# 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

1
by 2092000

2015-04-12T01:33:48+05:30

### This Is a Certified Answer

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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)