UTF-8 U+6211 U+662F wrote:Justin72835 wrote:You're definitely on the right track (I see that you already found the final velocities of the objects after the collision). Using the final velocity, you can find the kinetic energy of the object after the collision and set it equal to 1/2kx^2 and solve for x. This works because maximum compression occurs when the objects kinetic energy has been converted completely into potential energy.UTF-8 U+6211 U+662F wrote:1) kMv/(M+m) 2) 2kMv/(M+m)
[math]x_{inelastic}=Mv\sqrt{\frac{1}{k(M+m)}}[/math] [math]x_{elastic}=\frac{2Mv}{M+m}\sqrt{\frac{m}{k}}[/math]Edit: Sorry about the double post too.1) [math]Fd = \frac12(M+m)v_{final}^2[/math], where [math]v_{final}[/math] represents the speed of the cube after the collision happens. [math](kd)d = \frac12(M+m)v_{final}^2[/math] [math]kd^2 = \frac12(M+m)v_{final}^2[/math] [math]Mv = (M+m)v_{final}[/math] [math]v_{final} = \frac{Mv}{M+m}[/math] [math]kd^2 = \frac12(M+m)\left(\frac{Mv}{M+m}\right)^2[/math] [math]d^2 = \frac1{2k}(M+m)\left(\frac{Mv}{M+m}\right)^2[/math] [math]d^2 = \frac1{2k}\frac{(Mv)^2}{M+m}[/math] [math]d^2 = \frac{(Mv)^2}{2k(M+m)}[/math] [math]d = \sqrt{\frac{(Mv)^2}{2k(M+m)}}[/math] [math]d = Mv\sqrt{\frac1{2k(M+m)}}[/math] But this is slightly different from the posted answer, so where did I go wrong? 2) [math]kd^2 = \frac12mv_{final}^2[/math] [math]Mv = Mv_1 + mv_2[/math] [math]v = v_2 - v_1[/math] [math]v_1 = v_2 - v[/math] [math]Mv = M(v_2 - v) + mv_2[/math] [math]Mv = Mv_2 - Mv + mv_2[/math] [math]2Mv = v_2(M+m)[/math] [math]v_2 = \frac{2Mv}{M+m}[/math] [math]kd^2 = \frac12m(\frac{2Mv}{M+m})^2[/math] [math]d^2 = \frac1{2k}m(\frac{2Mv}{M+m})^2[/math] [math]d^2 = \frac{m(2Mv)^2}{2k(M+m)^2}[/math] [math]d = \frac{2Mv}{M+m}\sqrt{\frac{m}{2k}}[/math] Again, this answer is slightly different posted answer, so where did I go wrong?
You went wrong in assuming that the force applied by the spring remains constant. The work done by the spring isn't kd^2, it's 1/2kd^2 (potential energy of a spring). I had a really hard time understanding this when I first began learning about energy and started wondering where the '1/2' term in elastic potential energy (1/2kd^2) and kinetic energy (1/2mv^2) came from. For the first one, you can think of it like this: When the spring is uncompressed, it applies a force of 0; when it is fully compressed, it applies a force of kd. Thus the average force applied by the spring is 1/2kd. Plugging this into W=Fd gives W=1/2kd^2. You can also prove this using calculus but this is a far more intuitive way of thinking about it. The '1/2' term cancels out with the '1/2' from the kinetic energy equation, which is why my answer differs by a factor of root(1/2).
Note: it may seem like you need more information to solve this but if done correctly then a bunch of terms will cancel out (kind of like how mass always seems to cancel out in energy equations).
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