mathop {{v_1}}limits^ to જેટલા વેગથી ગતિ કરતો | vedclass

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Question

વેગમનના સરક્ષનના નિયમ અનુસાર 

હવે  $ \,\,\,_{{\upsilon _1}'}^ 	o \, \cdot _{{\upsilon _1}'}^ 	o =$ $( \,\,\,{\upsilon

Vedclass · Accepted Answer

$90$

Vedclass · Answer

વેગમનના સરક્ષનના નિયમ અનુસાર 

હવે  $ \,\,\,_{{\upsilon _1}'}^ 	o \, \cdot _{{\upsilon _1}'}^ 	o =$ $( \,\,\,{\upsilon _1}'{\upsilon _2}'\,\,)^.(\,\,\,{\upsilon _1}'{\upsilon _2}'\,\,)$

$	herefore \,\,{\upsilon _1}^2 = \,\,\,_{{\upsilon _1}'}^ 	o \, \cdot _{{\upsilon _1}'}^ 	o  + _{{\upsilon _1}'}^ 	o \, \cdot _{{\upsilon _2}'}^ 	o  + _{{\upsilon _2}'}^ 	o \, \cdot _{{\upsilon _1}'}^ 	o  + _{{\upsilon _2}'}^ 	o \, \cdot _{{\upsilon _2}'}^ 	o \,\,\,\,\,	herefore \,\,\,{\upsilon _1}^2 = \,\,\,{\upsilon _1}^{'\,2} + 2_{{\upsilon _1}'}^ 	o \, \cdot \,_{{\upsilon _1}'}^ 	o  + {\upsilon _2}^{'\,2}\,\,\,\,\,\,\,...\,\,...\,\,...\,\,(1)$

પરંતુ ઉર્જા સરક્ષણ ના નિયમ પરથી  $\frac{1}{2}m{\upsilon _1}^2 + 0 = \frac{1}{2}m{\upsilon _1}^{'2} + \frac{1}{2}m{\upsilon _2}^{'2}\,\,	herefore {\upsilon _1}^2\, = \,{\upsilon _1}^{'2} + {\upsilon _2}^{'2}$

આ કિમંત સમીકરણ (1) માં મુક્તા  $,\,{\upsilon _1}^2\, = \,{\upsilon _1}^2 + 2_{{\upsilon _1}'}^ 	o \, \cdot \,_{{\upsilon _2}'}^ 	o \,\,\,\,$

$\,\,	herefore \,\,\,\,2_{{\upsilon _1}'}^ 	o \, \cdot \,_{{\upsilon _2}'}^ 	o \, = 0\,\,\,\,	herefore \,\,\,{\upsilon _1}'{\upsilon _2}'\,\,\cos \,\,	heta \, = \,0\,\,\,	herefore \,\,\,\,\cos \,	heta \, = \,0\,\,\,	herefore \,\,\,	heta \, = \,{90^ \circ }$

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