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 by aurpeyz 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 by aurpeyz last updated on 28/Oct/20
$${thanks} \\ $$