AP Physics Vertical Spring Question

Question

A 5-kilogram block is fastened to an ideal vertical spring that has a 400 N/m spring constant. When the 5kg block is at rest on top of the spring, the spring is compressed to its equilibrium position at a distance of ∆x1 = 12.5 cm from the spring's equilibrium position. The 3.00 kg block is now released from rest 0.750 m above the 5.00 kg block, resulting in an inelastic collision between the two blocks. What is the amount of compression from the unstretched spring length when the two blocks reach the bottom of the oscillation?

I believe the amount of compression includes the 12.5cm from the 5kg block. The answer is in meters, but so far I have only calculated the velocity after the blocks touch. Any help is appreciated!

Answer 1

Let

k = 400 N/m be the spring constant,

m₁ = 5 kg be the mass of the block attached to the spring,

m₂ = 3 kg be the mass of the block that falls down,

Δx₁ = 12.5 cm = 0.125 m be the compression due to the block m₁,

Δx₂ (to be calculated) be the additional compression due to block m₂,

h = 0.750 m be the height from which m₂ falls on m₁,

g be gravitational acceleration.

The statement of the problem is unusual in a couple ways.

First is that all three quantities m₁, k, Δx₁ are given,

although they are related by the spring equation

m₁g = kΔx₁ (1)

But we can use it at least to see what approximation for g you are to use:

g = kΔx₁/m₁ = 400×0.125/5 = 10 m/s²

I am now going to calculate Δx₂, but remember that your answer should be Δx₁+Δx₂.

The solution is based on conservation of energy.

The potential energy of m₁, is converted to kinetic energy;

I assume that this is how you got the velocity after the blocks touch.

Continuing, when the bottom of the oscillation is reached,

the original potential energy got converted to the spring energy.

At the bottom of the oscillation the velocity is 0 and so is the kinetic energy.

It is convenient (but not necessary) to make the potential energy also 0 at the bottom of the oscillation.

For that we will use the bottom of the oscillation as the reference point for potential energy.

Thus at the beginning, the block m₁ has potential energy m₁gΔx₂, and

the block m₂ has potential energy m₂g(Δx₂+h).

There are two ways of solving it.

THE EASIER SOLUTION:

We can treat the combination of the spring together with m₁ as one spring

whose equilibrium position is at Δx₁. (This can be justified formally.)

Under that assumption, the spring energy at the bottom is

kΔx₂²/2

So the conservation of energy statement is

kΔx₂²/2 = m₂g(Δx₂+h) (2)

THE HARDER SOLUTION:

Use the original equilibrium position before compression by m₁.

In that case, the spring energy at the bottom is k(Δx₁+Δx₂)²/2, and

it came from the potential energy of m₁ and m₂, plus the initial spring energy

kΔx₁²/2.

So the conservation of energy statement is

k(Δx₁+Δx₂)²/2 = kΔx₁²/2 + m₂g(Δx₂+h) + m₁gΔx₂

Expand the left side

kΔx₁²/2 + kΔx₁Δx₂ + kΔx₂²/2 = kΔx₁²/2 + m₂g(Δx2+h) + m₁gΔx₂

Cancel equal terms

kΔx₁Δx₂ + kΔx₂²/2 = m₂g(Δx₂+h) + m₁gΔx₂

Apply the substitution (1)

m₁gΔx₂ + kΔx₂²/2 = m₂g(Δx₂+h) + m₁gΔx₂

Cancel equal terms

kΔx₂²/2 = m₂g(Δx₂+h)

which is the the same equation as (2).

Either way we get the quadratic equation

kΔx₂²/2 - m₂gΔx₂ - m₂gh = 0

It has the solutions

Δx₂ = (m₂g ± √((m₂g)² + 2km₂gh))/k = (3×10 ± √((3×10)² + 2×400×3×10×0.75))/400

Δx2 = 0.42 m

Δx2 = -0.27 m

The solution Δx₂ = 0.42 represent the bottom of the oscillation,

and the solution Δx₂ = -0.27 represents the top of the oscillation.

We get both extremes because the spring energy is the same at both extremes.