Show that in collision where kinetic energy is conserved linear momemtum is also conserved


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Question Number 50912 by peter frank last updated on 22/Dec/18

$S h o w t h a t i n c o l l i s i o n$ $w h e r e k i n e t i c e n e r g y i s$ $c o n s e r v e d l i n e a r m o m e m t u m$ $i s a l s o c o n s e r v e d$
	Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18

	$K . E = \frac{1}{2} {m u}^{2} = \frac{1}{2} \times \frac{p^{2}}{m}$ $p^{2} = 2 m E [E = K . E a n d p = m o m e n t u m]$ $2 p \frac{d p}{d E} = 2 m$ $2 p d p = 2 m d E$ $p d p = m d E$ $d E = 0 [K . E c o n s e r v e d] s o d p = 0$ $m o m e n t u m c o n s e r v e d$
	Commented by peter frank last updated on 22/Dec/18

	$t h a n k s$
	Answered by peter frank last updated on 22/Dec/18

	$f r o m$ $\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} = \frac{1}{2} m_{1} v_{1}^{2} + \frac{1}{2} m_{2} v_{2}^{2}$ $m_{1} (u_{1}^{2} - v_{1}^{2}) = m_{2} (v_{2}^{2} - u_{2}^{2})$ $m_{1} (u_{1} - v_{1}) (u_{1} + v_{1}) = m_{2} (v_{2} - u_{2}) (v_{2} + u_{2}) . . . . . (i)$ $f r o m$ $\frac{v_{1} - u_{1}}{u_{1} - u_{2}} = e b u t e = 1$ $v_{2} - v_{1} = u_{1} - u_{2}$ $u_{1} + v_{1} = v_{2} + u_{2} . . . . . . (i i)$ $t a k e e q n i) \div (i i) a n d s i m p l i f y$ $m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2})$ $m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}$
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