Final answer:
To find the velocity of block B when the energy stored in the spring bumpers is maximum, we can use the principle of conservation of momentum and energy. The velocity of block B is found to be 2.52 m/s.
Step-by-step explanation:
To find the velocity of block B when the energy stored in the spring bumpers is maximum, we can use the principle of conservation of momentum and energy. Since the collision is head-on and the surface is frictionless, the total momentum before the collision is equal to the total momentum after the collision.
Using the equation:
mAvAi + mBvBi = mAvAf + mBvBf
where: mA = 3.00 kg (mass of block A), mB = 9.00 kg (mass of block B), vAi = 5.00 m/s (initial velocity of block A), vBi = 0 m/s (initial velocity of block B), vAf = ? (final velocity of block A), vBf = ? (final velocity of block B)
Substituting the given values:
(3.00 kg)(5.00 m/s) + (9.00 kg)(0 m/s) = (3.00 kg)(vAf) + (9.00 kg)(vBf)
15.00 kg·m/s = 3.00 kg·vAf + 9.00 kg·vBf
Since the collision is elastic and the blocks have ideal spring bumpers, the total kinetic energy before the collision is equal to the total kinetic energy after the collision.
Using the equation:
0.5·mA·(vAi)^2 + 0.5·mB·(vBi)^2 = 0.5·mA·(vAf)^2 + 0.5·mB·(vBf)^2
Substituting the given values:
0.5·(3.00 kg)·(5.00 m/s)^2 + 0.5·(9.00 kg)·(0 m/s)^2 = 0.5·(3.00 kg)·(vAf)^2 + 0.5·(9.00 kg)·(vBf)^2
37.50 J = 4.50 J + 0.5·(9.00 kg)·(vBf)^2
33.00 J = 4.50 J + 4.50·(vBf)^2
28.50 J = 4.50·(vBf)^2
6.33 m2/s2 = (vBf)^2
vBf = ±2.52 m/s
Thus, the velocity of block B when the energy stored in the spring bumpers is maximum is 2.52 m/s.