1
$\begingroup$

This is problem #23 from the 2015 F=ma contest:

A 2.0 kg object falls from rest a distance of 5.0 meters onto a 6.0 kg object that is supported by a vertical massless spring with spring constant k = 72 N/m. The two objects stick together after the collision, which results in the mass/spring system oscillating. What is the maximum magnitude of the displacement of the 6.0 kg object from its original location before it is struck by the falling object?

The velocity of the 2kg block before impact is 10m/s, so the kinetic energy of the two objects together is 25J. This should translate to spring potential and gravitational potential:

$$\frac{1}{2}kx^2 +m_tgx=25$$ $$36x^2+80x-25=0$$

However, the solution says that $$m_tgx=20x$$ I thought that they might be using x as a different length. Could someone explain this?

$\endgroup$
  • $\begingroup$ Not clear to me what they mean by $x$. Your initial approach looks reasonable to me. $\endgroup$ – Floris Nov 23 '15 at 13:08
  • $\begingroup$ before it is stuck.???? or after it is struck.??? $\endgroup$ – Vinay5forPrime Nov 23 '15 at 14:39
1
$\begingroup$

By using work energy theorem it can be solved.

The velocity of the 2kg object till it reaches the 6kg object is given by $$\sqrt{2*g*5}=9.90m/s^2$$

apply the conservation of momentum for plastic impact. $$m_1u_1+m_2u_2=m_1m_2V$$ $$2*9.90+6*0=8*V$$ $$V=2.475m/s^2$$ work energy theorem $$\frac{1}{2}*8*2.475^2=\frac{1}{2}*72*(-x)^2+8*g*x$$ $$x=1.801472656m$$ or$$x=0.3777773442m$$ I hope it was helpful. Leave a comment.

$\endgroup$
  • $\begingroup$ What does x represent in your work-energy equation? We want to find "the maximum magnitude of the displacement of the 6.0 kg object from its original location before it is struck by the falling object". Do you know what this is? Also I got a different solution for the quadratic. $\endgroup$ – Andrew Wang Nov 23 '15 at 16:01
  • $\begingroup$ How can 6kg object move before being hit by the 2kg? That is why i have commented in the question. $\endgroup$ – Vinay5forPrime Nov 23 '15 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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