# Collision problem involving a mass moving at speed toward another stationary mass attached to spring

1. Oct 21, 2014

1. The problem statement, all variables and given/known data
Block 2 with mass 2kg is at rest on a surface an touching the end of relaxed spring with spring constant k=151 N/m . The other end of the spring is attached to a wall. Block 1 with a mass of 1.8kg and travelling with a speed of v=4.5m/s, collides with block 2, and the two blocks stick together.

When the blocks stop momentarily, what is the distance the spring is compressed?

See diagram --> http://s17.postimg.org/ry9qgk0sf/diagram22.jpg

EDIT: Sorry, forgot to mention the floors are frictionless!

2. Relevant equations
Conservation of linear momentum and conservation of mechanical energy

3. The attempt at a solution
First I set the problem up using conservation of mechanical energy

$KE_1 = U_s + KE_2 \\ \frac{1}{2} m_1 v_i^2 = \frac{1}{2} k x^2 + \frac{1}{2} (m_1 + m_2) v_f^2$

Then to find what the velocity on the right hand side of the equation I used the conservation of linear momentum

$m_1 v_i = (m_1 + m_2) v_f \\ v_f = \frac{m_1 v_i }{(m_1 + m_2)} \\ v_f = \frac{(1.8)(4.5)}{1.8+2} = 2.13 m/s$

Then used that in the first equation, of conservation of mechanical energy
$\frac{1}{2} m_1 v_i^2 = \frac{1}{2} k x^2 + \frac{1}{2} (m_1 + m_2) v_f^2 \\ \frac{1}{2} m_1 v_i^2 - \frac{1}{2} (m_1 + m_2) v_f^2 = \frac{1}{2} k x^2 \\ x=\sqrt{\frac{\frac{1}{2} m_1 v_i^2 - \frac{1}{2} (m_1 + m_2) v_f^2}{0.5k}} \\ x=\sqrt{\frac{\frac{1}{2} (1.8) (4.5)^2 - \frac{1}{2} (1.8+2) (2.13)^2}{(0.5)(151}} = 0.36m \\$

I am very unsure of my solution as I missed the two lectures covering the material this coursework is based on, so would appreciate any advice/feedback.

Last edited: Oct 21, 2014
2. Oct 21, 2014

### Staff: Mentor

Two blocks sticking together after collision means you have a totally inelastic collision. Note that kinetic energy is NOT conserved in inelastic collisions! So you need to "forget" anything you knew about the initial KE previous to the collision.

What you have, post collision, is the new KE that the combined blocks are carrying thanks to conservation of momentum allowing you to determine their post-collision speed. That's what you want to take into a conservation of energy scenario as the spring is compressed.

3. Oct 21, 2014

Ah right, ok, thanks for your help.

So going off that I calculated the velocity of the combined blocks correctly using the conservation of momentum does that mean then I can just do...
$\frac{1}{2}(m_1+m_2)v_f^2=\frac{1}{2}kx^2 \\ (m_1+m_2)v_f^2=kx^2 \\ x = \sqrt{\frac{(m_1+m_2)v_f^2}{k}}=\sqrt{\frac{(1.8+2)(2.13)^2}{151}}=0.34m$

?

Thanks :)

Last edited: Oct 21, 2014
4. Oct 21, 2014

### Staff: Mentor

Yup. Looks good.

5. Oct 22, 2014