A bullet of mass 0.012 kg and horizontal speed `70ms^(-1)` strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Estimate the loss in kinetic energy.

Explanation

`m_(b)=0.012kg,v_(ib)=70ms^(-1),m_(w)=0.4kg,h=?,DeltaKE=0`
Common speed `v=(m_(b)v_(ib))/(m_(b)+m_(w))=(0.012xx70)/(0.012+0.4)=2.04ms^(-1)`
Also `h=(v^(2))/(2g)=(2.04xx2.04)/(2xx10)=0.208m`
Loss in K.E. = Heat produced in the wood = `(1)/(2)((m_(b)m_(w))/(m_(b)+m_(w)))v_(ib)^(2)`
`=(1)/(2)xx((0.012xx0.4)/(0.412))xx70^(2)=28.54J`