Two blocks are resting on a horizontal, slippery surface. The blocks have the ma

Question

Two blocks are resting on a horizontal, slippery surface. The blocks have the masses block l=2m and block 2=m. The two blocks are loosely connected to a spring with spring constant k. The blocks are also connected to a string that ensures that they do not move in relation to each other. At time t = 0 the string is burned over. Starting position is shown in the figure below. At the moment the spring no longer has contact with the two blocks, the left block 1 has reached the speed v. Block 2 has not hit block 3 at this time. What is the speed of block 2 immediately after the blocks have lost contact with the spring? Please explain why. v/2 -v -2v 2v v

Explanation / Answer

Correct answer is option C) -2v . The solution is

initial momentum = 0(both blocks at rest)

Final moment should be equal to initial momentum since there is no external force acting in the system.

m1v1+m2v2=0

2mv+mv2=0

v2=-2v

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