- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- Chapter 4- Motion in a plane
- Chapter 5- Laws of motion
- Chapter 6- Work Energy and power
- Chapter 7- System of particles and Rotational Motion
- Chapter 8- Gravitation
- Chapter 9- Mechanical Properties Of Solids
- Chapter 10- Mechanical Properties Of Fluids
- Chapter 11- Thermal Properties of matter
- Chapter 12- Thermodynamics
- Chapter 13- Kinetic Theory
- Chapter 14- Oscillations
- Chapter 15- Waves

Chapter 1- Physical World |
Chapter 2- Units and Measurements |
Chapter 4- Motion in a plane |
Chapter 5- Laws of motion |
Chapter 6- Work Energy and power |
Chapter 7- System of particles and Rotational Motion |
Chapter 8- Gravitation |
Chapter 9- Mechanical Properties Of Solids |
Chapter 10- Mechanical Properties Of Fluids |
Chapter 11- Thermal Properties of matter |
Chapter 12- Thermodynamics |
Chapter 13- Kinetic Theory |
Chapter 14- Oscillations |
Chapter 15- Waves |

*In which of the following examples of motion, can the body be considered**approximately a point object:**(a) a railway carriage moving without jerks between two stations.**(b) a monkey sitting on top of a man cycling smoothly on a circular track.**(c) a spinning cricket ball that turns sharply on hitting the ground.**(d) a tumbling beaker that has slipped off the edge of a table.*

**Answer
1** :

( a ), ( b ) The size ofthe train carriage and the cap is very small as compared to the distancethey’ve travelled, i.e. the distance between the two stations and the length ofthe race track, respectively. Therefore, the cap and the carriage can beconsidered as point objects.

The size of thebasketball is comparable to the distance through which it bounces off afterhitting the floor. Thus, basketball cannot be treated as a point object.Likewise, the size of the bottle is comparable to the height of the chair fromwhich it drops. Thus, the bottle cannot be treated as a point object.

The position-time (x-t) graphs for two children A and B returning fromtheir school

O to their homes P and Q respectivelyare shown in Fig. 3.19. Choose the correct

entries in the brackets below ;

(a) (A/B) lives closer to the schoolthan (B/A)

(b) (A/B) starts from the schoolearlier than (B/A)

(c) (A/B) walks faster than (B/A)

(d) A and B reach home at the(same/different) time

(e) (A/B) overtakes (B/A) on the road(once/twice).

**Answer
2** :

(a) A lives closer tothe school than B, because A has to cover shorter distances [OP < OQ],

(b) A starts from school earlier than B, because for x= 0, t = 0 for A but forB, t has some finite time.

(c) The slope of B is greater than that of A, therefore B walks faster than A.

(d) Both A and B will reach their home at the same time.

(e) At the point of intersection, B overtakes A on the roads once.

A woman starts from her home at 9.00 am, walks at a speed of 5 km/h on a

straight road up to her office 2.5 kmaway, stays at the office up to 5.00 pm, and

returns home by an auto with a speed of25 km/h. Choose suitable scales and

plot the x-t graph of her motion.

**Answer
3** :

Distance till heroffice = 2.5 km.

Walking speed the woman= 5 km/h

Time taken to reach office while walking = (2.5/5 ) h=(1/2) h = 30 minutes

Speed of auto = 25km/h

Time taken to reachhome in auto = 2.5/25 = (1/10) h = 0.1 h = 6 minutes

In the graph, O istaken as the origin of the distance and the time, then at t = 9.00 am, x= 0

and at t = 9.30 am, x = 2.5 km

OA is the portion onthe x-t graph that represents her walk from home to the office. AB representsher time of stay in the office from 9.30 to 5. Her return journey isrepresented by BC.

A drunkard walking in a narrow lane takes 5 steps forward and 3 stepsbackwards, followed again by 5 steps forward and 3 steps backwards, and soon. Each step is 1m long and requires 1 s. Plot the x-t graph of his motion.Determine graphically and otherwise how long the drunkard takes to fall ina pit 13 m away from the start.

**Answer
4** :

The time taken to goone step is 1 second. In 5s he moves forward through a distance of 5m and thenin next 3s he comes back by 3m. Therefore, in 8s he covers 2m. So, tocover a distance of 8m he takes 32s. He must take another 5steps forward tofall into the pit. So, the total time taken is 32s + 5s = 37s to fall into apit 13 m away.

A jet airplane travelling at the speed of 500 km/h ejects its products ofcombustion at the speed of 1500 km/h relative to the jet plane. What isthe speed of the latter with respect to an observer on the ground?

**Answer
5** :

Speed of the jetairplane, V_{A}= 500 km/h

Speed at which thecombustion products are ejected relative to the jet plane, V_{B} –V_{A}

= – 1500 km/h

(The negative sign indicatesthat the combustion products move in a direction opposite to that of jet)

Speed of combustionproducts w.r.t. observer on the ground, V_{B} – 500 = – 1500

V_{B} = –1500 + 500 = – 1000 km/h

**Answer
6** :

The initial velocityof the car = u

Final velocity of thecar = v

Distance covered bythe car before coming to rest = 200 m

Using the equation,

v = u + at

t = (v – u)/a = 11.44sec.

Therefore, it takes11.44 sec for the car to stop.

Two trains A and B of length 400 m each are moving on two parallel trackswith a

uniform speed of 72 km h^{–1} inthe same direction, with A ahead of B. The driver of

B decides to overtake A and acceleratesby 1 m s^{–2}. If after 50 s, the guard of B just

brushes past the driver of A, what wasthe original distance between them?

**Answer
7** :

Length of the train Aand B = 400 m

Speed of both thetrains = 72 km/h = 72 x (5/18) = 20m/s

Using the relation, s= ut + (1/2)at^{2}

Distance covered bythe train B

S_{B} = u_{B}t+ (1/2)at^{2}

Acceleration, a = 1m/s

Time = 50 s

S_{B} =(20 x 50) + (1/2) x 1 x (50)^{2}

= 2250 m

Ditance covered by thetrain A

S_{A} = u_{A}t+ (1/2)at^{2}

Acceleration, a = 0

S_{A} = u_{A}t = 20 x 50 = 1000 m

Therefore, theoriginal distance between the two trains = S_{B} – S_{A} =2250 – 1000 = 1250 m

On a two-lane road, car A is travelling at a speed of 36 km/h. Two cars Band

C approach car A in opposite directionswith a speed of 54 km/h each. At a

certain instant, when the distance ABis equal to AC, both being 1 km, B decides

to overtake A before C does. Whatminimum acceleration of car B is required to

avoid an accident?

**Answer
8** :

The speed of car A =36 km/h = 36 x (5/8) = 10 m/s

Speed of car B = 54km/h = 54 x (5/18) = 15 m/s

Speed of car C =– 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are inopposite direction)

Relative speed of Aw.r.t C, V_{AC}= V_{A} – V_{B} = 10 – (-15) =25 m/s

Relative speed of Bw.r.t A, V_{BA} = V_{B} – V_{A} = 15 –10 = 5 m/s

Distance between AB =Distance between AC = 1 km = 1000 m

Time taken by the carC to cover the distance AC, t = 1000/V_{AC} = 1000/ 25 = 40 s

If a is the acceleration,then

s = ut + (1/2) at^{2}

1000 = (5 x 40) +(1/2) a (40) 2

a = (1000 – 200)/ 800= 1 m/s^{2}

Thus, the minimumacceleration of car B required to avoid an accident is 1 m/s^{2}

either direction every T minutes. A mancycling with a speed of 20 km h

direction A to B notices that a busgoes past him every 18 min in the direction of

his motion, and every 6 min in theopposite direction. What is the period T of the

bus service and with what speed(assumed constant) do the buses ply on the

road?

**Answer
9** :

Speed of each bus= V_{b}

Speed of the cyclist =V_{c }= 20 km/h

The relative velocityof the buses plying in the direction of motion of cyclist is V_{b} –V_{c}

The buses playing in the direction of motion of the cyclist go past him afterevery

18 minute i.e.(18/60) s.

Distance covered = (V_{b} –V_{c} ) x 18/60

Since the buses areleaving every T minutes. Therefore, the distance is equal to V_{b} x(T/60)

(V_{b} –V_{c} ) x 18/60 = V_{b} x (T/60) ——(1)

The relative velocityof the buses plying in the direction opposite to the motion of cyclist is V_{b} +V_{c}

The buses go past the cyclist every 6 minutes

18 minute i.e.(6/60) s.

Distance covered = (V_{b} +V_{c} ) x 6/60

Therefore, (V_{b} +V_{c} )x 6/60 = V_{b} x (T/60)——(2)

Dividing (2) by (1)

[(V_{b} –V_{c} ) x 18/60]/ [(V_{b} + V_{c} ) x6/60 ]= [V_{b} x (T/60)] /[V_{b} x (T/60)]

(V_{b} –V_{c} ) 18/(V_{b} +V_{c} ) 6 = 1

(V_{b} –V_{c} )3 = (V_{b} +V_{c} )

Substituting thevalue of V_{c}

(V_{b} –20 )3= (V_{b} + 20 )

3V_{b} –60 = V_{b} + 20

2V_{b} =80

V_{b} =80/2 = 40 km/h

To find the value ofT, substitute the values of V_{b} and V_{c} inequation (1)

(V_{b} –V_{c} ) x 18/60 = V_{b} x (T/60)

(40 – 20) x (18/60) =40 x (T/60)

T = (20 x 18) /40 = 9minutes

A player throws a ball upwards with an initial speed of 29.4 m/s.

(a) What is the direction ofacceleration during the upward motion of the ball?

(b) What are the velocity and accelerationof the ball at the highest point of its

motion?

(c) Choose the x = 0 m and t = 0 s tobe the location and time of the ball at its

highest point, vertically downwarddirection to be the positive direction of

the x-axis, and give the signs of position,velocity and acceleration of the ball

during its upward, and downward motion.

(d) To what height does the ball riseand after how long does the ball return to the

player’s hands? (Take g = 9.8 m s^{–2} andneglect air resistance).

**Answer
10** :

(a) The accelerationdue to gravity always acts downwards towards the centre of the Earth.

(b) At the highest point of its motion the velocity of the ball will be zerobut the acceleration due to gravity will be 9.8 m s^{–2 } actingvertically downward.

(c) If we consider the highest point of ball motion as x = 0, t = 0 andvertically downward direction to be +ve direction of the x-axis, then

(i) During upward motion of the ball before reaching the highest point position,x = +ve, velocity, v = -ve and acceleration, a = +ve.

(ii) During the downward motion of the ball after reaching the highest pointposition, velocity and acceleration all the three quantities are positive.

(d) Initial speed of the ball, u= -29.4 m/s

Final velocity of theball, v = 0

Acceleration = 9.8 m/s^{2}

Applying in theequation v^{2} – u^{2} = 2gs

0 – (-29.4)^{2} =2 (9.8) s

s = – 864.36/19.6 = –44.1

Height to which theball rise = – 44.1 m (negative sign represents upward direction)

Considering theequation of motion

v = u + at

0 = (-29.4) + 9.8t

t = 29.4/9.8 = 3seconds

Therefore, the totaltime taken for the ball to return to the player’s hands is 3 +3 = 6s

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