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## Homework Statement

In a frictionless plane, a block of mass [tex]M[/tex] is attached to a spring initially relaxed of constant [tex]k[/tex] which is attached to a wall. A piece of gum of mass [tex]m[/tex] is then horizontally thrown in direction to the block with velocity [tex]v_1[/tex] hitting it at time [tex]t = 0[/tex] in a perfectly inelastic collision. Find the expression for the displacement of the system at time [tex]t > 0[/tex]. The mass of the spring is not to be taken into consideration.

## Homework Equations

**1. SHM equation for the displacement of the system:**

[tex]x(t) = Asin(wt) + Bcos(wt) [/tex]

**2. SHM equation for the velocity of the system:**

[tex]\dot{x}(t) = Awcos(wt) - Bwsin(wt) [/tex]

Where,

[tex]w = \sqrt{\frac{k}{(M+m)}} [/tex]

**3. Equation of Conservation of Momentum for the system:**

[tex]p_{1} = p_{2},\ mv_1 = (m+M)v_2 [/tex]

## The Attempt at a Solution

First of all, from the third equation we know that

[tex]v_2 = \frac{mv_1}{(m+M)} [/tex]

The initial conditions for the system are easily found by using [tex]t = 0[/tex] at equations 1 and 2, so

[tex]x(0) = Asin(w \times 0) + Bcos(w \times 0) = B = 0[/tex]

[tex]\dot{x}(0) = v_2 = Awcos(w \times 0) - Bwsin(w \times 0) = Aw[/tex]

Where,

[tex]A = \frac{v_2}{w} [/tex]

[tex]A = \frac{mv_1}{(m+M)} \sqrt{\frac{m}{k}} [/tex]

So the equation for the displacement of the system at any given time(t) shall be

[tex]x(t) = \frac{mv_1}{(m+M)} \sqrt{\frac{m}{k}}sin(\sqrt{\frac{k}{m}}t) + Bcos(\sqrt{\frac{k}{m}}t) [/tex]

OK, so I know that 'A' shall be constant for any given time as it depends on the initial velocity of the system (which depends on the constant velocity of the piece of gum and the mass of the system) and on 'w' and both are constant. Now, how do I find 'B' for any time [tex] t > 0[/tex]? I know that at time [tex]t = 0[/tex] 'B' is zero but it won't necessarily be zero at [tex]t > 0[/tex] or am I wrong?

Thanks in advance.

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