Football Players: Solve Sliding Distance After Collision

  • Thread starter Thread starter Nightrider55
  • Start date Start date
AI Thread Summary
Fred, a 60 kg football player, collides with Brutus, a 120 kg player, and they slide after the impact. The calculations for the post-collision velocity and sliding distance were performed using conservation of momentum and kinematic equations, yielding a distance of 7.55 cm. Despite following the correct methodology, the answer was marked incorrect, leading to speculation about potential errors in the computer system's calculations or input tolerances. A comparison with another source indicated a different answer of 8.8 cm, prompting the instructor to contact tech support for clarification. These issues highlight the occasional discrepancies in computer-based grading systems compared to traditional instruction.
Nightrider55
Messages
18
Reaction score
0

Homework Statement


Fred (mass 60.0kg) is running with the football at a speed of 6.0m/s when he is met head-on by Brutus (mass 120kg), who is moving at 4.0m/s. Brutus grabs Fred in a tight grip, and they fall to the ground. How far do they slide?
Part A
The coefficient of kinetic friction between football uniforms and Astroturf is 0.3.
Give answer in cm


Homework Equations



MfVf + MbVb=(Mf+Mb)Vs
Kinematics

Ax=-mew x g



The Attempt at a Solution



I did the problem and doubled check my work and even had a friend check it and we couldn't find a mistake but I still get it wrong

To find the velocity after the collision I used the conservation of momentum equation above and calculated the "after impact) velocity to be -2/3 m/s. I then proceeded to find the acceleration and used Ax=-mew x g and got -2.94 m/s^2. From there I solved for time by plugging the velocity and acceleration into Vfx=Vix+Ax(delta T). I got the time for
the slide to be .22 seconds. I then plugged everything into the distance kinematics equation Xf=...and I got .075 m which I then converted to 7.55 cm, but it still says its wrong. Do you see any mistakes I made?
 
Physics news on Phys.org
Nightrider55 said:
To find the velocity after the collision I used the conservation of momentum equation above and calculated the "after impact) velocity to be -2/3 m/s. I then proceeded to find the acceleration and used Ax=-mew x g and got -2.94 m/s^2. From there I solved for time by plugging the velocity and acceleration into Vfx=Vix+Ax(delta T). I got the time for
the slide to be .22 seconds. I then plugged everything into the distance kinematics equation Xf=...and I got .075 m which I then converted to 7.55 cm, but it still says its wrong. Do you see any mistakes I made?

Is "it" that says the answer wrong a computer problem-system? If so, what units does it ask for the answer in? (Possibly #1 reason why the result of a correct method is rejected...) BTW, I agree with your solution (I used the "velocity-squared" equation,
(v_f)^2 = (v_i)^2 + 2a(delta_x), and found the same distance).
 
It wants it in cm. I don't get why it says I am wrong.
 
Not being there myself to see what the computer is doing, I can only be mystified. If you're sure you've used the right quantities for the problem, your method should give the correct answer.

It is conceivable that the computer system has a difficulty for this problem. The formula encoded for its calculation could be in error, or the tolerance for input answers might have been set to zero. (This happened once a ways back in a physics course here for one problem: good luck getting the computer's answer to sixteen decimal places...)

I would bring this up with the instructor and see if there is something set up wrong for that problem. It's not like that's never happened before. (Besides, I'm feeling cocky: I helped a student last week to get the instructor in a 5000-level stats course to re-examine their solution to an exam problem and find the error in the problem statement... ;-) )
 
Last edited:
I talked to my teacher about it and he came to the same answer when he did it. I did show answer on the website and they calculated 8.8 cm. I don't know how they got that :confused: He sent an e-mail to the website tech support to notify them of the error.

Thanks for the help!
 
Last edited:
I'm glad to hear the teacher agreed with us also! :) These difficulties do crop up from time to time with the computer-based systems; they aren't going to replace living instructors anytime soon...
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top