How far does arrow penetrate into board

1. Jan 20, 2008

sillybean

[SOLVED] How far does arrow penetrate into board

1. The problem statement, all variables and given/known data
An arrow is fired with a speed of 21.0 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 1560 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 block when this happens?
c)How far into the block does the arrow penetrate?

2. Relevant equations
x=Xi+Vi*t+ 1/2at^2
x=0+21m/s (1.04x10^-2 s)+ 1/2 (-1560m/s^2)(1.04*10^-2 s)^2
x=0.2184-0.0843648
x=0.134m

3. The attempt at a solution

I got parts a and b. But part c is giving me problem. I tried using the equation above to get the final position of the arrow but the answer was wrong. Am I using the wrong equation. Or is it a math error?

2. Jan 20, 2008

Staff: Mentor

Is the deceleration of the arrow (-1560 m/s^2) with respect to the block or some common reference point?

3. Jan 20, 2008

sillybean

Unfortunately, the problem doesn't specify that. I'm guessing since the problem gives the deceleration of the arrow while saying "during this process" it might be with respect to the block.

4. Jan 20, 2008

physixguru

if it is wid respect to the block...then use conservation of momentum ....

5. Jan 21, 2008

sillybean

ok. But just out of curiosity, what if it was just with respect to a common point. What would I do then?

6. Jan 21, 2008

sillybean

looking over conservation of momentum, i don't think that's how i need to solve the problem seeing as how I haven't been taught that. Besides that, I have no mass included in the problem.

I'm pretty sure I have to use one of the equations that relate velocity, time, position and acceleration. I used one of them( see above), but got the wrong answer.

7. Jan 21, 2008

amarett0

try using this formula
x = x0 + 1/2 (v0 + v) t

8. Jan 21, 2008

sillybean

that one worked. thanks amarett0!