(i) A body of mass 1.5 kg is allowed to fall freely under gravity. Find the momentum of the body 5 secs after it starts falling.
(ii) What will be the kinetic energy of the body at the same time?
(ili) Find the height at which the body will attain this kinetic energy when it falls freely.
(iv) What will be the velocity of the body on striking the ground?
(v) Find the ratio of P.E. to K.E. at a height of 62.5 m above the ground.
Answers
Answer:
1) GIVEN ;
m = 1kg
u =0
t = 5 sec
g = 10 m/s^2
v = ?
v = u+at
v = 0+10×5
v = 50 m/s
(1) momentum
= mv
= 1×50
= 50 kgm/s
(2) Kinetic energy
= 1/2 mv^2
=1/2 ×1 × 50 × 50
= 25×50
= 1250 Joule
2) a body may gain kinetic energy and potential energy simultaneously.
For example, a body is taken upward by continuously applying a force in the upward direction. Due to the upward force, the velocity (hence kinetic energy) of the body increases and as the body goes higher, its potential energy also increases.
But this system is not an isolated system as an external force is applied to it. Mechanical energy is conserved only when the forces acting on the system are conservative. Thus the reason is also correct and explains the assertion very well.
If conservative forces act on the system, then the mechanical energy of the system is conserved. Then the body does not gain kinetic energy and potential energy simultaneously.
3) Velocity after 5 secs , v=gt=10×5=50m/s
Mass of body , m=1kg
Momentum of body , p=mv=1×50=50kg m/s
Kinetic energy of body , K=21mv2=21×1×502=1250 J
4) For earth the escape velocity is 7 mi/sec or 11,000 m/s, 11 kilometers/sec or about 25,000 mph. An object given this velocity on the earth's surface will not return.
5) Total mechanical energy at height, H
EH=mgH
Let vh be velocity of the ball at height h(=43H)
∴ Total mechanical energy at height h,
Eh=mgh+21mvh2
According to law of conservation of mechanical energy,
EH=Eh;mgH=mgh+21mvh2
vh2=2g(H−h)
Required ratio of kinetic energy to potential energy at height h is
VhKh=mgh21mvh2=mgh21m2g(H−h)=(hH−1)=31.
Given: Mass of body = 1.5kg
Time = 5 seconds
To find: (i) momentum of the body after 5 seconds
(ii) kinetic energy of the body at 5 seconds
(iii) Height at which body will attain this kinetic energy
(iv) Velocity of the body on striking the ground
(v) ratio of potential and kinetic energy at a height of 62.5m above the ground.
Solution:
(i) Momentum is defined as the product of mass and velocity.
Since the body is falling freely, therefore initial velocity will be zero.
V = U + at ( v is the final velocity, u is the initial velocity, a is the acceleration, t is the time )
V = 0+ 10×5 = 50m/sec
Momentum = mass × velocity
= 1.5 × 50 = 75kgm/s
Therefore, the momentum of the body at 5 seconds is 75 kgm/s
(ii) Kinetic energy is the energy possessed by the body on the virtue of its motion, and its formula is 1/2mv².
Kinetic energy = 1/2 × 1.5 × (50)²
= 1875 J
Therefore, the kinetic energy will be 1875 J.
(iii) At that height, the potential energy of the body will be equal to the kinetic energy of the body.
Potential energy is the energy possessed by the body on the virtue of its position.
K.E = P.E
1/2mv² = mgh
1/2v² = gh
h = v²/2g
= (50²) / 2× 10
= 125m
Therefore, at a height of 125 m, the body will attain kinetic energy of 1875 J
(iv) The distance covered by the ball in the first 5 seconds.
v² - u² = 2as
(50)² - 0 = 2×10×s
s = 125m
Therefore, the total distance will be = 125 + 125 = 250m
Now, the velocity at which the body will strike the ground will be,
v² - u² = 2as
v² - 0 = 2×10×250
v² = 5000
v = 70.7 m/s
Therefore, the velocity of the body on striking the ground will be 70.7 m/s
(v) At 62.5m, the potential energy of the ball will be = mgh
= 1.5 × 10 × 62.5 = 937.5 J
The potential energy at 62.5m will be 937.5 J
The velocity of the ball at 62.5m, v² - u² = 2as
v² - 0 = 2×10×62.5
v = 35.35 m/s
Kinetic energy will be 1/2mv²
Kinetic energy = 1/2×1.5×(35.5)² J
= 945.18 J
The kinetic energy of the ball at 62.5m will be 945.18J
Ratio = potential energy/kinetic energy
= 937.5/945.18
= 0.99
Therefore, the ratio will be 0.99.