Question

A bullet of mass m moving horizontally strikes a block of wood of mass M lying at rest on a smooth horizontal surface. the fraction of kinetic energy lost by the bullet if it gets embedded in the block is

Answer 1

a bullet of mass m  has an initial velocity = v
initial kinetic energy of the bullet = 1/2  m v²

Wooden block :  mass  M , let the final velocity of the two bodies = V

       we apply the principle of conservation of momentum.
     m v + M * 0 = (M + m) V
               V =  m v / (M +m)

   The kinetic energy initial = 1/2 m v²
   the final kinetic energy = 1/2 (m+ M) V² = 1/2 * (m + M) [m v / (M+m) ]²
                     = 1/2 * m² v² / (M +m)

   loss of kinetic energy = 1/2 m v² [ 1 - m/(M+m) ] = 1/2 M m v² / (m+M)

Answer 2

The answer is provided in the attachment.

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