10 m/s ના વેગથી ગતિ કરતો 10 kg દળનો એક ગોળો ત | vedclass

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Question

${\upsilon _1}'\, = \,\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} \right)\,{\upsilon _1} + \,\left( {\frac{{2{m_2}}}{{{m_1} + {m_2}}}} \right

Vedclass · Accepted Answer

$6 m/s, 12 m/s$

Vedclass · Answer

${\upsilon _1}'\, = \,\left( {\frac{{{m_1} - {m_2}}}{{{m_1} + {m_2}}}} ight)\,{\upsilon _1} + \,\left( {\frac{{2{m_2}}}{{{m_1} + {m_2}}}} ight){\upsilon _2}$

$ = \,\,\,\left( {\frac{{10 - 5}}{{10 + 5}}} ight)10 + \left( {\frac{{2 	imes 5}}{{10 + 5}}} ight)4\,\,\, = \left( {\frac{5}{{15}}} ight)10 + \left( {\frac{{10}}{{15}}} ight)4\,\,\, = \frac{{10}}{3} + \frac{8}{3}\,\,\, = \,\frac{{18}}{3}\,\,\, = \,6\,m/s$

તેવી જ રીતે, ${\upsilon _2}'\, = \,\,\left( {\frac{{2{m_1}}}{{{m_1} + {m_2}}}} ight){\upsilon _1} + \,\,\left( {\frac{{{m_2} - {m_1}}}{{{m_1} + {m_2}}}} ight)\,{\upsilon _2}$ પરથી $\upsilon _{	ext{2}}^{	ext{'}}{	ext{  = 12 m/s}}$

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