- Chapter 1- Physical World
- Chapter 2- Units and Measurements
- Chapter 3- Motion in a Straight Line
- Chapter 4- Motion in a plane
- Chapter 5- Laws of motion
- 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 3- Motion in a Straight Line |
Chapter 4- Motion in a plane |
Chapter 5- Laws of motion |
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 |

** The sign of work done by a force on a body isimportant to understand. State carefully if the following quantities arepositive or negative:****(a) Work done by a man in liftinga bucket out of a well by means of a rope tied to the bucket,****(b) Work done by gravitationalforce in the above case,****(c) Work done by friction on abody sliding down an inclined plane,****(d) Work done by an applied forceon a body moving on a rough horizontal plane with uniform velocity,****(e) Work done by the resistiveforce of air on a vibrating pendulum in bringing it to rest.**

**Answer
1** :

Work done, W = T.S = Fs cos θ

(a) Work done ‘positive’, because force is acting in the direction ofdisplacement i.e., θ = 0°.

(b) Work done is negative, because force is acting against the displacementi.e., θ = 180°.

(c) Work done is negative, because force of friction is acting against thedisplacement i.e., θ= 180°.

(d) Work done is positive, because body moves in the direction of applied forcei.e., θ= 0°.

(e) Work done is negative, because the resistive force of air opposes themotion i.e., θ = 180°.

** A body of mass 2 kg initially at rest movesunder the action of an applied horizontal force of 7 N on a table withcoefficient of kinetic friction = 0.1. Compute the****(a) Work done by the appliedforce in 10 s****(b) Work done by friction in 10 s****(c) Work done by the net force onthe body in 10 s****(d) Change in kinetic energy ofthe body in 10 s and interpret your results.**

**Answer
2** :

(a) We know that U_{k} =frictional force/normal reaction

frictional force = U_{k} xnormal reaction

= 0.1 x 2 kg wt = 0.1 x 2 x 9.8 N = 1.96 N

net effective force = (7 – 1.96) N = 5.04 N

acceleration = 5.04/2 ms^{-2} =2.52 ms^{-2}

distance, s=1/2x 2.52 x 10 x 10 = 126 m

work done by applied force = 7 x 126 J = 882 J

(b) Work done by friction = 1.96 x 126 = -246.96 J

(c) Work done by net force = 5.04 x 126 = 635.04 J

(d) Change in the kinetic energy of the body

= work done by the net force in 10 seconds = 635.04 J (This is in accordancewith work-energy theorem).

**Answer
3** :

We know that total energy E = K.E. + P.E. or K.E. = E – P.E. andkinetic energy can never be negative.The object can not exist in the region,where its K.E. would become negative.

(a) In the region between x = 0 and x = a, potential energy is zero. So,kinetic energy is positive. In the region x > a, the potential energy has avalue greater than E. So, kinetic energy will be negative in this region. Thusthe particle cannot be present in the region x > a.

The minimum total energy that the particle can have in this case is zero.

(b) Here P.E. > E, the total energy of the object and as such the kineticenergy of the object would be negative. Thus object cannot be present in anyregion of the graph.

(c) Here x = 0 to x = a and x > b, the P.E. is more than E, so K.E. isnegative. The particle can not exist in these portions.

(d) The object can not exist in the region between x = -b/2 to x =-a/2 and x =-a/2 to x = -b/2 .Because in this region P.E. > E.

**Answer
4** :
Here, force constant k = 0.5 Nm^{-1} andtotal energy of particle E = 1J. The particle can go up to a maximum distance x_{m}, where its total energyis transformed into elastic potential energy.

**Answer
5** :
(a) Heat energy required for burning of casing ofrocket comes from the rocket itself. As

a result of work done against friction the kinetic energy of rocketcontinuously decreases – and this work against friction reappears as heatenergy.

(b) This is because gravitational force is a conservative force. Work done bythe gravitational ‘ force of the sun over a closed path in every complete orbitof the comet is zero.

(c) As an artificial satellite gradually loses its energy due todissipation against atmospheric resistance, its potential decreases rapidly. Asa result, kinetic energy of

satellite slightly increases i.e., its speed increases progressively.

(d) In Fig. (i), force is applied on the mass, by the man in vertically upwarddirection but distance is moved along the horizontal.

θ = 90°. W = F_{s} cos90° = zero

In Fig. (ii), force is applied along the horizontal and the distance moved isalso along the horizontal. Therefore, θ = 0°.

W = F_{s} cos θ =mg x s cos 0°

W = 15 x 9.8 x 2 x 1 = 294 joule.

Thus, work done in (ii) case is greater.

**Answer
6** :

(a) Potential energy of the body decreases because the body inthis case goes closer to the centre of the force.

(b) Kinetic energy, because friction does its work against the motion.

(c) Internal forces can not change the total or net momentum of a system. Hencethe rate of change of total momentum of many particle system is proportional tothe external force on the system.

(d) In an inelastic collision of two bodies, the quantities which do not changeafter the collision are the total kinetic energy/total linear momentum/ totalenergy of the system of two bodies.

** State if each of the following statements istrue or false. Give reasons for your answer.****(a) In an elastic collision oftwo bodies, the momentum and energy of each body is conserved.****(b) Total energy of a system isalways conserved, no matter what internal and external forces on the bodyare present.****(c) Work done in the motion of abody over a closed loop is zero for every force in nature.****(d) In an inelastic collision,the final kinetic energy is always less than the initial kinetic energy of thesystem.**

**Answer
7** :

(a) False, the total momentum and total energy of the system areconserved.

(b) False, the external force on the system may increase or decrease the totalenergy of the system.

(c) False, the work done during the motion of a body over a closed loop is zeroonly when body is moving under the action of a conservative force (such asgravitational or electrostatic force). Friction is not a conservative forcehence work done by force of friction (or work done on the body againstfriction) is not zero over a closed loop.

(d) True, usually in an inelastic collision the final kinetic energy is alwaysless than the initial kinetic energy of the system.

** Answer carefully, with reasons:****(a) In an elastic collision oftwo billiard balls, is the total kinetic energy conserved during the short timeof collision of the balls (i.e., when they are in contact)?****(b) Is the total linear momentumconserved during the short time of an elastic collision of two balls?****(c) What are the answers to (a)and (b) for an inelastic collision?****(d) If the potential energy oftwo billiard balls depends only on the separation distance between theircentres, is the collision elastic or inelastic? (Note, we are talking here of potentialenergy corresponding to the force during collision, not gravitational potentialenergy).**

**Answer
8** :

(a) In this case total kinetic energy is not conservedbecause when the bodies are in contact dining elastic collision even, thekinetic energy is converted into potential energy.

(b) Yes, because total momentum conserves as per law of conservation ofmomentum.

(c) The answers remain unchanged.

(d) It is a case of elastic collision because in this case the forces will beof conservative nature.

**A body is initiallyat rest. It undergoes a one-dimensional motion with constant acceleration. Thepower delivered to it at time t is proportional to****(i) t ^{1/2 }(ii)t (iii) t^{3/2} (iv) t^{2}**

**Answer
9** :

** **(ii)From v = u + at

v = 0 + at = at

As power, p = F x v

.’. p = (ma) x at = ma^{2}t

Since m and a are constants, therefore, p α t.

**Answer
10** :
(ii)p = force x velocity

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