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
