Dynamics of a particle moving in a straight line

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SUMMARY

The discussion centers on the dynamics of two particles, P and Q, with masses 2m and m respectively, connected by a string over a pulley. When released from a height of 2m, particle Q reaches a maximum height of 4.66m above the ground, but due to deceleration caused by tension, it ultimately achieves a height of 2.66m above its starting position. The calculations involve analyzing the forces acting on Q during its upward motion, factoring in gravitational acceleration and tension in the string.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Basic principles of mechanics involving pulleys
  • Knowledge of gravitational acceleration (9.81 m/s²)
  • Familiarity with kinematic equations for motion
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  • Study the principles of energy conservation in mechanical systems
  • Learn about the dynamics of connected particles in motion
  • Explore advanced kinematics involving acceleration and deceleration
  • Investigate the effects of tension in inextensible strings
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of particle motion in systems involving pulleys and tension.

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Hi all,

Sorry if I have posted in the wrong place. I am not to familiar with the forum titles and new setting. I have an argument and this is not homework.

Suppose two particles P and Q of masses 2m and m respectivily are attached to the ends of a light inextensible string which passes over a smooth fixed pulley. They both hang at a distance of 2m above the horizontal ground. The system is released from rest.

Given that particle Q does not reach the pulley, Find the greatest height that Q reaches above the ground.

I know how to find final velocities, decelerations etc through the system for each particle. I have got a total distance of 4.66m, however the answer is 2.66 which I am thinking is wrong.

Since both particles hung 2m above ground Q will be at a distance of 4m as soon as P hits the ground and then continue a small distance of 0.665m during the deceleration span due to the tension in the string.

Am I right.

Thank you.
 
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Your answer sounds correct to me.

As you point out, particle P hits the ground when particle Q is 2 meters above its starting position. That puts Q 4 meters above ground.

At that point Q had been accelerating through 2 meters with an upward acceleration of 1/3 gravity. It continued upward with a downward acceleration of 1 gravity. Equating the work done on particle Q in the two phases and solving, you get 2/3 of a meter of additional travel for a total of 2.66 meters travel ending 4.66 meters above ground.
 

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