Balistic Pendulum Equation Solution

[Woopy]

Homework Statement

An 8.0g bullet is fired into and becomes embedded in a 2.5kg block of wood at the end of a pendulum. The block of wood swing to a height of 6.0cm. (Watch Units)
a) What is the PE of the block & bullet at the top?
b) What is the KE at the bottom? (COE)
c) What is the velocity of the block & bullet right after the impact?
d) What is the pmomentum at the bottom?
e) What was the initial speed of the bullet? (COM)

Homework Equations

KE= PE
KE= 1/2 mv²
PE= mgh
P=mv
m₁v₁+m₂v₂=(m₁+m₂)v'

The Attempt at a Solution

a)PE = (.008kg + 2.5kg)(9.8 m/s²)(.06m) = 1.47 J (doesnt seem like a reasonable number to me)

b)KE = .5(.008kg + 2.5kg)(v)²

c) mgh=1/2mv²

2gh=v²
v=(square root)2gh
(square root) (2)(9.8m/s²)(.06m)=1.08 m/s

d)Does "bottom" mean not moving? if not, then
P=mv P=(.008kg+2.5kg)(1.08m/s)=2.71 kgm/s

e)(.008kg)(v₁) + (2.5kg)(v₂) = (.008kg + 2.5kg)v'
.008kg(v₁) + 2.5kg(v₂) = 2.508kg(v')

 

[Woopy]

figured it out

 

[turin]

EDIT: I was in the process of typing this up while you figured it out. You may still find my first comment useful, though.

Homework Statement

An 8.0g bullet is fired into and becomes embedded in a 2.5kg block of wood at the end of a pendulum. The block of wood swing to a height of 6.0cm. (Watch Units)
a) What is the PE of the block & bullet at the top?
b) What is the KE at the bottom? (COE)
c) What is the velocity of the block & bullet right after the impact?
d) What is the pmomentum at the bottom?
e) What was the initial speed of the bullet? (COM)

Homework Equations

KE= PE

This is too vague (i.e. generally untrue). I would advise against ever using this equation because it is imprecise and can cause confusion. I suggest the more explicit:

KEi+PEi=KEf+PEf

from which you can derive

KEi=PEf

by recognizing that KEf=0 (at the top of the swinging motion of the block) and PEi=0 (by simply choosing that as your zero point for PE, since gravitational PE is relative to your choice of zero).
Notice that the KE in this equation corresponds to a different situation than the PE, which I have indicated with the identifiers, "i" for initial and "f" for final.

Alternatively, try:

KEi+PEi=KEf+PEf
=>
KEi-KEf=PEf-PEi
=>
KEi=ΔPE
(is KEf=0)

a)PE = (.008kg + 2.5kg)(9.8 m/s²)(.06m) = 1.47 J (doesnt seem like a reasonable number to me)

Why does this seem unreasonable to you? I got the same thing, except for a sig fig issue. If you need to be picky about sig figs, then how many should you have?

b)KE = .5(.008kg + 2.5kg)(v)²

No. Use the hint, "(COE)". This is basically how you solved part (c), so you obviously know how to do this. Think of part (b) as an intermediate step between parts (a) and (c); I think that was the point of part (b).

d)Does "bottom" mean not moving?

Probably so, since it specifies "after the impact".

e)(.008kg)(v₁) + (2.5kg)(v₂) = (.008kg + 2.5kg)v'
.008kg(v₁) + 2.5kg(v₂) = 2.508kg(v')

Any guesses what is v₂ and v'?

 

[Woopy]

would PE be KE also, as part b as 1.47 J?

and for part e I ended up getting 339 m/s, which seems reasonable for a bullet

 

[turin]

would PE be KE also, as part b as 1.47 J?

Yes. And it looks like you're doing fine with these problems, but I still strongly suggest that you take heed of my first suggestion in my previous post. (That includes the use of more specific identification, such as "i" and "f".)

... for part e I ended up getting 339 m/s, which seems reasonable for a bullet

It seems reasonable to me, too. :)