Question

A block of mass m is held at rest while compressing an ideal spring by an amount x. The spring constant of the spring is k. The block has a mass M, where M > m. At time t = 0, the block is released. At time t = t1, the spring is no longer compressed and the block immediately collides with and sticks to another block of mass M. The blocks stick together and the two-block system moves with a constant speed v. Frictional effects are negligible. What is the value of t1?

1) t1 = sqrt((2m/k) + (2M/k))
2) t1 = sqrt((2m/k) - (2M/k))
3) t1 = sqrt((2M/k) - (2m/k))
4) t1 = sqrt((2M/k) + (2m/k))

Answer 1

The time t1 should be a quarter of the period of the simple harmonic motion, which is π√(m/k)/2. None of the given options match this result, indicating a possible error in the problem statement or the options.

Step-by-step explanation:

The value of t1, the time when the spring is no longer compressed, is related to the oscillation period of the mass-spring system. Since the spring compression is released at t = 0, and t1 is the time it takes for the spring to return to its equilibrium position and the block to collide with mass M, we must consider the period of a simple harmonic oscillator. The period T for a mass m attached to a spring with spring constant k is given by T = 2π√(m/k). Because we are looking for the time it takes to go from maximum compression to equilibrium position (which is a quarter of the period), t1 = T/4. Substituting in the equation for period we get t1 = (1/4) * 2π√(m/k) = (π√(m/k))/2. None of the provided options matches this value which indicates there might be a mistake in the problem statement or in the listed options.