| Topic: |
Science > Physics |
| User: |
"Bluethunder" |
| Date: |
05 Dec 2004 09:33:02 PM |
| Object: |
Help solve a problem |
I am asked to help with this problem but I'm stumped. The problem is from
Cutnell & Johnson Physics, 4E, Chapter 7, problem 35. Here goes the
problem:
A 0.01kg bullet traveling straight up at a speed of 750 m/s hits a 2kg block
of wood dropped from rest at the top of a building. The block of wood has
been falling for t seconds before the collision with the bullet. The block
of wood with the bullet imbedded in it as a result of the collision reverses
direction, rises and comes to a momentary halt at the top of the building.
Find t.
Could you help?
Ken
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| User: "AaronB" |
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| Title: Re: Help solve a problem |
06 Dec 2004 12:20:30 AM |
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"Bluethunder" <zdotcom@comcast.net> wrote in message news:<vMadnWJzJIuqTi7cRVn-rQ@comcast.com>...
I am asked to help with this problem but I'm stumped. The problem is from
Cutnell & Johnson Physics, 4E, Chapter 7, problem 35. Here goes the
problem:
A 0.01kg bullet traveling straight up at a speed of 750 m/s hits a 2kg block
of wood dropped from rest at the top of a building. The block of wood has
been falling for t seconds before the collision with the bullet. The block
of wood with the bullet imbedded in it as a result of the collision reverses
direction, rises and comes to a momentary halt at the top of the building.
Find t.
Could you help?
Ken
Well, I will warn you on the onset, most people here don't like
helping with homework problems unless you've demonstrated that you've
contributed a substantial amount of work toward a solution. I will
suggest that you try to show some of your work, because, in short
order, you will probably be flamed.
That said, I will give you a push in the right direction.
First, we have a slight problem of interpretation: was the bullet
fired at a speed of 750 m/s, or did it collide with the block at 750
m/s? Doesn't really make a difference in the long run, but it will
lead to slightly different solutions. I will assume the latter
(simpler) case, and leave it up to you to prove the other.
When two objects collide, momentum is conserved. You should know the
governing equation for that. This is a perfectly inelastic collision.
You should know how to deal with that as well. We do not know the
final speed of the block after it is hit, and we do not know the final
speed of the block before it was hit, but this is a function of t,
which is fine. We also know that the block rises up and just gets back
to the height of the building after the collision. What does that tell
you about the final speed of the block/bullet system? What does that
tell you about the time it takes?
A.
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| User: "Dave Baker" |
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| Title: Re: Help solve a problem |
05 Dec 2004 11:03:51 PM |
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Subject: Help solve a problem
From: "Bluethunder"
Date: 06/12/04 03:33 GMT Standard Time
Message-id: <vMadnWJzJIuqTi7cRVn-rQ@comcast.com>
I am asked to help with this problem but I'm stumped. The problem is from
Cutnell & Johnson Physics, 4E, Chapter 7, problem 35. Here goes the
problem:
A 0.01kg bullet traveling straight up at a speed of 750 m/s hits a 2kg block
of wood dropped from rest at the top of a building. The block of wood has
been falling for t seconds before the collision with the bullet. The block
of wood with the bullet imbedded in it as a result of the collision reverses
direction, rises and comes to a momentary halt at the top of the building.
Find t.
Could you help?
Seems very straightforward unless I'm missing something. If the bullet/block
combo is to ascend to the top of the building again then its upward speed after
impact must be the same as the downward speed the block had reached at impact.
Given that momentum is conserved and V is the block's downward speed at impact.
(0.01 x 750) - 2V = 2.01V
V = 1.8703 m/s
T = 1.8703 / 9.81 = 0.19 seconds
--
Dave Baker - Puma Race Engines (www.pumaracing.co.uk)
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| User: "Bluethunder" |
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| Title: Re: Help solve a problem |
06 Dec 2004 10:18:04 PM |
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I appologise for asking a straight forward question w/o showing some of my
work. The fact is, I keep going around in circles. I started trying to
solve the distance traveled y in a function of final velocity of the wooden
block vw just before it was hit:
vw**2 = 2 * 9.8 * y
y = vw**2 / 2 * 9.8
Then I tried to solve the velocity of the block based on conservation of
momentum
m1 * v1 - m2 * vw = (m1 + m2) * V where V is the combined
object's velocity as it is shot back up
At this point I realized I wasn't going to make it because I still have 2
variables vw and V.
Ken
"Bluethunder" <zdotcom@comcast.net> wrote in message
news:vMadnWJzJIuqTi7cRVn-rQ@comcast.com...
I am asked to help with this problem but I'm stumped. The problem is from
Cutnell & Johnson Physics, 4E, Chapter 7, problem 35. Here goes the
problem:
A 0.01kg bullet traveling straight up at a speed of 750 m/s hits a 2kg
block
of wood dropped from rest at the top of a building. The block of wood has
been falling for t seconds before the collision with the bullet. The
block
of wood with the bullet imbedded in it as a result of the collision
reverses
direction, rises and comes to a momentary halt at the top of the building.
Find t.
Could you help?
Ken
.
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| User: "Randy Poe" |
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| Title: Re: Help solve a problem |
06 Dec 2004 11:43:06 PM |
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Bluethunder wrote:
I appologise for asking a straight forward question w/o showing some
of my
work. The fact is, I keep going around in circles. I started trying
to
solve the distance traveled y in a function of final velocity of the
wooden
block vw just before it was hit:
vw**2 = 2 * 9.8 * y
y = vw**2 / 2 * 9.8
Then I tried to solve the velocity of the block based on
conservation of
momentum
m1 * v1 - m2 * vw = (m1 + m2) * V where V is the combined
object's velocity as it is shot back up
At this point I realized I wasn't going to make it because I still
have 2
variables vw and V.
You have one unknown: they're the same value.
If you give an object velocity v upward, how high will it
rise? Answer: until the kinetic energy 0.5*mv^2 is converted
into potential energy mgh.
If you drop an object starting at rest from height h, what
will its final velocity v be? Answer the potential energy
mgh is converted into kinetic energy 0.5*mv^2.
Note that in both cases, m cancels out from both sides. It
doesn't matter that you have a block going down and a block
plus bullet going up. The relation between v and h holds
independent of mass.
- Randy
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