Inelastic collision followed by circular motion

Thread starter Sal Coombs
Start date Nov 15, 2022
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Circular Circular motion Collision Inelastic Inelastic collision Motion

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After the inelastic collision, the masses travel at a speed of 2.25 m/s. The discussion explores the subsequent motion of these masses on a circular track. Three distinct possibilities for their behavior are proposed, although not explicitly listed in the thread. The focus remains on understanding the dynamics following the collision and the implications for circular motion. Further clarification on these possibilities is sought to deepen the analysis.

Nov 15, 2022

Sal Coombs

Homework Statement: A 3.0-kg mass is sliding on a horizontal frictionless surface with a speed of 3.0 m/s when it collides with a 1.0-kg mass initially at the bottom of a circular track. The masses stick together and slide up a frictionless circular track of radius 0.40 m. To what maximum height, h, above the horizontal surface (the original height of the masses) will the masses slide?

Relevant Equations: mv = mv Momentum
1/2mv^2 Kinetic Energy
mgh Potential Energy
(mv^2)/r Centripetal Force

Found the speed at which the masses will travel after their collision: 2.25m/s
Not sure what to do next...

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Nov 16, 2022

PeroK

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Sal Coombs said:

Found the speed at which the masses will travel after their collision: 2.25m/s
Not sure what to do next...

What happens when the masses follow the circular track?

Nov 16, 2022

haruspex

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To expand on @PeroK's question, there are three distinct possibilities. Can you list them?

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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