question_answer

A ball collides elastically with another ball of the same mass. The collision is oblique and initially one of the balls was at rest. After the collision, the two balls move with same speeds. What will be the angle between the velocity of the balls after the collision?

A)  \[{{30}^{o}}\]

B)  \[{{45}^{o}}\]

C)  \[{{60}^{o}}\]

D)  \[{{90}^{o}}\]

Correct Answer: B

Solution :

 Applying law of conservation of momentum along two axes of plane of motion, we get: \[{{V}_{3}}\] i.e.\[,\]                 \[MgC{{l}_{2}}\to 2NaOH\to Mg{{(OH)}_{2}}\downarrow +2NaCl\]                                                          ? (i) Also,    \[Mg{{(OH)}_{2}}\] \[M{{g}^{2+}}\] \[O{{H}^{-}}\] \ \[{{K}_{sp}}\] According to law of conservation of KE \[Mg{{(OH)}_{2}}\] \[Mg{{(OH)}_{2}}\,\underset{s}{\mathop{M{{g}^{2+}}}}\,+\underset{2s}{\mathop{2O{{H}^{-}}}}\,\] Or \[{{K}_{sp}}=4{{s}^{3}}\]                                         ?(iii) From Eqs. (ii) and (iii), we get          \[s=\sqrt[3]{\frac{12\times {{10}^{-12}}}{4}}=1.44\times {{10}^{-4}}\]                 \[\therefore \] \[\theta ={{45}^{o}}\]