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Geek Culture / Physics Question

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Robin
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Posted: 21st Oct 2004 03:48
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I was wondering if anyone here could just look over my answers to the question below and let me know if they're correct. I've got a feeling part A's wrong and if so, B,C & D will be wrong too :-( It's kinda important

A 50g bullet hits a 2Kg block of wood, suspended vertically from a string. The bullet penetrates 3cm into the wood and remains embedded as the block swings up through a vertical height of 30cm.

a) What was the speed of the bullet and block just after the collision?
b) What was the bullet's speed before the collision?
c) Estimate the average force exerted on the bullet by the block.
d) Calculate the kinetic energies before and just after the collision.
e) Was this collision elastic or inelastic?

My answers:

A) I got 2.4m/s (v=0m/s, a=-9.8m/s/s, s=0.3m, v^2 = u^2 + 2as)
B) Using the conservation of momentum rule:
m(1)u(1) + m(2)u(2) = m(3)v(3)
(0.05 x u) = (2.05 x 2.4)
u=98.4m/s
C) Using F=ma
v=0m/s u=98.4m/s s=0.03m
v^2 = u^2 + 2as
0 = (98.4)^2 + (2xax0.03)
a = 161376m/s/s
F = 0.05 x 161376 = 8068N
D) KE before collision:
KE = 0.5 x 0.05 x (98.4)^2 = 242J
KE after collision:
KE = 0.5 x 2.05 x (2.4)^2 = 5.9J
E) Inelastic, because the bullet gets embedded into the block of wood.

Thanks,

Robin

http://www.thegameszone.tk | robin@thegameszone.tk
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Neofish
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Posted: 21st Oct 2004 04:19
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Quote: "A 50g bullet hits a 2Kg block of wood, suspended vertically from a string. The bullet penetrates 3cm into the wood"


Quote: "a) What was the speed of the bullet and block just after the collision?"


Err technically it depends on the wood, the velocity of the bullet, the width of the wood (maybe) and other stuff like that...Anywho im not very good at physics like that...

I just noticed the second question

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Robin
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Posted: 21st Oct 2004 04:24 Edited at: 21st Oct 2004 04:25
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Yeah thats what I was thinking - if the question would have given you the time the bullet took to slow down or something else then it would be easier to work out - I'm assuming air resistance etc. can be ignored. I managed to get some answers but I don't know if they're right...

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Ian T
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Posted: 21st Oct 2004 04:33 Edited at: 21st Oct 2004 04:33
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Holy shtik. I should enjoy the relative non-mathiness of Chemistry while it lasts


Preparing for a month of insanity, inspiration and insight... again!
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Wiggett
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Posted: 22nd Oct 2004 21:01
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man physics sucks, why cant they just take it as it is, there is a bullet in the wood, ho wdid it get there? well obviously someone shot it out of a gun into the wood, how fast did it go? as fast as a speeding bullet. case closed. you think physics is hard try doing forensic physics, more bullets embedded in things like torso's and cars breaking through rails and flying over cliff faces. would be pretty fun if were not for the most boringest maths problems ever.


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Ilya
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Posted: 22nd Oct 2004 23:07
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I know simple physics from experience.

The default editor is fine.

Quote: "I've seen the word programming and I'm not sure what it means. Anybody please explain?"
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bitJericho
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Posted: 23rd Oct 2004 10:53
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eh.. robx... you're a friggin programmer^_^ This is stuff you *HAVE* to be able to do to make physics capable games

I loved my physics class And I hated chemistry Mostly I liked it because I could apply it to my programming..

Alas I have no answer for you, as you're missing some necessary details


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Wiggett
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Posted: 23rd Oct 2004 11:25
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hey if i whip out my work and formulae i could do this work, its jsut a matter of i don't want to so i will beg someon elike van b to do it for me


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pizzaman
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Posted: 23rd Oct 2004 12:35 Edited at: 23rd Oct 2004 12:38
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hi

My Answers (all to 3 significant figures)

A) 2.45 m/s

Gain in gravitational potental energy = Loss in kinetic energy
Therefore mgh = 0.5mv^2
v = sqrt(2gh)
v = sqrt(2 x 9.8 x 0.3)
v = 2.45

B) 101 m/s (Think your mistake was rounding error here)

m1.u1 + m2.u2 = m3.v
0.05 x u1 + 2 x 0 = 2.05 x 2.45
u1 = 5.03 / 0.05
u1 = 101

C) 81.8 N

a = ((v^2) - (u^2)) / (2s) Work out decceleration of bulletin wood
a = (0 - 101) / (2 x 0.03)
a = -1677

t = (v - u) / a Time it takes for bullet to deccelerate
t = (0 - 101) / 1677
t = 0.06

F = m(v - u) / t Average force produced by wood
F = 2(0 - 2.45) / 0.06
F = 81.8

D) 253 J (before) 6.03 J (After)

E = 0.5mv^2 + 0.5mv^2 E = mgh
E = 0.5 x 0.05 x 101^2 E = 2 x 9.8 x 0.3
E = 253 E = 6.03

E) Inelastic, because the bullet gets embedded into the block of wood.

Note my answers could be wrong, its been nearly a year and half since I've done physics, so done blame me if they are

Enjoy
pizzaman
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Robin
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Posted: 23rd Oct 2004 23:49
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Thank you so much!
Robin

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pizzaman
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Posted: 24th Oct 2004 04:45
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No probs, was kind of fun remembering how to do it. I take it your doing your A-levels? Second year?

Ah the good ol physics memories lol

Hope you do well with your studies
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Ilya
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Posted: 24th Oct 2004 05:33
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Quote: "A) 2.45 m/s

Gain in gravitational potental energy = Loss in kinetic energy
Therefore mgh = 0.5mv^2
v = sqrt(2gh)
v = sqrt(2 x 9.8 x 0.3)
v = 2.45

B) 101 m/s (Think your mistake was rounding error here)

m1.u1 + m2.u2 = m3.v
0.05 x u1 + 2 x 0 = 2.05 x 2.45
u1 = 5.03 / 0.05
u1 = 101

C) 81.8 N

a = ((v^2) - (u^2)) / (2s) Work out decceleration of bulletin wood
a = (0 - 101) / (2 x 0.03)
a = -1677

t = (v - u) / a Time it takes for bullet to deccelerate
t = (0 - 101) / 1677
t = 0.06

F = m(v - u) / t Average force produced by wood
F = 2(0 - 2.45) / 0.06
F = 81.8

D) 253 J (before) 6.03 J (After)

E = 0.5mv^2 + 0.5mv^2 E = mgh
E = 0.5 x 0.05 x 101^2 E = 2 x 9.8 x 0.3
E = 253 E = 6.03"


What?!??!?!
What's ^?
Do you get to use a calculator?

The default editor is fine.

Quote: "I've seen the word programming and I'm not sure what it means. Anybody please explain?"
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Neofish
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Posted: 24th Oct 2004 05:50
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^ is to the power of

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Ilya
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Posted: 24th Oct 2004 06:08 Edited at: 24th Oct 2004 06:09
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Oh.

What happened to little numbers?

The default editor is fine.

Quote: "I've seen the word programming and I'm not sure what it means. Anybody please explain?"
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Neofish
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Posted: 24th Oct 2004 06:51
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its harder to do, i dont know how to do it with HTML and it may be blocked (dunno, unless it uses the & and ; sys)

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bitJericho
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Posted: 24th Oct 2004 07:23
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..besides, it's the proper way to do powers when you're programming


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Neofish
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Posted: 24th Oct 2004 07:30
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was going to say tht but that would mean typing it

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Phaelax
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Posted: 24th Oct 2004 16:51
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Quote: "man physics sucks, why cant they just take it as it is, there is a bullet in the wood, "


Given the type of bullet(by looking at it) and figuring out the speed of the bullet, police could estimate where the shooter was at. It may not seem like it, but this is a very good and practical question that does have its usefullness.

also, the wood's density and size would help. plus you gotta consider string tension. Not sure if the length of the string would make a difference thought.

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Robin
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Posted: 25th Oct 2004 00:28
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Yeah I'm in my second year of A-levels It's not bad - my teacher pretty decent too.
btw. I think you forgot to square the 101 in part c)
Robin

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pizzaman
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Posted: 25th Oct 2004 06:44
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Your right I forgot to square 101 , altough it was quite late when I did those questions. Did my A-level in physics last year, got a B in the end so it wasn't too bad, however I wanted an A, o well thats how it goes I guess.
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Phaelax
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Posted: 25th Oct 2004 18:25
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I just remembered about this website. I visit it occationally.
http://physicsforums.com/

"eureka" - Archimedes
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