• # question_answer 22) The block of mass m moving on the frictionless horizontal surface collides with a spring of spring constant 'K' and compresses it by length L. The maximum momentum of the block after the .collision is :- A)  Zero             B)  $M{{L}^{2}}/R$C)  $\sqrt{MK}\,\,L$         D)  $K{{L}^{2}}/2M$

Correct Answer: C

Solution :

By come $\frac{1}{2}m{{v}^{2}}=\frac{1}{2}K{{L}^{2}}$ $\Rightarrow$ $m{{v}^{2}}=K{{L}^{2}}\Rightarrow (m{{v}^{2}})=mK{{L}^{2}}\Rightarrow mv=\sqrt{mK}\,L$

LIMITED OFFER HURRY UP! OFFER AVAILABLE ON ALL MATERIAL TILL TODAY ONLY!

You need to login to perform this action.
You will be redirected in 3 sec