A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.44 m long. The block is initially at rest. A bullet with mass 0.0134 kg is fired at the block with a horizontal velocity v0. The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m , the tension in the cord is 4.76 N .

What was the initial speed v0 of the bullet?

     v₀ = 240  m / s

Explanation:

This problem must be solved in two parts, first we must use the conservation of the moment, then the conservation of energy.

Let's start by applying moment conservation, to the system formed by the block and bullet, in this case the forces during the crash are internal and the moment is conserved

Instant starts. Before the crash

          p₀ = m v₀

Final moment. Right after the crash

          = (m + M) v

The moment is preserved

         Po =p_{f}

        M v₀ = (m + M) v

        v = m / (m + M) v₀        (1)

This is the speed with which the bullet block system comes out, now we can use energy conservation

Starting point. Right after the crash

         Em₀ = K = ½ (m + M) v²

Final point. Highest point of the path

          = U = (m + M) g y

         Em₀ = Em_{f}

         ½ (m + M) v² = (m + M) g y

          v = √2 g y                 (2)

We substitute 1 in 2

              m / (m + M) v₀ = √ 2gy

            v₀ = (m + M) / m √ 2gy

Let's calculate

           v₀ = (0.0134 +0.800) /0.0134    √ (2 9.8 0.8)

           v₀ = 240  m / s