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]]>**Circular Motion** HC Verma Exercise Questions Solutions Ch-7 Concept of Physics Vol-1 for ISC Class-11. Step by Step Solution of **Exercise **Questions of Ch-7 **Circular Motion **HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Ch-7 | Circular Motion |

Class | 11 |

Vol | 1st |

writer | H C Verma |

Book Name | Concept of Physics |

Topics | Solution of Exercise Questions |

Page-Number | 114, 115, 116 |

-: Select Topics :-

Exercise

(HC Verma Ch-7 Concept of Physics Vol-1 for ISC Class-11)

Page-114

Find the acceleration of the moon with respect to the earth from the following data : Distance between the earth and the moon = 3.85 × 10^{5} km and the time taken by the moon to complete one revolution around the earth = 27.3 days.

Distance between the Earth and the Moon :

r=3.85×10^{5}km =3.85×10^{8}m

Time taken by the Moon to revolve around the Earth :

T=27.3 days

= 24 × 3600 × 27.3s=2.36 × 10^{6}s

Velocity of the Moon :

Find the acceleration of a particle placed on the surface of the earth at the equator due to earth’s rotation. The diameter of earth = 12800 km and it takes 24 hours for the earth to complete one revolution about its axis.

Diameter of the Earth = 12800 km

So, radius of the Earth, R = 6400 km = 6.4 × 10^{6} m

Time period of revolution of the Earth about its axis :

A particle moves in a circle of radius 1.0 cm at a speed given by *v* = 2.0 *t* where *v* is cm/s and *t* in seconds.

(a) Find the radial acceleration of the particle at *t* = 1 s.

(b) Find the tangential acceleration at *t* = 1 s.

(c) Find the magnitude of the acceleration at *t* = 1 s.

Speed is given as a function of time. Therefore, we have:

*v* = 2*t*

Radius of the circle = *r* = 1 cm

At time *t *= 2 s, we get :

(a) Radial acceleration

A scooter weighing 150 kg together with its rider moving at 36 km/hr is to take a turn of a radius 30 m. What horizontal force on the scooter is needed to make the turn possible ?

Given:

Mass = m = 150 kg

Speed = v = 36 km/hr = 10 m/s

Radius of turn = r = 30 m

Let the horizontal force needed to make the turn be F. We have :

If the horizontal force needed for the turn in the previous problem is to be supplied by the normal force by the road, what should be the proper angle of banking?

Given:

Speed of the scooter = v = 36 km/hr = 10 m/s

Radius of turn = r = 30 m

Let the angle of banking be θ

We have :

A park has a radius of 10 m. If a vehicle goes round it at an average speed of 18 km/hr, what should be the proper angle of banking?

Given:

Speed of the vehicle = v = 18 km/h = 5 m/s

Radius of the park = r = 10 m

Let the angle of banking be θ

If the road of the previous problem is horizontal (no banking), what should be the minimum friction coefficient so that scooter going at 18 km/hr does not skid?

If the road is horizontal (no banking),

we have :

mv²/R = fs

N = mg

Here, f_{s} is the force of friction and N is the normal reaction.

If μ is the friction coefficient, we have :

Friction force = fs = *μ *N

So, mv²/R = *μ* mg

Here,

Velocity = v = 5 m/s

Radius = R = 10 m

A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?

Given:

Angle of banking = θ = 30°

Radius = r = 50 m

Assume that the vehicle travels on this road at speed v so that the friction is not used.

We get :

In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the centre at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes round the proton in a circle of radius 5.3 × 10^{−11} m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 × 10^{−31} kg and charge of the electron = 1.6 × 10^{−19} C.

Given :

Radius of the orbit of the ground state = r = 5.3 × 10^{−11} m

Mass of the electron = m = 9.1 × 10^{−31 }kg

Charge of electron = q = 1.6 × 10 ^{−19 }c

We know:

Centripetal force = Coulomb attraction

Therefore, we have :

** **

A stone is fastened to one end of a string and is whirled in a vertical circle of radius *R*. Find the minimum speed the stone can have at the highest point of the circle.

Let m be the mass of the stone.

Let v be the velocity of the stone at the highest point.

R is the radius of the circle.

Thus, in a vertical circle and at the highest point,

we have :

mv²/R = mg

⇒ v² = R g

⇒ v = √Rg

A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface?

Diameter of the fan = 120 cm

∴ Radius of the fan = r = 60 cm = 0.6 m

Mass of the particle = M = 1 g = 0.001 kg

Frequency of revolutions = n = 1500 rev/min = 25 rev/s

Angular velocity = ω = 2πn = 2π × 25 = 57.14 rev/s

Force of the blade on the particle:

F = Mrw^{2}

= (0.001) × 0.6 × (157.14)^{2}

=14.8 N

The moving fan exerts this force on the particle.

The particle also exerts a force of 14.8 N on the blade along its surface.

A mosquito is sitting on an L.P. record disc rotating on a turn table at revolutions per minute. The distance of the mosquito from the centre of the turn table is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than *π*^{2}/81. Take g =10 m/s^{2}.

r = 10 cm = 0 . 1 m

g = 10 m/ s²

It is given that the mosquito is sitting on the L.P. record disc.Therefore,we have : Friction force ≥ Centrifugal force on the mosquito

A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by he string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s^{2}.

Speed of the car = v = 36 km/hr = 10 m/s

Acceleration due to gravity = g = 10 m/s^{2 }

^{}

Let T be the tension in the string when the pendulum makes an angle θ with the vertical.

From the figure, we get :

(Page-115)

The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at this instant.

Given:

Mass of the bob = m = 100 gm = 0.1 kg

Length of the string = r = 1 m

Speed of bob at the lowest point in its path = 1.4 m/s

Let T be the tension in the string.

From the free body diagram,

we get :

^{2}/2 and SINθ ≈ θ for small θ.

Given:

Mass of the bob = m = 0.1 kg

Length of the circle = R = 1 m

Velocity of the bob = v = 1.4 m/s

Let T be the tension in the string when it makes an angle of 0.20 radian with the vertical.

From the free body diagram, we get :

Suppose the amplitude of a simple pendulum having a bob of mass m is θ_{0}. Find the tension in the string when the bob is at its extreme position.

Let T be the tension in the string at the extreme position.

Velocity of the pendulum is zero at the extreme position.

So, there is no centripetal force on the bob.

∴ T = mgcosθ_{0 }

_{}

A person stands on a spring balance at the equator.

(a) By what fraction is the balance reading less than his true weight?

(b) If the speed of earth’s rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?

(a)

Balance reading = Normal force on the balance by the Earth.

At equator, the normal force (N) on the spring balance :

N = mg − mω^{2}r

True weight = mg

Therefore, we have :

(b)

When the balance reading is half, we have :

A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?

Given:

Speed of vehicles = v = 36 km/hr = 10 m/s

Radius = r = 20 m

Coefficient of static friction = μ = 0.4

Let the road be banked with an angle θ

We have :

_{}

_{}

_{if R == N}

When the car travels at the maximum speed, it slips upward and μN_{1} acts downward.

Therefore we have :

Similarly, for the other case, it can be proved that :

Thus, the possible speeds are between 14.7 km/hr and 54 km/hr so that the car neither slips down nor skids up.

A motorcycle has to move with a constant speed on an over bridge which is in the form of a circular arc of radius R and has a total length L. Suppose the motorcycle starts from the highest point.(a) What can its maximum velocity be for which the contact with the road is not broken at the highest point? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road? (c) What maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?

R = Radius of the bridge

L = Total length of the over bridge

(a) At the highest point:

Let m be the mass of the motorcycle and v be the required velocity.

Suppose it loses contact at B.

So, it will lose contact at a distance *πR/3* from the highest point.

(c) Let the uniform speed on the bridge be v. The chances of losing contact is maximum at the end bridge. We have :

A car goes on a horizontal circular road of radius *R*, the speed increasing at a constant rate *dv/dt=a* . The friction coefficient between the road and the tyre is μ. Find the speed at which the car will skid.

Let v be the speed of the car.

Since the motion is non-uniform, the acceleration has both radial (a_{r}) and tangential (a_{t}) components.

a_{r}=v²/R

a_{t}=dv/dt=a

From free body diagram, we have :

A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is μ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip?

According to the question :

A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E.1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed? Take g = 10 m/s^{2}.

Given:

Radius of the curves = r = 100 m

Mass of the cycle = m = 100 kg

Velocity = v = 18 km/hr = 5 m/s

(b) At B and D, we have:

Tendency of the cycle to slide is zero.

So, at B and D, frictional force is zero.

At C, we have :

mgsinθ = f

(d) To find the minimum coefficient of friction, we have to consider a point where N is minimum or a point just before c .

In a children’s park a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod figure (7-E.2). Let the mass of each kid be 15 kg, the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.

Given :

Thus, the force of frictional on one of the kids is 10 π

A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is μ. Find the range of the angular speed for which the block will not slip.

When the bowl rotates at maximum angular speed, the block tends to slip upwards.

Also, the frictional force acts downward.

Here, we have:

Radius of the path that the block follow = r = Rsinθ

Let N_{1} be the normal reaction on the block and ω_{1} be the angular velocity after which the block will slip.

From the free body diagram-1, we get :

Also, the frictional force acts downward.

Here, we have:

Radius of the path that the block follow = r = Rsinθ

Let N

From the free body diagram-1, we get :

On solving the two equation, we get :

Let us now find the minimum speed (ω_{2}) on altering the direction of

Hence, the range of speed is between ω_{2} and ω_{1}.

A particle is projected with a speed *u* at an angle θ with the horizontal. Consider a small part of its path near the highest position and take it approximately to be a circular arc. What is the radius of this circular circle? This radius is called the radius of curvature of the curve at the point.

At the highest point, the vertical component of velocity is zero.

So, at the highest point, we have:

velocity = *v* = *u*cos*θ*

Centripetal force on the particle = mv²/r

At the highest point*,* we have :

mg=mv²/r

Here,* r *is the radius of curvature of the curve at the point.

What is the radius of curvature of the parabola traced out by the projectile in the previous problem at a point where the particle velocity makes an angle θ/2 with the horizontal?

Let *u* be the initial velocity and *v* be the velocity at the point where it makes an angle θ2 with the horizontal component remains unchanged

Therefore, we get :

A block of mass m moves on a horizontal circle against the wall of a cylindrical room of radius R. The floor of the room on which the block moves is smooth but the friction coefficient between the wall and the block is μ. The block is given an initial speed v_{0}. As a function of the speed v writes

(a) the normal force by the wall on the block,

(b) the frictional force by a wall, and

(c) the tangential acceleration of the block.

(d) Integrate the tangential acceleration to obtain the speed of the block after one revolution.

A block of mass ‘m’ moves on a horizontal circle against the wall of a cylindrical room of radius ‘R’

Friction coefficient between wall & the block is μ.

(Page-116)

A table with smooth horizontal surface is fixed in a cabin that rotates with a uniform angular velocity ω in a circular path of radius R( figure 7-E.3). A smooth groove AB of length L(<<R) is made the surface of the table. The groove makes an angle θ with the radius OA of the circle in which the cabin rotates. A small particle is kept at the point A in the groove and is released to move at the point A in the groove and is released to move along AB. Find the time taken by the particle to reach the point B.

Let the mass of the particle be m.

Radius of the path = R

Angular velocity = ω

Force experienced by the particle = mω^{2}R

The component of force mRω^{2} along the line AB (making an angle with the radius) provides the necessary force to the particle to move along AB.

A car moving at a speed of 36 km/hr is taking a turn on a circular road of radius 50 m. A small wooden plate is kept on the seat with its plane perpendicular to the radius of the circular road ( figure 7-E.4). A small block of mass 100 g is kept on the seat which rests against the plate. the friction coefficient between the block and the plate is. (a) Find the normal contact force exerted by the plate on the block. (b) The plate is slowly turned so that the angle between the normal to the plate and the radius of the road slowly increases. Find the angle at which the block will just start sliding on the plate.

Given :

Speed of the car = v = 36 km/h = 10 m/s

Radius of the road = r = 50 m

Friction coefficient between the block and the plate = μ = 0.58

Mass of the small body = m = 100 g = 0.1 kg

(a) Let us find the normal contact force (N) exerted by the plant of the block.

N=mv2r=0.1×10050=15=0.2

(b) The plate is turned; so, the angle between the normal to the plate and the radius of the rod slowly increases.

Therefore, we have :

A table with smooth horizontal surface is placed in a circle of a large radius R ( figure 7-E.5). A smooth pulley of small radius is fastened to the table. Two masses m and 2m placed on the table are connected through a string going over the pulley. Initially the masses are held by a person with the string along the outward radius and then the system is released from rest (with respect to the cabin). Find the magnitude of the initial acceleration of the masses as seen from the cabin and the tension in the string.

Let the bigger mass accelerates towards left with acceleration a.

Let T be the tension in the string and ω be the angular velocity of the table.

From the free body diagram, we have :

—: End of **Circular Motion ****Exercise Solutions** Concept of Physics:–

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]]>HC Verma Solutions Vol-1 Concept of Physics Bharti Bhawan Publishers for ISC Class-11 . There are various publications in Class 11th physics but HC Verma Bharti Bhawan Publishers is most famous among ISC Student. Visit official website CISCE for detail information about ISC Board Class-11 Physics HC Verma Solutions Vol-1 Concept of Physics Bharti Bhawan Publishers for ISC Class-11 […]

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]]>HC Verma Solutions Vol-1 Concept of Physics Bharti Bhawan Publishers for ISC Class-11 . There are various publications in Class 11th physics but HC Verma Bharti Bhawan Publishers is most famous among ISC Student. Visit official website CISCE for detail information about** ISC** Board **Class-11 Physics**

Board | ISC and Other |

Publications | Bharti Bhawan Publishers |

Subject | Physics |

Class | 11 |

Vol | 1st |

writer | H C Verma |

Book Name | Concept of Physics |

Topics | Solution of Question for Short Answer , Objective-I, Objective-II , Exercise |

Edition | 2020-2021 |

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**Chapter 1: Introduction to Physics**

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

Question for Short Answer

Objective-I

Objective-II

Exercise

HC Verma is one of the best books for studying physics concepts and practicing problems for ISC

HC Verma physics book is divided into two parts. Part-1 deals with the syllabus of class 11 and contain 22 chapters while part-2 deals with class 12 concepts and there are 25 chapters in total.

HC Verma book’s content is written in a simple language, concepts are explained in detail, amount of examples and variety of practice problems are present.

Harish Chandra Verma is a renowned professor at IIT-Kanpur working in the department of physics. He is also an experimental physicist and his field of research is in nuclear physics.

You can get HC Verma solutions with questions in proper format including **Question for Short Answer** , **Objective-I**, **Objective-II** and **Exercise** at www.icsehelp.com

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]]>Rest and Motions Exercise Questions Solutions HC Verma Ch-3 Concept of Physics Vol-1 for Class-11. Step by Step Solution of Exercise Questions for Ch-3 Rest and Motions Kinematics HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics. Rest and Motions Exercise Questions Solutions HC Verma Ch-3 Concept of Physics Vol-1 for […]

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]]>**Rest and Motions Exercise** Questions Solutions HC Verma Ch-3 Concept of Physics Vol-1 for Class-11. Step by Step Solution of **Exercise **Questions for Ch-3 **Rest and Motions **Kinematics** **HC Verma Concept of Physics . Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Ch-3 | Rest and Motions Kinematics |

Class | 11 |

Vol | 1st |

writer | H C Verma |

Book Name | Concept of Physics |

Topics | Solution of Exercise Questions |

Page-Number | 51, 52, 53, 54 |

-: Select Topics :-

Exercise (Currently Open)

(Page-51)

A man has to go 50 m due north, 40 m due east and 20 m due south to reach a field.

(a) What distance he has to walk to reach the field?

(b) What is his displacement from his house to the field?

(a) Distance travelled by the man = AB + BC + CD = 50 + 40 + 20 = 110 m

(b) AF = AB − BF = 50 − 20 = 30 m

Displacement = Final position − Initial position = AD

A particle starts from the origin, goes along the X-axis to the point (20 m, 0) and then return along the same line to the point (−20 m, 0). Find the distance and displacement of the particle during the trip.

Let the points be O(0,0), A(20 m, 0) and B(−20 m, 0).

(i) Distance travelled = OA + AB = 20 + 40 = 60 m

(ii) Displacement = OB = 20 m (in the negative direction)

It is 260 km from Patna to Ranchi by air and 320 km by road. An aeroplane takes 30 minutes to go from Patna to Ranchi whereas a delux bus takes 8 hours.

(a) Find the average speed of the plane.

(b) Find the average speed of the bus.

(c) Find the average velocity of the plane.

(d) Find the average velocity of the bus.

(a)

(b)

(c)

The plane moves in a straight path.

Average velocity = Average speed = 520 Km / h (Patna to Ranchi)

(d) Straight path distance from Patna to Ranchi = Displacement of the bus = 260 km

When a person leaves his home for sightseeing by his car, the meter reads 12352 km. When he returns home after two hours the reading is 12416 km

(a) What is the average speed of the care during this period?

(b) What is the average of velocity?

(a) Total distance covered = 12416 − 12352 = 64 km

(b) Because he returns to his house, the displacement is zero. So, the average velocity is zero

An athlete takes 2.0 s to reach his maximum speed of 18.0 km/h. What is the magnitude of this average acceleration.

Initial velocity, *u* = 0

Final velocity, *v* = 18 km/h = 5 m/s

Time interval, *t* = 2 s

Using

The speed of a car as a function of time is shown in the following figure. Find the distance travelled by the car in 8 seconds and its acceleration.

Figure-3-E.1

The acceleration of a cart started at t = 0, varies with time as shown in the following figure. Find the distance travelled in 30 seconds and draw the position-time graph

Figure-3-E.2

In the first 10 seconds,

At t = 10 s,

v = u + at = 0 + 5 × 10 = 50 ft/s

∴ From 10 to 20 seconds (∆t = 20 − 10 = 10 s), the particle moves with a uniform velocity of 50 ft/s.

Distance covered from t = 10 s to t = 20 s:

S_{2} = 50 × 10 = 500 ft

Between 20 s to 30 s, acceleration is constant, i.e., −5 ft/s^{2}.

At 20 s, velocity is 50 ft/s.

t = 30 − 20 = 10 s

Total distance travelled is 30 s:

S_{1} + S_{2} + S_{3}

= 250 + 500 + 250

= 1000 ft

The position–time graph:

Figure (3 -E.3) Shows the graph of velocity versus time for a particle going along the X-axis.

(a )Find the acceleration

(b) Find the distance travelled in 0 to 10s and

(c) Find the displacement in 0 to 10 s.

(a )

Slope of the *v–t *graph gives the acceleration.

Acceleration

(b)

Area in the *v–t* graph gives the distance travelled.

Distance travelled = Area of ∆ABC + Area of rectangle OABD

(c)

Displacement is the same as the distance travelled.

Displacement = 50 m

Figure (3 -E.4) shows the graph of the *x*-coordinate of a particle going along the X-axis as a function of time.

(a) Find the average velocity during 0 to 10 s,

(b) Find instantaneous velocity at 2, 5, 9 and 12s.

(a)

Displacement from* t* = 0 s to *t *= 10 s:

*x* = 100 m

Time = 10 s

Average velocity from 0 to 10 seconds,

(b)

Slope of the *x–t* graph gives the velocity.

At 2.5 s,

slope =

⇒ *v*_{inst} = 20 m/s

At 5 s, *v*_{inst} = 0.

At 8 s, *v*_{inst} = 20 m/s.

At 12 s, *v*_{inst} = −20 m/s.

From the velocity-time plot shown in figure,(3 -E.5) find the distance travelled by the particle during the first 40 seconds. Also find the average velocity during this period

Area shown in the *v–t* graph gives the distance travelled.

∴ Distance travelled in the first 40 seconds = Area of ∆OAB + Area of ∆BCD

As the displacement is zero, the average velocity is zero.

(Page-52)

figure (3 -E.6) shows x-t graph of a particle. Find the time t such that the average velocity of the particle during the period 0 to t is zero.

(3 -E.6)

Consider point B_{1} at 12 s.

At *t* = 0 s, S = 20 m and at *t* = 12 s, S = 20 m.

For the time interval 0–12, change in the displacement is zero.

Hence, the time is 12 seconds.

A particle starts from a point A and travels along the solid curve shown in figure (3 -E.7) . Find approximately the position B of the particle such that the average velocity between the position A and B has the same direction as the instantaneous velocity at B.

At position B , the instantaneous velocity of the particle has the direction along BC

An object having a velocity 4.0 m/s is accelerated at the rate of 1.2 m/s^{2} for 5.0 s. Find the distance travelled during the period of acceleration.

Given:

Velocity, *u* = 4.0 m/s

Acceleration, *a *= 1.2 m/s^{2}

Time, *t = *5.0 s

Distance travelled :

A person travelling at 43.2 km/h applies the brake giving a deceleration of 6.0 m/s^{2} to his scooter. How far will it travel before stopping?

Initial velocity, *u* = 43.2 km/h = 12 m/s

Final velocity, *v* = 0

Acceleration, *a* = −6 m/s^{2}

^{}

A train starts from rest and moves with a constant acceleration of 2.0 m/s^{2} for half a minute. The brakes are then applied and the train comes to rest in one minute. Find

(a) the total distance moved by the train

(b) the maximum speed attained by the train .

(c) the position(s) of the train at half the maximum speed.

(a)

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

Let the final velocity be *v* before the brakes are applied.

Now,

*t *= 30 s

*v *= *u* + *at*

*v* = 0 + 2 × 30

⇒ *v* = 60 m/s

^{}

When the brakes are applied:

*u*‘ = 60 m/s

*v*‘ = 0

*t* = 1 min = 60 s

Acceleration:

*s *= *s*_{1} +*s*_{2} = 1800 + 900 = 2700 m

⇒ *s* = 2.7 km

(b)

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

Let the final velocity be *v* before the brakes are applied.

Now,

*t *= 30 s

*v *= *u* + *at*

*v* = 0 + 2 × 30

⇒ *v* = 60 m/s

Maximum speed attained by the train,* v* = 60 m/s

(c)

Initial velocity, *u* = 0

Acceleration, *a* = 2 m/s^{2}

Let the final velocity be *v* before the brakes are applied.

Now,

*t *= 30 s

*v *= *u* + *at*

*v* = 0 + 2 × 30

⇒ *v* = 60 m/s

Half the maximum speed =60 / 2=30 m /s

When the train is accelerating with an acceleration of 2 m/s^{2}:

Distance,

⇒ *s = *225 m

When the train is decelerating with an acceleration of – 1 m/s^{2}:

Distance,

⇒ *s = *1350 m

Position from the starting point

Position from the starting point

= 900 + 1350 = 2250

= 2.25 km

A bullet travelling with a velocity of 16 m/s penetrates a tree trunk and comes to rest in 0.4 m. Find the time taken during the retardation.

Initial velocity, *u* = 16 m/s

Final velocity, *v* = 0

Distance, *s* = 0.4 m

Acceleration,

A bullet going with speed 350 m/s enters a concrete wall and penetrates a distance of 5.0 cm before coming to rest. Find the deceleration.

Initial velocity, *u *= 350 m/s

Final velocity, *v* = 0

Distance travelled by the bullet before coming to rest, *s* = 5 m

A particle starting from rest moves with constant acceleration. If it takes 5.0 s to reach the speed 18.0 km/h find

(a ) the average velocity during this period .

(b) distance travelled by the particle during this period.

(a )

Initial velocity of the particle, *u *= 0

Final velocity of the particle, *v *= 18 km/h = 5 m/s

Time, *t* = 5 s

Acceleration, *a* = (*v* − *u*) /*t*

⇒ *a* = (5 − 0) /5 = 1 m/s^{2}

Distance,

(b)

Initial velocity of the particle, *u *= 0

Final velocity of the particle,*v *= 18 km/h = 5 m/s

Time,*t* = 5 s

Acceleration,*a* = (*v* − *u*) /*t*

⇒*a* = (5 − 0) /5 = 1 m/s^{2}

Distance,

Final velocity of the particle,

Time,

Acceleration,

⇒

Distance,

Distance travelled,* s* = 12.5 m

A driver takes 0.20 s to apply the brakes after he sees a need for it. This is called the reaction time of the driver. If he is driving a car at a speed of 54 km/h and the brakes cause a deceleration of 6.0 m/s^{2}, find the distance travelled by the car after he sees the need to put the brakes on.

In the reaction time, the car moves with a constant speed of 54 km/h, i.e., 15 m/s.

Distance travelled in this time, *s*_{1} = 15 × 0.2 = 3 m

When the brakes are applied:

Initial velocity of the car, *u *= 15 m/s

Final velocity of the car, *v* = 0

Acceleration, *a* = −6 m/s^{2}

Distance,

Total distance, *s* = *s*_{1} +*s*_{2}

*⇒ s* = 3 + 18.75 = 21.75 m

*⇒ *22 m

Complete the following table:

Car Model |
Driver XReaction time 0.20 s |
Driver YReaction time 0.30 s |

A (deceleration on hard braking = 6.0 m/s^{2}) |
Speed = 54 km/h Braking distance a = …………Total stopping distance b = ………… |
Speed = 72 km/h Braking distance c = ………..Total stopping distance d = ………… |

B (deceleration on hard braking = 7.5 m/s^{2}) |
Speed = 54 km/h Breaking distance e = ………..Total stopping distance f = ………… |
Speed 72 km/h Braking distance g = ………….Total stopping distance h = ………… |

Braking distance: Distance travelled after the brakes are applied.

Total stopping distance = Braking distance + Distance travelled in the reaction time

**Case A:**

Deceleration = 6.0 m/s^{2}

**For driver X:**

Initial velocity, *u* = 54 km/h= 15 m/s

Final velocity, *v* = 0

Braking distance,

Distance travelled in the reaction time = 15 × 0.20 = 3 m

Total stopping distance,* b* = 19 + 3 = 22 m

**For driver Y:**

Initial velocity, *u* = 72 km/h= 20 m/s

Final velocity, *v* = 0

Braking distance,

Distance travelled in the reaction time = 20 × 0.30 = 6 m

Total stopping distance,*d* = 33 + 6 = 39 mCase B:

Deceleration = 6.0 m/s^{2}

Now, we have:

*e = *15 m

*f* = 18 m

*g = *27 m

*h* = 33 m

Total stopping distance,

Deceleration = 6.0 m/s

Now, we have:

Car Model |
Driver XReaction Time = 0.20 s |
Driver YReaction Time = 0.30 s |

A (deceleration on hard braking = 6.0 m/s^{2}) |
Speed = 54 km/h Braking distance , a = 19 mTotal stopping distance, b = 22 m |
Speed = 72 km/h Braking distance, c = 33 mTotal stopping distance, d = 39 m |

B (deceleration on hard braking = 7.5 m/s^{2}) |
Speed = 54 km/h Breaking distance, e = 15 mTotal stopping distance, f = 18 m |
Speed = 72 km/h Braking distance, g = 27 mTotal stopping distance, h = 33 m |

A police jeep is chasing a culprit going on a motorbike. The motorbike crosses a turning at a speed of 72 km/h. The jeep follows it at a speed of 90 km/h, crossing the turning ten seconds later than the bike. Assuming that they travel at constant speeds, how far from the turning will the jeep catch up with the bike?

Velocity of the police jeep,*v*_{p} = 90 km/h = 25 m/s

Velocity of the culprit riding the motor bike,*v*_{c} = 72 km/h = 20 m/s

Velocity of the culprit riding the motor bike,

In 10 seconds, the culprit reaches point B from point A.

Distance covered by the culprit:

*s *= *v _{c}t* = 20 × 10 = 200 m

At time *t* = 10 s, the police jeep is 200 m behind the culprit.

Relative velocity between the police jeep and the culprit:

25 − 20 = 5 m/s

In 40 seconds, the police jeep moves from point A to a distance *s*‘ to catch the culprit.

Here,

*s*‘ = *v _{p}t* = 25 × 40

⇒

Thus, the jeep will catch up with the bike 1.0 km away from the turning

A car travelling at 60 km/h overtakes another car travelling at 42 km/h. Assuming each car to be 5.0 m long, find the time taken during the overtake and the total road distance used for the overtake.

Velocity of the first car, *v*_{1}= 60 km/h = 16.7 m/s

Velocity of the second car, *v*_{2} = 42 km/h = 11.7 m/s

Relative velocity between the cars = (16.7 − 11.7) = 5 m/s

Distance travelled by the first car w.r.t. the second car = 5 + 5 = 10 m

Time,

t = s / v = 10 / 5s = 2 s

Distance covered by the first car w.r.t. the ground in 2 s = 16.7 × 2 = 33.4 m

The first car also covers a distance equal to its own length = 5 m

∴ Total road distance used for the overtake = 33.4 + 5 = 38 m

The first car also covers a distance equal to its own length = 5 m

∴ Total road distance used for the overtake = 33.4 + 5 = 38 m

A ball is projected vertically upward with a speed of 50 m/s. Find

(a) the maximum height.

(b) the time to reach the maximum height .

(c) the speed at half the maximum height. Take *g* = 10 m/s^{2}.

Given:

Initial speed of the ball*, u* = 50 m/s

Acceleration*, a *= −10 m/s^{2}

At the highest point, velocity*v *of the ball is 0.

Initial speed of the ball

Acceleration

At the highest point, velocity

Maximum height = 125 m

Given:

Initial speed of the ball*, u* = 50 m/s

Acceleration*, a *= −10 m/s^{2}

At the highest point, velocity*v *of the ball is 0.

Initial speed of the ball

Acceleration

At the highest point, velocity

Given:

Initial speed of the ball*, u* = 50 m/s

Acceleration*, a *= −10 m/s^{2}

At the highest point, velocity*v *of the ball is 0

Initial speed of the ball

Acceleration

At the highest point, velocity

A ball is dropped from a balloon going up at a speed of 7 m/s. If the balloon was at a height 60 m at the time of dropping the ball, how long will the ball take in reaching the ground?

Given:

Height of the balloon from the ground, *s* = 60 m

Balloon is moving upwards with velocity 7 m/s.

The balloon and the ball are moving upwards with the same speed.

When the ball is dropped, its initial velocity (*u*) is −7 m/s.

Acceleration due to gravity, *a* = *g* = 9.8 m/s^{2}

Using the equation of motion, we have:

Time taken by the ball to reach the ground = 4.3 s

A stone is thrown vertically upward with a speed of 28 m/s.

(a) Find the maximum height reached by the stone.

(b) Find its velocity one second before it reaches the maximum height.

(c) Does the answer of part (b) change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s ?

Given:

Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s

When the stone reaches the ground, its final velocity (v) is 0.

Also,

a = g = −9.8 m/s^{2} (Acceleration due to gravity)

Maximum height can be found using the equation of motion.

Thus, we have:

Given:

Initial velocity with which the stone is thrown vertically upwards, u = 28 m/s

When the stone reaches the ground, its final velocity (v) is 0.

Also,

a = g = −9.8 m/s^{2} (Acceleration due to gravity)

Total time taken by the stone to reach the maximum height:

As per the question, we need to find the velocity of the stone one second before it reaches the maximum height.

t’ = 2.85 − 1 = 1.85 s

Again, using the equation of motion, we get:

v’ = u + at’ = 28 − 9.8 × 1.85

⇒ v’ = 28 − 18.13 = 9.87 m/s

Hence, the velocity is 9.87 m/s

Given:

Initial velocity with which the stone is thrown vertically upwards, *u* = 28 m/s

When the stone reaches the ground, its final velocity (*v*) is 0.

Also,

*a* = *g = *−9.8 m/s^{2} (Acceleration due to gravity)

will not change, as after one second, the velocity becomes zero for any initial velocity and acceleration (*a* = − 9.8 m/s^{2}) remains the same. For any initial velocity more than 28 m/s, only the maximum height increases.

A person sitting on the top of a tall building is dropping balls at regular intervals of one second. Find the positions of the 3rd, 4th and 5th ball when the 6th ball is being dropped.

A person is releasing balls from a tall building at regular intervals of one second.

It means for each ball, the initial velocity *u* is 0.

Acceleration due to gravity, *a *= *g* = 9.8 m/s^{2}

When the 6^{th} ball is dropped, the 5^{th} ball moves for 1 second, the 4^{th} ball moves for 2 seconds and the 3^{rd}ball moves for 3 seconds.

Position of the 3^{rd} ball after *t =* 3 s:

Using the equation of motion, we get:

(Page-53)

A healthy young man standing at a distance of 7 m from a 11.8 m high building sees a kid slipping from the top floor. With what speed (assumed uniform) should he run to catch the kid at the arms height (1.8 m)?

Given:

Height of the building = 11.8 m

Distance of the young man from the building = 7 m

The kid should be caught over 1.8 m from ground.

As the kid is slipping, his initial velocity*u* is 0.

Acceleration,*a* = 9.8 m/s^{2}

Let*s* be the distance before which the kid has to be caught = 11.8 − 1.8 = 10 m

Using the equation of motion, we get:

Height of the building = 11.8 m

Distance of the young man from the building = 7 m

The kid should be caught over 1.8 m from ground.

As the kid is slipping, his initial velocity

Acceleration,

Let

Using the equation of motion, we get:

This is the time in which the man should reach the bottom of the building to catch the kid.

Velocity with which the man should run:

s / t = 7 / 1.42 = 4.9 m / s

An NCC parade is going at a uniform speed of 6 km/h through a place under a berry tree on which a bird is sitting at a height of 12.1 m. At a particular instant the bird drops a berry. Which cadet (give the distance from the tree at the instant) will receive the berry on his uniform?

Speed of the NCC cadets = 6 km/h = 1.66 m/s

Distance of the bird from the ground,*s* = 12.1 m

Initial velocity of the berry dropped by the bird,*u* = 0

Acceleration due to gravity,*a* = *g* = 9.8 m/s^{2}

Using the equation of motion, we can find the time taken*t* by the berry to reach the ground.

Thus

Distance of the bird from the ground,

Initial velocity of the berry dropped by the bird,

Acceleration due to gravity,

Using the equation of motion, we can find the time taken

Thus

Distance moved by the cadets = *v* × *t* = 1.57 × 1.66 = 2.6 m

Therefore, the cadet who is 2.6 m away from tree will receive the berry on his uniform.

Therefore, the cadet who is 2.6 m away from tree will receive the berry on his uniform.

A ball is dropped from a height. If it takes 0.200 s to cross the last 6.00 m before hitting the ground, find the height from which it was dropped. Take *g* = 10 m/s^{2}.

Given:

Distance travelled by the ball in 0.200 seconds = 6 m

Let:

Time, t = 0.200 s

Distance, s = 6 m

a = g = 10 m/s^{2} (Acceleration due to gravity)

Using the equation of motion, we get:

Let h be the height from which the ball is dropped.

We have:

u = 0 and v = 29 m/s

Now,

∴ Total height = 42.05 + 6 = 48.05 m ≈ 48 m

A ball is dropped from a height of 5 m onto a sandy floor and penetrates the sand up to 10 cm before coming to rest. Find the retardation of the ball is sand assuming it to be uniform.

A ball is dropped from a height of 5 m (s) above the sand level.

The same ball penetrates the sand up to 10 cm (s_{s}) before coming to rest.

Initial velocity of the ball, u = 0 And,

a = g = 9.8 m/s^{2} (Acceleration due to gravity)

Using the equation of motion, we get :

Thus, the time taken by the ball to cover the distance of 5 m is 1.01 seconds.

Velocity of the ball after 1.01 s:

*v* = *u* + *at*

⇒ *v* = 9.8 × 1.01 = 9.89 m/s

Hence, for the motion of the ball in the sand, the initial velocity *u*_{2}should be 9.89 m/s and the final velocity *v*_{2} should be 0.

*s*_{s} = 10 cm = 0.1 m

Again using the equation of motion, we get:

Hence, the sand offers the retardation of 490 m/s^{2}.

An elevator is descending with uniform acceleration. To measure the acceleration, a person in the elevator drops a coin at the moment the elevator starts. The coin is 6 ft above the floor of the elevator at the time it is dropped. The person observes that the coin strikes the floor in 1 second. Calculate from these data the acceleration of the elevator.

Given:

Distance between the coin and the floor of the elevator before the coin is dropped = 6 ft

Let *a *be the acceleration of the elevator*.*

It is given that the coin reaches the floor in 1 second. This means that the coin travels 6 ft distance.

The initial velocity is *u* for the coin and zero for the elevator.

Using the equation of motion, we get:

Equation for the coin :

Here,

*a*‘ = *g* − *a* ( *a*‘ is the acceleration felt by the coin.)

*g* = Acceleration due to gravity

*g* = 9.8 m/s^{2} = 32 ft/s^{2}

On substituting the values, we get

Hence, the acceleration of the elevator is 20 ft/s^{2}.

A ball is thrown horizontally from a point 100 m above the ground with a speed of 20 m/s. Find

(a) the time it takes to reach the ground,

(b) the horizontal distance it travels before reaching the ground,

(c) the velocity (direction and magnitude) with which it strikes the ground.

**(a)**

Given:

Speed of the ball, *u*_{x} = 20 m/s

Height from which the ball is dropped,* h* = 100 m

Let *t* be the time taken by the ball to reach the ground.

Using the equation of motion, we have:

Here,

Acceleration of gravity, *g* = 9.8 ms^{−2}

Vertical component of velocity, *u*_{y} = 0

Therefore, the time required by the ball to reach the ground is 4.5 seconds.

**(b)**

Given:

Speed of the ball, *u*_{x} = 20 m/s

Height from which the ball is dropped,* h* = 100 m

Horizontal distance travelled by the ball:

*x* = *u _{x}t *= 20 × 4.5 = 90 m

**(c)**

Given:

Speed of the ball, u_{x} = 20 m/s

Height from which the ball is dropped, h = 100 m

We know that horizontal velocity remains constant throughout the motion of the ball.

At A, v_{x} = 20 m/s.

v_{y} = u + gt = 0 + 9.8 × 4.5

⇒v_{y} = 44.1 m/s

Resultant velocity:

Therefore, the ball strikes the ground with a magnitude of velocity 49 m/s and the direction at an angle of 66° with the ground.

A ball is thrown at a speed of 40 m/s at an angle of 60° with the horizontal. Find

(a) the maximum height reached and

(b) the range of the ball. Take g = 10 m/s^{2}.

**(a)**

Given:

Initial speed of the ball, u = 40 m/s

Angle of projection of the ball with the horizontal, α = 60°

Also,

a = g = 10 m/s^{2 }

Maximum height reached by the ball:

**(b)**

Given:

Initial speed of the ball, *u* = 40 m/s

Angle of projection of the ball with the horizontal, α = 60°

Also,

*a* = g = 10 m/s^{2}

Horizontal range of the ball:

In a soccer practice session the football is kept at the centre of the filed 40 yards from the 10 ft high goalposts. A goal is attempted by kicking the football at a speed of 64 ft/s at an angle of 45° to the horizontal. Will the ball reach the goal post?

Given:

Height of the goalpost = 10 ft

The football is kept at a distance of 40 yards, i.e., 120 ft, from the goalpost.

Initial speed u with which the ball is hit = 64 ft/s

Acceleration due to gravity, a = g = 9.8 m/s^{2} = 32.2 ft/s^{2}

For the given question, 40 yards is the horizontal range (R).

Angle of projection, α = 45°

We know that the horizontal range is given by

R = u cos α(t)

= 6.86 ft < Height of the goalpost

Yes, the football will reach the goalpost

A popular game in Indian villages is *goli* which is played with small glass balls called golis. The goli of one player is situated at a distance of 2.0 m from the goli of the second player. This second player has to project his goli by keeping the thumb of the left hand at the place of his goli, holding the goli between his two middle fingers and making the throw. If the projected goli hits the goli of the first player, the second player wins. If the height from which the goli is projected is 19.6 cm from the ground and the goli is to be projected horizontally, with what speed should it be projected so that it directly hits the stationery goli without falling on the ground earlier?

Given:

Distance between the golis of the first and second players = 2.0 m = R = Horizontal range

Height h from which the goli is projected by the second player = 19.6 cm = 0.196 m

We know that the goli moves in projectile motion.

Acceleration due to gravity a = g = 9.8 m/s^{2}

The time in which the goli will reach the ground is given by the equation of motion.

As the initial velocity u_{y} in vertical direction is zero, we have:

Let us assume that the goli is projected with horizontal velocity u_{x} m/s

The horizontal range is given by

R = u_{x}t

Hence, if the second player projects the goli with a speed of 10 m/s, then his goli will hit the goli of the first player.

Figure (3−E8) shows a 11.7 ft wide ditch with the approach roads at an angle of 15° with the horizontal. With what minimum speed should a motorbike be moving on the road so that it safely crosses the ditch?

Assume that the length of the bike is 5 ft, and it leaves the road when the front part runs out of the approach road.

Given:

Width of the ditch = 11.7 ft

Length of the bike = 5 ft

The approach road makes an angle of 15˚ (α) with the horizontal.

Total horizontal range that should be covered by the biker to cross the ditch safely, *R* = 11.7 + 5 = 16.7 ft

Acceleration due to gravity, *a = g* = 9.8 m/s = 32.2 ft/s^{2}

We know that the horizontal range is given by

⇒ u =32ft/s

Therefore, the minimum speed with which the motorbike should be moving is 32 ft/s.

A person standing on the top of a cliff 171 ft high has to throw a packet to his friend standing on the ground 228 ft horizontally away. If he throws the packet directly aiming at the friend with a speed of 15.0 ft/s, how short will the packet fall?

Given:

Height (h) of the cliff = 171 ft

Horizontal distance from the bottom of the cliff = 228 ft

As per the question, the person throws the packet directly aiming to his friend at the initial speed (u) of 15.0 ft/s.

∴ θ = 37°

When the person throws the packet from the top of the cliff, it moves in projectile motion.

Let us take the reference axis at point A.

u is below the x-axis.

a = g = 32.2 ft/s^{2} (Acceleration due to gravity)

Using the second equation of motion, we get:

A ball is projected from a point on the floor with a speed of 15 m/s at an angle of 60° with the horizontal. Will it hit a vertical wall 5 m away from the point of projection and perpendicular to the plane of projection without hitting the floor? Will the answer differ if the wall is 22 m away?

Given:

Initial speed of the ball, u = 15 m/s

Angle of projection with horizontal, α = 60°

Distance of the wall from the point of projection = 5 m

a = g = 9.8 m/s^{2} (Acceleration due to gravity)

We know that the horizontal range for a projectile is given by

As the horizontal range of the projectile is 19.88 m, the ball will hit the wall 5 m away from the point of projection. If the wall is 22 m away from the point of projection, the ball will hit the wall because it is not in its horizontal range.

Find the average velocity of a projectile between the instants it crosses half the maximum height. It is projected with a speed* u* at an angle θ with the horizontal.

Given:

Initial velocity of the projectile = *u*

Angle of projection = *θ*

To find: Average velocity of the projectile

Average velocity

= Change in displacement / Time

Consider the projectile motion in the figure given below

By the symmetry of figure, it can be said that the line joining points A and B is horizontal.

So, there will be no effect of the vertical component of velocity of the projectile during displacement AB.

We know that the projectile moves at a constant velocity*u *cos *θ* in horizontal direction.

Hence, the average velocity of the projectile is*u *cos *θ**.*

So, there will be no effect of the vertical component of velocity of the projectile during displacement AB.

We know that the projectile moves at a constant velocity

Hence, the average velocity of the projectile is

A bomb is dropped from a plane flying horizontally with uniform speed. Show that the bomb will explode vertically below the plane. Is the statement true if the plane flies with uniform speed but not horizontally?

**Case-1 plane flying horizontally with uniform speed**

The plane is flying horizontally with a uniform speed. Therefore, the bomb also has the same speed.

Let the speed of the plane be represented by u.

Now, let t be the time taken by the bomb to reach the ground.

Distance travelled by the bomb in horizontal direction = ut

Both the plane and bomb are travelling in the same direction.

Distance travelled by the plane in the same time = ut

Hence, the bomb will explode vertically below the plane.

**Case-2** **plane flies with uniform speed but not horizontally**

When the plane is flying with a uniform speed but not horizontally:

Let us consider it will make an angle of projection θ along the horizontal direction.

So, both the plane and the bomb will be flying with the same angle of projection.

Therefore, both will have the same horizontal speed u cos θ, where u is the initial speed of the plane and the bomb.

When the bomb is released, the time taken by the bomb to reach the ground is t.

The distance travelled by the bomb and the plane will be u cos θt.

Hence, again the bomb will explode vertically below the plane.

**(i)**

During the motion of bomb, its horizontal velocity u remains constant and is the same as that of the plane at every point of its path.

Let the bomb reach the ground in time t.

Distance travelled in horizontal direction by the bomb = ut

Distance travelled in horizontal direction by the bomb is the same as that travelled by the plane.

So, the bomb will explode vertically below the plane.

**(ii)**

Let the plane move making an angle α with the horizontal.

Horizontal distance for both the bomb and the plane = u cos αt’

t’ = Time taken by the bomb to reach the ground

So, in this case also, the bomb will explode vertically below the plane.

A boy standing on a long railroad car throws a ball straight upwards. The car is moving on the horizontal road with an acceleration of 1 m/s^{2} and the projection velocity in the vertical direction is 9.8 m/s. How far behind the boy will the ball fall on the car?

Acceleration of the car = 1 m/s^{2}

Projection velocity of the ball (considered as a projectile) in the vertical direction = 9.8 m/s

Angle of projection, α = 90˚

Let u be the initial velocity of the car when the ball is thrown.

Both the car and the ball have the same horizontal velocity.

We know that the distance travelled by the ball in horizontal direction is given by -s_{ }= ut

Here, t is the time.

Also, the distance travelled by the car in horizontal direction is given by

Therefore, the ball drops 2 m behind the boy.

A staircase contains three steps each 10 cm high and 20 cm wide (figure 3−E9). What should be the minimum horizontal velocity of a ball rolling of the uppermost plane so as to hit directly the lowest plane?

Height of one step = 10 cm

Width of one step = 20 cm

Total height of the staircase = *y* = 30 cm

Total width of the staircase = *x* = 40 cm

To directly hit the lowest plane, the ball should just touch point E.

Let point A be the origin of reference coordinate.

Let *u* be the minimum speed of the ball.

We have:

*x* = 40 cm

*y* = −20 cm

*θ* = 0°

*g* = 10 m/s^{2} = 1000 cm/s^{2}

^{}

A person is standing on a truck moving with a constant velocity of 14.7 m/s on a horizontal road. The man throws a ball in such a way that it returns to the truck after the truck has moved 58.8 m. Find the speed and the angle of projection

(a) as seen from the truck,

(b) as seen from the road.

**a) as seen from the truck,**

Velocity of the truck = 14.7 m/s

Distance covered by the truck when the ball returns again to the truck = 58.8 m

Therefore, we can say that the time taken by the truck to cover 58.8 m distance is equal to the time of the flight of the truck.

Time in which the truck has moved the distance of 58.8 m:

Time (T) = s / v

= 58.8 / 14.7

= 4s

We consider the motion of the ball going upwards.

T = 4 s

Time taken to reach the maximum height when the final velocity v = 0:

T = 4 s

Time taken to reach the maximum height when the final velocity v = 0:

t = T / 2

= 4 / 2

= 2s

a = g = −9.8 m/s^{2} (Acceleration due to gravity)

∴ v = u − at

⇒ 0 = u + 9.8 × 2

⇒ u = 19.6 m/s

19.6 m/s is the initial velocity with which the ball is thrown upwards

∴ v = u − at

⇒ 0 = u + 9.8 × 2

⇒ u = 19.6 m/s

19.6 m/s is the initial velocity with which the ball is thrown upwards

** (b) as seen from the road.**

Velocity of the truck = 14.7 m/s

Distance covered by the truck when the ball returns again to the truck = 58.8 m

From the road, the motion of ball seems to be a projectile motion.

Total time of flight (T) = 4 seconds

Horizontal range covered by the ball in this time, R = 58.8 m

We know:

R = u cos αt

Here, α is the angle of projection.

Now,

u cos α = 14.7 …(i)

Now, take the vertical component of velocity.

Using the equation of motion, we get:

^{}

u= 14.7 /cos 53

= 24.42 m/s

=25 m/s (appr)

Therefore, when seen from the road, the speed of the ball is 25 m/s and the angle of projection is 53° with horizontal.

The benches of a gallery in a cricket stadium are 1 m wide and 1 m high. A batsman strikes the ball at a level one metre above the ground and hits a mammoth sixer. The ball starts at 35 m/s at an angle of 53° with the horizontal. The benches are perpendicular to the plane of motion and the first bench is 110 m from the batsman. On which bench will the ball hit?

Angle of projection of the ball, α = 53°

Width and height of the bench = 1 m

Initial speed of the ball = 35 m/s

Distance of the first bench from the batsman = 110 m

The batsman strikes the ball 1 m above the ground.

Let the ball land on the n^{th} bench.

∴ y = (n − 1) …(i)

And,

x = 110 + n – 1 = 110 + y

^{}

^{}

After Solving the equation, we will get:

y = 5

⇒ n − 1 = 5

⇒ n = 5 + 1

⇒ n = 6

Hence , The ball will hit the sixth bench of the gallery.

A man is sitting on the shore of a river. He is in the line of 1.0 m long boat and is 5.5 m away from the centre of the boat. He wishes to throw an apple into the boat. If he can throw the apple only with a speed of 10 m/s, find the minimum and maximum angles of projection for successful shot. Assume that the point of projection and the edge of the boat are in the same horizontal level.

Length of the boat = 1.0 m

Distance between the man and the centre of the boat (R) = 5.5 m

Initial speed (u) of throwing the apple by the man = 10 m/s

Acceleration due to gravity (g) = 10 m/s^{2}

We know that the horizontal range is given by

Similarly, for the end point of the boat, i.e., point C, we have:

Horizontal range (R) = 6 m

For a successful shot, the angle of projection α with initial speed 10 m/s may vary from 15° to 18° or from 71° to 75°. The minimum angle is 15° and the maximum angle is 75°, but there is an interval of 53° for which the successful shot is not allowed. We can show this by putting the successive value of α from 15° to 75°

(Page No-54)

A river 400 m wide is flowing at a rate of 2.0 m/s. A boat is sailing at a velocity of 10 m/s with respect to the water, in a direction perpendicular to the river.

(a) Find the time taken by the boat to reach the opposite bank.

(b) How far from the point directly opposite to the starting point does the boat reach the opposite bank?

Distance between the opposite shore of the river or width of the river = 400 m

Rate of flow of the river = 2.0 m/s

Boat is sailing at the rate of 10 m/s.

The vertical component of velocity 10 m/s takes the boat to the opposite shore. The boat sails at the resultant velocity v_{r}.

Time taken by the boat to reach the opposite shore:

Time = Distance / Time

= 400 / 10

= 40s

**(a)**

Distance the boat need to travel to reach the opposite shore

= 400 / sin(α)

= 407.9 m

Time= Total Distance / Total Velocity

T = 407.9 / 10.2

T = 40s

**(b)**

Note – from part (a) of Q-46

A swimmer wishes to cross a 500 m wide river flowing at 5 km/h. His speed with respect to water is 3 km/h.

(a) If he heads in a direction making an angle θ with the flow, find the time he takes to cross the river.

(b) Find the shortest possible time to cross the river.

**(a)**

** If he heads in a direction making an angle θ with the flow,**

Width of the river = 500 m

Rate of flow of the river = 5 km/h

Swimmer’s speed with respect to water = 3 km/h

As per the question, the swimmer heads in a direction making an angle *θ* with the flow.

We know that the vertical component of velocity 3 sin *θ* takes him to the opposite side of the river.

Distance to be travelled = 0.5 km

Vertical component of velocity = 3 sin *θ* km/h

Thus, we have:

**(b)**

**Shortest possible time to cover the river:**

Take *θ* = 90^{∘}

^{}

Hence, the required time is 10 minutes.

Consider the situation of the previous problem. The man has to reach the other shore at the point directly opposite to his starting point. If he reaches the other shore somewhere else, he has to walk down to this point. Find the minimum distance that he has to walk.

Width of the river = 500 m

Rate of flow of the river = 5 km/h

Swimmer’s speed with respect to water = 3 km/h

As per the question, the man has to reach the other shore at the point directly opposite to his starting point.

^{}

Horizontal distance is BD for the resultant velocity v_{r}.

x-component of the resultant velocity, R = 5 – 3 cos θ

Vertical component of velocity = 3 sin θ km/h

Time = Distance / Velocity

=0.5 h / 3 sinθ

This is the same as the horizontal component of velocity.

Thus, we have:

= 16 / 24

= 2 / 3 km

An aeroplane has to go from a point A to another point B, 500 km away due 30° east of north. A wind is blowing due north at a speed of 20 m/s. The air-speed of the plane is 150 m/s.

(a) Find the direction in which the pilot should head the plane to reach the point B.

(b) Find the time taken by the plane to go from A to B.

**(a)**

Distance between points A and B = 500 km

B from A is 30˚ east of north.

Speed of wind due north, v_{w} = 20 m/s

Airspeed of the plane, v_{a} = 150 m/s

Let R be the resultant direction of the plane to reach point B.

**(b)**

from part (a) of Q-49

Two friends A and B are standing a distance *x* apart in an open field and wind is blowing from A to B. A beat a drum and B hears the sound *t*_{1} time after he sees the event. A and B interchange their positions and the experiment is repeated. This time B hears the drum timer after he sees the event. Calculate the velocity of sound in still air v and the velocity of wind *u*. Neglect the time light takes in travelling between the friends.

Distance between A and B = x

Velocity of sound in air = v

Velocity of wind = u

**First Case:**

When A beats the drum from his original position:

Resultant velocity of sound = u + v

from (i) we get

Suppose A and B in the previous problem change their positions in such a way that the line joining them becomes perpendicular to the direction of wind while maintaining the separation x. What will be the time B finds between seeing and hearing the drum beating by A?

Distance between A and B = *x *

Let *v *be the velocity of sound in the direction along line AC.

Let *u** *be the velocity of air in the direction along line AB.

Angle between *v* and *u* =* θ* > π / 2

Resultant velocity of sound and air that will reach B =

Six particles situated at the corner of a regular hexagon of side* a* move at a constant speed v. Each particle maintains a direction towards the particle at the next corner. Calculate the time the particles will take to meet each other.

A regular hexagon has a side a. Six particles situated at the corners of the hexagon are moving with a constant speed v.

As per the question, each particle maintains a direction towards the particle at the next corner. So, particles will meet at centroid O of triangle PQR. Now, at any instant, the particles will form an equilateral triangle PQR with the same centroid O.

We know that P approaches Q, Q approaches R and so on.

Now, we will consider the motion of particle P. Its velocity makes an angle of 60˚.

This component is the rate of decrease of distance PO.

Relative velocity between P and Q:

—: End of **Rest and Motions Exercise** Questions Solutions HC Verma Vol-I :–

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]]>Obj-2 Rest and Motions Kinematics HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11. Solution of Objective -2 (MCQ-2) Questions of Ch-3 Rest and Motions Kinematics (Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics. Obj-2 (MCQ-2 ) Rest and Motions Kinematics HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11 Board ISC […]

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]]>**Obj-2 Rest and Motions** Kinematics HC Verma Solutions Ch-3 Vol-1 Concept of Physics for Class-11. **Solution of Objective -2 (MCQ-2) **Questions of Ch-3 **Rest and Motions** Kinematics** **(Concept of Physics) .Visit official Website CISCE for detail information about ISC Board Class-11 Physics.

Board | ISC and other board |

Publications | Bharti Bhawan Publishers |

Chapter-3 | Rest and Motions Kinematics |

Class | 11 |

Vol | 1st |

writer | H C Verma |

Book Name | Concept of Physics |

Topics | Solution of Objective-2 (MCQ-2) Questions |

Page-Number | 50 , 51 |

-: Select Topics :-

Question for Short Answer

Objective-I

Objective-II (Currently Open)

Exercise

(Page-50)

Consider the motion of the tip of the minute hand of a clock. In one hour

(a) the displacement is zero

(b) the distance covered is zero

(c) the average speed is zero

(d) the average velocity is zero

The options** (a)**** **and** (d) **are correct

**Explanation:**

Displacement is zero because the initial and final positions are the same.

A particle moves along the X-axis as x = u (t − 2 s) + a (t − 2 s)^{2}.

(a) the initial velocity of the particle is u

(b) the acceleration of the particle is a

(c) the acceleration of the particle is 2a

(d) at t = 2 s particle is at the origin.

The options** (c)**** **and** (d) **are correct

**Explanation:**

At t = 2 s,

x = u(2 s − 2 s) + a (2 s − 2 s)^{2} = 0 (origin)

x = u(2 s − 2 s) + a (2 s − 2 s)

Pick the correct statements:

(a) Average speed of a particle in a given time is never less than the magnitude of the average velocity.

(b) It is possible to have a situation in which

(c) The average velocity of a particle is zero in a time interval. It is possible that the instantaneous velocity is never zero is the interval.

(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed).

(d) The average velocity of a particle moving on a straight line is zero in a time interval. It is possible that the instantaneous velocity is never zero in the interval. (Infinite accelerations are not allowed).

The options** (a)**** , (b) **and** (c) **are correct

**Explanation:**

Average velocity = Displacement / Total time

Displacement ≤ Distance

∴ Average velocity ≤ Average speed In uniform circular motion, speed is constant but velocity is not.

In one complete circle of uniform motion, average velocity is zero. Instantaneous velocity is never zero in the interval.

An object may have

(a) varying speed without having varying velocity

(b) varying velocity without having varying speed

(c) nonzero acceleration without having varying velocity

(d) nonzero acceleration without having varying speed.

The options** (b) **and** (d) **are correct

**Explanation:**

Velocity and acceleration are vector quantities that can be changed by changing direction only (keeping magnitude constant).

Mark the correct statements for a particle going on a straight line:

(a) If the velocity and acceleration have opposite sign, the object is slowing down.

(b) If the position and velocity have opposite sign the particle is moving towards the origin.

(c) If the velocity is zero at an instant, the acceleration should also be zero at that instant.

(d) If the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval.

The options **(a), (b) **and** (d) **are correct

**Explanation:**

(a) Acceleration is given by

(b) If the position and velocity have opposite signs, the particle moves towards the origin. It can be explained by following figure

(c) If the velocity is zero in a certain time interval, then the change in the velocity in that time interval will also be zero. As acceleration is rate of change of velocity, it will also be zero at an instant in that time interval.

The velocity of a particle is zero at t = 0.

(a) The acceleration at t = 0 must be zero.

(b) The acceleration at t = 0 may be zero.

(c) If the acceleration is zero from t = 0 to t = 10 s, the speed is also zero in this interval.

(d) If the speed is zero from t = 0 to t = 10 s the acceleration is also zero in this interval.

The options **(b), (c) **and** (d) **are correct

**Explanation:**

(b) Acceleration will be zero only when the change in velocity is zero.

(c) Since the acceleration is zero from t = 0 s to t = 10 s, change in velocity is 0.

Velocity in this interval = Initial velocity = 0

Also,

Speed in this interval = Initial speed = 0

(d) From t = 0 s to t = 10 s, speed is zero.

Here, velocity is zero and initial velocity is zero.

So, the change in velocity is zero; i.e., acceleration is zero.

(c) Since the acceleration is zero from t = 0 s to t = 10 s, change in velocity is 0.

Velocity in this interval = Initial velocity = 0

Also,

Speed in this interval = Initial speed = 0

(d) From t = 0 s to t = 10 s, speed is zero.

Here, velocity is zero and initial velocity is zero.

So, the change in velocity is zero; i.e., acceleration is zero.

Mark the correct statements:

(a) The magnitude of the velocity of a particle is equal to its speed.

(b) The magnitude of average velocity in an interval is equal to its average speed in that interval.

(c) It is possible to have a situation in which the speed of a particle is always zero but the average speed is not zero.

(d) It is possible to have a situation in which the speed of the particle is never zero but the average speed in an interval is zero.

The option** (a) **is correct

**Explanation:**

The magnitude of the velocity of a particle is equal to its speed.

(a) Velocity being a vector quantity has magnitude as well as direction, and magnitude of velocity is called speed.

**(b) Average velocity** =

Total displacement / Total time taken

**Average speed** =

Total distance travelled / Total time taken

Distance ≥ Displacement

∴ Average speed ≥ Average velocity

The magnitude of average velocity in an interval is not always equal to its average speed in that interval.

(c) If speed is always zero, then the distance travelled is always zero. Hence, the total distance travelled and the average speed will be zero.

(d) If the speed of a particle is never zero, the distance travelled by the particle is never zero. Hence, the average speed will not be zero.

The velocity-time plot for a particle moving on a straight line is shown in the figure.

(a) The particle has a constant acceleration.

(b) The particle has never turned around.

(c) The particle has zero displacement.

(d) The average speed in the interval 0 to 10 s is the same as the average speed in the interval 10 s to 20 s.

The options **(a) **and** (d) **are correct

**Explanation:**

(a) The slope of the v–t graph gives the acceleration. For the given graph, the slope is constant. So, acceleration is constant.

(b) From 0 to 10 seconds, velocity is in positive direction and then in negative direction. This means that the particle turns around at t = 10 s.

(c) Area in the v–t curve gives the distance travelled by the particle.

Distance travelled in positive direction ≠ Distance travelled in negative direction

∴ Displacement ≠ Zero

(d) The area of the v–t graph from t = 0 s to t = 10 s is the same as that from t = 10 s to t= 20 s. So, the distance covered is the same. Hence, the average speed is the same.

In figure (3-Q5) shows the position of a particle moving on the X-axis as a function of time.

(a) The particle has come to rest 6 times.

(b) The maximum speed is at t = 6s.

(c) The velocity remains positive for t = 0 to t = 6s.

(d) The average velocity for the total period shown is negative.

The option **(a)**** **is correct

**Explanation:**

(a) The slope of the x–t graph gives the velocity. Here, 6 times the slope is zero. So, the particle has come to rest 6 times.

(b) As the slope is not maximum at t = 6 s, the maximum speed is not at t = 6 s.

(c) As the slope is not positive from t = 0 s to t = 6s, the velocity does not remain positive.

(d) Average velocity =

For the shown time (t = 6 s), the displacement of the particle is positive. Therefore, the average velocity is positive.

(Page-51)

The accelerations of a particle as seen from two frames S_{1} and S_{2} have equal magnitude 4 m/s^{2}.

(a) The frames must be at rest with respect to each other.

(b) The frames may be moving with respect to each other.

(c) The acceleration of S_{2} with respect to S_{1} may either be zero of 8 m/s^{2}.

(d) The acceleration of S_{2} with respect to S_{1} may be anything between zero and 8 m/s^{2}.

The option **(d)**** **is correct

**Explanation:**

—: End of **Obj-2 (MCQ-2) Rest and Motions** Kinematics HC Verma Solutions :–

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