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A and B are two identical balls. A moving with a speed of $${6}{m}/{s}$$, along the positive X-axis, undergoes a collision with B initially at rest. After collision, each ball moves along directions making angles of $$\pm{30}^{{\circ}}$$ with the X-axis. What are the speeds of A and B after the collision ? I s this collision perfectly eleastic ?
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## QuestionPhysicsClass 11

A and B are two identical balls. A moving with a speed of $${6}{m}/{s}$$, along the positive X-axis, undergoes a collision with B initially at rest. After collision, each ball moves along directions making angles of $$\pm{30}^{{\circ}}$$ with the X-axis. What are the speeds of A and B after the collision ? I s this collision perfectly eleastic ?

Here, $${u}={6}{m}{s}^{{-{1}}},\theta={30}^{{\circ}},\phi={30}^{{\circ}},\upsilon_{{{1}}}=?\upsilon_{{{2}}}=?$$
Using the law of conservation of linear momentum along X-axis
$${u}=\upsilon_{{{1}}}{\cos{\theta}}+\upsilon_{{{2}}}{\cos{\phi}}$$
$${6}=\upsilon_{{{1}}}{{\cos{{30}}}^{{\circ}}+}\upsilon_{{{2}}}{{\cos{{30}}}^{{\circ}}=}{\left(\upsilon_{{{1}}}+\upsilon_{{{2}}}\right)}\frac{{\sqrt{{{3}}}}}{{{2}}}$$ ...(i)
Using the law of conservation of linear momentum along Y-axis
$${0}=\upsilon_{{{1}}}{\sin{\theta}}-\upsilon_{{{2}}}{\sin{\phi}}=\upsilon_{{{1}}}{{\sin{{30}}}^{{\circ}}-}\upsilon_{{{2}}}{{\sin{{30}}}^{{\circ}}=}\frac{{\upsilon_{{{1}}}-\upsilon_{{{2}}}}}{{{2}}}\therefore\upsilon_{{{1}}}=\upsilon_{{{2}}}$$
From (i), $$\upsilon_{{{1}}}+\upsilon_{{{2}}}=\frac{{{12}}}{{\sqrt{{{3}}}}},{2}\upsilon_{{{1}}}=\frac{{{12}}}{{\sqrt{{{3}}}}},\upsilon_{{{1}}}=\frac{{{6}}}{{\sqrt{{{3}}}}}\times\frac{{\sqrt{{{3}}}}}{{\sqrt{{{3}}}}}={2}\sqrt{{{3}}}{m}{s}^{{-{1}}}$$
Hence $$\upsilon_{{{1}}}=\upsilon_{{{2}}}={2}\sqrt{{{3}}}{m}{s}^{{-{1}}}$$
As $$\theta+\phi\ne{90}^{{\circ}}\therefore$$ The collision is NOT perfectly elastic.