Answer:
a) v² = G M R³, b) T = 2π /
, c) 
Explanation:
a) The kinetic energy is
K = ½ m v²
to find the velocity let's use Newton's second law
F = m a
acceleration is centripetal
a = v² / R
force is the universal force of attraction
F = G m M / r²
we substitute
G m M R² = m v² R
v² = G M R³
the kinetic energy is
K = ½ m G M R³
b) angular and linear velocity are related
v = w R
w = v / R
w = 
w =
the angular velocity is related to the period
w = 2π / T
T = 2π / w
we substitute
T = 2π /
c) the angular moeomto is
L = m v r
L = m RA G M R³ R
L =
Given
m1(mass of the first object): 55 Kg
m2 (mass of the second object): 55 Kg
v1 (velocity of the first object): 4.5 m/s
v2 (velocity of the second object): ?
m3(mass of the object dropped): 2.5 Kg
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
270 = 62.5 v2
v2 = 4.32 m/s
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