Two objects slide over a frictionless horizontal surface. The first object, mass m_1 = 5 kg, is propelled with a speed u = 4.5 m/s towards the second object, mass m_2 = 5 kg, which is initially at rest. After the collision, both objects have velocities which are directed at θ = 60° on either side of the original line of motion of the first object. What can you say about the elasticity of collision?


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Question Number 25058 by Tinkutara last updated on 02/Dec/17

$T w o o b j e c t s s l i d e o v e r a f r i c t i o n l e s s$ $h o r i z o n t a l s u r f a c e . T h e f i r s t o b j e c t,$ $m a s s m_{1} = 5 k g, i s p r o p e l l e d w i t h a$ $s p e e d u = 4.5 m / s t o w a r d s t h e s e c o n d$ $o b j e c t, m a s s m_{2} = 5 k g, w h i c h i s$ $i n i t i a l l y a t r e s t . A f t e r t h e c o l l i s i o n,$ $b o t h o b j e c t s h a v e v e l o c i t i e s w h i c h a r e$ $d i r e c t e d a t θ = 60 ° o n e i t h e r s i d e o f$ $t h e o r i g i n a l l i n e o f m o t i o n o f t h e$ $f i r s t o b j e c t . W h a t c a n y o u s a y a b o u t$ $t h e e l a s t i c i t y o f c o l l i s i o n ?$
	Answered by jota+ last updated on 05/Dec/17

	$θ = 60$ $m_{1} u = m_{1} v_{1} \cos θ + m_{2} v_{2} \cos θ (1)$ $0 = m_{1} v_{1} \sin θ - m_{2} v_{2} \sin θ (2)$ $f r o m (2) p_{1} = m_{1} v_{1} = m_{2} v_{2} = p_{2}$ $t h e n$ $m_{1} u = 2 p_{1} \cos θ = p_{1}$ $T_{1} = \frac{1}{2} m_{1} u^{2} = \frac{p_{1}^{2}}{2 m_{1}}$ $T_{2} = \frac{p_{1}^{2}}{2 m_{1}} + \frac{p_{2}^{2}}{2 m_{2}}$ $Δ T = \frac{p_{2}^{2}}{2 m_{2}} \neq 0$ $T h e c o l l i s i o n i s i n e l a s t i c .$
	Commented by jota+ last updated on 05/Dec/17

	$N o . Δ T i s n o t z e r o$
	Commented by Tinkutara last updated on 05/Dec/17

	$A n s w e r g i v e n i s p e r f e c t l y e l a s t i c .$
	Commented by Tinkutara last updated on 05/Dec/17

	$Thank you Sir!$
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