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solve this problem

Créé le : 8 septembre 2025

Répondu en utilisant GPT-5 par Chat01

Question

image.pngimage.png

solve this problem

Python

python
from PIL import Image, ImageOps, ImageFilter, ImageEnhance, ImageStat, Image
im=Image.open('/mnt/data/image.png')
im.size

(695, 240)

Python

python
display(im.resize((1100,380)))

Python

python
display(im.crop((0,170,680,235)).resize((1000,200)))

Réflexion

Clarifying restitution formulas

Option (D) says the coefficient of restitution is "J/p + 1," and it looks like it's written with "J over p," and then the "+1" is below. This seems strange. Looking at option (C), I notice that the minus sign is at the baseline for the fraction. The directions of impulse are important too.

Analyzing momentum and impulses

A has momentum +p initially. During the impact, B gives impulse J to A, which would be -J, since it's opposite in direction. Option (A) claims the system's total momentum changes during the impact, but this is incorrect. The total momentum remains constant at p due to internal impulses canceling. Option (A) is false.

Regarding (B), since impulses are equal and opposite by Newton’s third law, it is correct.

Finding coefficient of restitution

Since B gives impulse J to A, A gives impulse J to B, which makes (B) true. Now, I need to find the coefficient of restitution. Using initial velocities and the changes in momentum, I get final velocities for A and B. The formula for restitution is e = (v2 - v1) / (u1 - u2), considering the sign change at collision.

Thought for 46s

Réponse

Answer: (B) and (C)

Let each mass be mm. Initially, AA has momentum p⇒uA=p/mp\Rightarrow u_A=p/m, BB is at rest uB=0u_B=0.
During the collision, BB imparts an impulse of magnitude JJ to AA (opposite to AA’s motion), so:

ΔpA=−J⇒pA′=p−J,ΔpB=+J⇒pB′=J.\Delta p_A=-J \quad\Rightarrow\quad p_A' = p-J,\qquad \Delta p_B=+J \quad\Rightarrow\quad p_B' = J.

Thus the system momentum is pA′+pB′=(p−J)+J=pp_A'+p_B'=(p-J)+J=p at all times; so (A) is false.

By Newton’s 3rd law, the impulse on BB due to AA has equal magnitude JJ (opposite direction) → (B) is true.

The coefficient of restitution:

e=vB′−vA′uA−uB=Jm−p−Jmpm=2J−pp=2Jp−1,e=\frac{v_B'-v_A'}{u_A-u_B} =\frac{\frac{J}{m}-\frac{p-J}{m}}{\frac{p}{m}} =\frac{2J-p}{p} =\frac{2J}{p}-1,

which matches (C). Statement (D) is false.

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