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A block of mass m executes SHM with amplitude A on a frictionless surface. At the instant when the block is at its negative extreme position (x = -A, velocity = 0), a bullet of mass m travelling with speed v0 strikes and embeds in the block (t = 0). Given that m*v0² = 2*k*A², the displacement x at any subsequent time t is:

1. x = -sqrt(2)*A*sin(sqrt(k/(2*m))*t + pi/4)
2. x = -sqrt(2)*A*sin(sqrt(k/(2*m))*t)
3. x = -sqrt(2)*A*sin(sqrt(k/(2*m))*t + pi/3)
4. x = -A*sin(sqrt(k/m)*t)

Correct answer: x = -sqrt(2)*A*sin(sqrt(k/(2*m))*t + pi/4)

Solution

Using momentum conservation the combined mass 2m has speed v0/2 at x = -A; the new amplitude is sqrt(2)*A and matching initial conditions (x=-A, v=v0/2) to x(t) = -sqrt(2)*A*sin(omega'*t + phi) gives phi = pi/4 when the bullet moves in the negative direction.

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