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.
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 1st object = p1=m1v1
Linear momentum of 2nd 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 mv2 = 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 mu2) /( ½ mv2) = ( ½ mu2) /{ ½ m(3v2)} =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 mv2= 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=p2 /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 106 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, p1 = m1v1
Momentum of second body, p2 = m2v2
As given: m1/m2 = ½ and p1 = p2
v1/v2 = m2/m1 = 2
Now kinetic energy, k = 1/2mv2
Ratio of kinetic energy, k1/k2 = m1v1 2/m2v2 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 mv2 = 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
K1 = K2
m1v1 2 = m2v22
as given, v1/v2 = 1/2 ⇒ v2/v1 = 2/1 ∴ m1/m2 = (v2/v1)2 = 4/1 or 4:1