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Maximum speed in the ensuing oscillations of 2 blocks on springs after they collide and stick together

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In the arrangement, two blocks attached at one end of each spring are held at rest. The other ends of the springs are attached to the walls. Relaxed length of each spring is l, distance between the walls is 2l, size of the blocks is negligible, and the floor is frictionless. Initially both the blocks are at distance 0.51l from the walls. When released, the block rush toward each other. collide head on and then stick to each other. Their maximum speed $v_m$ in the ensuing oscillatons is $\frac{l}{n}$$\sqrt{\frac{k}{m}}$ . Find the value of n .

My try :

But now how to proceed ?

asked Jun 15, 2018 in Physics Problems by koolman (4,286 points)
retagged Jun 15, 2018 by sammy gerbil

1 Answer

3 votes
Best answer

You have started correctly by writing the equations of motion $x(t)$. However, the blocks meet when $x_1+x_2=0$ not when $x_1+x_2=l$ (which is true in the initial position). This is because both $x_1, x_2$ are being measured outwards from the midpoint, so if the oscillations had the same frequency they would meet at the midpoint where $x_1=x_2=0$.

You might also observe that the oscillation frequencies are $\omega_1=\sqrt{\frac{k}{2m}}$ and $\omega_2=2\omega_1$. So the blocks meet when $$x_1+x_2=\frac{l}{2}\cos(\omega_1 t)+\frac{l}{2}\cos(2\omega_1 t) =l\cos(\frac32 \omega_1 t)\cos(\frac12\omega_1 t)=0$$ The faster oscillation becomes zero first, so the earliest time at which they meet is when $$\cos(\frac32 \omega_1 t)=0, \omega_1 t=\frac{\pi}{3}$$ The momenta just before collision are $$m_1u_1=(2m)\dot x_1=-(2m)\omega_1\frac{l}{2}\sin(\omega_1 t)=-m\omega_1 l \frac{\sqrt3}{2}$$ $$m_2u_2=m\dot x_2=-m(2\omega_1)\frac{l}{2}\sin(2\omega_1 t)=-m\omega_1 l \frac{\sqrt3}{2}$$ These momenta are equal and opposite (because $u_1, u_2$ are measured in opposite directions) so they cancel out. The kinetic energy of the blocks is completely wiped out by the collision. Only elastic potential energies remain at this point, which are $$\frac12 kx_1^2=\frac12 k \frac{l^2}{4}\cos^2(\omega_1 t)=\frac{kl^2}{32}$$ $$\frac12(2 k)x_2^2= k \frac{l^2}{4}\cos^2(2\omega_1 t)=\frac{2kl^2}{32}$$ The total elastic PE at this point decreases during the subsequent oscillation, becoming zero at the midpoint, at which KE is maximum. So $$\frac12 (3m)v_m^2=\frac{3kl^2}{32}$$ $$v_m=\frac{l}{4}\sqrt{\frac{k}{m}}$$ $$n=4$$

answered Jun 15, 2018 by sammy gerbil (28,896 points)
edited Jun 16, 2018 by sammy gerbil