A bullet of mass 50 g is fired into a wooden block of mass 2 kg resting on a smooth table surface. The bullet entres at `50 m*s^(-1)` and gets embedded in the block. Find the final velocity of the block. Find the initial and the final kinetic energy of the block -bullet system.

Explanation

Look at Fig.1.40. From the law of conservation of momentum,
`m_1v=(m_1+m_2)V`
or, `V=(m_1v)/(m_1+m_2)`….. (1)

Here, `m_1=50g =0.05kg , v=50 m*s^(-1) and m_2=2kg`.
`therefore` The final velocity of the block,
`V=(0.05xx50)/(0.05+2) =1.22 m*s^(-1)`
Initial kinetic energy of the system ,
`E_i=1/2xx0.05xx50^2=62.5J`.
Final kinetic energy ,
`E_f=1/2 xx(2+0.05)xx(1.22)^2=1.52 J`