10.0-g wad of sticky clay is hurled horizontally at a 110-g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?

 v₁ = 1146.6 m / s

Explanation:

This problem must be solved in two parts, a first part of the crash or another with energy after the crash.

Let's start by analyzing the crash.

Definition the system as formed by the clay block and the wooden block, in this case the moment is conserved because the forces are of action and reaction type

Let's use subscript 1 for the clay block and subscript 2 for the wood block

Initial. Before the crash

       p₀. = m₁ v₁

Final. Right after the crash

        = (m₁ + m₂) v

        p₀ = p_{f}

        m₁ v₁ = (m₁ + m₂) v

         v₁ = v (m₁ + m₂) / m₁

Now the body is formed by the joint of the plug + the wooden block

           W = ΔK

           -fr x = Kf - K₀

The final velocity is zero, so the final kinetic energy is zero

The equation for friction force is

             fr = μ N

From Newton's second law, on the horizontal surface

             N- W = 0

             N = (m₁ + m₂) g

             fr =  μ (m₁ + m₂) g

We substitute

          - μ (m₁ + m₂) g x = 0 - ½ (m₁ + m₂) v

           v = 2  μ g x

           v = 2 0.650 9.8 7.50

           v = 95.55 m / s

We substitute in the moment equation

           v₁ = v (m₁ + m₂) / m₁

           v₁ = 95.55 (10 + 110) / 10

           v₁ = 1146.6 m / s

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explanation:

where are the vectors ?