The block of  mass M  moving on  the frictionless horizontal  surface collides with the spring of spring constant  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)

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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