Solve Frictionless Drift Physics Problem: Angle of Inclination

  • Thread starter Thread starter KonigGeist
  • Start date Start date
  • Tags Tags
    Drift Frictionless
Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 2K views
KonigGeist
Messages
2
Reaction score
0

Homework Statement



A physics student playing with an air hockey table (a frictionless surface) finds that if she gives the puck a velocity of 5.70 m/s along the length (2.28 m) of the table at one end, by the time it has reached the other end the puck has drifted 5.78 cm to the right but still has a velocity component along the length of 5.70 m/s. She correctly concludes that the table is not level and correctly calculates its inclination from the above information. What is the angle of inclination?

Homework Equations


d = v * t
v2 = v02 + 2a(y-y0)


The Attempt at a Solution


A)Calculate the time using d = v * t:
t = d / v = 2.28 / 5.7 = 0.4s

B)Calculate the velocity of the drift using d = v * t:
v = d / t = 0.0578 / 0.4 = 0.1445m/s

C)Calculate the y-travel of the drift:
y = v2/2a = 0.14452 / 2(9.8) = 0.001065 m

D)Calculate angle a using tangent
sin(a) = 0.0578 / 0.001065
a = sin-1(0.001065 / 0.0578)


I've been working on this problem for several hours, but I haven't been able to get it. I have a feeling that I'm making a mistake somewhere, but I had two of my physics major friends look at it and they get the same answer as me.
 
Physics news on Phys.org
Welcome to PF!

Hi KonigGeist! Welcome to PF! :smile:
KonigGeist said:
B)Calculate the velocity of the drift using d = v * t:

No, you need to calculate the acceleration using one of the standard constant acceleration equations. :wink:
 
I see what you mean. I remember thinking about the acceleration when I first started, but I guess I forgot. :redface:
If I calculated the acceleration using x=x0 + v0t + 1 / 2at2:

a = 2x / t2 = 2(0.0578) / (0.4)2 = 0.7225 m / s2

Is this correct?
From here, how would I find the angle? Do I need to calculate the y distance, or can I relate the velocities, or something else entirely?