CBSE TEST PAPER-03 CLASS - XI PHYSICS (Work, Energy and Power) Topic: - Work, Energy and Power 1.

If two bodies stick together after collision will the collision be elastic or inelastic?

[1]

2.

When an air bubble rises in water, what happens to its potential energy?

[1]

3.

A spring is kept compressed by pressing its ends together lightly. It is then placed

[1]

A body is moving along Z – axis of a co – ordinate system is subjected to a constant  force F is given by ɵi, ɵj , kɵ F = −ɵi + 2 ɵj + 3kɵ N ,

[2]

nl

4.

od

in a strong acid, and released. What happens to its stored potential energy?

Where ɵi, ɵj , kɵ are unit vector along the x, y and z – axis of the system respectively

Z – axis? 5.

ow

what is the work done by this force in moving the body a distance of 4m along the

A ball is dropped from the height h1 and if rebounces to a height h2. Find the value

[2]

is

Th

State and prove work energy theorem analytically?

Fr ee

6.

D

of coefficient of restitution?

7.

An object of mass 0.4kg moving with a velocity of 4m/s collides with another object

[2] [2]

of mass 0.6kg moving in same direction with a velocity of 2m/s. If the collision is perfectly inelastic, what is the loss of K.E. due to impact?

8.

Prove that in an elastic collision in one dimension the relative velocity of approach

[3]

before impact is equal to the relative velocity of separation after impact?

9.

(a) Define potential energy. Give examples.

(b) Draw a graph showing variation of potential energy, kinetic energy and the total energy of a body freely falling on earth from a height h?

[5]

CBSE TEST PAPER-03 CLASS - XI PHYSICS (Work, Energy and Power) Topic: - Work, Energy and Power [ANSWERS] Ans1:

Inelastic collision.

Ans2:

Potential energy of an air bubble decreases because work is done by upthrust on the bubble.

(

)( )

Velocity of approach

W = 12 J

υ1 = 2 gh1

D

Ans5:

od

Sɵ = 4kɵ   W = F .S W = ɵi + 2 ɵj + 3kɵ . 4kɵ

nl

 F = −ɵi + 2 ɵj + 3kɵ N ,

(Ball drops form height h1)

Fr ee

Th

Velocity of separation = 2 gh2

is

Ans4:

The loss in potential energy appears as kinetic energy of the molecules of the cid.

ow

Ans3:

(Ball rebounds to height h2)

Coefficient of restitution

e=

υ2 2 gh2 = 2 gh1 υ1

e=

Ans6:

h2 h1

It states that work done by force acting on a body is equal to the change produced in its kinetic energy.  If F force is applied to move an object through a distance dS   Then dw = F .dS

 F = ma   dw = ma.d S  d v  dw = m .d S dt ds dυ dt dw = mυ dυ

dw = m

Integrating υ

w

u 2 υ

W=

mυ 2 mu 2 − 2 2

u

ow

V W =m 2

nl

o

od

∫ dW = W = ∫ mυ dυ

Hence W = Kf – Ki Where Kf and Ki are final and initial kinetic energy.

Ans7:

m1 = 0.4kg,

u1 = 4m/s,

m2 = 0.6kg

u2 = 2m/s.

1 1 m1u12 + m2u2 2 2 2 1 1 2 2 Ki = ( 0.4 ) × ( 4 ) + ( 0.6 ) × ( 2 ) 2 2 Ki = 4.4 J

Fr ee

Th

Ki =

is

D

Total K.E. be fore collision

Since collision is perfectly inelastic

υ=

m1u1 + m2u2 = 2.8m / s m1 + m2

Total K.E. after collision

1 ( m1 + m2 )υ 2 2 1 2 Kf = ( 0.4 + 0.6 ) × ( 2.8 ) 2 Kf =

Kf = 3.92 J

Loss in K.E. ( ∆K ) =Ki – Kf = 4.4 – 3.92 = 0.48J

Ans8:

m1u1 + m2u2 = m1υ1 + m2υ2

m1 ( u1 − υ1 ) = m2 (υ 2 − u2 ) − − − − − − − (1)

nl

K.E. also remains conserved.

od

According to law of conservation of linear momentum

Dividing (2) by (1)

is

u1 − u2 = υ2 − υ1

D

u1 − υ1 = υ2 + u2

ow

1 1 1 1 m1u12 + m2u2 2 = m1υ12 + m2υ2 2 2 2 2 2 2 2 2 2 m1 ( u1 − υ 1 ) = m2 (υ2 − u2 ) − − − − − −(2)

Fr ee

Th

I.e. Relative velocity of approach = Relative velocity of separation

Ans9:

(a) Potential energy is the energy possessed by a body by virtue of its position in a field or

due to change in its configuration example –

A gas compressed in a cylinder, A wound spring of a water, water raised to the overhead tank in a house etc.

(i) Gravitational potential energy decreases as the body falls downwards and is zero at the earth

(ii) Kinetic energy increases as the body falls downwards and is maximum when the body just strikes the ground. (iii) According to law of conservation of energy total mechanical (KE + PE) energy remains constant