prove-that-relative-velocity-is-reversed-by-a-head-on-collision-

Question Number 50914 by peter frank last updated on 22/Dec/18

p r o v e t h a t r e l a t i v e v e l o c i t y

i s r e v e r s e d b y a h e a d o n

c o l l i s i o n

Answered by tanmay.chaudhury50@gmail.com last updated on 22/Dec/18

f o r c o l l i s s i o n v e l o c i t y a r e {\vec{v}}_{1} = u_{1} \vec{i} a n d {\overset{⌣}{v}}_{2} = - u_{2} \vec{i}

r e l a t i v e v e l o c i t y o f {\vec{v}}_{1} w . r . t {\vec{v}}_{2} i s = {\vec{v}}_{1} - {\vec{v}}_{2}

= u_{1} \vec{i} + u_{2} \vec{i}

s o v e l o c i t y {\vec{v}}_{2} d i r e c t i i n r e v e r s e d a n d m o v i n g

a l o n g + v e x a x i s \dots

Commented by peter frank last updated on 22/Dec/18

t h a n k y o u

Answered by peter frank last updated on 22/Dec/18

f r o m c o n s e r v a t i o n o f

l i n e a r m o m e n t u m

m_{1} u_{1} + m_{2} u_{2} = m_{1} v_{1} + m_{2} v_{2}

m_{1} (u_{1} - v_{1}) = m_{2} (v_{2} - u_{2}) \dots \dots (i)

f r o m K . E c o n s e r v a t i o n

\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}) \dots . . (i i)

d i v i d e e q n (i i) b y e q n (i)

u_{1} + v_{1} = v_{2} + u_{2}

u_{1} - u_{2} = v_{2} - v_{1}

u_{2} - u_{1} = - (v_{2} - v_{1})