Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that there is no external torque on the system of two twins
so here we will use



Part b)
Since angular momentum is conserved here as there is no external torque
so we will have



Part c)
Work done by both of them = change in kinetic energy
so we have




Answer:
a) v = 0.8 m / s
, b)
= 0.777 m / s
, c) ΔK = 0.93 J
Explanation:
This exercise can be solved using the concepts of moment, first let's define the system as formed by the two blocks, so that the forces during the crash have been internal and the moment is conserved.
They give us the mass of block 1 (m1 = 100kg, its kinetic energy (K = 32 J), the mass of block 2 (m2 = 3.00 kg) and that it is at rest (v₀₂ = 0)
Before crash
po = m1 vo1 + m2 vo2
po = m1 vo1
After the crash
= (m1 + m2) 
a) The initial speed of the block of m1 = 100 kg, let's use the kinetic energy
K = ½ m v²
v = √2K / m
v = √ (2 32/100)
v = 0.8 m / s
b) The final speed,
p₀ =
m1 v₀1 = (m1 + m2) 
= m1 / (m1 + m2) v₀₁
The initial velocity is calculated in the previous part v₀₁ = v = 0.8 m / s
= 100 / (3 + 100) 0.8
= 0.777 m / s
c) The change in kinetic energy
Initial K₀ =
K₀ = 32 J
Final
= ½ (m1 + m2)
²
= ½ (3 + 100) 0.777²
= 31.07 J
ΔK =
- K₀
ΔK = 31.07 - 32
ΔK = -0.93 J
As it is a variation it could be given in absolute value
Part D
For this part s has the same initial kinetic energy K = 32 J, but it is block 2 (m2 = 3.00kg) in which it moves
d) we use kinetic energy
v = √ 2K / m2
v = √ (2 32/3)
v = 4.62 m / s
e) the final speed
v₀₂ = v = 4.62 m/s
p₀ = m2 v₀₂
m2 v₀₂ = (m1 + m2) 
= m2 / (m1 + m2) v₀₂
= 3 / (100 + 3) 4.62
= 0.135 m / s
f) variation of kinetic energy
= ½ (m1 + m2)
²
= ½ (3 + 100) 0.135²
= 0.9286 J
ΔK = 0.9286-32
ΔK = 31.06 J
The mass of a substance is given in atomic mass units and is calculated by adding the average atomic masses of all the atoms in the substance's chemical formula.
<h3>What empirical formula represents the total average atomic mass of every atom?</h3>
The Method The average atomic masses of all the atoms included in a formula's representation are added to get the mass of any molecule, formula unit, or ion. It has no bearing on the number of significant figures because the number of atoms is an exact quantity. One H2O molecule weighs 18.02 amu on average.
<h3>What connection exists between the empirical formula and the molecular formula?</h3>
You can determine the number of atoms of each element in a molecule using its molecular formula. These empirical formulations provide the most basic or reduced elemental ratio of a compound. The empirical formula and the molecular formula of a substance are same if the molecular formula can no longer be decreased.
To know more about atomic mass visit:-
brainly.com/question/17067547
#SPJ4