Determine mass of a glider on an air track

  • Thread starter Thread starter idlackage
  • Start date Start date
  • Tags Tags
    Air Mass Track
Click For Summary
SUMMARY

The mass of a glider on an air track can be determined using the principles of Newton's second law and kinematics. In this discussion, a 20.0 g mass is connected to the glider, and the acceleration is calculated as 0.975 m/s² using the velocities V1 = 0.143 m/s and V2 = 0.377 m/s over a time interval of Δt = 0.24 s. The net force equations reveal that the tension in the string and gravitational forces must be balanced to solve for the glider's mass, which was ultimately found to be 429 g when weighed. The initial confusion stemmed from misapplying the equations of motion and not considering the correct forces acting on the glider.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Basic kinematics, including velocity and acceleration calculations
  • Familiarity with tension forces in pulley systems
  • Knowledge of free body diagrams for analyzing forces
NEXT STEPS
  • Study the principles of free body diagrams to analyze forces in systems
  • Learn about tension in strings and its role in pulley systems
  • Explore advanced kinematics problems involving multiple masses
  • Investigate the effects of friction in real-world applications of Newton's laws
USEFUL FOR

Students in physics courses, educators teaching mechanics, and anyone interested in understanding dynamics involving pulleys and gliders.

idlackage
Messages
10
Reaction score
0
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.
 
Last edited by a moderator:
Physics news on Phys.org
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).
 
Last edited by a moderator:
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!
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
16K
Replies
1
Views
11K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 6 ·
Replies
6
Views
7K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
15K
  • · Replies 4 ·
Replies
4
Views
6K
  • · Replies 3 ·
Replies
3
Views
5K
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
6K