inelastic collision


by r16
Tags: collision, inelastic
r16
r16 is offline
#1
Apr24-07, 04:06 AM
P: 42
1. The problem statement, all variables and given/known data
A block with mass [itex]m_1 = 2 \textrm{kg}[/itex] slides along a frictionless table with a speed of 10 m/s. Directly in front of it and moving in the same direction is a block of mass [itex] m_2 = 5 \textrm{kg} [/itex] moving at 3 m/s. A massless spring with spring constant k = 1120 N/m is attached to the near side of [itex] m_2 [/itex]. When the blocks collide, what is the maximum compression of the spring? (Hint: At the moment of maximum compression of the spring, the two blocks move as one. Find the velocity by noting that the collision is completely inelastic to this point).


2. Relevant equations



3. The attempt at a solution
I tried doing this problem 2 ways, my way and then utilizing the hint (i got the same answer)

When they collide, block 1 will do work on block 2 until their kinetic energies are equal, and at this point the spring will be at it's maximum compression. So:

[tex] W = -\Delta U = \int^{x_f}_{x_0} \mathbf{F} \cdot dx = 1/2kx^2 [/tex]

[tex]KE_1 = (.5)(2)(10^2) = 100 \textrm{J}[/tex]
[tex]KE_2 = (.5)(5)(3^2) = 22.5 \textrm{J}[/tex]

Energy available for block 1 to do work on the spring:
[tex] KE_1 - KE_2 = \Delta U = 100 - 22.5 = 77.5 \textrm{J} = 1/2kx^2[/tex]
[tex] x= \sqrt{\frac{2U}{k}} = \sqrt{\frac{(2)(77.5)}{1120}} = .37\textrm{m} [/tex]

The book said the answer was .25m

So I went on to do it the second way, suggested by the hint:

[tex] m_1v_1 + m_2v_2 = (m_1 + m_2)v_3 [/tex]
[tex] v_3 = \frac{m_1v_1 + m_2v_2}{m_1 + m_2} = \frac{(2)(10) + (5)(3)}{2 + 5} = 3.5 \textrm{m/s}[/tex]

Now find the kinetic energy when at the instant when both blocks are moving as a unit:

[tex] KE= 1/2mv^2 = (.5)(2+5)(3.5^2) = 43 \textrm{J} [/tex]

Now energy is conserved both before and after the collision so:

[tex]E_f = KE_{\textrm{2 block system}} + U_{\textrm{spring}}[/tex]
[tex]E_i = KE_1 + KE_2 [/tex]
[tex] \Delta E = 0 = E_f - E_i [/tex]
[tex] E_f = E_i [/tex]
[tex] 43 + U = 100 + 22.5 [/tex]
[tex] U = 79.5 \textrm{J} =1/2kx^2 [/tex]
[tex]x = \sqrt{\frac{2U}{k}} = \sqrt{\frac{(2)(79.5)}{1120}} = .37 \textrm{m}[/tex]

I've gone over both ways and the math and physics looks pretty solid. Does anyone see an error in my physics / math?
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alphysicist
alphysicist is offline
#2
Apr24-07, 06:28 AM
HW Helper
P: 2,250
When you solved for v3, you divided by 10 instead of 7.
r16
r16 is offline
#3
Apr24-07, 07:20 AM
P: 42
Quote Quote by alphysicist View Post
When you solved for v3, you divided by 10 instead of 7.
Thanks that was it. This brings up a good point, which i've been doing alot lately. I do all of the major mathematics correctly, but I always make some stupid mistake and get the answer incorrect. How do everyone else deal with making those preventable errors?


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