A classical mechanics problem by Aditya Khatavkar
A bullet is fired at a speed of . It embeds itself in a block with mass hanging from a string. Mass of the bullet is . Find the maximum height (in ) achieved by the block (with the bullet embedded).
Assume that and the initial height of the block is .
The answer is 0.0173.
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1 solution
Let\quad 'm'\quad be\quad the\quad mass\quad of\quad the\quad bullet,\quad 'M'\quad be\quad the\quad mass\quad of\quad the\quad block,\quad 'v'\quad be\quad the\quad velocity\quad of\quad the\quad bullet\quad right\quad before\quad the\quad collision,\quad 'V'\quad be\quad the\quad velocity\quad of\quad the\quad \\ block\quad and\quad the\quad bullet\quad right\quad after\quad the\quad collision\quad (both\quad must\quad be\quad the\quad same\quad as\quad it\quad is\quad an\quad inelastic\quad collision),\quad and\quad 'h'\quad be\quad the\quad block\quad and\quad the\quad bullet's\\ maximum\quad height\quad achievable.\\ \\ \\ Total\quad momentum\quad just\quad before\quad the\quad collision\quad must\quad be\quad the\quad same\quad as\quad the\quad total\quad momentum\quad right\quad after\quad the\quad collision.\\ \\ \\ \therefore \quad mv\quad =\quad (m+M)V\\ \\ ⇒\quad V\quad =\quad \frac { mv }{ (m+M) } \\ \\ \\ The\quad kinetic\quad energy\quad of\quad the\quad block\quad and\quad the\quad bullet\quad right\quad after\quad the\quad collision\quad must\quad be\quad equal\quad to\quad the\quad potential\quad energy\quad of\quad the\quad block\quad and\quad the\quad bullet\\ when\quad they\quad are\quad at\quad their\quad maximum\quad height\quad achievable\quad (when\quad their\quad velocity\quad is\quad zero\quad m/s).\\ \\ \\ \therefore \quad \frac { 1 }{ 2 } (m+M){ (\frac { mv }{ m+M } ) }^{ 2 }\quad =\quad (m+M)gh\\ \\ ⇒\quad h\quad =\quad \frac { { (mv) }^{ 2 } }{ 2g{ (m+M) }^{ 2 } } \\ \\ \\ Plugging\quad in\quad the\quad values\quad for\quad 'm',\quad 'M',\quad 'v'\quad and\quad 'g',\quad we\quad get:\\ \\ \\ h\quad =\quad \frac { { (0.02*30) }^{ 2 } }{ 2*10*{ (0.02+1) }^{ 2 } } \quad =\quad \frac { 0.36 }{ 20.808 } \quad \~ \quad 0.173\quad meters