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Question Number 54289 by ajfour last updated on 01/Feb/19
Commented by ajfour last updated on 01/Feb/19
Find u such that upper block topples, if particle of mass m hits elastically. Also find where the particle hits the ground.
$${Find}\:{u}\:{such}\:{that}\:{upper}\:{block}\:{topples}, \\ $$$${if}\:{particle}\:{of}\:{mass}\:{m}\:{hits}\:{elastically}. \\ $$$$\boldsymbol{{Also}}\:\boldsymbol{{find}}\:\boldsymbol{{where}}\:\boldsymbol{{the}}\:\boldsymbol{{particle}} \\ $$$$\boldsymbol{{hits}}\:\boldsymbol{{the}}\:\boldsymbol{{ground}}. \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19
from youtube i saw this video...excellent explanation hence attached...
$${from}\:{youtube}\:{i}\:{saw}\:{this}\:{video}…{excellent}\:{explanation} \\ $$$${hence}\:{attached}… \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 01/Feb/19
Commented by mr W last updated on 01/Feb/19
u≥5(√((3μgb)/2)) ?
$${u}\geqslant\mathrm{5}\sqrt{\frac{\mathrm{3}\mu{gb}}{\mathrm{2}}}\:? \\ $$
Commented by ajfour last updated on 01/Feb/19
given u=(5/2)(√(μgb)) ; but just now i got the same result as yours Sir.
$${given}\:{u}=\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\mu{gb}}\:\:;\:{but}\:{just}\:{now}\:{i} \\ $$$${got}\:{the}\:{same}\:{result}\:{as}\:{yours}\:{Sir}. \\ $$
Answered by mr W last updated on 02/Feb/19
such that the upper block topples, the lower block must move at least a distance of s=2b after the hit. in this distance the friction force is f=μ(2m+4m)g=6μmg and the energy lost due to friction is fs, i.e. 6μmg×2b=12μmgb. that it to say the lower block must have a kinetic energy of at least 12μmgb after the hit. KE=(1/2)(4m)v^2 ≥12μmgb ⇒v=(√(6μgb))=velocity of lower block after hit let u_1 =velocity of particle after hit (←) u=u_1 +v since mu=−mu_1 +4mv, ⇒u=−u_1 +4v ⇒u=−(u−v)+4v ⇒2u=5v ⇒u=((5v)/2)≥(5/2)(√(6μgb))=5(√((3μgb)/2)) u_1 =u−v=((5v)/2)−v=((3v)/2)=3(√((3μgb)/2)) position where particle hits the ground: d=u_1 (√((2b)/g))=3(√(3μ))b
$${such}\:{that}\:{the}\:{upper}\:{block}\:{topples}, \\ $$$${the}\:{lower}\:{block}\:{must}\:{move}\:{at}\:{least} \\ $$$${a}\:{distance}\:{of}\:{s}=\mathrm{2}{b}\:{after}\:{the}\:{hit}. \\ $$$${in}\:{this}\:{distance}\:{the}\:{friction}\:{force} \\ $$$${is}\:{f}=\mu\left(\mathrm{2}{m}+\mathrm{4}{m}\right){g}=\mathrm{6}\mu{mg}\:{and}\:{the} \\ $$$${energy}\:{lost}\:{due}\:{to}\:{friction}\:{is}\:{fs}, \\ $$$${i}.{e}.\:\mathrm{6}\mu{mg}×\mathrm{2}{b}=\mathrm{12}\mu{mgb}.\:{that}\:{it}\:{to}\:{say} \\ $$$${the}\:{lower}\:{block}\:{must}\:{have}\:{a}\:{kinetic} \\ $$$${energy}\:{of}\:{at}\:{least}\:\mathrm{12}\mu{mgb}\:{after}\:{the}\:{hit}. \\ $$$${KE}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{m}\right){v}^{\mathrm{2}} \geqslant\mathrm{12}\mu{mgb} \\ $$$$\Rightarrow{v}=\sqrt{\mathrm{6}\mu{gb}}={velocity}\:{of}\:{lower}\:{block}\:{after}\:{hit} \\ $$$${let}\:{u}_{\mathrm{1}} ={velocity}\:{of}\:{particle}\:{after}\:{hit}\:\left(\leftarrow\right) \\ $$$${u}={u}_{\mathrm{1}} +{v} \\ $$$${since}\:{mu}=−{mu}_{\mathrm{1}} +\mathrm{4}{mv}, \\ $$$$\Rightarrow{u}=−{u}_{\mathrm{1}} +\mathrm{4}{v} \\ $$$$\Rightarrow{u}=−\left({u}−{v}\right)+\mathrm{4}{v} \\ $$$$\Rightarrow\mathrm{2}{u}=\mathrm{5}{v} \\ $$$$\Rightarrow{u}=\frac{\mathrm{5}{v}}{\mathrm{2}}\geqslant\frac{\mathrm{5}}{\mathrm{2}}\sqrt{\mathrm{6}\mu{gb}}=\mathrm{5}\sqrt{\frac{\mathrm{3}\mu{gb}}{\mathrm{2}}} \\ $$$$ \\ $$$${u}_{\mathrm{1}} ={u}−{v}=\frac{\mathrm{5}{v}}{\mathrm{2}}−{v}=\frac{\mathrm{3}{v}}{\mathrm{2}}=\mathrm{3}\sqrt{\frac{\mathrm{3}\mu{gb}}{\mathrm{2}}} \\ $$$${position}\:{where}\:{particle}\:{hits}\:{the}\:{ground}: \\ $$$${d}={u}_{\mathrm{1}} \sqrt{\frac{\mathrm{2}{b}}{{g}}}=\mathrm{3}\sqrt{\mathrm{3}\mu}{b} \\ $$
Commented by ajfour last updated on 01/Feb/19
i have got to know d= 6b(√(3μ)) .
$${i}\:{have}\:{got}\:{to}\:{know}\: \\ $$$$\:\:{d}=\:\mathrm{6}{b}\sqrt{\mathrm{3}\mu}\:\:. \\ $$
Commented by mr W last updated on 01/Feb/19
please check sir! h=b=((gt^2 )/2) d=u_1 t=u_1 (√((2b)/g))=3(√(((3μgb)/2)×((2b)/g)))=3(√(3μ))b
$${please}\:{check}\:{sir}! \\ $$$${h}={b}=\frac{{gt}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${d}={u}_{\mathrm{1}} {t}={u}_{\mathrm{1}} \sqrt{\frac{\mathrm{2}{b}}{{g}}}=\mathrm{3}\sqrt{\frac{\mathrm{3}\mu{gb}}{\mathrm{2}}×\frac{\mathrm{2}{b}}{{g}}}=\mathrm{3}\sqrt{\mathrm{3}\mu}{b} \\ $$
Commented by ajfour last updated on 01/Feb/19
sorry sir, it was asked if velocity of particle were double of previously found min. velocity, where would it hit the ground ?
$${sorry}\:{sir},\:{it}\:{was}\:{asked}\:{if}\:{velocity} \\ $$$${of}\:{particle}\:{were}\:{double}\:{of}\:{previously} \\ $$$${found}\:{min}.\:{velocity},\:{where}\:{would}\:{it} \\ $$$${hit}\:{the}\:{ground}\:? \\ $$
Commented by ajfour last updated on 01/Feb/19
As per my posted question , all your results and explanation, i believe are correct, Sir.Thanks again!
$${As}\:{per}\:{my}\:{posted}\:{question}\:,\:{all}\:{your} \\ $$$${results}\:{and}\:{explanation},\:{i}\:{believe} \\ $$$${are}\:{correct},\:{Sir}.{Thanks}\:{again}! \\ $$
Answered by ajfour last updated on 01/Feb/19
mu=4mV−mv u=V+v ⇒ u= 4V+V−u ⇒ V = ((2u)/5) (1/2)(4m)V^( 2) ≥ μ(6mg)(2b) ⇒ 2(((2u)/5))^2 ≥ 12μgb ⇒ u ≥ 5(√((3μgb)/2)) .
$${mu}=\mathrm{4}{mV}−{mv} \\ $$$${u}={V}+{v} \\ $$$$\Rightarrow\:\:{u}=\:\mathrm{4}{V}+{V}−{u} \\ $$$$\Rightarrow\:\:{V}\:=\:\frac{\mathrm{2}{u}}{\mathrm{5}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{4}{m}\right){V}^{\:\:\mathrm{2}} \geqslant\:\mu\left(\mathrm{6}{mg}\right)\left(\mathrm{2}{b}\right) \\ $$$$\Rightarrow\:\mathrm{2}\left(\frac{\mathrm{2}{u}}{\mathrm{5}}\right)^{\mathrm{2}} \:\geqslant\:\mathrm{12}\mu{gb} \\ $$$$\Rightarrow\:\:{u}\:\geqslant\:\mathrm{5}\sqrt{\frac{\mathrm{3}\mu{gb}}{\mathrm{2}}}\:. \\ $$
Commented by mr W last updated on 01/Feb/19
that is the correct result!
$${that}\:{is}\:{the}\:{correct}\:{result}! \\ $$

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