A spring has an unstretched length of L 0 = 0.39 m. A block with mass M = 0.4 kg

Question

A spring has an unstretched length of L0 = 0.39 m. A block with mass M = 0.4 kg is hung at rest from the spring, and the spring becomes L1 = 0.44 m long (see figure). Next the spring is stretched to a length of L2 = 0.47 m and the block is released from rest. Air resistance is negligible.

(a) How long does it take for the block to return to the equilibrium position, where the spring has length L1?


(b) Next the block is again positioned at rest, hanging from the spring (L1 = 0.44 m long) as shown in the figure below.

A bullet of mass m = 0.004 kg traveling at a speed of v = 250 m/s straight upward buries itself in the block, which then reaches a maximum height h above its original position. What is the speed of the block immediately after the bullet hits?

Explanation / Answer

a) the spring oscilates undergoing simple hormonic motion

time period T = 2*3.14 sqrt ( M/K)

where k = M * g/ L1-L0

T = 0.44879 sec

time taken is one forth of the total time

it takes T /4 sec to reach equilibrium

time taken = 0.11219 sec

B)

conserving momentum just after collision

M1 V1 = ( M1 +M2) V2

V2 = 2.475 m/sec

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