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A block of mass m executes SHM of amplitude A on a frictionless surface. At the instant it is at the negative extreme position, a bullet of mass m moving in the negative x-direction with speed v0 strikes it and gets embedded. The condition mv0² = 2kA² holds, where k is the spring constant. The displacement x as a function of time t after the collision is:

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

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

Solution

After the bullet embeds, the combined mass 2m has velocity -v0/2. The new angular frequency is omega = sqrt(k/(2m)). The new amplitude from energy: A'² = A² + (v0/2)²/omega² = 2A², so A' = sqrt(2)*A. Writing x = -sqrt(2)*A*sin(omega*t + phi) with initial conditions x(0)=-A and x'(0)=-v0/2 < 0 gives phi = pi/4, matching option A.

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