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Physics 17 Online
OpenStudy (anonymous):

I WILL give a Medal for the best response! A 2.00-kg object traveling east at 20.0 m/s collides with a 3.00-kg object traveling west at 10.0 m/s. After the collision, the 2.00-kg object has a velocity 5.00 m/s to the west. How much kinetic energy was lost during the collision? The correct answer is supposed to be 458 J. How can I solve this problem? I already tried using conservation of momentum, and conservation of energy, but I'm just not getting the right answer. Any help would be greatly appreciated.

OpenStudy (anonymous):

Energy is not conserved in such a collision. See if you can find the difference between total kinetic energy of the system before and after the collision has happened.

OpenStudy (anonymous):

I'm still stuck. Any more hints?

OpenStudy (anonymous):

Well, what answer are you getting, and what's the correct answer?

OpenStudy (anonymous):

Also, are you setting up the units correctly with the velocity? |dw:1384737173407:dw|

OpenStudy (anonymous):

The correct answer is supposed to be 458 J I tried so many times and got so many different answers and I don't remember what they were. I feel I'm using the wrong techniques!

OpenStudy (anonymous):

\[\begin{array}{l} v_{mi}=20m/s \\ v_{Mi}=-10m/s \\ \\ v_{mf} = -5m/s \\ v_{Mf} = ? \end{array}\]

OpenStudy (anonymous):

Using conservation of momentum, is the answer for \[v _{Mf}\] 6.67m/s?

OpenStudy (anonymous):

So you first solve for the final velocity of the larger mass, then take the kinetic energies of the two masses initially, and compare that to the kinetic energy of the two masses after the collision. \[ p_i=p_f\] \[mv_{mi}+Mv_{Mi}=mv_{mf}+Mv_{Mf}\] \[v_{Mf} = \frac{mv_{mi}+Mv_{Mi}-mv_{mf}}{M}=\frac{m(v_{mi}-v_{mf})+Mv_{Mi}}{M}\] \[v_{Mf}=\frac{(2kg)(25m/s)+(3kg)(-10m/s)}{3kg}\] \[v_{Mf}\approx 6.67 m/s\] yah

OpenStudy (anonymous):

Thank you sooo much! i was able to solve the question. Turns out I had the wrong sign for \[v _{mf}\]

OpenStudy (anonymous):

Hooray for sciencing with friends!! ^_^ Yeah, setting up your coordinate system in the problem helps tons for keeping track of those pesky signs ^_^ |dw:1384738108300:dw|

OpenStudy (anonymous):

I'll leave it open a little longer in case someone else also wants to give you a metal

OpenStudy (anonymous):

I'm going to post another question shortly. Maybe you can help me with it!

OpenStudy (anonymous):

Fo sho! ^^

OpenStudy (anonymous):

Okay, just posted it!

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