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**Physics of Impulsive Motion **

Recall from dynamics that the principle of impulse and momentum for a particle states that

(2â€“1)

where ^{N}G is the linear momentum of the particle as viewed by an observer in an inertial reference frame N. Suppose now that we consider the following system. A block of mass m is connected to a linear spring with spring constant K and unstretched length â„“_{0} and a viscous linear damper with damping coefficient c as shown in Fig. 2â€“1. The block is initially at rest (i.e., its initial velocity is zero) at its static equilibrium position (i.e., the spring is initially unstressed) when a horizontal impulse is applied. We are interested here in determining the velocity of the block immediately after the application of the impulse .

**Figure 2â€“1 **Block of mass m connected to linear spring and linear damper struck by horizontal impulse .

The solution of the above problem is found as follows. First, let F be the ground. Then,

choose the following coordinate system fixed in F:

Origin at block

when x = 0

E_{x} = To the left

E_{z} = Into page

E_{y} = E_{z} Ã— E_{x}

Then, the position of the block is given in terms of the displacement x as

r = x**E**_{x } (2â€“2)

Because {E_{x}, E_{y} , E_{z}} is a fixed basis, the velocity of the block in reference frame F is given as

Now because we are going to apply the principle of linear impulse and momentum to this problem, we do not need the acceleration of the block. Instead, we know that neither the spring nor the damper can apply an instantaneous impulse. Therefore, the only impulse applied to the system at t = 0 is that due to . Consequently, the external impulse acting on the system at t = 0 is

(2â€“4)

Furthermore, the linear momentum of the block the instant before the impulse is applied is zero (i.e., the block is initially at rest) while the linear momentum of the block the instant after the impulse is applied is given as

^{F}**G**â€² = m ^{F} **v** â€² = mvâ€²**E**_{x} (2â€“5)

Setting equal to ^{F}**G**â€² , we obtain

= mvâ€² â‰¡ mv(t = 0^{+}) (2â€“6)

Solving for v(t =0^{+}), we obtain

(2â€“7)

The result of this analysis shows that the response of a resting second-order linear system to an impulsive force is equivalent to giving the system the initial velocity shown in Eq. (2â€“7).

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