The particles of mass m1 and m2 and velocities u1 and u2 = ku (where  k not equal to 0) make an headon collition. If the initial kinetic energies of the two particles are equal and m1 comes to rest after collition, then calculate the value of  m1/m2  and u1/u2.

Asked by sudhanshubhushanroy | 25th Oct, 2017, 09:01: PM

Expert Answer:

From law of conservation of energy
 
begin mathsize 12px style 1 half m subscript 1 u subscript 1 superscript 2 space plus space fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction m subscript 2 u subscript 2 superscript 2 space equals space fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction m subscript 2 u subscript f space space end subscript superscript 2
h e n c e space m subscript 1 u subscript 1 superscript 2 space plus space m subscript 2 u subscript 2 superscript 2 space equals space m subscript 2 u subscript f space space end subscript superscript 2.................... e q n left parenthesis 1 right parenthesis end style
 
But initial kinetic energies of particles are equal.
 
begin mathsize 12px style m subscript 1 u subscript 1 superscript 2 space equals space m subscript 2 u subscript 2 superscript 2 space...................... e q n left parenthesis 2 right parenthesis end style
 
From law of conservation of momentum 
begin mathsize 12px style m subscript 1 u subscript 1 plus m subscript 2 u subscript 2 equals space m subscript 2 u subscript f space end subscript..................... e q n left parenthesis 3 right parenthesis end style
 
from eqn(3) final velocity u is given by 
begin mathsize 12px style u subscript f space equals space 1 over m subscript 2 open parentheses m subscript 1 u subscript 1 plus m subscript 2 u subscript 2 close parentheses space space........................ e q n left parenthesis 4 right parenthesis end style
By substituting ufrom eqn(4) in eqn(1) and after simplification
 
begin mathsize 12px style m subscript 1 m subscript 2 u subscript 1 superscript 2 space equals space m subscript 1 superscript 2 u subscript 1 superscript 2 space plus space 2 m subscript 1 m subscript 2 u subscript 1 u subscript 2
u sin g space e q n. left parenthesis 2 right parenthesis space a n d space f u r t h e r space s i m p l i f i c a t i o n

u subscript 2 open parentheses m subscript 2 space minus space m subscript 1 close parentheses space equals space 2 m subscript 1 u subscript 1 end style
 
begin mathsize 12px style H e n c e space u subscript 1 over u subscript 2 space equals space open parentheses fraction numerator m subscript 2 minus m subscript 1 over denominator 2 m subscript 1 end fraction close parentheses end style
 
 
from the above relation we can get mass ratio
 begin mathsize 12px style m subscript 1 over m subscript 2 space equals space fraction numerator u subscript 2 over denominator 2 u subscript 1 plus u subscript 2 end fraction end style

Answered by  | 27th Oct, 2017, 12:38: PM

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