Question Number 119648 by aurpeyz last updated on 26/Oct/20 | ||
$${Particles}\:{of}\:{mass}\:{m}_{\mathrm{1}} \:{and}\:{m}_{\mathrm{2}} \:\left({m}_{\mathrm{2}} >{m}_{\mathrm{1}} \right) \\ $$ $${are}\:{connected}\:{by}\:{a}\:{light}\:{inextensible} \\ $$ $${string}\:{passing}\:{over}\:{a}\:{smooth}\:{fixed}\: \\ $$ $${pulley}.\:{initially}\:{both}\:{masses}\:{hang} \\ $$ $${vertically}\:{with}\:{mass}\:{m}_{\mathrm{2}\:} {at}\:{a}\:{height} \\ $$ $${X}\:{above}\:{the}\:{floor}.\:{if}\:{the}\:{system}\:{is}\: \\ $$ $${released}\:{from}\:{rest}.\:{with}\:{what}\:{speed} \\ $$ $${will}\:{mass}\:{m}_{\mathrm{2}} \:{hit}\:{the}\:{floor}\:{and}\:{the} \\ $$ $${mass}\:{m}_{\mathrm{1}} \:{will}\:{rise}\:{a}\:{further}\:{distance}\:{of}\: \\ $$ $$\left[\frac{\left({m}_{\mathrm{2}} −{m}_{\mathrm{1}} \right){x}}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }\right]\:{after}\:{this}\:{occur}. \\ $$ | ||
Commented byaurpeyz last updated on 26/Oct/20 | ||
$${pls}\:{help} \\ $$ | ||
Answered by TANMAY PANACEA last updated on 26/Oct/20 | ||
$${m}_{\mathrm{2}} {g}−{T}={m}_{\mathrm{2}} {a} \\ $$ $${T}−{m}_{\mathrm{1}} {g}={m}_{\mathrm{1}} {a} \\ $$ $${a}=\frac{{g}\left({m}_{\mathrm{2}} −{m}_{\mathrm{1}} \right)}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} } \\ $$ $${v}^{\mathrm{2}} =\mathrm{0}^{\mathrm{2}} +\mathrm{2}{ax} \\ $$ $${v}=\sqrt{\mathrm{2}×\frac{{g}\left({m}_{\mathrm{2}} −{m}_{\mathrm{1}} \right)}{{m}_{\mathrm{1}} +{m}_{\mathrm{2}} }×{x}}\:\:\Lleftarrow{this}\:{is}\:{the}\:{speed} \\ $$ $${by}\:{which}\:{m}_{\mathrm{2}} \:{strike}\:{ground} \\ $$ $${pls}\:{chk} \\ $$ $$ \\ $$ | ||
Commented byaurpeyz last updated on 28/Oct/20 | ||
$${thanks} \\ $$ | ||