• # question_answer Two blocks A and B of masses m and 2m placed on a smooth surface are travelling in opposite directions with velocities of 6 m/s and 4 m/s respectively. A perfectly elastic spring is attached to block A. If after collision, velocity of A is $\frac{2}{3}\,\,m/s$ towards right , then velocity of block B would be          A) $\frac{4}{3}\,m/s$ towards left   B) $\frac{16}{3}\,m/s$  towards left C) $\frac{28}{3}\,m/s$ m/s towards left D)  4 m/s towards left

 [c] Collision is elastic. Hence, $V_{2}^{'}-V_{1}^{'}={{V}_{1}}-{{V}_{2}}$ or  $V_{2}^{'}-\frac{2}{3}=-6-4$ $P{{I}_{2}}=\frac{3}{2}\times 20=30cm$  $V_{2}^{'}=-10+\frac{2}{3}=-\frac{28}{3}m/s$ or velocity of B is ${{P}^{2}}=P_{1}^{2}+P_{2}^{2}+2{{P}_{1}}{{P}_{2}}\cos \theta$ towards left.