A solid block of mass m2 = 1.0 $kg$, at rest on a horizontal frictionless surface, is connected to a relaxed spring (with spring constant k = 260 N/m whose other end is fixed. Another solid block of mass m1 = 2.4 kg and speed v1 = 3.3 m/s collides with the 1.00 kg block. If the blocks stick together, what is their speed immediately after the collision?

What is the maximum compression of the spring?

To solve this problem, we can use the principle of conservation of momentum and the relationship between the maximum compression of a spring and the kinetic energy.

1. Conservation of momentum:
Before the collision, only the second block is stationary, so the momentum of the first block is given by:
m1 * v1 = (2.4 kg) * (3.3 m/s) = 7.92 kg·m/s

After the collision, the two blocks stick together, so their total momentum is conserved:
m1 * v1 + m2 * v2 = (m1 + m2) * vf

where vf is the final velocity of the two blocks stuck together.

2. Finding the final velocity:
Rearranging the equation above, we can solve for vf:
v2 = [(m1 + m2) * vf - m1 * v1] / m2

Plugging in the values:
v2 = [(2.4 kg + 1.0 kg) * vf - 2.4 kg * 3.3 m/s] / 1.0 kg
v2 = (3.4 kg * vf - 7.92 kg·m/s) / 1.0 kg

Since the two blocks stick together, their final velocity is the same, so we can rewrite the equation as:
v2 = (4.4 kg * vf - 7.92 kg·m/s) / 1.0 kg

3. Maximum compression of the spring:
To find the maximum compression of the spring, we need to relate it to the kinetic energy.

The initial kinetic energy of the two-block system is given by the equation:
KE_initial = (1/2) * m1 * v1^2

After the collision, the two blocks stick together and move with a final velocity vf. The total kinetic energy of the system is then given by:
KE_final = (1/2) * (m1 + m2) * vf^2

Since energy is conserved, we can equate the initial and final kinetic energies:
KE_initial = KE_final
(1/2) * m1 * v1^2 = (1/2) * (m1 + m2) * vf^2

We can solve this equation for vf:
vf^2 = (m1 * v1^2) / (m1 + m2)

Plugging in the values:
vf^2 = (2.4 kg * (3.3 m/s)^2) / (2.4 kg + 1.0 kg)
vf^2 = (2.4 kg * 10.89 m^2/s^2) / 3.4 kg
vf^2 = 7.7 m^2/s^2

Taking the square root of both sides gives us the final velocity vf:
vf = √(7.7 m^2/s^2)
vf ≈ 2.77 m/s

Now that we have the final velocity (vf), we can substitute it back into the equation from step 2 to find v2, the velocity immediately after the collision:
v2 = (4.4 kg * vf - 7.92 kg·m/s) / 1.0 kg
v2 = (4.4 kg * 2.77 m/s - 7.92 kg·m/s) / 1.0 kg
v2 ≈ 9.17 m/s

Therefore, the speed of the blocks immediately after the collision is approximately 9.17 m/s.

To find the maximum compression of the spring, we can use the formula for the potential energy stored in a spring:
PE_spring = (1/2) * k * x^2

where k is the spring constant and x is the maximum compression of the spring.

Since the kinetic energy is converted into potential energy in the spring, we can equate the initial kinetic energy of the two-block system to the potential energy stored in the spring at maximum compression:
KE_initial = PE_spring

Substituting in the variables:
(1/2) * m1 * v1^2 = (1/2) * k * x^2

Simplifying, we get:
x^2 = (m1 * v1^2) / k
x^2 = (2.4 kg * (3.3 m/s)^2) / 260 N/m
x^2 = (2.4 kg * 10.89 m^2/s^2) / 260 N/m
x^2 ≈ 0.10 m^2

Taking the square root of both sides gives us the maximum compression of the spring:
x = √(0.10 m^2)
x ≈ 0.32 m

Therefore, the maximum compression of the spring is approximately 0.32 m.