A) \[\frac{M{{L}^{2}}}{k}\]
B) zero
C) \[\frac{k{{L}^{2}}}{2M}\]
D) \[\sqrt{Mk}L\]
Correct Answer: D
Solution :
[d] 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.
You need to login to perform this action.
You will be redirected in
3 sec
