Question

A bullet of mass m, moving horizontally with speed u, meets a block of wood of mass M, travelling along the same line but in the opposite direction with speed ,U, and remains embedded in it. Show that the loss of kinetic energy is of the form 1/2kMm, where k is in terms of u, U, m, M and find the loss in KE when the bullet of mass 4.0g, travelling at 890ms^-1 hits a block of wood, mass 4.0kg moving at 7.0ms^-1

Answer 1

Step-by-step explanation:

First of all we shall calculate the velocity of composite mass . Let it be v . Applying law of conservation of momentum

mu - MU = ( m + M ) v

v = mu - MU / ( m + M )

loss of kinetic energy

= 1/ 2 mu² + 1/2 MU² - 1/2 ( M +m ) v²

= 1/ 2 mu² + 1/2 MU² - 1/2 ( M +m ) (mu - MU)² / ( m + M )²

= 1/ 2 mu² + 1/2 MU² - 1/2 (mu - MU)² / ( m + M )

= 1/2 [ m²u² + mMu² +mMU² + m²U² - m²u² - M²U² - 2 muMU ] / ( m + M )

= 1 / 2 [ mMu² + mMU² - 2 muMU ] / ( m + M )

= 1 / 2mM [ (u² + U² - 2 uU) / ( m + M )]

= 1/2 mM x k

where

k = [ (u² + U² - 2 uU) / ( m + M )]

Given

m = .004 kg

M = 4 kg

u = 890 ms⁻¹

U = 7 ms⁻¹

k = ( 890² + 7² - 2 x 890 x 7 ) / 4.004

= ( 792100 + 49 - 12460 ) / 4.004

= 194727.52

loss of kinetic energy

= 1/2 mM x k

= .5 x .004 x 4 x 194727.52

= 1557.82 J .