# Maximum compression of spring

An object with mass m1=0.4kg moves at v0=10 m/s toward a second object with mass m2 = 0.8kg. Attached to the second object is a spring with spring constant k = 200N/m, and natural length L0 = 0.1m. As the objects collide the spring is initially compressed. After compressing a maximum amount the spring then decompresses and the masses move apart. At the instant that the spring is compressed its maximum amount, the masses move with the same speed V. Determine the closest distant X, between the objects at this instant. Both momentum and mech energy are conserved.

M1=0.4kg
M2=0.8kg
V1=10 m/s
V2=0 (at rest)
K= 200N/m
L0=0.1m
X=?

2. Homework Equations
Vf=(m1v1+m2v2)/(m1+m2)

SPE=0.5Kx^2

KE=0.5mV^2

X=L-L0

3. The Attempt at a Solution
Vf=(.4)(10)/(1.2)=3.333

Alright so it's asking me maximum compression of the spring. and since they travel at the same speed later on I can count on them to have the masses combined at that moment right?

I used 1/2*m1v1^2=1/2*(m1+m2)*V^2 + 1/2*k*x^2

If i go with this formula, I'll get an X that is higher than the 0.1m of the natural length which doesnt make sense if it's compressing. (how can it compress more than 0.1m)

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PeterO
Homework Helper
An object with mass m1=0.4kg moves at v0=10 m/s toward a second object with mass m2 = 0.8kg. Attached to the second object is a spring with spring constant k = 200N/m, and natural length L0 = 0.1m. As the objects collide the spring is initially compressed. After compressing a maximum amount the spring then decompresses and the masses move apart. At the instant that the spring is compressed its maximum amount, the masses move with the same speed V. Determine the closest distant X, between the objects at this instant. Both momentum and mech energy are conserved.

M1=0.4kg
M2=0.8kg
V1=10 m/s
V2=0 (at rest)
K= 200N/m
L0=0.1m
X=?

2. Homework Equations
Vf=(m1v1+m2v2)/(m1+m2)

SPE=0.5Kx^2

KE=0.5mV^2

X=L-L0

3. The Attempt at a Solution
Vf=(.4)(10)/(1.2)=3.333

Alright so it's asking me maximum compression of the spring. and since they travel at the same speed later on I can count on them to have the masses combined at that moment right?

I used 1/2*m1v1^2=1/2*(m1+m2)*V^2 + 1/2*k*x^2

If i go with this formula, I'll get an X that is higher than the 0.1m of the natural length which doesnt make sense if it's compressing. (how can it compress more than 0.1m)

It can't compress that far, so the initial set-up is not plausible. I agree the "theoretical answer" is greater than 0.1 m.

10m/s is really fast for an object moving in the lab [you have to drop a brick 5m to get up to speeds like that]. I am sure that is mass 1 was only travelling at 1.0 m/s you would get both a plausible answer, and involving speeds and masses you might find in the lab.

Yeah I've confirmed it with someone else. The equation I used is correct. Either the spring constant is too small or/and the natural length is too small. The maximum the spring can absorb is even less than 1 J which leaves more than 19J left for the 2 masses KE to fill and can't be filled with 3.33 m/s alone for those two masses.

It's physically impossible and my teacher made a mistake IM SURE OF IT. It'll go to an answer greater than 0.1m. Which physically means there is no distance between them and the objects would be moving together with no space between them (meaning about roughly 5m/s for the two objects).

CWatters