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?
Best Answer
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.