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-r

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!