3. Two blocks with mass M and mass 3M that are sitting on a frictionless horizon

Question

3. Two blocks with mass M and mass 3M that are sitting on a frictionless horizontal surface, are pushed together and compress a spring that is mounted horizontally between them. The spring remains in palce because the compression leads to a sufficient amount of friction with the side walls of the blocks. It is important to understand that the spring is NOT attached to the blocks, in other words, the springs falls down as soon as the distance between the blocks exceeds the natural length of the spring. After the blocks are pushed together, a horizontal rope then secures the locks in place. Later the rope is cut with scissors, and the heavier block is observed to be launched with a speed 2 m/s in the positive x-direction.

Define the system and the interaction of interest (be careful not to overlook any forces!), and justify very precisely the principle(s) you apply to calculate the launching speed of the lighter block. In other words, do not just give an equation and the numerical answer, but explain with all the necessary detail why your are allowed to use this equation.

Thanks in advance!

Explanation / Answer

As no external force is being applied to the system.

So Momentum of the whole system will be conserved.

Now as soon as we cut the rope velocity attained by the heavier block with mass 3M is 2m/s in +X direction.

Momentum of heavier block = 6M

The initial momentum of the system was zero and this will be conserved as no external force is being applied. The friction force that is acting on block and spring is internal and hence NO EXTERNAL FORCE.

So after we cut the rope The lighter block will have momentum -6M

So velocity is 6M/M = -6 m/s in negative x direction.

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