Plastic impact into a mass–spring–damper (Part B): 22-Mec-A2 Kinematics and Dynamics of Machines, May 2018, Q6

Question 6: Plastic impact into a mass–spring–damper (Part B) (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. Incoming block $m=1\ \text{kg}$ at $v=10\ \text{m/s}$; target vibration mass $m=1\ \text{kg}$ (at rest) on spring $k=1000\ \text{N/m}$ and damper $c=10\ \text{N}\cdot\text{s/m}$. Perfectly plastic impact (they stick).

Find. The post-impact motion $x(t)$ (type, frequencies, decay, first peak).

Left: incoming block ($v=10$ m/s) strikes the mass–spring–damper and sticks. Right: the ensuing under-damped free response $x(t)=0.225\,e^{-2.5t}\sin(22.22\,t)$ m.

Approach. Conserve linear momentum through the (instantaneous) plastic impact to get the common velocity, then solve the free vibration of the combined mass on $k,c$ with that initial velocity and zero initial displacement.

Plastic impact — momentum conservation. During the impact the spring/damper impulse is negligible, so $$m\,v=(m+m)\,v_0\;\Rightarrow\;v_0=\frac{v}{2}=5\ \text{m/s},\qquad M=2m=2\ \text{kg}.$$
System parameters of the combined mass. $$\omega_n=\sqrt{\frac{k}{M}}=\sqrt{\frac{1000}{2}}=22.36\ \text{rad/s},\quad \zeta=\frac{c}{2\sqrt{kM}}=\frac{10}{2\sqrt{2000}}=0.112.$$ $\zeta<1\Rightarrow$ under-damped. Damped frequency: $$\omega_d=\omega_n\sqrt{1-\zeta^2}=22.36\sqrt{1-0.112^2}=22.22\ \text{rad/s}.$$
Free response with $x(0)=0,\ \dot x(0)=v_0$. $$x(t)=e^{-\zeta\omega_n t}\,\frac{v_0}{\omega_d}\sin(\omega_d t)=\boxed{0.225\,e^{-2.5\,t}\sin(22.22\,t)\ \text{m}},$$ with decay rate $\zeta\omega_n=c/2M=2.5\ \text{s}^{-1}$ (time constant $0.40$ s) and period $T_d=2\pi/\omega_d=0.283$ s.
First peak displacement. $\dot x=0$ at $t_p=\tfrac{1}{\omega_d}\tan^{-1}(\omega_d/\zeta\omega_n)=0.0657\ \text{s}$: $$x_{\max}=0.225\,e^{-2.5(0.0657)}\sin(22.22\cdot0.0657)=\boxed{0.19\ \text{m}}.$$

The stuck pair oscillates about the spring’s equilibrium at $\approx3.5\ \text{Hz}$, the amplitude decaying by $e^{-2.5t}$ — roughly a $54\%$ drop each cycle — and comes essentially to rest within about $2$ s.

Check: the source gives a single symbol $m=1$ kg and says “the two blocks stick”; both the striker and the vibration mass are taken as $m=1$ kg, giving $M=2$ kg and $v_0=5$ m/s. If instead only the vibration mass ($1$ kg) is intended to move after a striker of different mass, rescale $v_0$ by momentum accordingly; the response form is unchanged.

Quantity	Value
Post-impact velocity $v_0$ / mass $M$	$5\ \text{m/s}$ / $2\ \text{kg}$
$\omega_n,\ \zeta,\ \omega_d$	$22.36\ \text{rad/s},\ 0.112,\ 22.22\ \text{rad/s}$ (under-damped)
Response	$x(t)=0.225\,e^{-2.5t}\sin(22.22t)\ \text{m}$
First peak	$\approx0.19\ \text{m}$ at $t\approx0.066\ \text{s}$

Topic: Perfectly plastic impact followed by single-DOF under-damped free vibration.

Key relations: momentum $mv=(m+m)v_0$; $\omega_n=\sqrt{k/M}$; $\zeta=c/(2\sqrt{kM})$; $\omega_d=\omega_n\sqrt{1-\zeta^2}$; for $x(0)=0$, $x(t)=(v_0/\omega_d)e^{-\zeta\omega_n t}\sin\omega_d t$.

Why this works: the impact is so brief that the finite spring/damper forces impart negligible impulse, so momentum is conserved and sets the initial velocity; thereafter the merged mass is an ordinary damped oscillator released with that velocity from the equilibrium position.

Common pitfall: using the incoming velocity $10$ m/s (instead of the post-impact $5$ m/s) or the single mass $1$ kg (instead of $2$ kg) in the free-vibration solution; and forgetting energy is not conserved in a plastic impact (half the kinetic energy is lost).

Question 6 of 7: Plastic impact into a mass–spring–damper (Part B)

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