Can You Win the Puck Game at the Amusement Park?

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Homework Help Overview

The problem involves a puck being pushed up a frictionless ramp at an amusement park, where the goal is to determine if the puck can reach within 10 cm of the end of the ramp without going off. The puck is released with an initial speed and its speed is measured after traveling a certain distance.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the application of kinematic equations to determine acceleration and position over time. There is a focus on verifying calculations and addressing potential arithmetic errors that could affect the outcome.

Discussion Status

Some participants have confirmed the correctness of the original poster's approach while suggesting a review of arithmetic calculations. There is an ongoing exploration of whether the final distance calculation leads to winning the stuffed animal.

Contextual Notes

Participants express uncertainty regarding the implications of the results, particularly in relation to the winning condition of the game. There is mention of a perceived discrepancy between expected outcomes based on prior experiences and the current problem's requirements.

barthayn
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Homework Statement



One game at the amusement park has you push a puck up a long, frictionless ramp. You win a stuffed animal if the puck, at its highest point, comes to within 10 cm of the end of the ramp without going off. You give the puck a push, releasing it with a speed of 5.0m/s when it is 8.5m from the end of the ramp. The puck's speed after traveling 3.0m is 4.0 m/s. Are you a winner?

Homework Equations



vf^2 = vi^2 +2ad
x(t) = 1/2at^2 + vt

The Attempt at a Solution



a = (vf^2-vi^2)/2d
a = (16-25)/6
a = -3/2 m/s/s

x(t) = -3/2t^2 + 5t m
v(t) = -3t + 5 m/s
v(t) = 0 = -3t + 5
t = 5/3

x(5/3) = 26/5 m = 4.16m
df = 8.5 - 4.16
df = 4.3

Therefore, you are not a winner because you are off by a distance of 4.2m before you earn the stuffed animal at the amusement park.

Is this the correct method of doing this question. I feel that I am wrong because in high school it was always winning or going in the basket and not it is not. I believe it is unlikely that the answer will tell us that we are not a winner. Is there anything wrong with my logic process here?
 
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barthayn said:
a = (vf^2-vi^2)/2d
a = (16-25)/6
a = -3/2 m/s/s

This is correct.

barthayn said:
x(t) = -3/2t^2 + 5t m
v(t) = -3t + 5 m/s
v(t) = 0 = -3t + 5
t = 5/3
Check your relevant equations and your previous work. You have a slight arithmetical error here that makes a very large difference in your final solution.


Your process seems sound, just check your arithmetic. Post again if you want another double check.
 
Ignea_unda said:
This is correct.Check your relevant equations and your previous work. You have a slight arithmetical error here that makes a very large difference in your final solution.Your process seems sound, just check your arithmetic. Post again if you want another double check.

Thanks for pointing that out. I seen a two below the number and thought it was halved. It wasn't. I got it stopping 6.67cm away from the edge, and therefore I am the winner of the stuffed animal correct?
 
Yes, it looks like that is correct.
 

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