Question

Two blocks of masses m₁=1.7 kg and m₂=3.7 kg are put on a frictionless horizontal surface. A spring of stiffness k=970 N/m and negligible mass is placed between the blocks and they are pushed together to compress the spring by distance d from its equilibrium length. The configuration is fixed by a cord that holds the blocks. When the cord is burned, the first block moves to the left with a speed of v₁=1.9 m/s. Calculate the initial compression of the spring, d. 

Answer 1

$m_1 = 1.7kg;m_2=3.7kg;k= 970\frac{N}{m}$

$\text{Spring is compressed:}$

$v'_1= 0;v'_2= 0$

$\Delta x'= d$

$E = E'_k+E'_p$

$E'_k=\frac{m_1v_1'^2}{2}+\frac{m_2v_2'^2}{2}=0$

$E_p'=\frac{k\Delta x'^2}{2}= \frac{kd^2}{2}$

$E=\frac{kd^2}{2}$

$\text{Spring is released:}$

$v_1=1.9\frac{m}{s} ;v_2-?$

$\Delta x= 0$

$E = E_k+E_p$

$E_p=\frac{k\Delta x^2}{2}= 0$

$E_k=\frac{m_1v_1^2}{2}+\frac{m_2v_2^2}{2}$

$\text{by the law of conservation of momentum}$

$m_1v_1+m_2v_2 = m_1v'_1+m_2v'_2$

$m_1v_1+m_2v_2 = 0$

$v_2 = -\frac{m_1v_1}{m_2}= -\frac{1.7*1.9}{3.7}=-0.87$

$E_k=\frac{1.7*1.9^2}{2}+\frac{3.7*(-0.87)^2}{2}=4.47J$

$E = E_p+E_k = 4.47$

$E=\frac{kd^2}{2}$

$d = \sqrt{\frac{2E}{k}}= \sqrt{\frac{2*4.47}{970}}\approx 0.07m$

$\text{Answer: initial compression of the spring, d = 0.07m}$