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I tried to use the conservation of energy to solve this problem, here's what I tried to do:

$\require{enclose}$ $$\begin{align} \enclose{downdiagonalstrike} {\frac{1}{2}} m v^{2} &= \enclose{downdiagonalstrike} {\frac{1}{2}}Kx^{2} \\[1em] m v^{2}&=K x^{2} \\[1em] \frac{m v^{2}}{K}&=x^{2} \\ x&=\sqrt{\frac{m v^{2}}{K}} \end{align}$$

A block of mass M is initially at rest on a frictionless floor, as shown in the accompanying figure. The block, attached to a massless spring with spring constant k, is initially at its equilibrium position. An arrow with mass m and velocity v is shot into the block The arrow sticks in the block. What is the maximum compression of the spring?

The correct answer is E, but I need someone to explain it

enter image description here

  • A) $x=v \sqrt{\frac{k}{m}}$

  • B) $x=v \sqrt{\frac{m}{k}}$

  • C) $x=v \sqrt{\frac{m+M}{k}}$

  • D) $x=\frac{(m+M) v}{\sqrt{m k}}$

  • E) $x=\frac{m v}{\sqrt{(m+M) k}}$

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  • $\begingroup$ Can you provide your attempt to a solution and/or your thoughts? $\endgroup$ – zhutchens1 May 9 '19 at 1:35
  • $\begingroup$ I tried to use the conservation of energy to solve this problem but it didn't work though $\endgroup$ – John C. May 9 '19 at 1:46
  • $\begingroup$ That seems like a valid approach. You should edit the question to fully describe your line of thinking and your complete solution, so that we can fully address your question. $\endgroup$ – zhutchens1 May 9 '19 at 1:49
  • $\begingroup$ I have edited the post nw $\endgroup$ – John C. May 9 '19 at 2:02
  • $\begingroup$ It looks like your answer is the same as (b). I think you are correct. $\endgroup$ – zhutchens1 May 9 '19 at 3:18
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You forgot about conservation of momentum in your formula so momentum before collision is $m.v$ and after collision is $(m+M)v_2$ by conservation of momentum $v_2=\frac{m.v}{m+M}$ since arrow stuck to the system new mass is$m+M$ substituting these in conservation of energy you get your E answer

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