Blocks A (mass 3.00 kg ) and B (mass 9.00 kg ) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 5.00 m/s . The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of block A. Find the velocity of block B when the energy stored in the spring bumpers is maximum.

An head on collision is aka elastic collision and in elastic collision momentum and energy is conserved that is the total momentum before collision = total momentum after collision

Given

Mass of block A mA = 3kg, initial Velocity of block A vA1 = 5m/s, Final velocity = vA2

Mass of block B mB = 9kg, Initial velocity of block B vB1 = 0m/s, Final velocity = vB2

Therefore we have the equation

1)mAvA1 + mBvB1 = mAvA2 + mBvB2

Also in elastic collision between 2 objects, the relative velocities before and after the collision have the same magnitude but opposite direction

vA1 - vB1 = vB2 - vA2

vA2 = vB2 - vA1 when vB1 = 0

Substitute into the equation we have

(3*5)+(9*0) = 3*(vB2-vA1) + (9*vB2)

15 = 3vB2 - 3vA1 +9vB2

15=3vB2 - (3*5) + 9vB2

15+15 = 12vB2

vB2 = 2.5m/s