Kinematics problem from a competition: Will the 2 sliding boxes collide?

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Homework Help Overview

The discussion revolves around a kinematics problem involving two sliding boxes and whether they will collide based on their respective distances traveled. The problem involves analyzing the motion of the boxes under the influence of friction.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the distances traveled by each box and the conditions necessary for a collision to occur. There is a focus on the equations derived for the distances and the implications of these equations regarding the collision condition.

Discussion Status

Some participants have confirmed the calculations presented by the original poster and have reiterated the condition for collision. There is an ongoing exploration of the implications of the derived condition and its relation to the problem's options.

Contextual Notes

There is mention of potential issues with the problem's construction, particularly regarding the clarity and quality of the multiple-choice options provided in the competition context.

Jim Alexandridis
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Homework Statement
two small boxes of rectangular parallelepiped shape are launched simultaneously against each other, with velocities of meters 𝑢1=𝑢 and 𝑢2=2𝑢, from positions A and B of a straight and horizontal trajectory 𝑂𝑥 lying in a plane. The distance between them is 𝛢𝛣=𝑠 and the coefficient of friction between each body and the surface is μ. Boxes will collide if:
a)u≥2√(μgs)/3
b)u≥1√(μgs)/3
c)u≥2√(μgs)/5
d)u≥3√(μgs)/5

This problem was on an greek physics competition on 10th grade and there is a disagreement about the answer, so i would like your opinion depending on my answer
Relevant Equations
Kinematics:
V=v0+at
Dx=v0t+1/2at²
Smax=v²/2a
The distance covered by the first box is :s1max=v²/2|a|=v²/2μg where a=-μg by second newtons law
Similarly S2max=(2v)²/2|a|=4v²/2μg
It gas to be s1max+s2max≥S => v²/2a +4v²/2a ≥s => 5v²≥2aS =>v²≥ 2μgS/5=> v≥√(2μgs/5)
But this is in the possible solution, am I wrong somewhere? I appreciate your help
 
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Hi @Jim Alexandridis and welcome to PF.

The first box travels distance ##s_1=\dfrac{u^2}{2\mu g}##
The second box travels distance ##s_2=\dfrac{(2u)^2}{2\mu g}=\dfrac{4u^2}{2\mu g}=4s_1.##
The two boxes together travel distance ##s_1+s_2=5s_1=\dfrac{5u^2}{2\mu g}##
So what must be true for a collision to take place?

Sometimes multiple choice problems are not very well constructed.
 
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kuruman said:
Hi @Jim Alexandridis and welcome to PF.

The first box travels distance ##s_1=\dfrac{u^2}{2\mu g}##
The second box travels distance ##s_2=\dfrac{(2u)^2}{2\mu g}=\dfrac{4u^2}{2\mu g}=4s_1.##
The two boxes together travel distance ##s_1+s_2=5s_1=\dfrac{5u^2}{2\mu g}##
So what must be true for a collision to take place?

Sometimes multiple choice problems are not very well constructed.
Isn’t that the same as the OP did in post #1?
 
haruspex said:
Isn’t that the same as the OP did in post #1?
Yes. It's reassurance to a new user that when you're right, you're right.
 
I believe that for the collision to happen it has to be :

##S_{1}+S_{2}\ge S##
 
Jim Alexandridis said:
I believe that for the collision to happen it has to be :

##S_{1}+S_{2}\ge S##
That is correct. As you already posted, this condition leads to $$u \geq \sqrt{\frac{2\mu~g~s}{5}}$$which is not one of the options (a) - (d). What you can do with this information is up to you. It ranges from doing nothing to bringing it to the attention of someone who is in a position to improve the quality control of exams like this.
 
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kuruman said:
That is correct. As you already posted, this condition leads to $$u \geq \sqrt{\frac{2\mu~g~s}{5}}$$which is not one of the options (a) - (d). What you can do with this information is up to you. It ranges from doing nothing to bringing it to the attention of someone who is in a position to improve the quality control of exams like this.
It might be a "copy/paste" issue. For example, in some other forum, I've just posted in LaTeX:
1712846272189.png

but after "edit and save" it has been converted to this:

1712846529653.png
 

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