0
$\begingroup$

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) $x=\frac{m v}{\sqrt{(m+M) k}}$

$\endgroup$

closed as off-topic by JMac, ZeroTheHero, Bob D, Qmechanic May 9 at 5:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – JMac, ZeroTheHero, Bob D, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

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

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

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.