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  • Need clarity, kindly explain! The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant Kand compresses it by length L.The maximum momentum of the block after co

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

Option 1)

Zero

 Option 2)

\frac{ML^{2}}{K}
Option 3)

\sqrt{MK}L

 Option 4)

\frac{KL^{2}}{2M}

Answers (1)

best_answer

As we learnt in

If only conservative forces act on a system, total mechnical energy remains constant -

K+U=E\left ( constant \right )

\Delta K+\Delta U=0

\Delta K=-\Delta U

-

 

 Let initial velocity of block is v then initial energy =\frac{1}{2}mv^{2}

Final potential energy =\frac{1}{2}kx^{2}

From energy conservation 

\frac{1}{2}mv^{2}=\frac{1}{2}kx^{2}

\therefore\ v=\sqrt{}\frac{k}{m}.x\ \because x=L

\Rightarrow   \Rightarrow\ maximum\ v =(\frac{\sqrt{} k}{m})L

Therefore maximum momentum = mv

    =\sqrt{}{mkL}

Correct answer is 3

 


Option 1)

Zero

Option 2)

\frac{ML^{2}}{K}

Option 3)

\sqrt{MK}L

Option 4)

\frac{KL^{2}}{2M}

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