Is the Calculation of Height in a Pulley System Accurate?

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The discussion centers on a physics problem involving a mass system connected by a pulley, where a mass m falls onto a stationary scale pan of mass m, affecting the motion of a mass of 3m. The original poster calculated that the mass of 3m rises to a height of 4h/5, but the expected answer is h/5. Key points include the need to correctly account for the momentum and forces acting on the system, particularly distinguishing between the moving and stationary masses during the collision. Clarifications were provided regarding the total mass involved in the calculations, emphasizing that only the falling mass m was initially moving. The conversation highlights the importance of accurately applying physics principles to solve problems involving pulleys and mass interactions.
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Homework Statement


An object of mass 3m is connected by means of a light, inextensible string to a scale pan of mass m. The object rests on the ground. The string passes over a smooth pulley and the scale pan hangs suspended. An object of mass m, falling from rest a distance h above the scale pan, lands on it and does not bounce. Show that the mass of 3m will rise to a height h/5.

Sorry but there's no diagram given! The answer I got was 4h/5 instead! :(

Homework Equations





The Attempt at a Solution


To find the speed of the scale pan before the string becomes taut:
mgh=0.5u^2
u=√(2gh)

To find the speed of the scale pan/speed at which the 3m object jerks in motion:
Take upwards direction as negative:
-I = 2mv-2m√(2gh) (this is the change in momentum of the scale pan)
-I = -3mv

Equate both to get:
-3mv=2mv-2m√(2gh)
-5mv=-2m√(2gh)
v=2√(2gh)/5

This speed, v is the initial speed of the upward motion of the 3m mass.
u=2√(2gh)/5
v=0 (mass of 3m rise to a height of h/5 and stop)
To find acceleration, a:

For the 3m mass (it's going upwards):
T-3mg=3ma
For the scale pan+object of mass m (going downwards):
2mg-T=2ma

Equate both equations:
T=3ma+3mg
T=2mg-2ma
3ma+3mg=2mg-2ma
5ma=-mg
5a=-g
a=-g/5

Using an equation of linear motion:
v^2=u^2+2as
0=(2√(2gh)/5)^2 + 2(-g/5)s
s=4h/5

Did I do something wrong in the middle? Was I on the right track in the first place? Thanks! :D :smile:
 
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hi jiayingsim123! :smile:

(try using the X2 button just above the Reply box :wink:)
jiayingsim123 said:
-I = 2mv-2m√(2gh) (this is the change in momentum of the scale pan)

no, the 2m√(2gh) is wrong …

only the mass m was moving, the pan was stationary! :redface:

(btw, i find this sort of problem easier if i pretend everything is in a straight line …

in this case, a mass m collides with a stationary mass 4m :wink:)​
 
Hi tiny-tim, thanks for helping out again!
But I thought the mass m drops onto the scale pan, meaning both of it will move downwards as a single entity and therefore the total mass is 2m? So isn't it a mass of 2m colliding with a stationary mass of 3m?
 
jiayingsim123 said:
… isn't it a mass of 2m colliding with a stationary mass of 3m?

no!

there's 4m of stationary mass, isn't there?? :smile:

(you're being confused by the fact that the 4m is in two parts)

only a mass of m was originally moving, and 4m wasn't :wink:
 
Okay thanks for clearing that up, tiny-tim! You're really awesome! :)
 
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