# What's the maximum compression of the spring? [closed]

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

• 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) $$x=\frac{m v}{\sqrt{(m+M) k}}$$

• Can you provide your attempt to a solution and/or your thoughts? – zhutchens1 May 9 '19 at 1:35
• I tried to use the conservation of energy to solve this problem but it didn't work though – John C. May 9 '19 at 1:46
• 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. – zhutchens1 May 9 '19 at 1:49
• I have edited the post nw – John C. May 9 '19 at 2:02
• It looks like your answer is the same as (b). I think you are correct. – zhutchens1 May 9 '19 at 3:18

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