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An 8.0-g bullet is shot into a 4.0-kg block, at rest on a frictionless horizontal surface (see the figure). The bullet remains lodged in the block. The block moves into an ideal massless spring and compresses it by 8.7 cm. The spring constant of he spring is 2400 N/m. The initial velocity of the bullet is closest to..."

What I did first was find the amount of force the bullet+block exert on the spring ($208.8 \text{ N}$). I don't know what to do after that. I know I have to find final velocity, and then do $m_1v_1 = (m_1+m_2)v_2$, and solve for $v_1$, but how would I do that? At first I thought I could do $F = ma$, but that didn't work. Then I tried to do $F = \frac{dm}{dt} + \frac{d}{dt}(mv)$, but I don't know the time of the collision, so I wouldn't be able to use that, correct?

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closed as off-topic by ACuriousMind, Kyle Kanos, John Rennie, Bill N, Ryan Unger Oct 26 '15 at 23:28

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  • $\begingroup$ Note that you appear to try to use Newton's second law in this form: $F = dp/dt$. Two comments for you to keep in mind for the future. First, you did the derivative incorrectly. But more importantly: Newton's second law does not apply to systems whose mass is changing. $dm/dt$ must be zero. This little fact is often not given enough emphasis. $\endgroup$ – garyp Oct 26 '15 at 14:32
  • $\begingroup$ This is a standard momentum-energy drill problem. Try energy conservation after the collision, and solve the process "backward" (compression, collision, shooting). $\endgroup$ – Bill N Oct 26 '15 at 20:52
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In general, the force on the spring isn't going to be a helpful quantity. However, you can calculate how much energy is stored in the spring when it is compressed. From this energy, you should be able to calculate the energy of the block bullet system after impact, and then the velocity $v_2$.

You should be able to do this using only conservation laws (energy and momentum), and shouldn't need to using any of your force equations. Work backwards, starting with the final state of the system.

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  • $\begingroup$ Thank you very much! I will accept this answer as soon as it lets me! So this was more of a conservation of energy problem, than a conservation of momentum. At least until the last step. $\endgroup$ – Mahmud Assamaray Oct 26 '15 at 14:15
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    $\begingroup$ Glad I could help. For many problems, you have to use both depending on the situation, Since the spring is presumably attached to something rigid, conservation of momentum won't work there. Also, since the collision between the bullet and block is inelastic, conservation of energy won't be used in that case. You just need to consider the interaction at each step of the problem to see which method is appropriate. $\endgroup$ – tmwilson26 Oct 26 '15 at 14:18

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