Homework Help: Two blocks collide with spring in-between.

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1. Apr 11, 2016

Jose Bouza

1. The problem statement, all variables and given/known data
Block 1 (mass 2 kg) is moving rightward at 10 m/s and block 2 (mass 5kg kg) is moving rightward at 3 m/s. The surface is frictionless, and a spring with spring constant of 1120 N/m is fixed on the left side on block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

2. Relevant equations

3. The attempt at a solution
I realize after the fact that this could probably be solved very easily with conservation of energy and/or momentum but I tried a different route and I'm confused as to why it gave me the wrong answer. What I did was use F=ma to set up a differential equation for the velocity of Block 1 in terms of its compression of the block x.
Since $$F_1 = -kx$$ the velocity at any position x is given by the solution to the equation
$$\frac{dv}{dx}m = -kx$$
which is
$$v_1(x) = v_i1 -\frac{kx^2}{2m_1}$$
By newtons third law the velocity of the second block at any position x should thus be
$$v_2(x) = v_i2 +\frac{kx^2}{2m_2}$$
If we set these equal to each other, plug in the relevant variables and solve for x we get a solution x = 0.1336m. When checked against the solution manual this is wrong though, where did I go wrong? Thanks.

2. Apr 11, 2016

BvU

Hello Jose,

I think it goes wrong at the very first $F_1 = -kx$. This is true for a spring with the other end fixed, but here the other end moves, so you have something with $F_1 = -k(x_1 - x_2 - C)$ with C something like $x_1-x_2$ when both blocks touch the end of the spring (without compressing it).

3. Apr 11, 2016

Jose Bouza

Yup you're right. When I add this into my solution I end up with a system of equations that I cant seem to solve for x2-x1 so I guess the energy method is the only one that works here. Thanks anyways.