Answer:
Here, \[{{M}_{1}}=1kg\],
\[{{M}_{2}}=2kg\]
\[m=50g=\frac{50}{100}kg=\frac{1}{20}kg\]
Let \[\upsilon \] = initial
velocity of the bullet
\[{{\upsilon }_{1}}\]= velocity of the bullet after piercing through the first block
V = velocity of each block when hit by the bullet. Fig,
3(HT).5.
As the bullet penetrates first
block, as per the principle of conservation of linear momentum
\[m\upsilon ={{M}_{1}}V+m{{\upsilon }_{1}}\] ???.
(i)
Again, applying the same
principle, when the bullet is embedded in second block:
\[m{{\upsilon }_{1}}=\left( {{M}_{2}}+m \right)V\]
\[{{\upsilon }_{1}}=\frac{\left(
{{M}_{2}}+m \right)V}{m}=\frac{\left( 2+0.05 \right)}{0.05}V=41V\]
From \[0.05\upsilon
=1V+0.05\times {{\upsilon }_{1}}=V+0.05\times 41V=3.05V\]
\[\upsilon
=\frac{3.05V}{0.05}=61V\]
%age loss in initial velocity of
bullet = \[\frac{\left( \upsilon -{{\upsilon }_{1}} \right)\times
100}{\upsilon }=\frac{\left( 61V-41V \right)}{61V}\]
\[=\frac{20}{21}\times
100=32.8%\]
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