Maximum spring compression and final velocities of two blocks colliding with spring

[physicsss]

A 2.0 kg block slides along a frictionless tabletop at 6.0 m/s toward a second block (at rest) of mass 4.5 kg. A coil spring, which obeys Hooke's law and has spring constant k = 850 N/m, is attached to the second block in such a way that it will be compressed when struck by the moving block.

(a) What will be the maximum compression of the spring?

(b) What will be the final velocities of the blocks after the collision?
(Assume the initial direction of the 2.7 kg block is positive.)


(c) Is the collision elastic? (Compare energy values to the nearest joule.)

I don't know how I should treat the spring attached to the mass...and does the 4.5kg move when the first block hits it?

 

[Tide]

You can use energy and momentum conservation to solve this problem.

Initially, the kinetic energy and momentum are all carried by the first block. When contact is made via the spring, both blocks will have momentum and kinetic energy PLUS they will have potential energy due to the spring. In the energy conservation equation you can eliminate one of the speeds by using the momentum conservation relation.

This gives a relation between the amount of compression and the speed of one of the blocks (it's quadratic!) and you can determine the maximum compression either with calculus (if you've had that) or by using the properties of quadratic equations (parabolas).

Actually, you use the energy conservation equation to find the speed that gives the minimum compression and use that result to find the maximum compression.

If you are familiar with the "center of mass frame" I think it would be a lot easier to do the problem in that frame - if not then nevermind! :-)

 

[wais]

(a) To find the maximum compression of the spring, we can use the conservation of energy principle. Initially, the system has only kinetic energy, which is given by KE = 1/2mv^2. The total kinetic energy of the system is then 1/2(2.0 kg)(6.0 m/s)^2 = 36 J. When the first block collides with the second block, the spring will absorb some of this kinetic energy and store it as potential energy. The maximum compression of the spring will occur when all the kinetic energy is converted into potential energy. This can be represented by the equation 1/2kx^2 = 36 J, where k is the spring constant and x is the maximum compression of the spring. Solving for x, we get x = √(2(36 J)/850 N/m) = 0.272 m.

(b) To find the final velocities of the blocks after the collision, we can use the conservation of momentum principle. Initially, the total momentum of the system is 2.0 kg x 6.0 m/s = 12 kg m/s. After the collision, the total momentum will still be conserved, but it will be divided between the two blocks. This can be represented by the equation m1v1 + m2v2 = m1v1' + m2v2', where m1 and m2 are the masses of the two blocks, v1 and v2 are their initial velocities, and v1' and v2' are their final velocities. Plugging in the values, we get (2.0 kg)(6.0 m/s) + (4.5 kg)(0 m/s) = (2.0 kg)v1' + (4.5 kg)v2'. Solving for v1' and v2', we get v1' = 2.25 m/s and v2' = 1.33 m/s. Therefore, the final velocities of the blocks after the collision are 2.25 m/s for the first block and 1.33 m/s for the second block.

(c) To determine if the collision is elastic, we can compare the initial kinetic energy with the final kinetic energy. The initial kinetic energy was 36 J, and the final kinetic energy is (1/2)(2.0 kg)(2.25 m/s)^2 + (1/