A 0.9-kg block (mass m) and a second block (mass M) are both initially at rest on a frictionless inclined plane. Mass M rests against a spring that has a force constant of 12.2 kN/m. The distance d along the plane between the two blocks is 3.64 m. The 0.9-kg block is released, making an elastic collision with the larger block. The 1.00-kg block then rebounds a distance of 2.72 m back up the inclined plane. The block of mass M momentarily comes to rest 4.20 cm from its initial position. Find M.

To solve this problem, we need to apply the principles of conservation of momentum and conservation of energy.

1. Conservation of Momentum:
Before the collision, both blocks are initially at rest, so the initial momentum is zero. After the collision, the smaller block rebounds and moves in the opposite direction, while the larger block moves forward. The total momentum before and after the collision should still be zero, as no external forces act on the system.

Let's assign velocities to the smaller block and the larger block after the collision as v1 and v2 respectively. Using the conservation of momentum equation:

Initial momentum = Final momentum
0 = (0.9 kg)(-v1) + M(v2)

We can rearrange the equation to solve for v2 in terms of v1:
v2 = (0.9 kg)(v1) / M ---- Equation (1)

2. Conservation of Energy:
The smaller block rebounds a distance of 2.72 m back up the inclined plane. We can use the conservation of energy equation to determine the velocity of the smaller block just before it rebounds.

The potential energy gained by the smaller block is given by:
Potential Energy = (0.9 kg)(9.8 m/s^2)(2.72 m) = 23.0592 J

This potential energy is converted into kinetic energy when the block reaches the bottom of the inclined plane:
Kinetic Energy = (1/2)(0.9 kg)(v1^2)

Equating the potential energy to the kinetic energy:
23.0592 J = (1/2)(0.9 kg)(v1^2)
v1^2 = (2 * 23.0592 J) / 0.9 kg
v1 = sqrt((2 * 23.0592 J) / 0.9 kg) ---- Equation (2)

3. Find M:
We can substitute the value of v1 from equation (2) into equation (1) and solve for M:

v2 = (0.9 kg)(v1) / M
v2 = (0.9 kg)(sqrt((2 * 23.0592 J) / 0.9 kg)) / M
v2 = sqrt(2 * 23.0592 J / 0.9 kg) / M

We also know that the larger block momentarily comes to rest 4.20 cm from its initial position. We can use this information to find the velocity v2 using the equation for the potential energy of the spring:

Potential Energy = (1/2)(12.2 kN/m)(0.042 m)^2
Potential Energy = 0.0121 kJ = (1/2)(M)(v2^2)

Solving for v2:
v2^2 = (2 * 0.0121 kJ) / M
v2 = sqrt((2 * 0.0121 kJ) / M) ---- Equation (3)

Now we can equate the values of v2 from equation (1) and equation (3):

sqrt(2 * 23.0592 J / 0.9 kg) / M = sqrt((2 * 0.0121 kJ) / M)

Squaring both sides and rearranging, we get:

2 * 23.0592 J / 0.9 kg = 2 * 0.0121 kJ / M^2
23.0592 J / 0.9 kg = 0.0121 kJ / M^2
M^2 = (0.0121 kJ / (23.0592 J / 0.9 kg))
M^2 = (0.0121 kJ * 0.9 kg) / 23.0592 J
M^2 = 0.011289 kJ kg / J
M = sqrt(0.011289 kJ kg / J)

Now we can calculate the value of M using the above equation.