A 3.0- block slides along a frictionless tabletop at 8.0 toward a second block (

Question

A 3.0- block slides along a frictionless tabletop at 8.0 toward a second block (at rest) of mass 4.5 . A coil spring, which obeys Hooke's law and has spring constant = 900 , is attached to the second block in such a way that it will be compressed when struck by the moving block.

What will be the final velocities of the blocks after the collision?

A 3.0- block slides along a frictionless tabletop at 8.0 toward a second block (at rest) of mass 4.5 . A coil spring, which obeys Hooke's law and has spring constant = 900 , is attached to the second block in such a way that it will be compressed when struck by the moving block. What will be the final velocities of the blocks after the collision?

Explanation / Answer

The maximum compression is when all the initial kinetic energy is converted to potential energy. KE is (1/2)mv^2 PE of compression is (1/2)kx^2. So you need to set these equal. What you need to know is v, the speed that the 2nd block is going to be moving when struck. It sounds to me like this is an elastic collision, though the wording is not completely clear on that point. So first you solve the equations of elastic collision to get the velocity of the 4.5 kg block after the collision. Then you set KE = PE and solve for x.

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