Question-82726

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} \dots . . (ii)

v = gtsin θ_{2} + Vcos θ_{2} \dots . . (iii)

m_{1} ucos θ_{1} + (M + m_{1} + m_{2}) V

= m_{2} vcos θ_{2} \dots . . (iv)

using (i), (ii), (iii), (iv)

V is obtained \dots

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

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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