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Question Number 82726 by mr W last updated on 23/Feb/20

Commented by mr W last updated on 23/Feb/20

$m a s s m_{1} a n d m_{2} a r e r e l e a s e d f r o m t h e$ $t o p o f a w e d g e s i m u t a n e o u s l y . a s s u m e$ $t h a t m a s s m_{1} r e a c h e s t h e b o t t u m o f$ $t h e w e d g e f i r s t l y . f i n d t h e v e l o c i t y o f$ $t h e w e d g e a t t h i s m o m e n t .$ $a l l s u r f a c e s a r e f r i c t i o n l e s s .$
Commented by TawaTawa last updated on 23/Feb/20

$Sir please i need help, i have post the question several time .$ $i repost now Q 82729 . Or the question is not correct so that i can stop$ $posting it . Please help . Thanks sir .$
Commented by ajfour last updated on 23/Feb/20

Commented by TawaTawa last updated on 23/Feb/20

$Sir, i have typed the question . please check . God bless you sir .$ $Q 82729 .$
Commented by ajfour last updated on 23/Feb/20

Commented by ajfour last updated on 23/Feb/20

$a_{1} = gsin θ_{1} - Acos θ_{1}$ $a_{2} = gsin θ_{2} + Acos θ_{2}$ $\frac{h}{\sin θ_{1}} = \frac{1}{2} (gsin θ_{1} - Acos θ_{1}) t^{2}$ $\Rightarrow \frac{2 h}{\sin θ_{1}} = {gt}^{2} \sin θ_{1} - Vtcos θ_{1} . . (i)$ $[from above eq . we get t (V)]$ $u = (gsin θ_{1} - Acos θ_{1}) t$ $= gtsin θ_{1} - Vcos θ_{2} . . . . . (ii)$ $v = gtsin θ_{2} + Vcos θ_{2} . . . . . (iii)$ $m_{1} ucos θ_{1} + (M + m_{1} + m_{2}) V$ $= m_{2} vcos θ_{2} . . . . . (iv)$ $using (i), (ii), (iii), (iv)$ $V is obtained . . .$
Commented by mr W last updated on 23/Feb/20
$wonderful solution sir! we can get V=((8gh)/(cos θ_1 {((4 tan θ_1 (M+m_1 sin^2 θ_1 +m_2 sin^2 θ_2 ))/(m_2 sin 2θ_2 −m_1 sin 2θ_1 ))−1}^2 −1))$
$w o n d e r f u l s o l u t i o n s i r!$ $w e c a n g e t$ $V = \frac{8 g h}{\cos θ_{1} {\frac{4 \tan θ_{1} (M + m_{1} \sin^{2} θ_{1} + m_{2} \sin^{2} θ_{2})}{m_{2} \sin 2 θ_{2} - m_{1} \sin 2 θ_{1}} - 1}^{2} - 1}$
Commented by mr W last updated on 24/Feb/20

${w h a t}^{'} s t h e c o n d i t i o n t h a t m_{1} r e a c h e s$ $t h e b o t t o m f i r s t l y ?$
Commented by ajfour last updated on 24/Feb/20

$If t_{1} < t_{2}, m_{1} reaches first .$ $a_{1} = gsin θ_{1} - Acos θ_{1}$ $a_{2} = gsin θ_{2} + Acos θ_{2}$ $\frac{1}{2} a_{1} t_{1}^{2} = \frac{h}{\sin θ_{1}}, \frac{1}{2} a_{2} t_{2}^{2} = \frac{h}{\sin θ_{2}}$ $t_{1} = \sqrt{\frac{2 h}{a_{1} \sin θ_{1}}}, t_{2} = \sqrt{\frac{2 h}{a_{2} \sin θ_{2}}}$ $N_{1} = m_{1} gcos θ_{1} + m_{1} Asin θ_{1}$ $N_{2} = m_{2} gcos θ_{2} - m_{2} Asin θ_{2}$ $N_{2} \sin θ_{2} - N_{1} \sin θ_{1} = MA$ $A is determined from above 3 eqs .$ $hence we have a_{1}, a_{2} .$ $Then the condition can be$ $expressed . .$
Commented by mr W last updated on 24/Feb/20

$c l e a r! t h a n k s!$
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