Question-225661

Question Number 225661 by mr W last updated on 05/Nov/25

Commented by mr W last updated on 05/Nov/25

Commented by fantastic last updated on 05/Nov/25

i have a feeling that the block M will acheive its max v at 2nd phase when the pellete will start to go left after going to the right What do you think sir

$${i}\:{have}\:{a}\:{feeling}\:{that} \\ $$$${the}\:{block}\:{M}\:{will}\:{acheive} \\ $$$${its}\:{max}\:{v}\:{at}\:\mathrm{2}{nd}\:{phase} \\ $$$${when}\:{the}\:{pellete}\:{will} \\ $$$${start}\:{to}\:{go}\:{left}\:{after} \\ $$$${going}\:{to}\:{the}\:{right} \\ $$$${What}\:{do}\:{you}\:{think}\:{sir} \\ $$

Commented by mr W last updated on 06/Nov/25

$${maybe}.\:{what}'{s}\:{your}\:{exact}\:{answer}? \\ $$

Commented by mahdipoor last updated on 06/Nov/25

my final ode eq is : (A=v_M ^. ) A(((M+m)/(mr)))+ω^2 cosθ+αsinθ=0 A((M/(mr)))tanθ+(g/r)+ω^2 sinθ−αcosθ=0 ω=θ^. , α=θ^(..) this eq must solve from { ((θ=90 , ω^2 =((2g)/r) , α=0)),((v_M =0 , A=0)) :} to θ=90 again . v_M at this is v_(max)

$$\mathrm{my}\:\mathrm{final}\:\mathrm{ode}\:\mathrm{eq}\:\mathrm{is}\::\:\:\:\left(\mathrm{A}=\overset{.} {\mathrm{v}}_{\mathrm{M}} \right) \\ $$$$\mathrm{A}\left(\frac{\mathrm{M}+\mathrm{m}}{\mathrm{mr}}\right)+\omega^{\mathrm{2}} \mathrm{cos}\theta+\alpha\mathrm{sin}\theta=\mathrm{0} \\ $$$$\mathrm{A}\left(\frac{\mathrm{M}}{\mathrm{mr}}\right)\mathrm{tan}\theta+\frac{\mathrm{g}}{\mathrm{r}}+\omega^{\mathrm{2}} \mathrm{sin}\theta−\alpha\mathrm{cos}\theta=\mathrm{0} \\ $$$$\omega=\overset{.} {\theta}\:\:\:,\:\:\:\alpha=\overset{..} {\theta} \\ $$$$\mathrm{this}\:\mathrm{eq}\:\mathrm{must}\:\mathrm{solve}\:\mathrm{from}\:\begin{cases}{\theta=\mathrm{90}\:,\:\omega^{\mathrm{2}} =\frac{\mathrm{2g}}{\mathrm{r}}\:\:,\:\:\alpha=\mathrm{0}}\\{\mathrm{v}_{\mathrm{M}} =\mathrm{0}\:\:,\:\mathrm{A}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{to}\:\theta=\mathrm{90}\:\:\mathrm{again}\:.\:\mathrm{v}_{\mathrm{M}} \:\mathrm{at}\:\mathrm{this}\:\mathrm{is}\:\mathrm{v}_{\mathrm{max}} \\ $$

Commented by mr W last updated on 07/Nov/25

$${thanks}\:{sir}! \\ $$$${to}\:{exactly}\:{solve}\:{this}\:{problem}\:{is}\:{not} \\ $$$${possible},\:{due}\:{to}\:{this}\:{d}.{e}.\:{as}\:{you}\: \\ $$$${showed},\:{which}\:{can}\:{not}\:{be}\:{solved}. \\ $$$${but}\:{fortunately}\:{the}\:{question}\:{only}\: \\ $$$${requests}\:{the}\:{maximum}\:{velocity}.\: \\ $$$${then}\:{we}\:{don}'{t}\:{need}\:{to}\:{solve}\:{this}\: \\ $$$${hard}\:{d}.{e}.\:{that}\:{means}\:{we}\:{can}\:{find} \\ $$$${out}\:{the}\:{maximum}\:{velocity},\:{but}\:{we} \\ $$$${can}\:{not}\:{find}\:{out}\:{at}\:{what}\:{time}\:{and} \\ $$$${at}\:{what}\:{position}\:{of}\:{the}\:{block}\:{this} \\ $$$${happens}. \\ $$

Answered by mr W last updated on 07/Nov/25

from t=0 to t_1 the block M doesn′t move, because a motion of it to the left is prevented by the wall. when the pellet reaches the bottom of the cavity, its normal reaction force will push the block to the right. upon now the block can start to move. at t=t_1 : velocity of pellet: v_0 =(√(2gr)) velocity of block: 0 upon now the block starts to move to the right, away from the wall. at some time t=t_2 : the block has maximum velocity V_(max) , when the pellet is at the lowest position. MV_(max) −mv=mv_0 ⇒v=((MV_(max) )/m)−v_0 ((MV_(max) ^2 )/2)+((mv^2 )/2)=((mv_0 ^2 )/2) MV_(max) ^2 +m(((MV_(max) )/m)−v_0 )^2 =mv_0 ^2 ⇒V_(max) =((2(√(2gr)))/(1+(M/m))) ✓ ⇒v=(1−(2/(1+(M/m))))(√(2gr))

$${from}\:{t}=\mathrm{0}\:{to}\:{t}_{\mathrm{1}} \:{the}\:{block}\:{M}\:{doesn}'{t} \\ $$$${move},\:{because}\:{a}\:{motion}\:{of}\:{it}\:{to}\:{the} \\ $$$${left}\:{is}\:{prevented}\:{by}\:{the}\:{wall}.\:{when} \\ $$$${the}\:{pellet}\:{reaches}\:{the}\:{bottom}\:{of}\:{the} \\ $$$${cavity},\:{its}\:{normal}\:{reaction}\:{force} \\ $$$${will}\:{push}\:{the}\:{block}\:{to}\:{the}\:{right}.\:{upon} \\ $$$${now}\:{the}\:{block}\:{can}\:{start}\:{to}\:{move}. \\ $$$$ \\ $$$${at}\:{t}={t}_{\mathrm{1}} : \\ $$$${velocity}\:{of}\:{pellet}:\:{v}_{\mathrm{0}} =\sqrt{\mathrm{2}{gr}} \\ $$$${velocity}\:{of}\:{block}:\:\mathrm{0} \\ $$$${upon}\:{now}\:{the}\:{block}\:{starts}\:{to}\:{move}\:{to} \\ $$$${the}\:{right},\:{away}\:{from}\:{the}\:{wall}. \\ $$$$ \\ $$$${at}\:{some}\:{time}\:{t}={t}_{\mathrm{2}} : \\ $$$${the}\:{block}\:{has}\:{maximum}\:{velocity}\:{V}_{{max}} , \\ $$$${when}\:{the}\:{pellet}\:{is}\:{at}\:{the}\:{lowest} \\ $$$${position}. \\ $$$${MV}_{{max}} −{mv}={mv}_{\mathrm{0}} \:\Rightarrow{v}=\frac{{MV}_{{max}} }{{m}}−{v}_{\mathrm{0}} \\ $$$$\frac{{MV}_{{max}} ^{\mathrm{2}} }{\mathrm{2}}+\frac{{mv}^{\mathrm{2}} }{\mathrm{2}}=\frac{{mv}_{\mathrm{0}} ^{\mathrm{2}} }{\mathrm{2}} \\ $$$${MV}_{{max}} ^{\mathrm{2}} +{m}\left(\frac{{MV}_{{max}} }{{m}}−{v}_{\mathrm{0}} \right)^{\mathrm{2}} ={mv}_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\Rightarrow{V}_{{max}} =\frac{\mathrm{2}\sqrt{\mathrm{2}{gr}}}{\mathrm{1}+\frac{{M}}{{m}}}\:\:\checkmark \\ $$$$\Rightarrow{v}=\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{1}+\frac{{M}}{{m}}}\right)\sqrt{\mathrm{2}{gr}} \\ $$

Commented by mr W last updated on 07/Nov/25

Commented by mr W last updated on 07/Nov/25

at some time t=t_3 , the pellet is at its highest position, it has the same velocity as the block: V_m (M+m)V_m =mv_0 ⇒V_m =(v_0 /(1+(M/m)))=((√(2gr))/(1+(M/m)))=(V_(max) /2) (((M+m)V_m ^2 )/2)=mg(r−h) ⇒h=(r/(1+(m/M)))<r

$${at}\:{some}\:{time}\:{t}={t}_{\mathrm{3}} ,\:{the}\:{pellet}\:{is}\:{at} \\ $$$${its}\:{highest}\:{position},\:{it}\:{has}\:{the}\:{same} \\ $$$${velocity}\:{as}\:{the}\:{block}:\:{V}_{{m}} \\ $$$$\left({M}+{m}\right){V}_{{m}} ={mv}_{\mathrm{0}} \\ $$$$\Rightarrow{V}_{{m}} =\frac{{v}_{\mathrm{0}} }{\mathrm{1}+\frac{{M}}{{m}}}=\frac{\sqrt{\mathrm{2}{gr}}}{\mathrm{1}+\frac{{M}}{{m}}}=\frac{{V}_{{max}} }{\mathrm{2}} \\ $$$$\frac{\left({M}+{m}\right){V}_{{m}} ^{\mathrm{2}} }{\mathrm{2}}={mg}\left({r}−{h}\right) \\ $$$$\Rightarrow{h}=\frac{{r}}{\mathrm{1}+\frac{{m}}{{M}}}<{r} \\ $$

Commented by mr W last updated on 07/Nov/25