible value, and the string breaks suddenly, which of the following correctly describes the trajectory of the stone after the string breaks:

(a) the stone moves radially outwards,

(b) the stone flies off tangentially from the instant the string breaks,

(c) the stone flies off at an angle with the tangent whose magnitude depends on the speed of the particle ?

Ans: (b) The stone will go in the direction of the velocity at the time the string breaks. The direction of the velocity vector is tangential to the path of the stone at that time, according to Newton's first law of motion. As a result, as soon as the string snaps, the stone will fly off in a tangential direction.

 Solve the Following Questions                            2 Mark

1. Give the magnitude and direction of the net force acting on

(a) A drop of rain falling down with constant speed.

(b) A kite skillfully held stationary in the sky.

Ans: 

(a) As the raindrop is falling with a constant speed, so its acceleration will be $0$. As the force acting on a particle is given by, so the net force acting on the rain drop will be $0$.

(b) As the kite is held stationary, so by Newton's first law of motion, the algebraic sum of forces acting on the kite is zero.

2. Two blocks of masses ${{m}_{1}},{{m}_{2}}$ are connected by light spring on a smooth horizontal surface. The two masses are pulled apart and then released. Prove that the ratio of their acceleration is inversely proportional to their masses.

Ans: Due to inertia, the mass of the two bodies tries to expand, and the acceleration will act in the opposite direction as it shrinks.

So let us assume that the ${F_1}$ and ${F_2}$ be the forces acting in opposite directions due to masses ${m_1}$ and ${m_2}$. 

Thus ${F_1} + {F_2} = 0$

${m_1}{a_1} + {m_2}{a_2} = 0$

${m_1}{a_1} =  - {m_2}{a_2}$

$\dfrac{{{a_1}}}{{{a_2}}} =  - \dfrac{{{m_1}}}{{{m_2}}}$

Hence the above is proved.

3. A shell of mass 0.020 kg is fired by a gun of mass 100 kg. If the muzzle speed of the shell is $80\;{\text{m}}{{\text{s}}^{ - 1}}$ , what is the recoil speed of the gun?

Ans: From the question, we have the shell having mass 0.020 kg and is fired by a gun having the mass 100 kg. We need to find the recoil speed of the gun, when the muzzle speed is given as $80\;{\text{m}}{{\text{s}}^{ - 1}}$.

Since, the momentum before firing $ = 0$

As the momentum after firing = momentum of (bullet + gun). 

Therefore, momentum after firing will be $= {m_b}{v_b} - {m_g}{v_g}$

As per the law of conservation of linear momentum:

$\Rightarrow 0={{m}_{b}}{{v}_{b}}-{{m}_{g}}{{v}_{g}}\text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{  }\!\!~\!\!\text{ }\Rightarrow {{m}_{b}}{{v}_{b}}={{m}_{g}}{{v}_{g}}\text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ }\Rightarrow {{v}_{g}}=\frac{{{m}_{b}}{{v}_{b}}}{{{m}_{g}}}$$\Rightarrow {{v}_{g}}=\frac{{{m}_{b}}{{v}_{b}}}{{{m}_{g}}}\text{ }\!\!~\!\!\text{ }\Rightarrow {{v}_{g}}\text{ }\!\!~\!\!\text{ }=\frac{0.02\times 80}{100}\text{ }\!\!~\!\!\text{ }\Rightarrow {{v}_{g}}=0.016\ \text{m}{{\text{s}}^{-1}}$

4. A force is being applied on a body but it causes no acceleration. What possibilities may be considered to explain the observation?

Ans: (1) If the force is a deforming force, no acceleration is produced.

(2) Internal force is incapable of causing acceleration.

5. Forces of 16N and 12N are acting on a mass of 200kg in mutually perpendicular directions. Find the magnitude of the acceleration produced?

Ans: In the given question, we have the force of $16N$ and $12N$ given and they are acting on a mass of $200kg$ in mutually perpendicular directions. We need to find the magnitude of the acceleration produced.

 $F = \sqrt {F_1^2 + F_2^2 + 2{F_1}{F_2}\cos \theta }$

Since, the forces are in mutually perpendicular directions. Therefore, $\theta = 90^\circ$. Hence, the force will become:

$F = \sqrt {F_1^2 + F_2^2} \quad$

Now substituting the values, we get

                                             \[\text{ }\!\!~\!\!\text{ F=}\left( \sqrt{\text{16}{{\text{)}}^{\text{2}}}\text{+(12}{{\text{)}}^{\text{2}}}} \right)\text{ }\!\!~\!\!\text{ =20}\text{N}\]

Hence, the magnitude of the acceleration will be:

$a=\frac{F}{m}\text{ }\!\!~\!\!\text{ }a=\frac{20}{200}\text{ }\!\!~\!\!\text{ }a=0.1\text{m}{{\text{s}}^{-2}}$

6. An elevator weighs 3000 kg. What is its acceleration when the tension supporting cable is 33000 N. Given that ${g = 9.8 m^{ - 2}}$ .

Ans: From the question, we have an elevator having the weighs given as 3000 kg. We need to find the acceleration, if the tension in the supporting cable is given as 33000 N.

Net upward force on the Elevator $F$ is equal to $=T\text{ }mg\,because\,F=ma$

$\Rightarrow ma = T - mg$

$\Rightarrow T = m(a + g)$

$\Rightarrow T = 33000\;{\text{N}} = 3000(a + 9.8)$

\[a=\frac{33000-3000\times 9.8}{3000}\text{ }\!\!~\!\!\text{  }\!\!~\!\!\text{ }=1.2\ \text{m}{{\text{s}}^{-2}}\]

7. Write two consequences of Newton's second law of motion?

Ans: 

1. It demonstrates that only when force is added to the motion is it accelerated. 

2. It introduces the notion of a body's inertial mass.

8. A bird is sitting on the floor of a wire cage and the cage is in the hand of a boy. The bird starts flying in the cage. Will the boy experience any change in the weight of the cage?

Ans: When the bird begins to fly within the cage, the weight of the bird is no longer felt since the air inside is in direct touch with ambient air, making the cage look lighter.

9. Why does a cyclist lean to one side, while going along curve? In what direction does he lean?

Ans: A cyclist leans while riding along a curve because a component of the ground's natural response supplies him with the centripetal force he needs to turn.

He must lean inwards from his vertical posture, towards the circular path's centre.

10. How does banking of roads reduce wear and tear of the tyres?

Ans: When a curving road is unbanked, the centripetal force is provided by friction between the tyres and the road. Friction must be increased, resulting in wear and tear. When the curving road is banked, however, a component of the ground's natural response supplies the necessary centripetal force, reducing tyre wear and tear.

11. A monkey of mass 4 kg climbs on a rope which can stand a maximum tension 600 N. In which of the following cases will the rope break? The monkey (a) climbs up with an acceleration of  $6\;{\text{m}}{{\text{s}}^{ - 2}}$ (b) climbs down with an acceleration of $4\;{\text{m}}{{\text{s}}^{ - 2}}$ (c) climbs up with a uniform seed of 5 ms (d) falls down the rope freely under gravity. Take ${\text{g}} = 10\;{\text{m}}{{\text{s}}^{ - 2}}$ and ignore the mass of the rope.

Ans: Here, from the question we have:

$m = 40\;{\text{kg}},\;\,T = 600\;{\text{N}}$ (max tension rope can hold)

If the response (R) is greater than the tension, the rope will break $\left( T \right)$ 

(a) $a = 6\;{\text{m}}{{\text{s}}^{ - 2}}$

$R = m(g + a) = 40(10 + 6) = 640\;{\text{N}}$ (Rope will break)

(b) $a = 4\;{\text{m}}{{\text{s}}^{ - 2}}$

$R = m(g - a) = 40(10 - 6) = 240\;{\text{N}}$ (Rope will not break)

(c) $v = 5\;{\text{m}}{{\text{s}}^{ - 1}}$ (constant $a = 0$)

$R = mg = 40 \times 10 = 400\;{\text{N}}$ (Rope will not break)

(d) $a = g;\;R = m(g - a) = m(g - g)$

$R = 0$ $({\text{Rope}}$ will not break)

Since, the rope only breaks when the monkey climbs up with an acceleration of $6\;{\text{m}}{{\text{s}}^{ - 2}}$.

Hence, option (a) will be correct.

12. A soda water bottle is falling freely. Will the bubbles of the gas rise in the water of the bottle?

Ans: As the water in a freely falling bottle is in a state of weight – lessening. So no up thrust force occurs on the bubbles and bubbles will not ascend in the water.

13. Two billiard balls each of mass 0.05 kg moving in opposite directions with speed 6m/s collide and rebound with the same speed. What is the impulse imparted to each ball due to other.

Ans: From the question, we have the initial momentum to the ball $A = 0.05(6) = 0.3\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$

Because the speed of ball $A$ is reversed when it collides, its ultimate momentum is: $A = 0.05( - 6) =  - 0.3\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$

Impulse imparted to ball ${\text{A}} =$ change in momentum of ball $A =$ final momentum – initial momentum $=  - 0.3 - 0.3 =  - 0.6\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$.

14. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei, the products must be emitted in opposite directions.

Ans: Total momentum is conserved according to the conservation of linear momentum concept.

Before disintegration linear momentum will be equal to zero.

After disintegration linear momentum $= {m_1}\overrightarrow {{v_1}}  + {m_2}\overrightarrow {{v_2}}$

$\Rightarrow {m_1}{v_1} + {m_2}{v_2} = 0 \Rightarrow {v_2} =  - \dfrac{{{m_1}{v_1}}}{{{m_2}}}$

15. Explain why passengers are thrown forward form their seats when a speeding bus stops suddenly.

Ans: When a fast bus comes to a complete stop, the bottom half of the body in touch with the seat comes to a complete halt, while the upper section of the passengers' bodies prefer to retain their uniform motion. As a result, the passengers are pushed forward.

16. A rocket with a lift-off mass 20,000 kg is blasted upwards with an initial acceleration of $5.0\;{\text{m}}{{\text{s}}^{ - 2}}.$ Calculate the initial thrust (force) of the blast.

Ans: From the question, we have:

The mass of the rocket given as $m = 20,000\;{\text{kg}}$

Initial acceleration given as $a = 5\;{\text{m}}{{\text{s}}^{ - 2}}$

Acceleration due to gravity given as, $g = 10\;{\text{m}}{{\text{s}}^{ - 2}}$

Using Newton's second law of motion, the net force (thrust) acting on the rocket is given by the relation:

$F - mg = ma$

$F = m(g + a)$

$F = 20000 \times (10 + 5)$

$F = 20000 \times 15$ 

$F = 3 \times {10^5}\;{\text{N}}$

17. A bob of mass $0.1\;{\text{kg}}$ hung from the ceiling of a room by a string $2\;{\text{m}}$ long is set into oscillation. The speed of the bob at its mean position is $1\;{\text{m}}{{\text{s}}^{ - 1}}.$ What is the trajectory of the bob if the string is cut when the bob is (a) at one of its extreme positions, (b) at its mean position.

Ans: (a) Vertically downward

(b) Parabolic path

(a) The bob's velocity is $0$ at its most extreme position. The bob will fall vertically to the ground if the string is severed at this point.

(b) The bob's velocity is $1\;{\text{m}}{{\text{s}}^{ - 1}}$ at its average location. This velocity runs perpendicular to the arc produced by the oscillating bob. The bob will follow a projectile path with only the horizontal component of velocity if it is severed at the mean location. As a result, it will take a parabolic course.

18. Two billiard balls each of mass $0.05\;{\text{kg}}$ moving in opposite directions with speed 6 ms-1collide and rebound with the same speed. What is the impulse imparted to each ball due to the other?

Ans: As from the question, we have the mass of each ball given as $0.05\;{\text{kg}}$

Initial velocity of each ball will be $= 6\;{\text{m}}{{\text{s}}^{ - 1}}$

Magnitude of the initial momentum of each ball given as, ${p_i} = 0.3\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$

The balls alter their directions of motion after colliding, but their velocity magnitudes do not change.

Final momentum of each ball given as, ${p_f} =  - 0.3\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$

Each ball's impulse equals a change in the system's momentum.

$= {p_f} - {p_i}$

$=  - 0.3 - 0.3 =  - 0.6\;{\text{kg}}\;{\text{m}}{{\text{s}}^{ - 1}}$

The negative indication implies that the balls are receiving opposite-direction shocks.

19. A train runs along an unbanked circular track of radius $30\;{\text{m}}$ at a speed of $54\;{\text{km}}{{\text{h}}^{ - 1}}$. The mass of the train is ${10^6}\;{\text{kg}}$. What provides the centripetal force required for this purpose - The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Ans: In the given question, we have the radius of the circular track given as, $r = 30\;{\text{m}}$

Speed of the train is given as, $v = 54\;{\text{km}}{{\text{h}}^{ - 1}} = 15\;{\text{m}}{{\text{s}}^{ - 1}}$

Mass of the train is given as, $m = {10^6}\;{\text{kg}}$

The lateral thrust of the rail on the wheel provides the centripetal force. The wheel exerts an equal and opposite force on the rail, according to Newton's third law of motion. The wear and damage of the rail are caused by this reaction force.

Since, the angle of banking $\theta$, is associated to the radius $(r)$ and speed $(v)$ by the relation:

$\tan \theta  = \dfrac{{{v^2}}}{{rg}}$

$= \dfrac{{{{(15)}^2}}}{{30 \times 10}} = \dfrac{{225}}{{300}}$

$\theta  = {\tan ^{ - 1}}(0.75) = {36.87}^{{}^\circ}$

Therefore, the angle of banking is about ${36.87}^{{}^\circ}$.

20. A constant retarding force of $50\;{\text{N}}$ is applied to a body of mass $20\;{\text{kg}}$ moving initially with a speed of $15\;{\text{m}}{{\text{s}}^{ - 1}}$. How long does the body take to stop?

Ans: As in the given question, we have the retarding force given as, $F =  - 50\;{\text{N}}$

Mass of the body is given as, $m = 20\;{\text{kg}}$

Initial velocity of the body is given as, $u = 15\;{\text{m}}{{\text{s}}^{ - 1}}$

Final velocity of the body is given as, $v = 0$

The acceleration created in the body may be estimated using Newton's second rule of motion:$F = ma$

$- 50 = 20 \times a$

$\therefore a = \dfrac{{ - 50}}{{20}} =  - 2.5\;{\text{m}}{{\text{s}}^{ - 2}}$

The time it takes for the body to come to rest may be estimated using the first equation of motion:

i.e. $v = u + at$

$\therefore t = \dfrac{{ - u}}{a} = \dfrac{{ - 15}}{{ - 2.5}}$

$t = 6\;{\text{s}}$

21. A nucleus is at rest in the laboratory frame of reference. Show that if it disintegrates into two smaller nuclei the products must move in opposite directions.

Ans: Let us assume that $m,{m_1}$, and ${m_2}$ represent the parent nucleus's and the two daughter nuclei's respective masses. The parent nucleus is in a state of dormancy.

Therefore, the initial momentum of the system (parent nucleus) will be = 0

Also let us assume that ${v_1}$ and ${v_2}$ will be the particular velocities of the daughter nuclei having masses $m$ and $2m$.

After disintegration, the system's total linear momentum $= {m_1}{v_1} + {m_2}{v_2}$

As per the law of conservation of momentum the total initial momentum will be equal to total final momentum.

$0 = {m_1}{v_1} + {m_2} + {v_2}$

${v_1} = \dfrac{{ - {m_2}{v_2}}}{{{m_1}}}$

The negative indication here implies that the parent nucleus pieces are moving in opposing directions.

22. A shell of mass $0.020\;{\text{kg}}$ is fired by a gun of mass $100\;{\text{kg}}$. If the muzzle speed of the shell is $80\;{\text{m}}{{\text{s}}^{ - 1}}$, what is the recoil speed of the gun?

Ans: In the given question, we have the mass of the gun given as, $M = 100\;{\text{kg}}$

Mass of the shell is given as, $m = 0.020\;{\text{kg}}$

Muzzle speed of the shell is given as, $v = 80\;{\text{m}}{{\text{s}}^{ - 1}}$

Recoil speed of the gun will become $V$

Since the gun and the shell, both are at rest initially.

Therefore, initial momentum of the system will be and

Final momentum of the system will be $= mv - MV$

Because the shell and the gun are pointing in opposing directions, the negative sign emerges.

As per the law of conservation of momentum the total initial momentum will be equal to total final momentum.

$mv - MV = 0$

$\therefore V = \dfrac{{mv}}{M}$

$= \dfrac{{0.020 \times 80}}{{100 \times 1000}} = 0.016\;{\text{m}}{{\text{s}}^{ - 1}}$

Solve the Following Questions                                 3 Mark

1. A train runs along an unbanked circular bend of radius $30 \mathrm{~m}$ at a speed of $54~\text{kmh}{{\text{r}}^{-1}}$. The mass of the train is $106 \mathrm{~kg}$. What provides the necessary centripetal force required for this purpose? The engine or the rails? What is the angle of banking required to prevent wearing out of the rail?

Ans: From the question, we have the radius of circular bend given as, $r=30~\text{m}$

$\text{Speed of train }=v=54\text{km}{{\text{h}}^{-1}}=54\times \dfrac{5}{18}=15~\text{m}{{\text{s}}^{-1}}$

Mass of train given as, $m={{10}^{6}}~\text{kg}$ 

Then we need to find the angle of banking $\theta$.

(1) The centripetal force is generated by the lateral force exerted by rails on the train's wheels. 

(2) The centripetal force is provided by the lateral thrust by the outer rail.

(3) According to Newton's third law of motion, the train exerts (i.e., causes) an equal and opposite thrust on the outer rail causing its wear and tear.

Therefore, the angle of baking:

\[\tan \theta =\frac{{{v}^{2}}}{rg}=\frac{{{(15)}^{2}}}{30\times 9.8}\]

$\theta =37.4$

2. A block of mass $15 \mathrm{~kg}$ is placed on a long trolley. The coefficient of static friction between the block and the trolley is $0.18$. The trolley accelerates from rest with $0.5$ $\mathrm{ms}^{-2}$ for $20 \mathrm{~s}$ and then moves with uniform velocity. Discuss the motion of the block as viewed by (a) a stationary observer on the ground, (b) an observer moving with the trolley.

Ans:

(a) Mass of the block is given as, $m=15 \mathrm{~kg}$

Coefficient of static friction is given as, $\mu=0.18$

Acceleration of the trolley is given as, $a=0.5~\text{m}{{\text{s}}^{-2}}$

According to Newton's second law of motion, the force $F$ exerted on the block by the trolley's motion is given by the relationship:

$\Rightarrow F=m a=15 \times 0.5=7.5 \mathrm{~N}$

This force is applied in the trolley's forward motion.

The block and the trolley have a static friction force of:

$\Rightarrow f =\mu m g$

$\Rightarrow f =0.18 \times 15 \times 10=27 \mathrm{~N}$

The applied external force is larger than the static friction force between the block and the trolley. As a result, the block will appear to be at rest to a ground observer.

There will be no applied external force while the trolley moves at a constant speed. In this case, the only force acting on the block is friction.  

(b) When moving with the trolley, a spectator experiences some acceleration. This is a non-inertial frame of reference situation. A pseudo force of equal size opposes the frictional force pushing on the trolley backward. This force, on the other hand, works in the opposite direction. For the spectator moving with the trolley, the trolley will appear to be at rest.

3. What is the acceleration of the blocks? What is the net force on the block $P$? What force does $P$ apply on $Q$. What force does $Q$ apply on $R$ ? (Given: $\tan \theta=0.7653$)

Ans: If $a$ is the acceleration

Then $\mathrm{F}=(3 \mathrm{~m}) \mathrm{a}$

$a=F / 3 m$

(1) Net force on $\mathrm{P}$

$F_{1}=m a=m \times \dfrac{F}{3 m}$

$F_{1}=F / 3$

(2) Force applied on $Q$

$\mathrm{F}_{2}=(\mathrm{m}+\mathrm{m}) \mathrm{a}$

\[\begin{align} & {{F}_{2}}=2m\times a \\ & =2m\times \dfrac{F}{3m} \end{align}\]

\[{{F}_{2}}=\dfrac{2F}{3}\]

(3) Force applied on $\mathrm{R}$ by $Q$

\[\begin{align} & {{F}_{3}}=m\times a \\  & =m\times \dfrac{F}{3m} \end{align}\]

${{F}_{3}}=\dfrac{F}{3}$

4. How is centripetal force provided in case of the following?

(i) Motion of planet around the sun,

(ii) Motion of moon around the earth.

(iii) Motion of an electron around the nucleus in an atom.

Ans: 

(i) The centripetal force is provided by the gravitational force acting on the earth and the sun.

(ii) Centripetal force is provided by the earth's gravitational attraction on the moon.

(iii) The centripetal force is provided by the electrostatic attraction between the electron and the proton.

5. State Newton's second, law of motion. Express it mathematically and hence obtain a relation between force and acceleration.

Ans: According to Newton's second law, the rate of change of momentum is precisely proportional to the force.\[\text{ i}\text{.e}\text{. F}\alpha \text{ rate of change of momentum }\left( \dfrac{d\overrightarrow{p}}{dt} \right)\]

\[\vec{F}=k\dfrac{d\overrightarrow{p}}{dt}\ \because (\vec{P}=m\vec{v})\]

\[\Rightarrow \vec{F}=km\dfrac{d\overrightarrow{v}}{dt}\]

\[\Rightarrow \vec{F}=km\overrightarrow{a}\,\,\,\,(\text{In S}\text{.I}\text{. unit k}=1\text{)}\]

\[\Rightarrow \vec{F}=km\overrightarrow{a}\]

6. A railway car of mass \[20\text{ tonnes}\] moves with an initial speed of $54 \mathrm{~km} / \mathrm{hr}$. .On applying brakes, a constant negative acceleration of $0.3 \mathrm{~m} / \mathrm{s}^{2}$ is produced.

(i) What is the breaking force acting on the car?

(ii) In what time it will stop?

(iii) What distance will be covered by the car before if finally stops?

Ans: $m=20$ tonnes $=20\times 1000~\text{kg, }u=54\text{kmh}{{\text{r}}^{-1}}=15~\text{m}{{\text{s}}^{-1}}$

$a=-0.3~\text{m}{{\text{s}}^{-2}},\,\,\vartheta =0$

(a) $\mathrm{F}=\mathrm{ma}$

$F=20000 \times(-0.3)$

$\mathrm{F}=-6000 \mathrm{~N}$

(b) $v=u+a t$

$v-u=a t$

$t=\dfrac{v-u}{a}=\dfrac{0-15}{-0.3}$

$\mathrm{t}=50 \mathrm{~s}$

(c) $v^{2}-v^{2}=2$ as

$(0)^{2}-(15)^{2}=2(-0.3) s$

$\mathrm{S}=375 \mathrm{~m}$

7. What is meant by coefficient of friction and angle of friction? Establish the relation between the two? OR

A block of mass $10 \mathrm{~kg}$ is sliding on a surface inclined at an angle of $30^{\circ}$ with the horizontal. Calculate the acceleration of the block. The coefficient of kinetic friction between the block and the surface is $0.5$.

Ans: The contact between the outcome of limiting friction and normal reaction is known as the angle of friction.

$\tan \alpha=\dfrac{F s}{R}$ , will be with the normal reaction.

Coefficient of static friction

The limiting value of static frictional force is proportion to the normal reaction is given by:${{F}_{s}}\prec R\Rightarrow {{F}_{s}}={{\mu }_{s}}R$

Or ${{\mu }_{s}}=\dfrac{{{F}_{s}}}{R}\,\,\,\,\,\,\,\,\,\,......(2)$

From (1) & (2)

${{\mu }_{s}}=\tan \alpha =\dfrac{{{F}_{s}}}{R}$

OR

A block of mass $10 \mathrm{~kg}$ is sliding on a surface inclined at an angle of $30{}^\circ $ with the horizontal.

Calculate the acceleration of the block. The coefficient of kinetic friction between the block

and the surface is $0.5$.

$m=10~\text{kg,}\quad \theta ={{30}^{{}^\circ }},\,\,\,{{\mu }_{k}}=0.5$

$a=g(\sin \theta-\mu k \cos \theta)$

$a=9.8\left(\sin 30^{\circ}-0.5 \cos 30^{\circ}\right)$

$a=9.8(-.5-0.5 \times 0.866)$

$\mathrm{a}=0.657 \mathrm{~m} / \mathrm{s}^{2}$

8. State and prove the principle of law of conservation of linear momentum?

Ans: Law of conservation of momentum states that unless an external force is applied, the two or more objects acting upon each other in an isolated system, the total momentum of the system remains constant. This also means that the total momentum of an isolated system before is equal to the total momentum of the isolated system after.

The law of conservation of linear momentum applies if no external force occurs on the system. The entire momentum of the system remains unchanged.

i.e. if ${{\vec{F}}_{ext}}=0$then $\vec{P}=$ constant

Impulse experienced by ${{m}_{1}}=\overrightarrow{{{F}_{12}}}\Delta t={{m}_{1}}\overrightarrow{{{u}_{1}}}-{{m}_{1}}\overrightarrow{{{v}_{1}}}$

Impulse experienced by \[{{m}_{2}}=\overrightarrow{{{F}_{21}}}\Delta t={{m}_{2}}\overrightarrow{{{u}_{2}}}-{{m}_{2}}\overrightarrow{{{v}_{2}}}\]

According to Newton's third law

$\overrightarrow{{{F}_{12}}}=-\overrightarrow{{{F}_{21}}}$

$\Rightarrow {{m}_{1}}\overrightarrow{{{u}_{1}}}-{{m}_{1}}\overrightarrow{{{v}_{1}}}=-\left( {{m}_{2}}\overrightarrow{{{u}_{2}}}-{{m}_{2}}\overrightarrow{{{v}_{2}}} \right)$

${{m}_{1}}\overrightarrow{{{u}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}}={{m}_{1}}\overrightarrow{{{u}_{1}}}+{{m}_{2}}\overrightarrow{{{v}_{2}}}$

As a result, the momentum gained by one ball is lost by the other. As a result, linear momentum is conserved.

9. A particle of mass $0.40 \mathrm{~kg}$ moving initially with constant speed of $10 \mathrm{~m} / \mathrm{s}$ to the north is subject to a constant force of $8.0 \mathrm{~N}$ directed towards south for $30 \mathrm{~s}$. Take at that instant, the force is applied to be $t=0$, and the position of the particle at that time to be $x=0$, predict its position at $t=-5\text{s},\,25\text{s},30\text{s}$?

Ans: $\mathrm{m}=0.40 \mathrm{~kg}$

$u=\text{l}0~\text{m}{{\text{s}}^{-1}}$ due North

$\mathrm{F}=-8.0 \mathrm{~N}$

$a=\dfrac{F}{m}=\dfrac{-8.0}{0.40}$

$a=-20~\text{m}{{\text{s}}^{-2}}$ 

(1) At $t=-5 \mathrm{~s}$

$x=u t=10 \times(-5)$

$x=-50 m$

(2) At $t=25 \mathrm{~s}$

$x=u t+\dfrac{1}{2} a t^{2}$

$x=10 \times 25+(-20)(25)^{2}$

$\mathrm{x}=-6000 \mathrm{~m}$

(3) At $\mathrm{t}=30 \mathrm{~s}$

$x_{1}=u t+\dfrac{1}{2} a t^{2}$

$x_{1}=10 \times 30+\dfrac{1}{2}(-20)(30)^{2}$

$x_{1}=-8700 \mathrm{~m}$

(4) At $\mathrm{t}=30 \mathrm{~s}$

$v=u+a t$

$v=10+(-20)(30)$

$v=-590~\text{m}{{\text{s}}^{-1}}$

$\therefore$ Motion from 30 s to $100 \mathrm{~s}$

$x_{2}=u t=-590 \times 70$

$x_{2}=-41300 \mathrm{~m}$

$\therefore$ Total distance $\mathrm{x}=\mathrm{x}_{1}+\mathrm{x}_{2}$

$\mathrm{x}=-50000 \mathrm{~m}$

10. A block of mass $25 \mathrm{~kg}$ is raised by a $50 \mathrm{~kg}$ man in two different ways as shown in Fig.5.19. What is the action on the floor by the man in the two cases? If the floor yields to a normal force of $700 \mathrm{~N}$, which mode should the man adopt to lift the block without the floor yielding?

Ans:

$750 \mathrm{~N}$ and $250 \mathrm{~N}$ in the respective cases; Method (b)

From the question, we have the mass of the block I given as $m=25 \mathrm{~kg}$

Mass of the man given as $M=50 \mathrm{~kg}$

Acceleration due to gravity is given as, $\text{g}=10~\text{m}{{\text{s}}^{-2}}$

Force applied on the block is given as, $F=25 \times 10=250 \mathrm{~N}$

Weight of the man is given as, $W=50 \times 10=500 \mathrm{~N}$ 

Case (a): When the man directly raises the block.

In this situation, the man exerts an upward push. This makes him appear heavier.

$\therefore$ Action on the floor by the man$\operatorname{man}=250+500=750 \mathrm{~N}$

Case (b): The man uses a pulley to lift the block.

In this situation, the man exerts a downward force. His apparent weight is reduced as a result of this.

$\therefore$ Action on the floor by the man$\operatorname{man}=500-250=250 \mathrm{~N}$

If the floor can withstand a normal force of $700\text{N}$, the man should use the second approach to lift the block more readily while using less force.

11. 

(a) State Impulse-momentum theorem?

(b) A ball of mass $0.1 \mathrm{~kg}$ is thrown against a wall. It strikes the wall normally with a velocity of $30~\text{m}{{\text{s}}^{-1}}$ and rebounds with a velocity of $20~\text{m}{{\text{s}}^{-1}}$. Calculate the impulse of the force exerted by the ball on the wall.

Ans: 

(a) The impulse-momentum theorem generally states that the impulse applied to a body is equal to the change in momentum of that body.

Impulse $\Rightarrow m(u-v)=\overrightarrow{{{P}_{2}}}-\overrightarrow{{{P}_{1}}}$

(b) \[m=0.1~\text{kg, }v=30~\text{m}{{\text{s}}^{-1}}\text{, }u=-20~\text{m}{{\text{s}}^{-1}}\]

Impulse, 

\[\begin{align} & \overrightarrow{{{P}_{2}}}-\overrightarrow{{{P}_{1}}}=mv-mu \\  & =m\left( v-u \right) \\  & =m\left( 30-\left( -20 \right) \right) \\  & =m\left( 30+20 \right) \\  & =\left( 0.1\times 50 \right)\,\text{Ns} \\  & =5\,\text{Ns}  \end{align}\]

12. Ten one rupee coins are put on top of one another on a table. Each coin has a mass $m\,\text{kg}$. Give the magnitude and direction of

(a) The force on the $7^{\text {th }}$ coin (counted from the bottom) due to all coins above it.

(b) The force on the $7^{\text {th }}$ coin by the eighth coin and

(c) The reaction of the sixth coin on the seventh coin.

Ans: 

(a) The force on $7^{\text {th }}$ coin is due to weight of the three coins lying above it. 

Therefore, 

\[\begin{align} & F=(3~\text{m})\text{kg} \\ & F=(3\text{mg})\text{N} \\ \end{align}\]

Here $\mathrm{g}$ is acceleration due to gravity. This force acts vertically downwards.

(b) The eighth coin is already weighed down by the weight of the two coins above it, as well as its own. As a result, the force on the $7^{\text {th }}$ coin owing to the $8^{\text {th }}$ coin is equal to the sum of the two forces, i.e.

\[\begin{align} & F=2~m+m \\ & =(3~m)\text{kg} \\ & F=(3mg)\text{N} \end{align}\]

The force acts in a vertical downward direction. 

(c) The sixth coins is under the weight of four coins above it 

Reaction, 

\[\begin{align} & R=-F=-4~m(\text{kg}) \\ & F=-(4mg)\text{N} \end{align}\]

The -ve symbol denotes that the reaction is vertically upwards.

13. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with a constant speed,

(b) a cork of mass $10 \mathrm{~g}$ floating on water,

(c) a kite skillfully held stationary in the sky,

(d) a car moving with a constant velocity of $30~\text{km}{{\text{h}}^{-1}}$ on a rough road,

(e) a high-speed electron in space far from all material objects, and free of electric and magnetic fields.

Ans:

(a) Zero net force

The raindrops are falling at a steady rate. As a result, the acceleration is zero. The net force acting on the rain drop is zero, according to Newton's second law of motion. 

(b) Zero net force

The cork's weight is acting downward. The buoyant force exerted by the water in an upward direction balances it. As a result, there is no net force acting on the floating cork. 

(c) Zero net force

In the sky, the kite is stationary, i.e. it is not moving at all. As a result, according to Newton's first rule of motion, there is no net force acting on the kite.  

(d) Zero net force

The car is going at a consistent speed down a bumpy route. As a result, it has no acceleration. There is no net force operating on the car, according to Newton's second law of motion. 

(e) Zero net force

All fields have no affect on the high-speed electron. As a result, there is no net force acting on the electron.

14. A pebble of mass $0.05 \mathrm{~kg}$ is thrown vertically upwards. Give the direction and magnitude of the net force on the pebble,

(a) during its upward motion,

(b) during its downward motion,

(c) at the highest point where it is momentarily at rest. Do your answers change if the pebble was thrown at an angle of $45^{\circ}$ with the horizontal direction?

Ignore air resistance.

Ans: $0.5 \mathrm{~N}$, in vertically downward direction, in all cases

Gravitational acceleration always operates downward, regardless of the direction of motion of an item. In all three scenarios, the gravitational force is the only force acting on the stone. Newton's second law of motion gives its magnitude as: $F=m \times a$

Where,

$F=$ Net force

$m=$ Mass of the pebble $=0.05 \mathrm{~kg}$

$a=\text{g}=10~\text{m}{{\text{s}}^{-2}}$

$\therefore F=0.05 \times 10=0.5 \mathrm{~N}$

In all three circumstances, the net force acting on the stone is $0.5\,\text{N}$, and this force acts downward. 

The horizontal and vertical components of velocity will be present if the stone is thrown at an angle $45^{\circ}$to the horizontal. Only the vertical component of velocity becomes zero at the highest point. Throughout its travel, however, the stone will have a horizontal component of velocity. The net force applied on the stone is unaffected by this component of velocity. 

15. A constant force acting on a body of mass $3.0 \mathrm{~kg}$ changes its speed from $2.0 \mathrm{~m} \mathrm{~s}^{-1}$ to $ 3.5 \mathrm{~m} \mathrm{~s}^{-1}$ in $25 \mathrm{~s}$. The direction of the motion of the body remains unchanged. What is the magnitude and direction of the force?

Ans: $0.18 \mathrm{~N}$; in the direction of motion of the body

Mass of the body is given as, $m=3 \mathrm{~kg}$

Initial speed of the body is given as, $u=2~\text{m}{{\text{s}}^{-1}}$

Final speed of the body is given as, $v=3.5~\text{m}{{\text{s}}^{-1}}$

Time is given as, $t=25 \mathrm{~s}$

The acceleration (a) produced in the body can be estimated using the first equation of motion:$v=u+a t$

$\therefore a=\dfrac{v-u}{t}$

$=\dfrac{3.5-2}{25}=\dfrac{1.5}{25}=0..06~\text{m}{{\text{s}}^{-1}}$

Newton's second law of motion states that force is equal to:

$F=m a$

$=3 \times 0.06=0.18 \mathrm{~N}$

The net force acting on the body is in the direction of its motion since the application of force does not affect the direction of the body.

16. A body of mass $5 \mathrm{~kg}$ is acted upon by two perpendicular forces $8 \mathrm{~N}$ and $6 \mathrm{~N}$. Give the magnitude and direction of the acceleration of the body.

Ans: In the given question, we have the mass of the body given as, $m=5 \mathrm{~kg}$.

It is acting on two perpendicular forces, given as $8 \mathrm{~N}$ and $6 \mathrm{~N}$.

The following is a representation of the situation:

The resultant of two forces is given as:

$R=\sqrt{(8)^{2}+(-6)^{2}}=\sqrt{64+36}=10 N$

$\theta$ is the angle made by $R$ with the force of $8 \mathrm{~N}$

$\therefore \theta ={{\tan }^{-1}}\left( \dfrac{-6}{8} \right)=-36.87{}^\circ $

The –ve sign indicates that $\theta$ will be in the clockwise direction with respect to the force having the magnitude $8 \mathrm{~N}$.

The acceleration (a) of a body is provided by Newton's second law of motion:

$F=m a$

$\therefore a=\dfrac{F}{m}=\dfrac{10}{5}=2~\text{m}{{\text{s}}^{-2}}$

17. A stone of mass $0.25 \mathrm{~kg}$ tied to the end of a string is whirled round in a circle of radius $1.5 \mathrm{~m}$ with a speed of 40 rev./min in a horizontal plane. What is the tension in the string? What is the maximum speed with which the stone can be whirled around if the string can withstand a maximum tension of $200 \mathrm{~N}$ ?

Ans:

Mass of the stone is given as, $m=0.25 \mathrm{~kg}$

Radius of the circle is given as, $r=1.5 \mathrm{mt}$

Number of revolution per second is given as, $n=\dfrac{40}{60}=\dfrac{2}{3} r p s$

Since, Angular velocity is given by the formula, $\omega=\dfrac{v}{r}=2 \pi n \ldots(i)$

The tension $T$ in the string provides the centripetal force for the stone, i.e.,

$T={{F}_{Centripetal}}$

$=\dfrac{m v^{2}}{r}=m r \omega^{2}=m r(2 \pi n)^{2}$

$=0.25 \times 1.5 \times\left(2 \times 3.14 \times \dfrac{2}{3}\right)^{2}$

$=6.57 \mathrm{~N}$

Since, the maximum tension in the string, $T_{\max }=200 \mathrm{~N}$

${{T}_{\max }}=\dfrac{m{{v}^{2}}_{\max }}{r}$

$\therefore v_{\max }=\sqrt{\dfrac{T_{\max } \times r}{m}}$

$=\sqrt{\dfrac{200 \times 1.5}{0.25}}$

$=\sqrt{1200}=34.64~\text{m}{{\text{s}}^{-1}}$

Therefore, the maximum speed of the stone will be equal to $34.64~\text{m}{{\text{s}}^{-1}}$.

18. Figure 5.18 shows a man standing stationary with respect to a horizontal conveyor belt that is accelerating with $1 \mathrm{~m} \mathrm{~s}^{-2} .$ What is the net force on the man? If the coefficient of static friction between the man's shoes and the belt is $0.2$, up to what acceleration of the belt can the man continue to be stationary relative to the belt? \[\left( \text{Mass}\,\text{of}\,\text{the}\,\text{man}=65\,\text{kg} \right)\]

Ans: From the question, we have the mass of the man is given as, $m=65\,\text{kg}$

Acceleration of the belt is given as, $a=1~\text{m}{{\text{s}}^{-2}}$

Coefficient of static friction is given as, $\mu=0.2$

Newton's second law of motion gives the net force $F$ operating on the man as:${{F}^{\prime }}_{net}=ma=65\times 1=65~\text{N}$

The individual will remain motionless in relation to the conveyor belt until his net force is equal to or less than the frictional force $f_{z}$, exerted by the belt, i.e.,

$F_{net}^{\prime }={{f}_{z}}$

$m a^{\prime}=\mu m g$

$\therefore {{a}^{\prime }}=0.2\times 10=2~\text{m}{{\text{s}}^{-2}}$

As a result, the maximum belt acceleration at which the guy may remain immobile is $2~\text{m}{{\text{s}}^{-2}}$.

19. A stream of water flowing horizontally with a speed of $15 \mathrm{~ms}^{-1}$ gushes out of a tube of cross-sectional area $10^{-2} \mathrm{~m}^{2}$, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?

Ans: Since, the speed of the water stream is given as, $v=15~\text{m}{{\text{s}}^{-1}}$ 

Cross-sectional area of the tube is given as, $A=10^{-2} \mathrm{~m}^{2}$ 

Therefore, the volume of water coming out from the pipe per second will be: $V=Av=15\times {{10}^{-2}}~{{\text{m}}^{3}}{{\text{s}}^{-1}}$

As the density of water is, \[p={{10}^{3}}~\text{kg}{{\text{m}}^{-3}}\]

Flow rate of water via the pipe in gallons per second $=p\times V=150~\text{kg}{{\text{s}}^{-1}}$ 

When the water hits the wall, it does not bounce back. As a result, Newton's second rule of motion gives the force exerted by the water on the wall as:

$F=$ Rate of change of momentum $=\dfrac{\Delta P}{\Delta t}$ $=\dfrac{m v}{t}$

$=150 \times 15=2250 \mathrm{~N}$

20. An aircraft executes a horizontal loop at a speed of $720~\text{km}{{\text{h}}^{-1}}$ with its wings banked at $15{}^\circ .$ What is the radius of the loop?

Ans: 

From the question, it is given that the peed of the aircraft, $v=720~\text{km}{{\text{h}}^{-1}}=720\times \dfrac{5}{18}=200~\text{m}{{\text{s}}^{-1}}$

Also the acceleration due to gravity, $\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$ and the angle of banking, $\theta=15^{\circ}$ 

Therefore, for radius $r$, of the loop, we have the relation:

$\tan \theta=\dfrac{v^{2}}{r g}$

$r=\dfrac{v^{2}}{g \tan \theta}$

$=\dfrac{200 \times 200}{10 \times \tan 15}=\dfrac{4000}{0.268}$

$=14925.37 \mathrm{~m}$

$=14.92 \mathrm{~km}$

Solve the Following Questions                            4 Mark

1. Give the magnitude and direction of the net force acting on a stone of mass $0.1 \mathrm{~kg}$,

(a) just after it is dropped from the window of a stationary train,

(b) just after it is dropped from the window of a train running at a constant velocity of $36 \mathrm{~km} / \mathrm{h}$

(c) just after it is dropped from the window of a train accelerating with $1 \mathrm{~m} \mathrm{~s}^{-2}$,

(d) lying on the floor of a train which is accelerating with $1 \mathrm{~m} \mathrm{~s}-2$, the stone being at rest relative to the train. Neglect air resistance throughout.

Ans: 

(a) $1 \mathrm{~N}$; vertically downward

From the question, we have the mass of the stone given as, $m=0.1 \mathrm{~kg}$

Acceleration of the stone is given as, $a=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$

The net force exerted on the stone, according to Newton's second law of motion, is

$F=m a=m \mathrm{~g}$

$=0.1 \times 10=1 \mathrm{~N}$

Gravitational acceleration always works in the downward direction. 

(b) $1 \mathrm{~N}$; vertically downward

The train is travelling at a constant speed. As a result, its acceleration in the horizontal direction, where it is moving, is zero. As a result, there is no horizontal force acting on the stone. 

The net force acting on the stone is due to gravity's acceleration, and it is always vertically downward. This force has a magnitude of $1 \mathrm{~N}$.

(c) $1 \mathrm{~N}$; vertically downward

It is given that the train is accelerating at the rate of $1 \mathrm{~m} / \mathrm{s}^{2}$.

Hence, the net force acting on the stone will be equal to, $F^{\prime}=m a=0.1 \times 1=0.1 \mathrm{~N}$

This force has a horizontal component to it. The horizontal force $F'$, no longer acts on the stone when it is dropped. This is due to the fact that the force acting on a body at any one time is determined by the current circumstance rather than previous ones.

As a result, the net force acting on the stone is determined only by gravity's acceleration.$F=m g=1 \mathrm{~N}$

This force acts vertically downward.

(d) $0.1 \mathrm{~N} ;$ in the direction of motion of the train

The typical reaction of the floor balances the weight of the stone. The train's horizontal motion is the only source of acceleration.

Acceleration of the train, $a=0.1 \mathrm{~m} / \mathrm{s}^{2}$

The net force acting on the stone will be directed in the train's direction of travel. Its magnitude is given by:

$F=m a$ 

$=0.1 \times 1=0.1 \mathrm{~N}$

2. The driver of a three-wheeler moving with a speed of $36 \mathrm{~km} / \mathrm{h}$ sees a child standing in the middle of the road and brings his vehicle to rest in $4.0 \mathrm{~s}$ just in time to save the child. What is the average retarding force on the vehicle? The mass of the three wheeler is $400 \mathrm{~kg}$ and the mass of the driver is $65 \mathrm{~kg}$.

Ans: From the question, it is given that the initial speed of the three-wheeler, $u=36 \mathrm{~km} / \mathrm{h}=10 \mathrm{~m} / \mathrm{s}$

Also the final speed of the three-wheeler is given as, $v=0 \mathrm{~m} / \mathrm{s}$

Time, $t=4 \mathrm{~s}$

Mass of the three-wheeler is given as, $m=400 \mathrm{~kg}$

Mass of the driver is given as, $m^{\prime}=65 \mathrm{~kg}$

Therefore, the total mass of the system, $M=400+65=465 \mathrm{~kg}$

So, with the use of first law of motion, the acceleration $(a)$ of the three-wheeler can be calculated by using the formula: $v=u+a t$

$\therefore a=\dfrac{v-u}{t}=\dfrac{0-10}{4}=-2.5 \mathrm{~m} / \mathrm{s}^{2}$

The negative indication implies that the three-velocity wheeler's is decreasing over time.

The net force operating on the three-wheeler may be estimated using Newton's second law of motion:

$F=M a$

$=465 \times(-2.5)=-1162.5 \mathrm{~N}$

The minus symbol shows that the force is acting in the opposite direction of motion of the three-wheeler.

3. A body of mass $0.40 \mathrm{~kg}$ moving initially with a constant speed of $10 \mathrm{~ms}$ -1to the north is subject to a constant force of $8.0 \mathrm{~N}$ directed towards the south for $30 \mathrm{~s}$. Take the instant the force is applied to be $t=0$, the position of the body at that time to be $x=0$, and predict its position at $t=-5 \mathrm{~s}, 25 \mathrm{~s}, 100 \mathrm{~s}$.

Ans: From the question, we have the value given as the mass of the body, $m=0.40 \mathrm{~kg}$

Initial speed of the body is given as, $u=10 \mathrm{~m} / \mathrm{s}$ due north

Force acting on the body is given as, $F=-8.0 \mathrm{~N}$

Therefore, the acceleration produced in the body, $a=\dfrac{F}{m}=\dfrac{-8.0}{0.40}=-20 \mathrm{~m} / \mathrm{s}^{2}$

(i) At $t=-5 \mathrm{~s}$

Acceleration, $a^{\prime}=0$ and $u=10 \mathrm{~m} / \mathrm{s}$

$s=u t+\dfrac{1}{2} a^{\prime} t^{2}$

$=10 \times(-5)=-50 \mathrm{~m}$

(ii) At $t=25 \mathrm{~s}$

Acceleration, $a^{\prime \prime}=-20 \mathrm{~m} / \mathrm{s}^{2}$ and $u=10 \mathrm{~m} / \mathrm{s}$

$s^{\prime}=u t^{\prime}+\dfrac{1}{2} a^{\prime \prime} t^{2}$

$=10 \times 25+\dfrac{1}{2} \times(-20) \times(25)^{2}$

$=250+6250=-6000 \mathrm{~m}$

(iii) At $t=100 \mathrm{~s}$

For $0 \leq t \leq 30 \mathrm{~s}$

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

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

$s_{1}=u t+\dfrac{1}{2} a^{n} t^{2}$

$=10 \times 30+\dfrac{1}{2} \times(-20) \times(30)^{2}$

$=300-9000$

$=-8700 \mathrm{~m}$

For 30 < t  \[\le 100~\text{s}\]

According to the first equation of motion, for $t=30 \mathrm{~s}$, final velocity will be calculated by the formula:

$v=u+a t$

$=10+(-20) \times 30=-590 \mathrm{~m} / \mathrm{s}$

Velocity of the body after $30 \mathrm{~s}=-590 \mathrm{~m} / \mathrm{s}$

For motion between $30 \mathrm{~s}$ to $100 \mathrm{~s}$, i.e., in $70 \mathrm{~s}$ :

$s_{2}=v t+\dfrac{1}{2} a^{n} t^{2}$

$=-590 \times 70=-41300 \mathrm{~m}$

$=-590 \times 70=-41300 \mathrm{~m}$

$\therefore$ Total distance, ${{s}^{\prime \prime }}={{s}_{1}}+{{s}_{2}}=-8700-41300=-50000\,\text{m}$

4. A truck starts from rest and accelerates uniformly at $2.0 \mathrm{~ms}^{-2} .$ At $t=10 \mathrm{~s}$, a stone is dropped by a person standing on the top of the truck ( $6 \mathrm{~m}$ high from the ground). What are the (a) velocity, and (b) acceleration of the stone at $t=11 \mathrm{~s}$ ? (Neglect air resistance.)

(a) $22.36 \mathrm{~m} / \mathrm{s}$, at an angle of $26.57{}^\circ $ with the motion of the truck

(b) $10 \mathrm{~m} / \mathrm{s}^{2}$

Ans: 

(a) In the given question, we have the initial velocity of the truck as, $u=0$

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

Time, $t=10 \mathrm{~s}$

The final velocity is provided by the first equation of motion:

$v=u+a t$

$=0+2 \times 10=20 \mathrm{~m} / \mathrm{s}$

The final velocity of the truck and the stone is $20 \mathrm{~m} / \mathrm{s}$.

At $t=11 \mathrm{~s}$, the horizontal component $\left(v_{x}\right)$ of velocity, in the absence of air resistance, the rate of change of velocity remains constant, i.e.$v_{x}=20 \mathrm{~m} / \mathrm{s}$

The first equation of motion gives the vertical component of the stone's velocity as: 

${{v}_{y}}=u+{{a}_{y}}\vartriangle t$

Where, $\vartriangle t=11-10=1~\text{s}$ and $a_{y}=\mathrm{g}=10 \mathrm{~m} / \mathrm{s}^{2}$

$\therefore v_{y}=0+10 \times 1=10 \mathrm{~m} / \mathrm{s}$

The resultant velocity $(v)$ of the stone is given as:

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}$

$=\sqrt{20^{2}+10^{2}}=\sqrt{400+100}$

$=\sqrt{500}=22.36 \mathrm{~m} / \mathrm{s}$

Assume that the resultant velocity makes an angle $\theta $ with the horizontal component of velocity.$v_{x}$

$\therefore \tan \theta=\left(\dfrac{v_{y}}{v_{z}}\right)$

$\theta=\tan ^{-1}\left(\dfrac{10}{20}\right)$

$=\tan ^{-1}(0.5)$

$=26.57^{\circ}$

(b) The horizontal force acting on the stone is zero when it is dropped from the truck. The stone, however, continues to move due to gravity's pull. As a result, the stone's acceleration is $10\,\text{m/}{{\text{s}}^{2}}$ and it operates vertically downward.

5. Figure 5.16 shows the position-time graph of a particle of mass 4kg. What is the (a) force on the particle for t<0, 0<t<4s, 0<t<4s ? (b) impulse at t=0 and t=4s (Consider one-dimensional motion only).

Ans:

(a) For $t<0$

The location of the particle is coincident with the time axis, as can be seen in the graph. It means that the particle's movement in this time interval is zero. As a result, there is no force acting on the particle.

For $t>4 \mathrm{~s}$

The location of the particle in the provided graph is parallel to the time axis, as can be seen. It means the particle is resting at a distance of from the origin. As a result, there is no force acting on the particle.

For 0 < t < 4

The provided position-time graph exhibits a constant slope, as can be seen. As a result, the particle's acceleration is zero. As a result, there is no force acting on the particle. 

(b) At $t=0$

Impulse = Change in momentum

$=m v-m u$

Mass of the particle is given as, $m=4 \mathrm{~kg}$

Initial velocity of the particle, $u=0$

Final velocity of the particle will be, $v=\dfrac{3}{4} \mathrm{~m} / \mathrm{s}$

$\therefore$ Impulse $=4\left(\dfrac{3}{4}-0\right)=3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

At $t=4 \mathrm{~s}$

Initial velocity of the particle, $u=\dfrac{3}{4} \mathrm{~m} / \mathrm{s}$

Final velocity of the particle, $v=0$

$\therefore$ Impulse $4\left(\dfrac{3}{4}-0\right)=3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

6. Two bodies of masses $10 \mathrm{~kg}$ and $20 \mathrm{~kg}$ respectively kept on a smooth, horizontal surface are tied to the ends of a light string. A horizontal force $F=600 \mathrm{~N}$ is applied to (i) A, (ii) B along the direction of string. What is the tension in the string in each case?

Ans:

From the question, we have the horizontal force given as, $F=600 \mathrm{~N}$

Mass of body A given as, $m_{1}=10 \mathrm{~kg}$

Mass of body B given as, $m_{2}=20 \mathrm{~kg}$

Therefore, the total mass of the system will become, $m=m_{1}+m_{2}=30 \mathrm{~kg}$

The acceleration produced in the system may be estimated using Newton's second rule of motion:$F=m a$

$\therefore a=\dfrac{F}{m}=\dfrac{600}{30}=20 \mathrm{~m} / \mathrm{s}^{2}$

When force $F$ is applied on body A:

The motion equation may be expressed as follows:

$F-T=m_{1} a$

$\therefore T=F-m_{1} a$

$=600-10 \times 20=400 \mathrm{~N} \ldots$ (i)

When force $F$ is applied on body $\mathrm{B}$ :

The motion equation may be expressed as follows:

$F-T=m_{2} a$

$T=F-m_{2} a$

\[\therefore T=600-20\times 20=200~\text{N}\]

7. A batsman deflects a ball by an angle of $45{}^\circ $ without changing its initial speed which is equal to $54 \mathrm{~km} / \mathrm{h}$. What is the impulse imparted to the ball? (Mass of the ball is $0.15\,\text{kg}$.)

Ans: The given situation can be represented as shown in the following figure.

Where,

AO will be the incident path of the ball and OB will be the path followed by the ball after deflection.

\[\angle \text{AOB}\] will be the angle between the incident and the deflected paths of the ball and is equal to ${{45}^{{}^\circ }}$.

$\angle \mathrm{AOP}=\angle \mathrm{BOP}=22.5^{\circ}=\theta$

Let us assume that the initial and final velocities of the ball $=v$

Horizontal component of the initial velocity will be $=v \cos \theta$ along $\mathrm{RO}$

Vertical component of the initial velocity will be$=v \sin \theta$ along PO

Horizontal component of the final velocity will be$=v \cos \theta$ along $\mathrm{OS}$

Vertical component of the final velocity = $v\sin \theta $ along $\text{OP}$

There is no change in the horizontal components of velocities. The vertical components of velocities are in a clockwise orientation.

$\therefore$ Impulse imparted to the ball = Change in the linear momentum of the ball

$=m v \cos \theta-(-m v \cos \theta)$

$=2 m v \cos \theta$

Mass of the ball, $m=0.15 \mathrm{~kg}$

Velocity of the ball, $v=54 \mathrm{~km} / \mathrm{h}=15 \mathrm{~m} / \mathrm{s}$

$\therefore$ Impulse $=2 \times 0.15 \times 15 \cos 22.5^{\circ}=4.16 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

8. Figure 5.17 shows the position-time graph of a body of mass $0.04 \mathrm{~kg}$. Suggest a suitable physical context for this motion. What is the time between two consecutive impulses received by the body? What is the magnitude of each impulse?

Ans:

A ball bounces back and forth between at $x=0$ and $x=2 \mathrm{~cm}$; after every $2 \mathrm{~s}$, the ball receives an impulse of magnitude $0.08 \times 10^{-2} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ from the walls.

A body changes its direction of motion every $2 \mathrm{~s}$, as seen in the graph. In terms of physical representation, imagine a ball rebounding Aback and forth between two fixed walls located at $x=0$ and $x=2 \mathrm{~cm}$. The ball collides with a wall every $2 \mathrm{~s}$ because the slope of the $x-t$ graph flips after every $2 \mathrm{~s}$. As a result, every$2 \mathrm{~s}$, the ball receives an impulse.

Mass of the ball, $m=0.04 \mathrm{~kg}$

The ball's velocity is determined by the graph's slope. We may compute initial velocity using the graph as follows: 

$u=\dfrac{(2-0) \times 10^{-2}}{(2-0)}=10^{-2} \mathrm{~m} / \mathrm{s}$

Velocity of the ball before collision will be, $u=10^{-2} \mathrm{~m} / \mathrm{s}$

Velocity of the ball after collision will be, $v=-10^{-2} \mathrm{~m} / \mathrm{s}$

(As the ball reverses its direction of motion, the negative sign appears.)

Since, magnitude of impulse is equal to change in momentum.

$=|m v-m u|$

$=|0.04(v-u)|$

$=\mid 0.04\left(-10^{-2}=10^{-2} \mid\right.$

$=0.08 \times 10^{-2} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

9. A stone of mass m tied to the end of a string revolves in a vertical circle of radius R. The net forces at the lowest and highest points of the circle directed vertically downwards are: (Choose the correct alternative)

Ans: (a) The following graphic depicts the free body diagram of the stone at its lowest position. 

The net force exerted on the stone at this moment, according to Newton's second law of motion, is equal to the centripetal force, i.e.

$F_{n e t}=T=m g=\dfrac{m v_{1}^{2}}{R} \ldots(i)$

Here, $v_{1}=$ Velocity at the lowest point

The following graphic depicts the stone's free body diagram at its highest position.

We may calculate the following using Newton's second rule of motion:

$T+m g=\dfrac{m v_{2}^{2}}{R} \ldots$ (ii)

Where, $v_{2}=$ Velocity at the highest point

The net force operating at the lowest and highest locations is $(T-m g)$ and $(T+m g)$, as shown by equations (i) and (ii), respectively.

10. A disc revolves with a speed of $33 \dfrac{1}{3}$ rev/min, and has a radius of $15 \mathrm{~cm}$. Two coins are placed at $4 \mathrm{~cm}$ and $14 \mathrm{~cm}$ away from the centre of the record. If the co-efficient of friction between the coins and the record is $0.15$, which of the coins will revolve with the record?

Ans: Coin placed at $4 \mathrm{~cm}$ from the centre 

Mass of each coin $=m$

Radius of the disc, $r=15 \mathrm{~cm}=0.15 \mathrm{~m}$ 

Frequency of revolution, $v=33 \dfrac{1}{3} \mathrm{rev} / \mathrm{min}=\dfrac{100}{3 \times 60}=\dfrac{5}{9} \mathrm{rev} / \mathrm{s}$

Coefficient of friction, $\mu=0.15$

The coin with a friction force higher than or equal to the centripetal force supplied by the rotation of the disc will rotate with the disc in the present circumstance. If this isn't the case, the coin will fall out of the disc.

When the coin placed at $4 \mathrm{~cm}$ :

Radius of revolution, $r^{\prime}=4 \mathrm{~cm}=0.04 \mathrm{~m}$ 

Angular frequency, $\omega =2\pi v=2\times \dfrac{22}{7}\times \dfrac{5}{9}=3.49\ {{\text{s}}^{-1}}$

Frictional force, $f={{\mu }_{mg}}=0.15\times m\times 10=1.5\ \text{mN}$

Centripetal force on the coin:

$=0.49 \mathrm{~m} \mathrm{~N}$

As$f>{{F}_{cent.}}$, along with the record, the coin will spin.

Coin placed at $14 \mathrm{~cm}$ :

Radius, $r^{\prime \prime}=14 \mathrm{~cm}=0.14 \mathrm{~m}$

Angular frequency, $\omega=3.49 \mathrm{~s}^{-1}$ 

Frictional force, $f=1.5 \mathrm{~m} \mathrm{~N}$ 

Centripetal force is given as:

${{F}_{\text{cent}\text{. }}}=m{{r}^{\prime \prime }}{{\omega }^{2}}$

$=m \times 0.14 \times(3.49)^{2}=1.7 \mathrm{~m} \mathrm{~N}$

Since $f < {F_{\text {cert }}$, the coin will slip from the surface of the record.

11. A $70 \mathrm{~kg}$ man stands in contact against the inner wall of a hollow cylindrical drum of radius $3\,\text{m}$ rotating about its vertical axis with$\text{200 rev}/\text{min}$. The coefficient of friction between the wall and his clothing is $0.15$. What is the minimum rotational speed of the cylinder to enable the man to remain stuck to the wall (without falling) when the floor is suddenly removed?

Ans: In the given question, we have the mass of the man given as,$m=70 \mathrm{~kg}$

Radius of the drum given as, $r=3 \mathrm{~m}$

Coefficient of friction, $\mu=0.15$ 

Frequency of rotation, $v=200\,\text{rev}/\text{min}=\dfrac{200}{60}=\dfrac{10}{3}\text{rev}/\text{s}$

The normal force provides the essential centripetal force for the rotation of the man$({{F}_{N}})$. 

The man adheres to the drum's wall while the floor spins. As a result, the frictional force $\left( f=\mu {{F}_{N}} \right)$ acting upward balances the weight of the man $mg$ acting downward.

As a result, the man won't fall until: 

mg<f

$m g<\mu F_{N}=\mu m r \omega 2$

$g<\mu r \omega^{2}$

$\omega>\sqrt{\dfrac{g}{\mu r}}$

The minimum angular speed is calculated as follows:

$\omega_{\min }=\sqrt{\dfrac{g}{\mu r}}$

$=\sqrt{\dfrac{10}{0.15\times 3}}=4.71\,\text{rad}{{\text{s}}^{-1}}$

Solve the Following Questions                            5 Mark

1. (a) Define impulse. State its S.I. unit?

(b) State and prove impulse momentum theorem?

Ans: (a) Impulsive forces are those that are applied repeatedly over a short period of time.

Impulse $\mathrm{I}=\mathrm{F} \times \mathrm{t}$

Unit $=\text{NS}$

The term "impulse" refers to a vector quantity that is oriented along the average force$\vec{F} a v$.

(b) The change in momentum of the body is equal to the force's impulse. Newton's second law is that:

$\vec{F}=\dfrac{d \vec{p}}{d t}$

or $d \vec{p}=\vec{F} d t$

At $t=0 \vec{P}=\vec{P}_{1}$ and at $t=t \vec{P}=\vec{P}_{2}$

\[\int_{\overrightarrow{{{P}_{1}}}}^{\overrightarrow{{{P}_{2}}}}{dP}=\int_{v}^{t}{F}dt\]

${{\vec{P}}_{2}}-{{\vec{P}}_{1}}=Ft$

${{\vec{P}}_{2}}-{{\vec{P}}_{1}}=I$

$[\because F t=I($Impulse$)]$

2. A man of mass $70 \mathrm{~kg}$ stands on a weighing scale in a lift which is moving

(a) upwards with a uniform speed of $10 \mathrm{~m} \mathrm{~s}^{-1}$,

(b) downwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$,

(c) upwards with a uniform acceleration of $5 \mathrm{~m} \mathrm{~s}^{-2}$.

What would be the readings on the scale in each case?

(d) What would be the reading if the lift mechanism failed and it hurtled down freely under gravity?

Ans: (a) In the question, we have the mass of the man given as, $m=70 \mathrm{~kg}$

Acceleration given as, $a=0$

We may express the equation of motion as: using Newton's second law of motion.

$R-m g=m a$

Where, $ma$ is the net force acting on the man.

As the lift moves at a constant pace, acceleration $a=0$

$\therefore R=m \mathrm{~g}$

$=70 \times 10=700 \mathrm{~N}$

$\therefore$ Taking a weighing scale reading $=\dfrac{700}{g}=\dfrac{700}{10}=70 \mathrm{~kg}$

(b) Mass of the man, $m=70 \mathrm{~kg}$

Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ downward

We may express the equation of motion as: using Newton's second law of motion.

$R+m \mathrm{~g}=m a$

$R=m(g-a)$

$=70(10-5)=70 \times 5$

$=350 \mathrm{~N}$

$\therefore$ Reading on the weighing scale $=\dfrac{350}{g}=\dfrac{350}{10}=35 \mathrm{~kg}$

(c) Mass of the man, $m=70 \mathrm{~kg}$

Acceleration, $a=5 \mathrm{~m} / \mathrm{s}^{2}$ upward

We may express the equation of motion as: using Newton's second law of motion.

$R-m g=m a$

$R=m(\mathrm{~g}+a)$

$=70(10+5)=70 \times 15$

$=1050 \mathrm{~N}$

$\therefore$ Reading on the weighing scale $=\dfrac{1050}{g}=\dfrac{1050}{10}=105 \mathrm{~kg}$

(d) When the lift is allowed to move due to gravity, acceleration $a=\mathrm{g}$

We may express the equation of motion as: using Newton's second law of motion.

$R+m \mathrm{~g}=m a$

$R=m(g-a)$

$=m(\mathrm{~g}-\mathrm{g})=0$

$\therefore$ Reading on the weighing scale $=\dfrac{0}{g}=0 \mathrm{~kg}$

The man will experience weightlessness.

3. Two masses $8 \mathrm{~kg}$ and $12 \mathrm{~kg}$ 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.

Ans:

The following diagram shows how the given system with two masses and a pulley may be represented:

So, the smaller mass is given as, $m_{1}=8 \mathrm{~kg}$

Larger mass is given as, $m_{2}=12 \mathrm{~kg}$

Therefore, tension in the string $=T$

Mass $m_{2}$, owing to its weight, moves downward with acceleration $a$, and mass $m_{1}$ moves upward.

The system of each mass is subjected to Newton's second law of motion:

For mass $m_{1}:$

The motion equation may be expressed as follows:

$T-m_{1} \mathrm{~g}=m a \ldots(i)$

For mass $m_{2}:$

The motion equation may be expressed as follows:

$m_{2} g-T=m_{2} a \ldots(i i)$

Now on adding equations (i) and (ii), we get:

$\left(m_{2}-m_{1}\right) g=\left(m_{1}+m_{2}\right) a$

$\therefore a=\left( \dfrac{{{m}_{2}}-{{m}_{1}}}{{{m}_{1}}+{{m}_{2}}} \right)g$

$=\left(\dfrac{12-8}{12+8}\right) \times 10=\dfrac{4}{20} \times 10=2 \mathrm{~m} / \mathrm{s}^{2}$

As a result, the acceleration of the masses is $2 \mathrm{~m} / \mathrm{s}^{2}$.

When we replace the value $a$ in equation (ii), we get:

$m_{2} g-T=m_{2}\left(\dfrac{m_{2}-m_{1}}{m_{1}+m_{2}}\right) g$

$T=\left(m_{2}-\dfrac{m_{2}^{2}-m_{1} m_{2}}{m_{1}+m_{2}}\right) g$

$=\left(\dfrac{2 m_{1} m_{2}}{m_{1}+m_{2}}\right) g$

$=\left(\dfrac{2 \times 12 \times 8}{12+8}\right) \times 10$

$=\dfrac{2 \times 12 \times 8}{20} \times 10=96 \mathrm{~N}$

As a result, the string tension is $96 \mathrm{~N}$.

4. Explain why

(a) a horse cannot pull a cart and run in empty space,

(b) passengers are thrown forward from their seats when a speeding bus stops suddenly,

(c) it is easier to pull a lawn mower than to push it,

(d) a cricketer moves his hands backwards while holding a catch.

Ans: (a) A horse pushes the earth backward with some force in order to drive a waggon. The horse's feet, in turn, are subjected to an equal and opposite reaction force from the ground. The horse moves forward as a result of this response force.

There is no response force in an empty area. As a result, a horse can't pull a cart and run in the open. 

(b) When a fast bus comes to a quick stop, the lower section of a passenger's body in touch with the seat comes to an abrupt stop. The top part, on the other hand, prefers to stay in motion (as per the first law of motion). As a result, the upper body of the passenger is pushed forward in the direction of the bus's movement. 

(c) A force at an angle is applied to a lawn mower as it is being pulled, as indicated in the diagram below.

This applied force's vertical component acts upward. This decreases the mower's effective weight.

When pulling a lawn mower, on the other hand, a force is delivered at an angle, as indicated in the diagram below.

The vertical component of the applied force operates in this situation in the direction of the mower's weight. This raises the mower's effective weight.

Because the lawn mower's effective weight is lower in the first scenario, pulling the lawn mower is easier than pushing it. 

(d) The equation of motion, according to Newton's second law of motion, is:

$F=m a=m \dfrac{\Delta v}{\Delta t} \ldots(i)$

Where,

$F=$ As he catches the ball, the cricketer feels a braking force.

$m=$ Mass of the ball

$\Delta t=$ Time of impact of the ball with the hand

From equation I it can be shown that the impact force is inversely proportional to the impact time, i.e.

$F \propto \dfrac{1}{\Delta t} \ldots \ldots(i i)$

Equation (ii) demonstrates that as the time of impact rises, the force received by the cricketer reduces, and vice versa.

A cricketer extends his hand backward while taking a catch to lengthen the time of impact$\vartriangle t$. As a result, the stopping force decreases, protecting the cricketer's hands from being injured.

5. A helicopter of mass $1000 \mathrm{~kg}$ rises with a vertical acceleration of $15 \mathrm{~ms}^{-2} .$ The crew and the passengers weigh $300 \mathrm{~kg}$. Give the magnitude and direction of the

(a) force on the floor by the crew and passengers,

(b) action of the rotor of the helicopter on the surrounding air,

(c) force on the helicopter due to the surrounding air.

Ans: 

(a) Mass of the helicopter is given as, ${{m}_{\text{h}}}=1000~\text{kg}$

Mass of the crew and passengers is given as, $m_{p}=300 \mathrm{~kg}$

Therefore, the total mass of the system, $m=1300 \mathrm{~kg}$

As the acceleration of the helicopter is given as, $a=15 \mathrm{~m} / \mathrm{s}^{2}$

The reaction force $R$ exerted on the system by the floor may be computed using Newton's second equation of motion:

$R-m_{\mathrm{p}} \mathrm{g}=m a$

$=m_{p}(g+a)$

$=300(10+15)=300 \times 25$

$=7500 \mathrm{~N}$

The response force will likewise be directed upward because the chopper is accelerating vertically upward. As a result, the force exerted on the floor by the crew and passengers is $7500~\text{N}$, directed downward, according to Newton's third law of motion. 

(b) The reaction force $R$ experienced by the helicopter may be computed using Newton's second equation of motion as follows:

$=32500 \mathrm{~N}$

The helicopter is being pushed higher by the reaction force of the surrounding air. As a result, the rotor's action on the surrounding air will be $32500\,\text{N}$, directed downward, according to Newton's third law of motion. 

(c) The surrounding air exerts a force of $32500\,\text{N}$on the helicopter, which is directed upward.

6. Ten one-rupee coins are put on top of each other on a table. Each coin has a mass $\text{m}$. Give the magnitude and direction of

(a) the force on the $7^{\text {th }}$ coin (counted from the bottom) due to all the coins on its top,

(b) the force on the $7^{\text {th }}$ coin by the eighth coin,

(c) the reaction of the $6^{\text {th }}$ coin on the $7^{\text {th }}$ coin.

Ans:

(a) The weight of the three coins on top of the seventh coin exerts force on it.

So, let us assume that the weight of one $=3 \mathrm{mg}$

Therefore, the weight of three coins $=3 \mathrm{mg}$

As a result, the three coins on top of the $7^{\text {th }}$ coin exert $3\text{mg}$ of force on it. This force works in a vertical downward direction. 

(b) Because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top, the eighth coin exerts force on the seventh coin.

Therefore, the weight of the eighth coin will be equal to $\operatorname{coin}=m \mathrm{~g}$.

Weight of the ninth and tenth coin will also be the same i.e. $\operatorname{coin}=m \mathrm{~g}$

Therefore, the total weight of these three coins $=3 \mathrm{mg}$

Hence, the force exerted on the $7^{\text {th }}$ coin by the ${{8}^{\text{th }}}$coin is $3\text{mg}$. This force acts vertically downward.

(c) Because of the weight of the four coins \[\left( {{7}^{\text{th }}},{{8}^{\text{th }}},{{9}^{\text{th }}},{{10}^{\text{th }}} \right)\] on top, the ${{6}^{\text{th}}}$ coin suffers a downward pull.

Therefore, $4 \mathrm{mg}$will be the total downward force experienced by $6^{\text {th }}$coin

As per Newton's third law of motion, the $6^{\text {th }}$ coin will produce an equal reaction force on the $7^{\text {th }}$ coin, but in the opposite direction. Hence, the reaction force of the $6^{\text {th }}$coin on the $7^{\text {th }}$ coin is of magnitude $4 \mathrm{mg}$. This force acts in the upward direction.

7. A monkey of mass \[40~\text{kg}\] climbs on a rope (Fig. 5.20) which can stand a maximum tension of $600 \mathrm{~N}$. In which of the following cases will the rope break: the monkey

(a) climbs up with an acceleration of $6 \mathrm{~m} \mathrm{~s}^{-2}$

(b) climbs down with an acceleration of $4 \mathrm{~m} \mathrm{~s}^{-2}$

(c) climbs up with a uniform speed of $5 \mathrm{~m} \mathrm{~s}^{-1}$

(d) falls down the rope nearly freely under gravity?

(Ignore the mass of the rope).

Ans:

Case (a)

From the given question, we have the information given as mass of the monkey, $m=40 \mathrm{~kg}$

Acceleration due to gravity is given as, $g=10 \mathrm{~m} / \mathrm{s}$

Maximum tension that the rope can bear is given as, $T_{\max }=600 \mathrm{~N}$

Acceleration of the monkey is given as, $a=6 \mathrm{~m} / \mathrm{s}^{2}$ upward

So, the equation of motion can be written by using the Newton’s second law of motion:

$T-m \mathrm{~g}=m a$

$\therefore T=m(g+a)$

$=40(10+6)$

$=640 \mathrm{~N}$

As $T>T_{\max }$, in this situation, the rope will break.

Case (b)

Now in this case, the acceleration of the monkey, $a=4 \mathrm{~m} / \mathrm{s}^{2}$ acting downward.

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

$m g-T=m a$

$\therefore T=m(\mathrm{~g}-a)$

$=40(10-4)$

$=240 \mathrm{~N}$

As T < Tmax , in this situation, the rope will not break.

Case (c)

In this case the monkey is climbing with a uniform speed of $5 \mathrm{~m} / \mathrm{s}$. Hence, the acceleration will become zero, i.e.,

$a=0 .$

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

$T-mg=ma$

$T-m g=0$

$\therefore T=m g$

$=40 \times 10=400 \mathrm{~N}$

As T<T in this situation, the rope will not break.

Case (d)

When a monkey will fall freely under the presence of the gravity, the acceleration will become equal to the gravitational acceleration. 

i.e., $a=\mathrm{g}$

Here we will use the Newton’s second law of motion to write the equation of motion. So it can be written as:

$m g-T=m g$

$\therefore T=m(g-g)=0$

As T<Tmax , in this situation, the rope will not break.

8. Two bodies $A$ and $B$ of masses $5 \mathrm{~kg}$ and $10 \mathrm{~kg}$ in contact with each other rest on a table against a rigid wall (Fig. 5.21). The coefficient of friction between the bodies and the table is $0.15$. A force of $200 \mathrm{~N}$ is applied horizontally to $A$. What are (a) the reaction of the partition (b) the action-reaction forces between $A$ and $B$ ? What happens when the wall is removed? Does the answer to (b) change, when the bodies are in motion? Ignore the difference between ${{\mu }_{s}}$ and $\mu_{\mathrm{k}}$.

Ans:

(a) In the given question, the mass of body A is given as, $m_{A}=5 \mathrm{~kg}$

Mass of body B is given as, $m_{\mathrm{B}}=10 \mathrm{~kg}$

Applied force, $F=200 \mathrm{~N}$

Coefficient of friction, $\mu_{s}=0.15$

The friction force is calculated using the following formula:$f_{s}=\mu\left(m_{\mathrm{A}}+m_{\mathrm{B}}\right) \mathrm{g}$

$=0.15(5+10) \times 10$

$=1.5 \times 15=22.5 \mathrm{~N}$ in leftward

Net force acting on the partition $=200-22.5=177.5 \mathrm{~N}$ rightward

The reaction force of the partition will be in the opposite direction as the net applied force, according to Newton's third law of motion.

Hence, the reaction of the partition will be $177.5 \mathrm{~N}$, in the leftward direction.

(b) Force of friction on mass A:

$f_{\mathrm{A}}=\mu m_{\mathrm{A}} \mathrm{g}$

$=0.15 \times 5 \times 10=7.5 \mathrm{~N}$ leftward

Net force exerted by mass $\mathrm{A}$ on mass $\mathrm{B}=200-7.5=192.5 \mathrm{~N}$ rightward

According to Newton's third rule of motion, mass B will exert an equivalent amount of reaction force on mass A, i.e., $192.5 \mathrm{~N}$ acting leftward.

The two bodies will move in the direction of the applied force after the wall is removed.

On the moving system, there is a net force$=177.5 \mathrm{~N}$

As a result, the system of acceleration's equation of motion may be represented as:

Net force $=\left(m_{A}+m_{B}\right) a$

$\therefore a=\dfrac{\text { Net force }}{m_{A}+m_{\mathrm{B}}}$

$=\dfrac{177.5}{5+10}=\dfrac{177.5}{15}=11.83 \mathrm{~m} / \mathrm{s}^{2}$

Net force causing mass $\mathrm{A}$ to move:

$F_{A}=m_{A} a=5 \times 11.83=59.15 \mathrm{~N}$

Net force exerted by mass $\mathrm{A}$ on mass $\mathrm{B}=192.5-59.15=133.35 \mathrm{~N}$

The direction of motion will be affected by this force. According to Newton's third law of motion, mass B will exert an equal amount of force on mass A, i.e., $133.3 \mathrm{~N}$, acting in the opposite direction of motion.

9. The rear side of a truck is open and a box of $40 \mathrm{~kg}$ mass is placed $5 \mathrm{~m}$ away from the open end as shown in Fig. 5.22. The coefficient of friction between the box and the surface below it is 0.15. On a straight road, the truck starts from rest and accelerates with $2 \mathrm{~m} \mathrm{~s}-2 .$ At what distance from the starting point does the box fall off the truck? (Ignore the size of the box).

Ans: In the give question, we have the mass of the box given as, $m=40 \mathrm{~kg}$

Coefficient of friction is given as, $\mu=0.15$

Initial velocity is given as, $u=0$

Acceleration is given as, $a=2 \mathrm{~m} / \mathrm{s}^{2}$

Distance of the box from the end of the truck is given as, $s^{\prime}=5 \mathrm{~m}$

The force on the box generated by the truck's accelerated speed is described by Newton's second law of motion:

$F=m a$

$=40 \times 2=80 \mathrm{~N}$

A reaction force of $80 \mathrm{~N}$ is operating on the box in the rearward direction, according to Newton's third law of motion. The force of friction $f$ occurring between the box and the truck's floor opposes the box's backward motion. The following factors contribute to this force:

$f=\mu m g$

$=0.15 \times 40 \times 10=60 \mathrm{~N}$

$\therefore$ On the block, the net force is:

$F_{\text {mat }}=80-60=20 \mathrm{~N}$ acting in backward.

The box's rearward acceleration is calculated as follows:

$a_{h x k}=\dfrac{F_{m t}}{m}=\dfrac{20}{40}=0.5 \mathrm{~m} / \mathrm{s}^{2}$

Time $t$ may be computed using the second equation of motion as follows:

$s^{\prime}=u t+\dfrac{1}{2} a_{k x c i} t^{2}$

$5=0+\dfrac{1}{2} \times 0.5 \times t^{2}$

$\therefore t=\sqrt{20}\ \text{s}$

As a result, after $\sqrt{20} s$, the box will fall from the truck.

The below formula will give the distance $s$ travelled by the vehicle in$\sqrt{20} \mathrm{~s}$:

$s=u t+\dfrac{1}{2} a t^{2}$

$=0+\dfrac{1}{2} \times 2 \times(20)^{2}$

$=20 \mathrm{~m}$

10. You may have seen in a circus a motorcyclist driving in vertical loops inside a 'death-well' (a hollow spherical chamber with holes, so the spectators can watch from outside). Explain clearly why the motorcyclist does not drop down when he is at the uppermost point, with no support from below. What is the minimum speed required at the uppermost position to perform a vertical loop if the radius of the chamber is $25 \mathrm{~m}$ ?

Ans:

Because both the force of normal response and the weight of the biker act downward and are balanced by the centripetal force, a motorcyclist does not fall at the top point of a vertical loop in a death-well. The scenario is depicted in the diagram below.

The total of the normal force $({{F}_{N}})$ and the force due to gravity$\left( {{F}_{g}}={{m}_{g}} \right)$ acting on the biker is the net force.

The centripetal acceleration $\left( a \right)$ equation of motion may be expressed as:

$F_{\text {net }}=m a c$

${{F}_{N}}+{{F}_{g}}=m{{a}_{c}}$

${{F}_{N}}+mg=\dfrac{m{{v}^{2}}}{r}$

The motorcyclist's speed provides a normal reaction. At the slowest possible speed $\left( {{v}_{min}} \right),{{F}_{N}}=0$

$mg=\dfrac{m{{v}_{\min }}^{2}}{r}$

$\therefore v_{\min }=\sqrt{r g}$

$=\sqrt{25 \times 10}=15.8 \mathrm{~m} / \mathrm{s}$

11. A thin circular loop of radius $R$ rotates about its vertical diameter with an angular frequency $\omega$. Show that a small bead on the wire loop remains at its lowermost point for $\omega \leq \sqrt{g / R}$. What is the angle made by the radius vector joining the centre to the bead with the vertical downward direction for $\omega=\sqrt{2 g / R}$ ? Neglect friction.

Ans:

Let us assume that the radius vector joining the bead with the centre is making an angle $\theta$, in the vertically downward direction.

OP $=R=$will be the radius of the circle

$N=$ Normal reaction

The vertical and horizontal force equations can be expressed as follows:

$m g=N \cos \theta \ldots(i)$

$m l \omega^{2}=N \sin \theta \ldots(i i)$

In $\Delta O P Q$, we have:

$\sin \theta=\dfrac{l}{R}$

$l=R \sin \theta \ldots$ (iii)

Putting equation (iii) in equation (ii), we get:

$m(R \sin \theta) \omega^{2}=N \sin \theta$

$m R \omega^{2}=N$

Putting equation (iv) in equation (i), we get:

$m g=m R \omega^{2} \cos \theta$

$\cos \theta=\dfrac{g}{R \omega^{2}} \ldots(v)$

Since $\cos \theta \leq 1$, the bead will remain at its lowermost point for $\dfrac{g}{R \omega^{2}} \leq 1$, i.e., for $\omega \leq \sqrt{\dfrac{g}{R}}$ 

For $\omega=\sqrt{\dfrac{2 g}{R}}$ or $\omega^{2}=\dfrac{2 g}{R} \quad \ldots \ldots(v i)$

On equating equations $(v)$ and $(v i)$, we get:

$\dfrac{2 g}{R}=\dfrac{g}{R \cos \theta}$

$\Rightarrow \cos \theta =\dfrac{1}{2}$

$\therefore \theta=\cos ^{-1}(0.5)=60^{\circ}$

Important Questions for CBSE Class 11 Physics Chapter 5 (Law of Motion) - Benefits of Studying the Important Questions and Answers

The CBSE Class 11 students are advised to include this study material of important questions and answers to get the following benefits:

1. Important questions and answers make the students aware of the important topics in the chapter.

2. The chapter might seem lengthy for some students, in that case, they can always avail themselves of these important questions and answers. 

3. Students can also use this study material at the time of revising this chapter. This will help them to cover the important points right before their exam.

4. Laws of Motion can seem complicated for some students, in that case, they can only study the important questions and answers before the Physics exam.

Conclusion  

The CBSE Class 11 Physics Chapter 5 - Law of Motion covers fundamental concepts that form the basis of classical mechanics. Understanding these principles is crucial for building a strong foundation in physics. The chapter introduces Newton's laws of motion, emphasizing their application in real-world scenarios. By mastering these concepts, students can analyze the motion of objects, calculate forces, and solve practical problems. Important questions in this chapter provide invaluable practice and reinforce students' understanding of the topic. Regular practice of these questions enhances problem-solving skills, critical thinking, and conceptual clarity, ultimately preparing students for higher-level physics topics and various competitive exams.

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