Answer:
launching speed of the lighter block = -6 m/s
Explanation:
We are given;
Mass of light block; M
Mass of heavy block; 3M
Speed of launched block: v = 2m/s
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
