JEE Mains · Physics · STD 11 - 5. work,energy,power and collision
Two particles \(A\) and \(B\) of equal mass \(M\) are moving with the same speed \(v\) as shown in the figure . They collide completely inelastically and move as a single particle \(C\). The angle \(\theta \) that the path of \(C\) makes with the \(X-\) axis is given by
- A \(\tan \theta = \frac{{\sqrt 3 + \sqrt 2 }}{{1 - \sqrt 2 }}\)
- B \(\tan \theta = \frac{{\sqrt 3 - \sqrt 2 }}{{1 - \sqrt 2 }}\)
- C \(\tan \theta = \frac{{1 - \sqrt 2 }}{{\sqrt 2 \left( {1 + \sqrt 3 } \right)}}\)
- D \(\tan \theta = \frac{{1 - \sqrt 3 }}{{1 + \sqrt 2 }}\)
Answer & Solution
Correct Answer
(A) \(\tan \theta = \frac{{\sqrt 3 + \sqrt 2 }}{{1 - \sqrt 2 }}\)
Step-by-step Solution
Detailed explanation
For particle \(C,\) According to law of conservation of linear momentum, verticle component, \(2\,mv'\,\sin \,\theta = mv\sin {60^ \circ } + mv\,\sin {45^ \circ }\)…
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