Question 6 of 7: Plastic impact on a mass–spring–damper

Question 6 (Part B): Plastic impact on a mass–spring–damper (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Falling block $m_b=0.25m=0.5$ kg, initial speed $2$ m/s downward, drop height $10$ m. Primary system $m=2$ kg, $k=2000$ N/m, $c=20$ N·s/m, initially at rest. Perfectly plastic (stick) collision.

Find. the ensuing motion $x(t)$ after impact.

Approach. Get the block’s impact speed by energy; apply impulse–momentum through the (instantaneous) plastic collision to get the common post-impact velocity; then solve the underdamped free-vibration initial-value problem about the shifted equilibrium.

1. Impact speed of the block. $$v_{\text{imp}}=\sqrt{v^2+2gh}=\sqrt{2^2+2(9.81)(10)}=14.15\ \text{m/s (down)}.$$
2. Plastic collision (momentum). The spring/damper impulses are negligible during the instantaneous impact, so $$v_0=\frac{m_b\,v_{\text{imp}}}{m+m_b}=\frac{0.5(14.15)}{2.5}=\boxed{2.83\ \text{m/s (down)}}.$$ Half the kinetic energy is lost in the perfectly plastic collision.
3. Post-impact system parameters. The stuck mass is $M=2.5$ kg, so $$\omega_n=\sqrt{k/M}=28.3\ \text{rad/s},\quad \zeta=\frac{c}{2\sqrt{kM}}=0.141\ (\text{underdamped}),\quad \omega_d=\omega_n\sqrt{1-\zeta^2}=28.0\ \text{rad/s}.$$
4. Initial conditions about the new equilibrium. Adding the block lowers the static equilibrium by $\delta=m_b g/k=2.45$ mm. Measuring $x$ (down positive) from the new equilibrium, at $t=0$ the mass is at the old rest point, i.e. $x_0=-\delta=-2.45$ mm and $\dot x_0=+2.83$ m/s.
5. Ensuing motion. The underdamped free response is $$x(t)=e^{-\zeta\omega_n t}\!\left[x_0\cos\omega_d t+\frac{\dot x_0+\zeta\omega_n x_0}{\omega_d}\sin\omega_d t\right],$$ $$\boxed{x(t)=e^{-4t}\big[-0.0025\cos(28.0\,t)+0.101\sin(28.0\,t)\big]\ \text{m}.}$$ The mass oscillates with an initial amplitude $\approx0.10$ m at $\omega_d=28$ rad/s ($f_d=4.46$ Hz), decaying as $e^{-4t}$ (time constant $0.25$ s), and settles at the new equilibrium $2.45$ mm below the original rest position.