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Question Number 98355 by aurpeyz last updated on 13/Jun/20

Commented by aurpeyz last updated on 13/Jun/20

$pls help me$
	Answered by mr W last updated on 13/Jun/20

	Commented by mr W last updated on 14/Jun/20

	$m_{1} = 0.4 g$ $m_{2} = 1.6 g$ $v_{2} = 4.4 m / s$ $m_{1} v_{1} \sin α = m_{2} v_{2} \sin β . . . (i)$ $m_{1} v_{1} \cos α + m_{2} v_{2} \cos β = m_{1} u_{1} . . . (i i)$ $\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2} . . . (i i i)$ $w e h a v e t h r e e e q u a t i o n s f o r t h r e e$ $u n k n o w n s : u_{1}, v_{1} a n d β .$ $f r o m (i) :$ $\sin β = \frac{m_{1} v_{1} \sin α}{m_{2} v_{2}}$ $f r o m (i i) :$ $\cos β = \frac{m_{1} (u_{1} - v_{1} \cos α)}{m_{2} v_{2}}$ $\sin^{2} β + \cos^{2} β = 1 :$ $\frac{m_{1}^{2} v_{1}^{2} \sin^{2} α + m_{1}^{2} {(u_{1} - v_{1} \cos α)}^{2}}{m_{2}^{2} v_{2}^{2}} = 1$ $u_{1}^{2} + v_{1}^{2} - 2 u_{1} v_{1} \cos α = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{2} . . . (i v)$ $f r o m (i i i) :$ $u_{1}^{2} - v_{1}^{2} = (\frac{m_{2}}{m_{1}}) v_{2}^{2} . . . (v)$ $l e t X = u_{1} + v_{1}, Y = u_{1} v_{1}$ ${(u_{1} + v_{1})}^{2} - 2 u_{1} v_{1} (1 + \cos α) = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{2}$ $X^{2} - 2 (1 + \cos α) Y = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{2} = 16 v_{2}^{2} . . . (v i)$ ${(u_{1} + v_{1})}^{2} {(u_{1} - v_{1})}^{2} = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{4}$ $X^{2} (X^{2} - 4 Y) = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{4} = 16 v_{2}^{4} . . . (v i i)$ $w e g e t$ $Y = u_{1} v_{1} = 89.39$ $X = u_{1} + v_{1} = 19.33$ $u_{1}, v_{1} a r e r o o t s o f$ $x^{2} - 19.33 x + 89.39 = 0$ $\Rightarrow x_{1} = u_{1} = 11.67 m / s$ $\Rightarrow x_{2} = v_{1} = 7.66 m / s$ $β o = \sin^{- 1} \frac{m_{1} v_{1} \sin α}{m_{2} v_{2}} = 19.48 °$
	Commented by 1549442205 last updated on 14/Jun/20

	$we infer \tan β = \frac{v_{1} \sin α}{u_{1} - v_{1} \cos α} (1)$ $In above formula known v_{1} = 4.4, α = 50 °$ $Dividing (4 i) by (5 i) we get \frac{u_{1}^{2} + v_{1}^{2} - {2 u}_{1} v_{1} \cos α}{u_{1}^{2} - v_{1}^{2}} = \frac{m_{2}}{m_{1}} = 4$ $which gives us the quadratic with respect to u_{1}$ ${3 u}_{1}^{2} + {2 u}_{1} v_{1} \cos α - {5 v}_{1}^{2} = 0 (2)$ $which has unique root u_{1} = 4.8153 (m / s)$ $Replace into (1) we obtain :$ $\tan β = 1.69629 \Rightarrow β \approx 59 ° 28^{'}$
	Commented by aurpeyz last updated on 13/Jun/20

	$no v 1$
	Commented by aurpeyz last updated on 13/Jun/20

	$is this the final answer Sir ?$
	Commented by mr W last updated on 13/Jun/20

	$i g o t t h e s a m e r e s u l t a s g i v e n .$ $i {d i d n}^{'} t w r i t e a l l s t e p s f o r t h e s o l u t i o n$ $o f e q u a t i o n s . {t h a t}^{'} s n o t d i f f i c u l t . y o u$ $c a n d o i t b y y o u r s e l f .$
	Commented by 1549442205 last updated on 14/Jun/20

	$great work, thank prof . sir . I did a mistake$
	Commented by mr W last updated on 14/Jun/20

	$i m a d e t h e s a m e m i s t a k e a t f i r s t a s$ $y o u c a n s e e i n m y d i a g r a m, b e c a u s e i$ ${d i d n}^{'} t n o t i c e t h a t m_{2} > m_{1} . w h e n$ $m_{2} > m_{1}, m_{1} m u s t m o v e b a c k w a r d s$ $a f t e r c o l l i s i o n, i . e . t h e a n g l e 50 ° i s$ $b a c k w a r d s, i n t h e d i a g r a m α \neq 50 °,$ $b u t α = 130 °, t h e n y o u c a n g e t a p r o p e r$ $s o l u t i o n f o r t h e e q u a t i o n s .$
	Commented by aurpeyz last updated on 14/Jun/20

	${(u_{1} + v_{1})}^{2} {(u_{1} - v_{1})}^{2} = {(\frac{m_{2}}{m_{1}})}^{2} v_{2}^{4}$ $how did you get this Sir ?$
	Commented by mr W last updated on 14/Jun/20

	$s q u a r i n g e q n . (v)$
	Commented by aurpeyz last updated on 14/Jun/20

	$pls how do you get u 1 without knowing V 1 ? pls$
	Commented by mr W last updated on 14/Jun/20

	$y o u c a n s o l v e (v i) a n d (v i i) t o g e t$ $X a n d Y w i t h$ $X = u_{1} + v_{1} a n d Y = u_{1} v_{1}$ $a f t e r y o u h a v e g o t X a n d Y, y o u c a n$ $g e t u_{1} a n d v_{1} : t h e y a r e j u s t t h e r o o t s$ $o f x^{2} - X x + Y = 0 .$
	Commented by aurpeyz last updated on 14/Jun/20

	$i will work on that Sir . this is the best platform for learning sincerely$
	Commented by mr W last updated on 14/Jun/20

	$j u s t t r y s i r! g o o d l u c k!$ $i s h o w e d o n l y o n e o f p o s s i b l e w a y s$ $w h i c h i t h i n k i s g o o d f o r b e g i n e r s t o$ $u n d e r s t a n d .$ $m a y b e t h e r e a r e o t h e r b e t t e r w a y s .$
	Commented by aurpeyz last updated on 14/Jun/20

	$pls sir . explain how you solve (vi) and (vii) to get X and Y . thanks alot Sir$
	Commented by aurpeyz last updated on 14/Jun/20

	$hurray . i have gotten it Sir . thanks alot$
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