How far into the block does the arrow penetrate?

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Homework Help Overview

The problem involves an arrow penetrating a block of Styrofoam while both the arrow and the block are accelerating. The scenario includes calculating the time it takes for the arrow to stop relative to the block, the common speed of both after the arrow stops, and the distance the arrow penetrates into the block.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the relationship between the deceleration of the arrow and the acceleration of the block, suggesting the use of equations to find the time until they reach the same final velocity. Some participants share their calculations for parts (a) and (b) and express uncertainty about how to approach part (c).

Discussion Status

Some participants have made progress on parts (a) and (b) and are seeking confirmation of their calculations. There is an ongoing exploration of how to tackle part (c), with hints being offered to simplify the approach. Multiple interpretations of the problem are being considered.

Contextual Notes

Participants are encouraged to show their work and reasoning before receiving assistance, adhering to forum guidelines. There is a focus on collaborative problem-solving without providing direct answers.

Zippy
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An arrow is fired with a speed of 20.0 m/s at a block of Styrofoam resting on a smooth surface. The arrow penetrates a certain distance into the block before coming to rest relative to it. During this process the arrow's deceleration has a magnitude of 1550 m/s^2 and the block's acceleration has a magnitude of 450 m/s^2.
a) how long does it take for the arrow to stop moving with respect to the block?
b) what is the common speed of the arrow and the block when this happens?
c) how far into the block does the arrow penetrate?
 
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You know that the block is accelerating as the arrow is decelerating, and at some time t they reach the same final velocity v. Use this to write down two equations involving initial velocity, acceleration or deceleration, and time. The left hand side of the two equations is v, you can forget about that for now. But that means the right hand side of the two equations equal one another, and you can use that to work out t.

That gives you a) and you can work out the rest from there.
 
Welcome to PF, Zippy.

You need to show what you've done, or your ideas on how to solve the problem, before we can help you.
 
neutrino: sorry, we overlapped there.

zippy: the idea is to give you help and encouragement, but not to do it all for you.
 
Thanks for the help. I think I've got (a) and (b), but I'm not sure how to approach (c).

This is what I've got so far:

(a) vi1+a1t=vi2+a2t
20-1550t=0+450t
2000t=20
t=0.01s

(b) v=vi2+a2t
v=0+450*0.01
v=4.5 m/s

Can you tell me if these are correct?
 
Farsight said:
neutrino: sorry, we overlapped there.
That's not a problem. :)

zippy: the idea is to give you help and encouragement, but not to do it all for you.
That's right. :D
 
Could someone give me a hint for (c)? I really have no idea where to start.
 
For (c), if I start plugging in numbers into x=(v^2-vi^2)/(2a), I get x1=0.1225 m and x2=0.0225 m. The difference is 0.1 m, so is that how far the arrow penetrates the block?
 
You can make c) simpler. Ask yourself how much the arrow decelerated, in what time, and at what rate.
 
  • #10
Hey Zippy, I followed the same reasoning you did and got the same answer, but could I get some confirmation if it's right or if there was some flaw in the reasoning?
 

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