Solve Frictionless Drift Physics Problem: Angle of Inclination

[KonigGeist]

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.

 

[tiny-tim]

Welcome to PF!

Hi KonigGeist! Welcome to PF!

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.

 

[KonigGeist]

I see what you mean. I remember thinking about the acceleration when I first started, but I guess I forgot.
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?

 

[tiny-tim]

Hi KonigGeist!

Yes. that looks ok.

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

 

[rwisz]

Can you help me?



Hello there,

Thank you for sharing your attempt at solving this problem. It seems like you have the right approach, but there are a couple of mistakes in your calculations.

First, in part B, when calculating the velocity of the drift, you should use the total distance traveled by the puck (5.78 cm) instead of the length of the table (2.28 m). So the correct calculation should be v = 0.0578 / 0.4 = 0.1445 m/s.

Second, in part C, when calculating the y-travel of the drift, you have used the wrong formula. The correct formula to use is y = v^2 / 2a, where v is the velocity of the drift and a is the acceleration due to gravity (9.8 m/s^2). So the correct calculation should be y = (0.1445)^2 / (2 * 9.8) = 0.001065 m.

Finally, in part D, you are using the wrong trigonometric function. The correct formula to use is tan(a) = 0.0578 / 0.001065, which gives us an angle of approximately 3.08 degrees.

I hope this helps you in solving the problem. Remember to always double-check your calculations and use the correct formulas. Good luck!