Elastic Collision on a Frictionless Track

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SUMMARY

The discussion focuses on calculating the maximum height to which a block of mass M1 (5.00 kg) rises after an elastic collision with a stationary block of mass m2 (10 kg) on a frictionless track. The key equations involved are the conservation of momentum and kinetic energy: p1 + p2 = p1' + p2' and 0.5mv2 + 0.5mv2 = 0.5mv2' + 0.5mv2+. The initial attempt yielded a height of 5 meters, which was deemed incorrect. The correct approach involves using the conservation laws to determine the new velocity of M1 post-collision, leading to the accurate calculation of the maximum height.

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komarxian
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Homework Statement


Consider a frictionless track as shown (I will attach an image in a follow up post). A block of mass M1= 5.00 kg is released from point A. It makes a head-on elastic collision at B with a block of mass m2 = 10 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

Homework Equations


p1+p2 = p1' + p2'
.5mv2+.5mv2=.5mv2'+.5mv2+

The Attempt at a Solution


Okay, so I tried mixing the Kinetic energy with some momentum equations and also potential energy because KE=-U, but I feel like I probably totally screwed up, and when I tried to use kinematics it didn't work out very well.
The answer I got was 5 meters, but I do not know what the correct answer is. (I will post an image with my work as well, because personally I find it hard to read typed math and prefer pictures, and someone else might too) Thank you so much to anyone who can help!
 
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ImageUploadedByPhysics Forums1453425941.973192.jpg

The top is the image that goes with the problem. Below is my work.
ImageUploadedByPhysics Forums1453425995.806060.jpg
 
Clearly 5m must be too much.
Looks like you found the collision velocity to be 10m/s, which is right if you are taking g as 10m/s2.
Couldn't follow your working after that. You need to use conservation of momentum and of energy to find the new velocity of the first mass.
Or, if you have been taught it, you can use "Newton's experimental law" with R=1. This can be derived from those two conservation equations.
 

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