Question Number 28585 by Tinkutara last updated on 27/Jan/18
Answered by ajfour last updated on 27/Jan/18
$${collision}\:{for}\:{m}_{\mathrm{1}} \:{is}\:{elastic}. \\ $$$${after}\:{collision},\:{velocity}\:{of}\:{m}_{\mathrm{1}} \\ $$$${is}\:{reversed};\:{kinetic}\:{energy}\: \\ $$$${is}\:{regathered}\:{as}\:{spring}\:{comes}\:{to} \\ $$$${relaxed}\:{length} \\ $$$$\:\:\:{K}_{{i}} ={m}_{\mathrm{1}} {gh}\:\:\:\:\: \\ $$$${m}_{\mathrm{1}} \:{still}\:{rises}\:{further}\:{by}\:'\boldsymbol{{x}}'\:{till}\:{it}\:{stops} \\ $$$${just}\:{when}\:{m}_{\mathrm{2}} \:{breaks}\:{off}\:{the} \\ $$$${surface},\:{so}\:\:\:\:\:\boldsymbol{{kx}}=\boldsymbol{{m}}_{\mathrm{2}} \boldsymbol{{g}}\:\:\:...\left({i}\right) \\ $$$$\:\:\:\:\:\boldsymbol{{K}}_{\mathrm{1}} =\boldsymbol{{K}}_{{f}} +\bigtriangleup\boldsymbol{{U}}_{{g}} +\bigtriangleup\boldsymbol{{U}}_{{spring}} \\ $$$$\:\:\:\:\:{m}_{\mathrm{1}} {gh}=\mathrm{0}+{m}_{\mathrm{1}} {gx}+\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:={x}\left({m}_{\mathrm{1}} {g}+\frac{{kx}}{\mathrm{2}}\right) \\ $$$${using}\:\left({i}\right) \\ $$$$\:\:\:\:\:\:\boldsymbol{{h}}=\frac{\boldsymbol{{m}}_{\mathrm{2}} \boldsymbol{{g}}}{\boldsymbol{{k}}}\left(\mathrm{1}+\frac{\boldsymbol{{m}}_{\mathrm{2}} }{\mathrm{2}\boldsymbol{{m}}_{\mathrm{1}} }\right)\:. \\ $$
Commented by Tinkutara last updated on 27/Jan/18
Why collision of m1 is elastic?
Commented by ajfour last updated on 27/Jan/18
$${only}\:{conservative}\:{forces}\:{acted} \\ $$$${on}\:{it}\:. \\ $$
Commented by Tinkutara last updated on 28/Jan/18
But what is the datum assumed here?
Commented by ajfour last updated on 28/Jan/18
in simpler words please ..
Commented by Tinkutara last updated on 28/Jan/18
What is the reference line assumed for calculating potential energy?
Commented by ajfour last updated on 28/Jan/18
$${with}\:{m}_{\mathrm{2}} \:{at}\:{ground}\:{level}\:{the} \\ $$$${level}\:{of}\:{m}_{\mathrm{1}} \:{when}\:{spring}\:{comes} \\ $$$${to}\:{natural}\:{length}\:{after}\:{collision} \\ $$$${is}\:{taken}\:{to}\:{be}\:{the}\:{reference}\:{level} \\ $$$${of}\:{gravitational}\:{potential}\:{energy}. \\ $$
Commented by Tinkutara last updated on 28/Jan/18
$${But}\:{then}\:{they}\:{should}\:{be}\:−{m}_{\mathrm{1}} {gh}\:{and} \\ $$$$−{m}_{\mathrm{1}} {gx}.\:{Or}\:{not}? \\ $$
Commented by ajfour last updated on 28/Jan/18
$${No},\:{spring}\:{is}\:{being}\:{stretched}\:{after} \\ $$$${this}\:{due}\:{to}\:{momentum}\:{of}\:{m}_{\mathrm{1}} . \\ $$$${m}_{\mathrm{1}} \:{is}\:{rising}\:{up}. \\ $$
Commented by Tinkutara last updated on 28/Jan/18
But potential energy when measured downwards should be taken negative or not?
Commented by ajfour last updated on 28/Jan/18
$${K}_{\mathrm{0}} \:{of}\:{m}_{\mathrm{1}} =−\bigtriangleup{U} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=−\left(−{m}_{\mathrm{1}} {gh}\right)\:={m}_{\mathrm{1}} {gh} \\ $$$${just}\:{before}\:{m}_{\mathrm{2}} \:{hits}\:{the}\:{ground}: \\ $$$${after}\:{this}\:{there}\:{is}\:{m}_{\mathrm{2}} \:{collides} \\ $$$${with}\:{ground}\:{and}\:{looses}\:{all}\:{its} \\ $$$${kinetic}\:{energy}.\:{m}_{\mathrm{1}} \:{compresses} \\ $$$${the}\:{spring}\:{and}\:{thereafter},\:{by}\:{the} \\ $$$${time}\:{springs}\:{regains}\:{its}\:{original} \\ $$$${length}\:{m}_{\mathrm{1}} \:{regathers}\:{its}\:{Kinetic} \\ $$$${energy}.\:{the}\:{level}\:{of}\:{m}_{\mathrm{1}} \:{at}\:{this} \\ $$$${level}\:{is}\:{now}\:{taken}\:{to}\:{be}\:{initial} \\ $$$${position}\:{to}\:{apply}\:{work}\:{energy} \\ $$$${theorwm}.\:{K}_{\mathrm{1}} \:{of}\:{m}_{\mathrm{1}} \:={K}_{\mathrm{0}} ={m}_{\mathrm{1}} {gh}. \\ $$$${m}_{\mathrm{1}} \:{rises}\:{beyond}\:{this}\:{point}\:{and} \\ $$$${by}\:{the}\:{time}\:{its}\:{speed}\:{gets}\:{zero}, \\ $$$${m}_{\mathrm{2}} \:{is}\:{breaks}\:{contact}. \\ $$$${work}-{energy}\:{eq}.\:{for}\:{m}_{\mathrm{1}} \:{is} \\ $$$${W}_{{ext}} ={K}_{\mathrm{2}} −{K}_{\mathrm{1}} +\bigtriangleup{U}_{{g}} +\bigtriangleup{U}_{{spring}} \\ $$$$\mathrm{0}=\left(\mathrm{0}−{m}_{\mathrm{1}} {gh}\right)+\left({m}_{\mathrm{1}} {gx}−\mathrm{0}\right)+\left(\frac{\mathrm{1}}{\mathrm{2}}{kx}^{\mathrm{2}} −\mathrm{0}\right) \\ $$$${spring}\:{force}\:{finally}\:{is}\:{kx}={m}_{\mathrm{2}} {g} \\ $$$${hence} \\ $$$${m}_{\mathrm{1}} {gh}={x}\left({m}_{\mathrm{1}} {g}+\frac{{kx}}{\mathrm{2}}\right) \\ $$$${h}={x}\left(\mathrm{1}+\frac{{kx}}{\mathrm{2}{m}_{\mathrm{1}} {g}}\right)\:=\:\frac{\boldsymbol{{m}}_{\mathrm{2}} \boldsymbol{{g}}}{\boldsymbol{{k}}}\left(\mathrm{1}+\frac{\boldsymbol{{m}}_{\mathrm{2}} }{\mathrm{2}\boldsymbol{{m}}_{\mathrm{1}} }\right)\:. \\ $$
Commented by Tinkutara last updated on 29/Jan/18
Thank you very much Sir! I got the answer.