Two blocks sliding and the collision

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SUMMARY

The discussion revolves around a physics problem involving two blocks on a frictionless track, where a 1 kg block (Block 1) collides inelastically with a stationary 5 kg block (Block 2) on a 30-degree incline. The calculations utilize the energy conservation equation and the inelastic collision equation to determine the maximum distance "s" that Block 2 travels up the incline. The correct answer for "s" is established as 0.54 meters, with critical corrections made regarding the velocity calculations, specifically ensuring that the velocity of Block 1 is treated as negative due to its direction.

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Homework Statement



Two blocks are on a frictionless track with a vertical, quarter circle, and 30 degree incline. Released from rest at point A, 1 kg smaller block 1 slides to B and makes an inelastic collision with 5 kg stationary larger block 2. The collision sents block 2 up the ramp. The smaller block slides back up along the track after collision reach highest point C. If angle locates point C is 60 degrees, find max distance "s" block 2 travel on the incline.
[PLAIN]http://img704.imageshack.us/img704/752/picturefv.jpg

Homework Equations


Energy equation : K1+ Ug1+ UE1+Work= K2+Ug2+UE2
Inelastic collision equation: m1v1+ m2v2= m1v1' + m2v2'


The Attempt at a Solution



1. I found the velocity for slider 1 BEFORE collision using energy equation.
2. I then use the inelastic collision equation and found V'1(at B) + 5V'2(at B) = 7.668. Which is the equation for velocity of slider 1 and 2 at point B after they collide.
3. Then I used energy equation again for B to C motion by slider 1 AFTER collision. I found V'1( at B) = 14.7. I plug this in the step 2 result and found V'2 (at B), which is the velocity of slider 2 at Point B AFTER collision.
4. Then I used energy equation for motion B to D for slider 2 AFTER collision. D being furthest part on the incline slider 2 travels on. Here I find "s" being 0.202 meters.
However, the answer is 0.54 meters, I am not sure what I am doing wrong, I doubled checked my steps and my calculations, too.

Hopefully someone could help, thanks!
 
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I tried the question and I get 0.54m as well. Exactly actually. I think your mistake is here "V'1( at B) = 14.7". I get sqrt(1.5g) = 3.834. Did you forget to square root it here? Also, don't forget V1 should be plugged in as negative since it's moving to the left.
 
jdc15 said:
I tried the question and I get 0.54m as well. Exactly actually. I think your mistake is here "V'1( at B) = 14.7". I get sqrt(1.5g) = 3.834. Did you forget to square root it here? Also, don't forget V1 should be plugged in as negative since it's moving to the left.

I tried it again with cleaner procedures and found that it should be sqrt 14.7. However, I forgot to make V1 negative this time. Good thing I checked this and with your help I changed V1 to negative and guess what, I finally found the correct answer.
Thank you so much! This was the only one I'm stuck on for my test tomorrow and now I am all set!
 

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