JEE Mains · Physics · STD 11 - 4.1 newtons laws of motion
Particle \(A\) of mass \(m _{1}\) moving with velocity \((\sqrt{3} \hat{i}+\hat{j})\, ms ^{-1}\) collides with another particle \(B\) of mass \(m _{2}\) which is at rest initially. Let \(\overrightarrow{ V }_{1}\) and \(\overrightarrow{ V }_{2}\) be the velocities of particles \(A\) and \(B\) after collision respectively. If \(m _{1}=2\, m _{2}\) and after collision \(\overrightarrow{ V }_{1}=(\hat{ i }+\sqrt{3} \hat{ j })\, ms ^{-1},\) the angle between \(\overrightarrow{ V }_{1}\) and \(\overrightarrow{ V }_{2}\) is\(......^o\)
- A \(60\)
- B \(15\)
- C \(-45\)
- D \(105\)
Answer & Solution
Correct Answer
(D) \(105\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{ v }_{01}=(\sqrt{3 \hat{ i }}+\hat{ j }) m / s\) \(\overrightarrow{ v }_{02}=\overrightarrow{0}\) \(m _{1}=2 m _{2}\) After collision, \(\overrightarrow{ v }_{1}=(\hat{ i }+\sqrt{3} \hat{ j }) m / s\) \(\overrightarrow{ v }_{2}=?\) Applying conservation of…
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