Answer:
a) 5.197rev/s
b) Kf/Ki =2.28
Explanation:
a) Angular momentum of the system L = Iw
ButLi=Lf
Kiwi =Ifwf
wf = (Ii/If)will = (4.65/3.4)×3.8=5.197rev/s
b)Kinetic energy KE= 0.5Iw^2
Ki = 0.5Iiwi^2
Kf=0.5Ifwf^2
Kf/Ki = Ifwf/Iiwi
Kf/Ki = (4.65/3.4))(5.197/3.8)
Kf/Ki = 1.22(1.368)^2
Kf/Ki = 2.28
After the collision the magnitude of the momentum of the system is Mv
Given:
mass of 1st object = M
speed of 1st object = v
mass of 2nd object = M
speed of 2nd object = 0
To Find:
magnitude of the momentum after collision
Solution: Product of the mass of a particle and its velocity. Momentum is a vector quantity; i.e., it has both magnitude and direction. Isaac Newton's second law of motion states that the time rate of change of momentum is equal to the force acting on the particle.
Applying conservation of linear momentum
Mv + M(0) = 2MV
Mv = 2MV
V = v/2
So, after collision momentum is
p = 2MV = 2xMxv/2 = Mv
So, after collision momentum is Mv
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Answer:
V is greater
Explanation:
because v intial at that time V final is the that speed which it is going at that time
Answer: 
Explanation:
If we make an analysis of the net force 
of the rock that was thrown upwards, we will have the following:
(1)
Where:
is the force with which the rock was thrown
is the weight of the rock
Being the weight the relation between the mass 
of the rock and the acceleration due gravity 
:
(2)
(3)
Substituting (3) in (1):
(4)
(5) This is the net Force on the rock
On the other hand, we know this force is equal to the multiplication of the mass with the acceleration, according to Newton's 2nd Law:
(6)
Finding the acceleration 
:
(7)
(8)
Finally:
