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## Q. 3.13

Figure 3.21 shows two masses $m_{1} \text{ and } m_{2}$ where $m_{1} = 5 \ kg \text{ and } m_2 = 1 \ kg.$
The mass $m_{1}$ which is initially at rest is supported by a spring and a damper.

The stiffness coefficient of the spring k is assumed to be 1000 N/m, and the damping coefficient c is assumed to be 10 N · s/m. The mass $m_2$ is assumed to move with a constant velocity $v_2 = −5 \ m/s.$ Assuming the coefficient of restitution e = 0.9, determine the displacement equations of the two masses after impact.

## Verified Solution

The conservation of momentum and the restitution conditions are
$m_{1}v^{\prime}_{1} + m_{2}v^{\prime}_{2} = m_{1}v_{1} + m_{2}v_{2}$

$v^{\prime}_{2}− v^{\prime}_{1} = e(v_{1} − v_{2})$

where $v_{1} = 0 \text{ and } v_{2}$ = −5 m/s. Therefore, the conservation of momentum and restitution equations can be written as
$5v^{\prime}_{1} + v^{\prime}_{2}$ = (5)(0) + (1)(−5)

$v^{\prime}_{2} − v^{\prime}_{1}$ = 0.9(0 − (−5))

which yield

$5v^{\prime}_{1} + v^{\prime}_{2} = −5, \quad \quad v^{\prime}_{1} − v^{\prime}_{2}$ = −4.5

By adding these two equations, one obtains $6v^{\prime}_{1} = −9.5,$ or
$v_{1}$ = −1.5833 m/s
and
$v^{\prime}_{2} = v^{\prime}_{1}$ + 4.5 = −1.5833 + 4.5 = 2.9167 m/s
The equations of motion of the two masses after impact can be written as

$m_{1} \ddot{x_{1}} + c \dot{x_{1}} + kx_{1}$ = 0
$m_{2} \ddot{x_{2}}$ = 0

The solutions of these two equations can be determined as
$x_{1} = Xe ^{−ξωt} \sin(ω_dt + \phi)$

$x_{2} = A_{1}t + A_{2}$

where $X, \phi, A_1, \text{ and } A_2$ are constants that can be determined using the initial conditions $x_{1 0} = 0, \dot{x}_{1 0} = v^{\prime}_{1}$ = −1.5833 m/s, $x_{2 0} = 0, \dot{x}_{2 0} = v^{\prime}_{2}$ = 2.9167 m/s. The damping factor ξ , natural frequency ω, and the damped natural frequency $ω_d$ are given by

ξ = $\frac{c}{C_{c}} = \frac{c}{2√km} = \frac{10}{2√(1000)(5)}$ = 0.0707

ω = $\sqrt{\frac{k}{m}} = \sqrt{\frac{1000}{5}}$ = 14.142 rad/s

$ω_{d} = ω \sqrt{1 − ξ^{2}}$ = 14.1066 rad/s

By using the initial conditions one can show that the displacements of the two masses as functions of time can be written as

$x_1 = −0.1122e^{−0.9998t} \sin(14.1066t)$
$x_2 = 2.9167t$

The results presented in Fig. 3.22, which shows $x_1 \text{ and } x_2$ as functions of time, clearly demonstrate that the two masses do not encounter a second impact.