Solve Frictionless Drift Physics Problem: Angle of Inclination

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A physics student analyzes the motion of a puck on a frictionless air hockey table, discovering it drifts 5.78 cm to the right while maintaining a velocity of 5.70 m/s over a distance of 2.28 m. To solve for the angle of inclination, she calculates the time taken for the puck to travel the length of the table as 0.4 seconds. The drift velocity is found to be 0.1445 m/s, and the acceleration due to the drift is calculated as 0.7225 m/s². The discussion emphasizes the need to relate this acceleration to the gravitational component acting on an inclined plane to determine the angle. The problem-solving process highlights the importance of correctly applying physics equations to find the solution.
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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.
 
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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?
 
Hi KonigGeist! :wink:

Yes. that looks ok. :smile:

Now compare that acceleration with the component of gravity down a slope. :wink:
 
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