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तीन टकराने वाले कणों के द्रव्यमान क्रमशः 10, 20 तथा 40 ग्राम हैं तथा प्रत्येक कण के वेग का मान क्रमशः X,Y तथा Z अक्षों के अनदिश 20 सेमी./से. है।

Solved397 viewsSeptember 29, 2022भौतिक विज्ञान

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admin 56.29K September 29, 2022 0 Comments

प्रश्न : तीन टकराने वाले कणों के द्रव्यमान क्रमशः 10, 20 तथा 40 ग्राम हैं तथा प्रत्येक कण के वेग का मान क्रमशः X,Y तथा Z अक्षों के अनदिश 20 सेमी./से. है। यदि अन्योन्य क्रिया के बलों द्वारा तीसरा कण विरामावस्था में आ जाता है तथा दूसरे कण का वेग (10 j+5k) सेमी./से. हो जाता है तो पहले कण के वेग की गणना करो।

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Question is closed for new answers.

admin Selected answer as best September 29, 2022

1 Answer

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admin 56.29K Posted September 29, 2022 0 Comments

हल / उत्तर : प्रश्नानुसार, m₁ = 10 ग्राम, m₂ = 20 ग्राम, m₃ = 40 ग्राम

U₁ = 20 i सेमी./से., u₂ = 20 j सेमी./से.

सब्सक्राइब करे youtube चैनल

U₃ = 20 सेमी./से.

V₃ = 0,V₂ = (10i + 5k) सेमी./से.

V₁ =?

कुल प्रारम्भिक संवेग = m₁ u₁ + m₂ u₂ + m₃ u₃

= 200i + 400j + 800k

‘कुल अन्तिम संवेग = m₁v₁ +m₂v₂ + m₃v₃

=10 v₁ +20 (10j + 5k) + 40 x 0

= 10V₁ + 200j + 100k

संवेग संरक्षण के नियम से

M₁ u₁ + m₂ u₂ + m₃ u₃ = m₁ v₁ + m₂ v₂ + m₃ v₃

200i +400j +800k = 10v₁ + 200j + 100k

10v₁ = 200i +400j +800k – 200j -100k

200 i +200j +700k

V₁ = 20 i + 20 j +70k सेमी/से

हिंदी माध्यम नोट्स

Class 6

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Class 7

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Class 7

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admin Selected answer as best September 29, 2022

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