Wondering if you can explain how equation 4-22 was determined as well as complet

Question

Wondering if you can explain how equation 4-22 was determined as well as completing the problem if not just the explanation of equation 4-22 is fine.

Example 4-8 Consider the mechanical system shown in Figure 4-14. A bullet of mass m is shot into a block of mass M (where M > m). Assume that when the bullet hits the block, it becomes embedded there. Determine the response (displacement x) of the block after it is hit by the bullet. The displacement x of the block is measured from the equilibrium position before the bullet hits it. Suppose that the bullet is shot at t = 0-and that the initial velocity of the bullet is v(0-). Assuming the following numerical values for M, m, b, k, and v(0-), draw a curve x(t) versus t: M = 50 kg, m = 0.01 kg, b 100 N-s/m k-2500 N/m, u(0-)-800 m/s The input to the system in this case can be considered an impulse, the magnitude of which is equal to the rate of change of momentum of the bullet. At the instant the bullet hits the block, the velocity of the bullet becomes the same as that of the block, since the bullet is assumed to be embedded in it. As a result, there is a sudden change in the velocity of the bullet. [See Figure 4-15(a).] Since the change in the velocity of the bullet occurs instantaneously, v has the form of an impulse. (Note that v is negative.) Fort > 0, the block and the bullet move as a combined mass M + m. The equa- tion of motion for the system is (M+m)2 + bk + kx = F(t) (4-22)

Explanation / Answer

sol. it is a spring damper model.

as per no slip condition, as bullet hit the the block, at the point to hit the velocity of bullet becomes equal to the velocity of block(i.e, v =0), soall the kinetic energy gets transform in to a pressure energy, which in turn creats the deflection in spring damper system uder the combined mass od bullet and the block.

and hence equation 4.22 can be written by considering total mas as m+M.

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