Question 3.113: As shown in the figure, the weight W1 strikes W2 from a heig...

The drop of weight W_{1} converts potential energy, W_{1} h, to kinetic energy \frac{1}{2} \frac{W_{1}}{g} v_{1}^{2} .

Equating these provides the velocity of W_{1} at impact with W_{2}.

W_{1} h=\frac{1}{2} \frac{W_{1}}{g} v_{1}^{2} \quad \Rightarrow \quad v_{1}=\sqrt{2 g h}                     (1)

Since the collision is inelastic, momentum is conserved. That is, \left(m_{1}+m_{2}\right) v_{2}=m_{1} v_{1} , where v_{2} is the velocity of W_{1}+W_{2} after impact. Thus

\frac{W_{1}+W_{2}}{g} v_{2}=\frac{W_{1}}{g} v_{1} \quad \Rightarrow \quad v_{2}=\frac{W_{1}}{W_{1}+W_{2}} v_{1}=\frac{W_{1}}{W_{1}+W_{2}} \sqrt{2 g h}                                (2)

The kinetic and potential energies of W_{1}+W_{2} are then converted to potential energy of the spring. Thus,

Substituting in Eq. (1) and rearranging results in

\delta^{2}-2 \frac{W_{1}+W_{2}}{k} \delta-2 \frac{W_{1}^{2}}{W_{1}+W_{2}} \frac{h}{k}=0                                   (3)

Solving for the positive root (see discussion on p. 192)

\delta=\frac{1}{2}\left[2 \frac{W_{1}+W_{2}}{k}+\sqrt{4\left(\frac{W_{1}+W_{2}}{k}\right)^{2}+8 \frac{W_{1}^{2}}{W_{1}+W_{2}} \frac{h}{k}}\right]                               (4)

W_{1}=40  N , W_{2}=400  N , h=200  mm , k=32  kN / m =32  N / mm .