How Does Projectile Mass Affect Exit Speed in Physics?

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Homework Help Overview

The discussion revolves around the effect of projectile mass on exit speed from a toy cannon, specifically examining how varying mass influences the velocity of projectiles when subjected to a constant force along the barrel.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between force, mass, and acceleration, questioning how the time spent in the barrel affects the final speed of different masses. There is discussion about comparing accelerations and how final speed relates to distance and acceleration.

Discussion Status

Participants are actively engaging with the problem, raising questions about the assumptions made regarding time and acceleration. Some guidance has been offered regarding the use of equations, but no consensus or resolution has been reached yet.

Contextual Notes

The problem is framed as a multiple-choice question, with specific answer options provided, but the original poster notes a lack of explanation from the source material regarding the correct answer.

dtl42
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[SOLVED] SAT Subject Test in Physics

Homework Statement


Assume that every projectile fired from a toy cannon experiences force F along the length of the barrel. If a projectile of mass M leaves the barrel with speed, V, at what speed will a projectile of mass 2M leave the barrel with?


Homework Equations


F=ma, a=(Vf-Vo)/t


The Attempt at a Solution


The force F, produces acceleration of V/t for mass m, so the same force should produce F=(2m)(v/t), so the velocity equals v/2?

This is a multiple choice question with choices A) v/2, B)v/sqrt(2), C)v, D)2v, E)4v

The book says the answer is B, but gives no explanation.
 
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dtl42 said:
The force F, produces acceleration of V/t for mass m, so the same force should produce F=(2m)(v/t), so the velocity equals v/2?
Careful: The time in the barrel is not the same for the two projectiles.
 
Oh, that's true, so how can I get around that?
 
dtl42 said:
Oh, that's true, so how can I get around that?
How do the accelerations compare? How does final speed (at the end of the barrel) depend on distance and acceleration?
 
The final speed is Acceleration/Distance, and we know the distance must be the same for both objects. Can we rearrange the v=a/d and substitute it into F=ma?
 
Wait, v doesn't equal a/d, I don't know what I'm thinking...
 
Oh, I found out how to do it, just use v^2=2ad and work it out.

EDIT: Sorry for the obnoxious triple post.
 

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