Determine mass of a glider on an air track

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The discussion revolves around calculating the mass of a glider on an air track connected to a 20.0 g mass via a pulley. The user initially misapplied the equations of motion, leading to an incorrect mass calculation of 20 g for the glider. After clarification, it was noted that vertical forces cancel out, and only horizontal forces, specifically tension, should be considered. The correct approach resolved the issue, confirming the calculations were valid when focusing on the appropriate forces. The user successfully understood the problem and expressed gratitude for the assistance received.
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Determine mass of a glider on an air track [solved]

Homework Statement



A 20.0 g mass is stringed to a glider on an air track through a pulley (https://www.physicsforums.com/attachment.php?attachmentid=30478&stc=1&d=1291794499"). I'm supposed to find the mass of said glider, assuming that the air track and pulley are frictionless and the string is massless.

Data from a graph:

V1 = 0.143 m/s
V2 = 0.377 m/s
Δt = 0.24 s

Homework Equations


a = (V2 – V1) / Δt
Fnet = ma

The Attempt at a Solution



a = (V2 – V1) / Δt
= (0.377 – 0.143) / 0.24
= 0.975 m/s2

Mass 1 (https://www.physicsforums.com/attachment.php?attachmentid=30479&stc=1&d=1291794499"):

Fnet = ma
Fn1 + Fg1 + T1 = -m1a
-m1g + T1 = -m1a
-9.8m1 + T1 = -0.975m1
+T = 8.825m1

Mass 2 (https://www.physicsforums.com/attachment.php?attachmentid=30480&stc=1&d=1291794499"):

Fg2 + T2 = m2a
+m2g – T2 = +m2a
(20.0)(9.80) – T2 = 20.0(0.975)
196 – T2 = 19.5
-T2 = -176.5

Final Calculations:

+ T = 8.825m1
(+) – T = -179.84
---------------------
0 = 8.825m1 – 179.84
8.825m1 = 179.84
m1 = 179.84/8.825
= 20 g

When the glider was weighted on a scale, it was said to be 429 g. So I have no idea what I'm doing wrong--am I getting the mass of the "20.0 g mass" here, or is this number just coincidental and irrelevant? Are the equations used incorrect? Thanks for any help.
 
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idlackage said:

Homework Statement



Mass 1 (https://www.physicsforums.com/attachment.php?attachmentid=30479&stc=1&d=1291794499"):

Fnet = ma
Fn1 + Fg1 + T1 = -m1a
-m1g + T1 = -m1a
-9.8m1 + T1 = -0.975m1
+T = 8.825m1


You do not have to add in Fn1 and Fg1 because they cancel each other out + they are in the vertical direction. You only need to take into account the horizontal forces(which in this case is the tension from the string).
 
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The tensions are equal
 
Kikora said:
You do not have to add in Fn1 and Fg1 because they cancel each other out + they are in the vertical direction. You only need to take into account the horizontal forces(which in this case is the tension from the string).

It works now, thank you so much!
 
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