question_answer

Two wooden blocks of masses 1 kg and 2 kg are separated by a certain distance. A bullet of mass 50 g fired from a gun pierces through the block of mass 1 kg and then stopped in the second block. After the impact of the bullet, both blocks start moving with the same speed. Calculate the percentage loss in the initial velocity of the bullet when it is in between the two blocks.

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         = \[\frac{\left( \upsilon -{{\upsilon }_{1}} \right)\times 100}{\upsilon }=\frac{\left( 61V-41V \right)}{61V}\]                                                                                 \[=\frac{20}{21}\times 100=32.8%\]