question_answer

A wooden block of mass M rests on a horizontal surface. A bullet of mass m moving in the horizontal direction strikes and gets embedded in it. The combined system covers a distance x on the surface. If the coefficient of friction between wood and the surface is \[\mu \], the speed of the bullet at the time of striking the block is (where m is mass of the bullet)

A) \[\sqrt{\frac{2\,Mg}{\mu m}}\]               

B) \[\sqrt{\frac{2\,\mu mg}{Mx}}\]  

C) \[\sqrt{2\mu gx}\,\left( \frac{M+m}{m} \right)\]    

D) \[\sqrt{\frac{2\mu mx}{M+m}}\]

Correct Answer: C

Solution :

   \[\,Let\text{ }speed\text{ }of\text{ }the\text{ }bullet\text{ }=\text{ }v\] \[Speed\text{ }of\text{ }the\text{ }system\text{ }after\text{ }the\text{ }collision\text{ }=V\] \[\operatorname{By}\,\,conservation of momentum\,\,m\nu =\left( m+M \right)V\]\[\Rightarrow \,\,\,V=\frac{m\nu }{M+m}\] So the initial K.E. acquired by the system \[\Rightarrow \,\,\frac{1}{2}(M+m){{V}^{2}}\] \[\Rightarrow \,\,\frac{1}{2}(m+M){{\left( \frac{m\nu }{M+m} \right)}^{2}}\] \[=\,\,\frac{1}{2}\,\,\frac{{{m}^{2}}{{\nu }^{2}}}{(m+M)}\] This kinetic energy goes against friction work done by friction \[=\mu R\times x=m(m+M)g\,\,\times \,\,x\] By the law of conservation of energy \[\frac{1}{2}\frac{{{m}^{2}}{{\nu }^{2}}}{(m\,\,+\,\,M)}\,\,=\,\mu (m+M)\,g\,\,\times \,\,x\] \[\Rightarrow \,\,{{\nu }^{2}}=2\mu gx{{\left( \frac{m+M}{m} \right)}^{2}}\] \[\therefore \,\,\,v=\sqrt{2\mu gx}\,\left( \frac{M+m}{m} \right)\]