- #1
FaraDazed
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Homework Statement
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 traveling 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!
Homework Equations
Conservation of linear momentum and conservation of mechanical energy
The Attempt at a Solution
First I set the problem up using conservation of mechanical energy
[itex]
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
[/itex]
Then to find what the velocity on the right hand side of the equation I used the conservation of linear momentum
[itex]
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
[/itex]
Then used that in the first equation, of conservation of mechanical energy
[itex]
\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 \\
[/itex]
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.
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