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Question Number 56585 by Tinkutara last updated on 18/Mar/19

Commented by Tinkutara last updated on 18/Mar/19

Commented by mr W last updated on 19/Mar/19

$v_{0} = s p e e d o f r o c k e t a t t = 0 .$ $s p e e d o f b a l l 1 = v_{1} = v_{0} + 0.3$ $s p e e d o f b a l l 2 = v_{2} = v_{0} - 0.2$ $s p e e d o f r o c k e t a t t :$ $v_{R} = v_{0} + 2 t$ $l e t v_{R} = v_{1}$ $v_{0} + 2 t = v_{0} + 0.3$ $\Rightarrow t = 0.15 s$ $i . e . a f t e r 0.15 s e c o n d s t h e w a l l o f c h a m b e r$ $a n d b a l l 1 a r e t o g e t h e r . i n t h i s t i m e$ $t h e b a l l 1 h a s m o v e d 0.15 (v_{0} + 0.3) a n d$ $t h e b a l l 2 h a s m o v e d 0.15 (v_{0} - 0.2)$ $\Rightarrow 0.15 (v_{0} + 0.3) - 0.15 (v_{0} - 0.2) = 0.075 m$ $i . e . a t t = 0.15, t h e d i s t a n c e b e t w e e n$ $b a l l 1 a n d b a l l 2 i s 4 - 0.075 = 3.925 m .$ $a f t e r t = 0.15 s t h e b a l l 1 m o v e s w i t h$ $t h e r o c k e t, {l e t}^{'} s s a y b o t h b a l l s c o l l i d e$ $a f t e r t i m e t, t h e n$ $(v_{0} + 0.3) t + \frac{2 t^{2}}{2} - (v_{0} - 0.2) t = 3.925$ $0.5 t + t^{2} - 3.925 = 0$ $\Rightarrow t = \frac{- 0.5 + \sqrt{0 . 5^{2} + 4 \times 3.925}}{2} \approx 1.75 s$ $0.15 + 1.75 = 1.90 s$ $i . e . b o t h b a l l s c o l l i d e i n 1.90 s .$
Commented by Tinkutara last updated on 19/Mar/19
Thank you so much Sir! Awesome explanation!
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