# Totally inelastic collision

1. Dec 11, 2005

### suspenc3

a block of mass m1 = 2kg slides along a frictionless table with a speed of 10m/s. Directly in front of it, and moving in the same direction is a block of mass m2 = 5kg moving at 3.0m/s. A massless spring with a spring constant of k = 1120N/m is attached to the near side of m2. When the blocks collide what is the max compression of the spring?

Ok, this would be treated as a totally inelastic collision because at the point of max compression, the two masses will be moving together as one, right?

my question is how do I incorporate the spring constant into the forumlas?

Last edited: Dec 11, 2005
2. Dec 11, 2005

### dicerandom

I think you need to use both energy and momentum in this situation. Use conservation of momentum to determine the velocity of the m1-m2 system at the point of maximum compression, then calculate the difference in kinetic and potential energies in the initial and final situations to determine how much potential energy is stored in the spring and thus how far it is compressed.

3. Dec 11, 2005

### suspenc3

yes...That makes sence..heres what I did

$$p_1_i + p_2_i = p_2_f$$
$$m_1v_1_i + m_2v_2_i = (m_1 + m_2)V$$

Solve for V:

$$V = 5m/s$$

and then:

$$K_i + U_i = K_f + U_f$$
$$1/2m_1v_1^2 + 1/2m_2v_2^2 + 0 = 1/2(m_1 + m_2)V^2 + 1/2kx^2$$

solve for x yields 0.05m or 5cm...but in the book it should be 25cm

What did I do wrong?

4. Dec 11, 2005

### dicerandom

Looks like it's just a numerical/algebraic error. Your final velocity is correct and your equation relating the energies is correct, double check your calculation when solving for x.