A) Zero
B) \[\frac{M{{L}^{2}}}{K}\]
C) \[\sqrt{MK}\,L\]
D) \[\frac{K{{L}^{2}}}{2M}\]
Correct Answer: C
Solution :
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 \[\frac{1}{2}M{{v}^{2}}=\frac{1}{2}K{{L}^{2}}\] \[\therefore \] \[v=\sqrt{\frac{K}{M}}L\] Where v is the velocity of block by which it collides with spring. So, its maximum momentum \[P=Mv=M\sqrt{\frac{K}{M}}\,L\] = \[\sqrt{MK}\,L\] After collision the block will rebound with same linear momentum.
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