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JEE Mains · Physics · STD 11 - Newtons laws of motion
  • JEE Mains 2020
  • Hard
  • Numerical

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\(......^o\)

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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 linear momentum,

\(m _{1} \overrightarrow{ v }_{01}+ m _{2} \overrightarrow{ v }_{0 2}= m _{1} \overrightarrow{ v }_{1}+ m _{2} \overrightarrow{ v }_{2}\)

\(2 m_{2}(\sqrt{3 \hat{i}}+\hat{j})+0=2 m_{2}(\hat{i}+\sqrt{3} \hat{j})+m_{2} \vec{v}_{2}\)

\(\overrightarrow{ v }_{2}=2(\sqrt{3 \hat{ i }}+\hat{ j })-2(\hat{ i }+\sqrt{3} \hat{ j })\)

\(=2(\sqrt{3 \hat{1}}-\hat{j})+2(\hat{i}-\sqrt{3} \hat{j})\)

\(\overrightarrow{ v }_{2}=2(\sqrt{3}-1)(\hat{ i }-\hat{ j })\)

for angle between \(\overrightarrow{ v }_{1} \& \overrightarrow{ v }_{2}\)

\(\cos \theta=\frac{\overrightarrow{ v }_{1} \cdot \overrightarrow{ v }_{2}}{\overrightarrow{ v }_{1} \overrightarrow{ v }_{2}}=\frac{2(\sqrt{3}-1)(1-\sqrt{3})}{2 \times 2 \sqrt{2}(\sqrt{3}-1)}\)

\(\cos \theta=\frac{1-\sqrt{3}}{2 \sqrt{2}} \Rightarrow \theta=105^{\circ}\)

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