• # question_answer Two identical balls A and B are released from the positions shown in figure. They collide elastically on horizontal portion$M/N$. All surfaces are smooth. The ratio of heights attained by A and B after collision will be   A) $1:4$ B) $2:1$                            C) $4:13$ D) $2:5$

 [c] After collision, balls exchange their velocities $\therefore$ ${{V}_{A}}=\sqrt{2gh}$  and  ${{V}_{B}}=\sqrt{2g(4h)}=2\sqrt{2gh}$ Height gained by A will be  ${{h}_{A}}=h$ But path of B will be first straight line and then parabolic as shown in figure After calculation, we can show that ${{h}_{B}}=\frac{13}{4}h$ $\therefore$      $\frac{{{h}_{A}}}{{{h}_{B}}}=\frac{4}{13}$