• # question_answer 3) The block of mass (M) moving on the frictionless horizontal surface collides with the spring of spring constant K and compresses it by length L. The maximum momentum of the block after collision is A)  0                                B)  $\frac{M{{L}^{2}}}{K}$C) $\sqrt{(MK)}L$                      D)  $\frac{K{{L}^{2}}}{2M}$

When block of mass M collides with the spring its kinetic energy gets converted into elastic potential energy of the spring. From the law of conservation of energy $\frac{1}{2}m{{v}^{2}}=\frac{1}{2}K{{L}^{2}},$$\therefore \,v=\sqrt{\frac{K}{M}}L.$so, Its maximum momentum, $P=Mv=M\sqrt{\frac{K}{M}}L=\sqrt{MK}.L.$ Hence, the correction option is [c].