Question-90923

Question Number 90923 by ajfour last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

A b a l l o f m a s s m h i t s h o r i z o n t a l l y

a n d e l a s t i c a l l y t h e c e n t r e o f a

h i n g e d s q u a r e p l a t e o f s i d e s a n d

m a s s M . F i n d v e l o c i t y v o f b a l l

j u z t a f t e r i m p a c t .

Answered by mr W last updated on 26/Apr/20

I = \frac{{M s}^{2}}{3}

b e f o r e t h e h i t :

E = \frac{1}{2} {m u}^{2}

a f t e r t h e h i t :

E = \frac{1}{2} {m v}^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} {m u}^{2}

\Rightarrow v^{2} + \frac{{M s}^{2}}{3 m} ω^{2} = u^{2}

I ω - \frac{s}{2} m v = \frac{s}{2} m u

\Rightarrow ω = (u + v) \frac{3 m}{2 M s}

v^{2} + {(u + v)}^{2} \frac{3 m}{4 M} = u^{2}

3 m (u + v) = 4 M (u - v)

\Rightarrow v = \frac{4 M - 3 m}{4 M + 3 m} u (b a c k w a r d s +)

Commented by ajfour last updated on 27/Apr/20

V e r y N i c e a n d c o r r e c t, S i r!

Facebook Tweet Pin