question_answer

A bullet of mass 0.02 kg travelling horizontally with velocity \[250\text{ }m{{s}^{-1}}\]strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40 m. The coefficient of sliding friction of the rough surface is \[\left( g=9.8\text{ }m{{s}^{-2}} \right)\]

A)  0.75

B)  0.61

C)  0.51   

D)  0.30  

Correct Answer: C

Solution :

Alter  impact  the bullet   and   block move together and   comes to rest after \[m=0.2\text{ }kg\] covering a distance of \[40\text{ }m\]. By conservation of momentum \[{{m}_{1}}{{u}_{1}}+{{m}_{2}}{{u}_{2}}={{m}_{1}}{{v}_{1}}+{{m}_{2}}{{v}_{2}}\] or  \[0.02\times 250+0.23\times 0=0.02v+0.23v\] \[5+0=v(0.25)\] \[\frac{500}{25}=v=20m{{s}^{-1}}\] Now, by conservation of energy \[\frac{1}{2}M{{v}^{2}}=\mu R.d\] or \[\frac{1}{2}\times 0.25\times 400=\mu \times 0.25\times 9.8\times 40\] \[\Rightarrow \] \[\mu =\frac{200}{9.8\times 40}=0.51\]