question_answer

A bullet of mass 0.01kg and travelling at a speed of 500 m/s strikes a block of mass 2kg which is suspended by a string of length 5 m. The centre of gravity of the block is found to rise a vertical distance of 0.1 m, Fig. 4.47. What is the speed of the bullet after it emerges from the block? \[\left( g=9.8\,m/{{s}^{2}} \right)\].

Answer:

                Here, m = 0.01 kg, \[\upsilon =500\,m/s\] \[M=2kg,\,l=5m,\,h=0.1m,g=9.8\,m/{{s}^{2}}\] Let V be the velocity acquired by the blocks. \[\therefore \]   \[\frac{1}{2}M{{V}^{2}}=Mgh\] \[V=\sqrt{2gh}=\sqrt{2\times 9.8\times 0.1}=1.4m/s\] If is speed of bullet on emerging out of block, Fig. 4.47, then by law of conservation of momentum, we get \[m\upsilon +M\times 0=MV+m\upsilon '\] or \[\upsilon '=\frac{m\upsilon -MV}{m}=\frac{0.01\times 500-2\times 1.4\,}{0.01}\] \[\therefore \]   \[\upsilon '=220\,m/s\]