Is the Calculation of Height in a Pulley System Accurate?

[jiayingsim123]

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

 

[tiny-tim]

hi jiayingsim123!

(try using the X² button just above the Reply box )

-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!

(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 )​

 

[jiayingsim123]

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?

 

[tiny-tim]

… isn't it a mass of 2m colliding with a stationary mass of 3m?

no!

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

(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

 

[jiayingsim123]

Okay thanks for clearing that up, tiny-tim! You're really awesome! :)