Extra
score Questions

Q. Can any object have mechanical energy
even if its momentum is zero? Explain.

Ans: Yes, mechanical energy comprises
of both potential energy and kinetic energy. Zero momentum means that
velocity is zero. Hence, there it no kinetic energy but the object may
possess potential energy.

Q. Can kinetic energy of a body be negative?

Ans: No, It is because mass and velocity cannot ne negative

Q. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Ans: A freely falling object just before hitting the ground has maximum kinetic energy. After falling, it rolls on the rough ground and finally comes to rest. The kinetic energy of the object is used up in doing work against friction; which finally appears as heat energy.

Ans: No, It is because mass and velocity cannot ne negative

Q. A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?

Ans: A freely falling object just before hitting the ground has maximum kinetic energy. After falling, it rolls on the rough ground and finally comes to rest. The kinetic energy of the object is used up in doing work against friction; which finally appears as heat energy.

Q. A light and a heavy object have the same
momentum find out the ratio of their kinetic energies.Which
one has a larger kinetic energy?

Linear momentum of 1

^{st}object = p1=m1v1
Linear momentum of 2

^{nd}object = p2=m2v2
Given, p1 > p2
---------------------------------(i)

Þ m1v1 > m2v2

But, m1<m2 (A light and a heavy
object) Þ v1 > v2 ------------(ii)

Ke = ½ mv

^{2}= ½ m vx v =1/2 pv
From (i)and (ii) p1v1 >
p2v2 Þ ½ p1v1 > ½ p2v2 Þ KE1> KE2

Q. A rocket is moving up with a velocity v. If the
velocity of this rocket is suddenly tripled, what will be the ratio of two
kinetic energies?

Ans: Initial KE/Final KE = ( ½ mu

^{2}) /( ½ mv^{2}) = ( ½ mu^{2}) /{ ½ m(3v^{2})} =1:9
Q. Give one example each of potential energy (I)
due to position (ii) due to shape.

Ans: (i) Potential energy due to position: Water
stored in dam has potential energy.

(ii) Potential energy due to shape: In a toy
car, the wound spring possesses potential energy, and as the spring is released,
its potential energy changes into kinetic energy due to which the car moves.

Q. What kind of energy transformation takes place
when a body is dropped from a certain height?

Ans: When a body falls, its potential energy
gradually gets converted into kinetic energy. On reaching the ground, the
whole of the potential energy of the body gets converted into kinetic energy.

Q. A man drops a stone of 200g from a height of 5m. What
is its kinetic energy when it reaches the ground? What is its speed before it
hits the ground?

Ans: Using
the equation of motion,

u =0m/s ; v = ? s = 5m

V2=u2 + 2as

v2= o + 2 x 9.8 x5 =9.9m/s

This is the speed with which body hits the ground.

Kinetic energy of the body before hitting the ground is,

m = 200g = 0.2kg

KE = ½ mv

^{2}= ½ x 0.2 x9.9x9.9 =9.8j
Q. Two bodies have same momentum. Which will have greater kinetic
energy- heavier body or lighter body?

Let mass of light body be 'm1 ' & heavy body be 'm2 '

Given that m1V1=m2v2

But, KE=p

^{2}/2m
Since as p is constant K E is inversely proportional
to the mass of the object

Hence light body has greater kinetic energy

Q. An electric bulb of 60w is used for 6h per day .Calculate the
units of energy consumed in one day by the bulb.

p =w/t or, p =E/t

E = wt = 60w x 6h = 60w x 6 x
3600 sec = 1.3 x 10

^{6}J
Q. A boy of mass 50kg runs up to a stair case of 45 steps in 9s. If
the height of a step is 15cm, find his power. (g= 10m/s2)

Ans: h will be the net height
attained by him once he puts all the steps i.e. (45 x 0.15) = 6.75 /m

Power = w/t =PE/t = mgh/t = = (50x10x6.75)/9 = 375 J/s

Q. Two particles of masses 1g and 2g have equal momentum. Find the
ratio between their kinetic energies?

Momentum of first body, p

Momentum of second body, p

As given: m

v

Now kinetic energy, k = 1/2mv

Ratio of kinetic energy, k

_{1}= m_{1}v_{1}Momentum of second body, p

_{2}= m_{2}v_{2}As given: m

_{1}/m_{2}= ½ and p_{1}= p_{2}v

_{1}/v_{2}= m_{2}/m_{1}= 2Now kinetic energy, k = 1/2mv

^{2}Ratio of kinetic energy, k

_{1}/k_{2}= m_{1}v_{1}^{2}/m_{2}v_{2}^{2}= 2/1
Q. If the K.E. of a object increased by 300% ,Find the % increased
in momentum of the body .

Ans: E = ½ mv

^{2 }=^{ }p2/2m
Þ √2mE=p

New E’ = E + 300%of E = 4E

New, p’ = √2m4E =2 √2mE

Percentage increase in the momentum = (p’ - p
)/p x 100%

= (2 √2mE - √2mE)/ √2mE x 100% =100%

Q. A 400gm bag is lifted from
ground and kept on the table at a height of 2 m .The time taken to do so is 4s.
(Assume g=10m/s)

Ans: As
we know that work dome against the gravitational force

= mgh 400/1000gm x 10 x 2 = 8 J Now, Power =w/t = 8/4= 2 J/sec

Q.Two bodies I and II have the same kinetic energy. Their
velocities v1/ v2 are in the ratio of 1:2.

Since Two bodies I and II have the same kinetic energy: comparing
the KE of both the bodies

K

_{1}= K_{2}
m

_{1}v_{1}^{2}= m_{2}v_{2}^{2 }
as given, v1/v2 = 1/2 ⇒
v2/v1 = 2/1 ∴ m1/m2 = (v2/v1)2 = 4/1 or 4:1

**For further study**

9th Work and Energy
NCERT Exercise Solution

IX Work Power and Energy
Solution of important Questions Part-1 View

9th Work, Energy And
Power-CBSE Questions with solution Part-2 View

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