Linear Vibrations Lab

Worked examples › Example 4

Example 4Impact: a block sticks to a damped carriage

impactdamped free vibrationinitial conditions

Problem

Block A (mass $2m$) strikes the stationary block B (mass $m$) with speed $v$ and sticks to it. B is attached to the wall by a spring $k$ and a damper $c$. All surfaces are smooth. $x$ is measured to the left from B's rest position.

Use $v=30$ m/s, $k=3000$ N/m, $c=30$ N·s/m, $m=10$ kg.

Find: the time history $x(t)$ after A sticks to B.

InteractiveThe system in motion

Displacements in the drawing are scaled up so the motion is visible; the plot shows the true values.

x(t)Decay envelope

Solution

Every solution follows the same five steps: coordinates, free-body diagrams, kinetics, kinematics, solve. Try the problem first, then open one step at a time to check your work.

Step 1 · Choose a coordinate system

Let $x$ be the position of B (and later of A + B), positive to the left, measured from B's rest position where the spring is unstretched. Take $t=0$ at the instant of impact.

Step 2 · Draw the free-body diagrams
During impact (very short)BAF̂F̂spring and damper forces are finite:no impulse during the impactAfter impact (A stuck to B)3mkxcẋ3mgN+x
Left: the equal and opposite contact impulse between A and B. Right: the combined body after impact, with x positive to the left.
Step 3 · Kinetics (Newton–Euler equations)

During impact (A and B together): only the internal contact impulse acts, so linear momentum is conserved:

$$2m\,v+m(0)=(2m+m)\,V_0$$

After impact (combined body, mass $3m$):

$$3m\ddot x=-kx-c\dot x$$
Step 4 · Kinematics (constraints)

A sticks to B, so afterward they share one coordinate: $x_A=x_B=x$. The impact is instantaneous, so B has not moved yet:

$$x(0)=0,\qquad \dot x(0)=V_0=\tfrac23v=20\ \text{m/s}$$
Step 5 · Solve

$3m\ddot x+c\dot x+kx=0$ with $\omega_n=\sqrt{k/3m}=\sqrt{3000/30}=10$ rad/s and $\zeta=\dfrac{c}{2\sqrt{k(3m)}}=\dfrac{30}{600}=0.05$.

Underdamped: $\omega_d=\omega_n\sqrt{1-\zeta^2}=9.987$ rad/s and $x=e^{-\zeta\omega_nt}(A\cos\omega_dt+B\sin\omega_dt)$.

$x(0)=0\Rightarrow A=0$. $\;\dot x(0)=B\omega_d=V_0\Rightarrow B=20/9.987=2.003$ m.

$$x(t)=2.00\,e^{-0.5t}\sin(9.99\,t)\ \text{m}$$

One third of the kinetic energy is lost in the sticking collision.

Answer

$$x(t)=2.00\,e^{-0.5t}\sin(9.99\,t)\ \text{m}$$

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