Two blocks with mass M and 3M on a horizontal frictionless surface are pushed together and compress a spring of negligible mass that is squeezed between them. The spring remains in place because the compression leads to a sufficient amount of friction with the sides of the blocks. It is important to understand that the spring is NOT attached to the blocks, in other words, that it will fall 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 blocks in place.Later the rope is cut with scissors and the heavier block is launched with a speed of 2 m/s in the positive x-direction.

Define very precisely the system, the interaction(s) and the interaction time(s) of interest, and justify very precisely the principle(s) you apply to calculate the launching speed of the lighter block.

We are told that the two blocks are sitting on the horizontal frictionless surface. Thus, we can say that no external force is being applied on the system and so, the momentum of the whole system is conserved accordingly to that condition.

We are also told that when the rope is cut with scissors, that the heavier block attains the speed of 2 m/s in the positive x-direction which is horizontal direction.

We know that formula for momentum is; M = mass x velocity.

Thus, the momentum of the heavier block is calculated as;

M_1 = 3M × 2

M_1 = 6M kg.m/s

Since no external force is applied on the object, the initial momentum will be zero.

Hence, to conserve the system, the momentum of the lighter block will be equal and opposite to the momentum of heavier block.

So, momentum of lighter block is;

M_2 = -6M kg.m/s

Since mass of lighter block is M and formula for momentum = mass x velocity.

Thus;

-6M = Mv

Where v is speed of lighter block.

So, v = -6M/M

v = -6 m/s