Two blocks of clay, one of mass 100 kg and one of mass 3.00 kg, undergo a completely inelastic collision. before the collision one of the blocks is at rest and the other block is moving with kinetic energy 32.0 j the 3.00 ke block in intially at rest and the 1.00 ke block is moving, what is the initial speed of the 100 ke block? express your answer with the appropriate units value units submit part 8 what is the common nal speed of the two block? express your answer with the appropriate units value sub request answe part how much kinetic energy is lost in the collision? express your answer with the appropriate units. |ak|- value units submit request answer part d if the 100 kg block is initially at rest and the 300 kg block is moving, what is the initial speed of the 3.00 kg block? express your answer with the appropriate units. na obi? url value units submit request answer part e what is the common final speed of the two blocks in case of the 1.00 kg block is initially at rest? express your answer with the appropriate units. 13- value units submit request answer part f how much kinetic energy is lost in the collision in case of the 100 kg block is initially at rest? express your answer with the appropriate units o ? value submit reguet 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

part 1: which substance has a higher solubility?

a

smallest amount left undissolved atall temps in table shown most has dissolved high sol'by