Two blocks of masses 6 kg and 5.5 kg are
placed on a horizontal, frictionless surface. a
light spring is attached to one of them, and
the blocks are pushed together with the spring
between them. a cord holding them together
is burned, after which the block of mass 5.5 kg
moves to the right with a speed of 6.8 m/s.
what is the velocity of the other mass in
m/s?
answer in units of m/s

When you squish the spring, you put some energy into it, and after the cord
burns and they go boing in opposite directions, that energy that you stored
in the spring is what gives the blocks their kinetic energy.

But linear momentum still has to be conserved.  It was zero while they were
tied together and nothing was moving, so it has to be zero after they both
take off.

Momentum = (mass) x (velocity)

After the launch, the 5.5-kg moves to the right at 6.8 m/s,
so its momentum is
                               (5.5 x 6.8) = 37.4 kg-m/s to the right.

In order for the total momentum to be zero, the other block has to
carry the same amount of momentum in the opposite direction.

               M x V = (6 x speed) = 37.4 kg-m/s to the left.

Divide each side by  6 :      Speed = 37.4 / 6 =  6.2333... m/s left

(That number is  (6 and 7/30) m/s .)

Ihope you get it right bro