Question-22956

Question Number 22956 by Physics lover last updated on 24/Oct/17

Commented by Physics lover last updated on 24/Oct/17

f i n d a n g u l a r v e l o c i t y a f t e r

c o l l i s i o n a n d l o s s i n k i n e t i c

e n e r g y ?

Commented by ajfour last updated on 24/Oct/17

Commented by ajfour last updated on 24/Oct/17

ω l / 2 = v_{1} - V \dots (i)

c o n s e r v i n g l i n e a r m o m e n t u m

(2 m) V + {m v}_{1} = m v

\Rightarrow 2 V + v_{1} = v \dots (i i)

c o n s e r v i n g a n g u l a r m o m e n t u m

a b o u t a p o i n t o n t a b l e j u s t b e l o w

t h e r i g h t e n d o f s t i c k (∙) .

(\frac{2 {m l}^{2}}{12}) ω - (2 m) V (\frac{l}{2}) = 0

\Rightarrow ω l = 6 V \dots (i i i)

u s i n g t h i s i n (i)

v_{1} - V = 3 V o r v_{1} = 4 V

u s i n g t h i s i n (i i)

6 V = v o r V = \frac{v}{6}

t h e n v_{1} = 4 V = \frac{2 v}{3}

ω l = 6 V = v

ω = v / l .

K_{i} = \frac{1}{2} {m v}^{2}

K_{f} = \frac{1}{2} (2 m) V^{2} + \frac{1}{2} (\frac{2 {m l}^{2}}{12}) ω^{2} +

+ \frac{1}{2} {m v}_{1}^{2}

= \frac{{m v}^{2}}{36} + \frac{{m v}^{2}}{12} + \frac{2 {m v}^{2}}{9}

= \frac{12 {m v}^{2}}{36} = \frac{{m v}^{2}}{3}

l o s s i n K = K_{i} - K_{f}

= \frac{{m v}^{2}}{2} - \frac{{m v}^{2}}{3} = \frac{{m v}^{2}}{6} .

Commented by Physics lover last updated on 25/Oct/17

t h a n k d y o u v e r y m u c h, M r A j f o u r .

Answered by ajfour last updated on 24/Oct/17

ω = \frac{v}{l}; l o s s i n K = \frac{{m v}^{2}}{6} .

Commented by Physics lover last updated on 24/Oct/17

y e s s i r!!