Two blocks with mass M and 3M that are sitting on a frictionless horizontal surf

Question

Two blocks with mass M and 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 place bcause the compression leads to 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 spring falls down as soon as the distance betwen the blocks exceeds the natural length of the spring. After the blocks are pushed togother, a horizontal rope then secures the blocks in place. Later the rope is cut, with scissors, and the heavier block is observed to be launched with a speed of 2m/s in the positive x-direction.

Define the system and the interaction of interest(be careful no to overlook any forces!), and justify very precisely the priciples you apply to calculate the speed of the lighter block. Explain with all the necessary details why you are allowed to use the equation.

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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