Two blocks, each with a mass m = 0.400 kg, can slide without friction on a horizontal surface. Initially, block 1 is in motion with a speed $v$ = 1.20 m/s; block 2 is at rest. When block 1 collides with block 2, a spring bumper on block 1 is compressed. Maximum compression of the spring occurs when the two blocks move with the same speed, v/2 = 0.600 m/s. If the maximum compression of the spring is 1.77 cm, what is its force constant?

To find the force constant of the spring, we can use the principle of conservation of mechanical energy.

1. First, let's find the initial kinetic energy of block 1 before the collision.
The formula for kinetic energy is KE = (1/2)mv^2, where m is the mass and v is the velocity.
Given that the mass of block 1 is 0.400 kg and its velocity is 1.20 m/s, the initial kinetic energy is:
KE1 = (1/2)(0.400 kg)(1.20 m/s)^2.

2. Next, let's find the final kinetic energy of the system when the two blocks move with a common speed of v/2 = 0.600 m/s.
Since the blocks move with the same speed, the final kinetic energy can be calculated as:
KE2 = (1/2)(2m)(0.600 m/s)^2, where 2m represents the total mass of the system (both blocks).

3. The work done by the spring is equal to the change in kinetic energy between the initial and final states of the system.
Therefore, the work done by the spring is:
W = KE2 - KE1.

4. The work done by the spring is given by the formula W = (1/2)kx^2, where k is the force constant of the spring and x is the maximum compression.
Given that W is equal to KE2 - KE1 and the maximum compression x is 1.77 cm (0.0177 m), we can set up the following equation:
(1/2)k(0.0177 m)^2 = KE2 - KE1.

5. Now, we can solve the equation to find the force constant k.
Rearrange the equation as follows:
k = (2/m)(KE2 - KE1) / (0.0177 m)^2.

6. Substitute the given values and calculate the force constant.
Substitute the values into the equation and calculate:
k = (2/0.400 kg)(KE2 - KE1) / (0.0177 m)^2.

By plugging in the values into the equation, you will be able to find the force constant k.