Motion of masses connected by a string over a pulley

Click For Summary

Homework Help Overview

The discussion revolves around the motion of two particles connected by a string over a pulley, specifically focusing on the dynamics of the system as one particle falls and impacts the ground. The problem involves concepts from mechanics, including kinematics and forces acting on the masses.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the acceleration of the system and the resulting speeds of the particles at various points in the motion. There are attempts to apply kinematic equations to find speeds and distances traveled.
  • Some participants question the assumptions made about momentum conservation during the interaction between the masses when the string becomes taut.
  • There is discussion about the impulse experienced by the falling mass and the implications of the system's behavior when one mass hits the ground.
  • Participants express uncertainty about the effects of the string's tension and the nature of the interaction between the two masses.

Discussion Status

The discussion is ongoing, with various interpretations being explored regarding the mechanics of the system. Some participants have offered insights into the application of momentum conservation, while others are clarifying the implications of the string's inextensibility and the resulting forces. There is no explicit consensus yet, but the dialogue is productive.

Contextual Notes

Participants are navigating the complexities of the problem, including the effects of external forces and the definitions of momentum exchange in the context of the system's dynamics. The original poster's assumptions and the problem's constraints are under examination.

  • #31
ii) Impulse on ground when mass B impacts: 2mΔv ?
 
Physics news on Phys.org
  • #32
neilparker62 said:
ii) Impulse on ground when mass B impacts: 2mΔv ?

Yes, sorry. The mass should be 2m.For (v) :
Distance of B = twice the distance when it accelerates upwards + decelerate to stop (when A hits the ground)

Speed of B when A hits the ground
v^2 = (\frac{1}{3} \sqrt{\frac{2gh}{3}})^2 + 2 . \frac{g}{3} . 2h
v^2 = \frac{38gh}{27}

Distance traveled by B when decelerating:
v^2 = u^2 - 2gd
0 = \frac{38gh}{27} - 2gd
d = \frac{19h}{27}

So total distance
= 4h + 2 . \frac{19h}{27}
= \frac{146h}{27}

For (vi), I think I need to use formula of sum to infinity of geometric progression but I can not find the ratio. The distance traveled by B:
a. when released from rest until hits the ground = h
b. between first hit and second hit = \frac{146h}{27}
c. between second hit and third hit (repeating the same process as (v)) = \frac{1298h}{243}

How to find the total distance?

Thanks
 
Last edited:

Similar threads

Replies
34
Views
3K
Replies
40
Views
3K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 38 ·
2
Replies
38
Views
2K
  • · Replies 1 ·
Replies
1
Views
991
Replies
24
Views
4K
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
15
Views
2K