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Blocks (mass 2.00 kg) and (mass 6.00 kg) move on a frictionless, horizontal surface. initially, block is at rest and block is moving toward it at the blocks are equipped with ideal spring bumpers, as in example 8.10 (section 8.4). the collision is head-on, so all motion before and after the collision is along a straight line.

(a) find the maximum energy stored in the spring bumpers and the velocity of each block at that time.
(b) find the velocity of each block after they have moved apart.

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Final answer:

To calculate the maximum energy stored in spring bumpers and the velocities at that time, conservation of momentum is used to find the common velocity at maximum compression. Post-separation velocities following an elastic collision are determined using kinetic energy and momentum conservation equations.

Step-by-step explanation:

The student's question appears to be about calculating the maximum energy stored in spring bumpers during a collision between two masses, and determining the velocities of the masses both at the moment when the energy is maximum and after they have separated post-collision, in a scenario where momentum and energy conservation principles are applied.

Maximum Energy Stored and Velocities at Maximum Energy

The maximum energy stored in the spring bumpers occurs when the relative velocity between the two blocks is zero. This happens at the moment of maximum compression of the spring when both blocks momentarily move together with the same velocity due to the conservation of momentum. The velocity of each block at this time can be found using the equation mAvA + mBvB = (mA + mB)vcommon, where vcommon is the common velocity at maximum compression. The energy stored can then be calculated using the kinetic energy formula ½(mA + mB)vcommon².

Velocities After Separation

After the blocks have moved apart, assuming the collision to be elastic, the velocities can be determined using the conservation of kinetic energy and conservation of momentum. The final velocities can be found using the equations mAvA² + mBvB² = mAv'A² + mBv'B² and mAvA + mBvB = mAv'A + mBv'B, allowing for the calculation of the post-collision velocities v'A and v'B.

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