Question

Jack D. Ripper flipped out after missing a Must-Do-It question for the third time on his Minds On Physics assignment. Outraged by the futility of his efforts, he flings a 4.0-gram pencil across the room. The pencil lodges into a 221.0-gram Sponge Bob doll which is at rest on a countertop. Once in motion, the pencil/doll combination slide a distance of 11.9 cm across the countertop before stopping. The coefficient of friction between the doll and the countertop is 0.325. Determine the speed at which the pencil is moving prior to striking Sponge Bob.

Answer 1

the two equations to use here are\[F=ma\]and\[v_f^2=v_0^2+2ad\](for some reason people always forget that kinematic equation I wrote, but it's so useful)

Answer 2

but would a be 9.8m/s^2? It travels horizontally, so I thought gravity didn't matter...

Answer 3

first, how are we going to represent the first event: the transfer of momentum from the pencil to the pencil-spongebob comination?

Answer 4

p=mv...
p=(0.225kg)(v)

Answer 5

ok I took the srong approach
this is an energy problem...

Answer 6

wrong*

Answer 7

no it's a momentum problem.

http://www.physicsclassroom.com/calcpad/momentum/problems.cfm

number 28

Answer 8

really?
because this is what happened to me when I tried it as a momentum problem\[Ft=J=\Delta p\]the initial momentum of the object is\[p_0=(m_1+m_2)v_0\]and the only force acting is friction\[f=\mu N=\mu (m_1+m_2)g\]since the object comes to a rest we have that\[p_0-ft=0\]but since we don't have the time we need the fact that the force of friction acting over that distance will remove all the kinetic energy from the object, so we need to use energy it seems

Answer 9

so in term of energy that would be\[\frac12(m_1+m_2)v_0^2-fd=0\]and now we can solve for vo, from which we can go back to momentum and get v0 for the pencil from before the collision

Answer 10

so we use a combination of energy and momentum I think here

Answer 11

I'm not sure how u came to the conclusion that initial momentum minus Ft=0. But i'm with you so far...and i'm not sure, maybe we could use energy.

Answer 12

it's impulse
I'm not saying the initial momentum is zero (though I could since I can change reference frames at any time)
I'm just saying the force times time is change in momentum
in this case the object ends at rest, so when the impulse is applied the momentum goes to zero

but as I said, we don't know the time, so go the other way I wrote about

Answer 13

let's just take it from the sponge-pencil moment of combination...
what is the kinetic energy of that system?

Answer 14

0J

Answer 15

no, the object is moving
we won't have a number as an answer
(really you should wait and put numbers in at the end when possible)

Answer 16

we don't know speed though. so KE=1/2mv^2

Answer 17

where m is really two masses
and what is the force of friction?

Answer 18

i got 0.717 for the force of friction.

Answer 19

in symbols?

Answer 20

let's try to save all numbers until the very end if we can

Answer 21

Ff=μN
Ff=0.325*2.205
Ff=0.717

Answer 22

how about just leaving it as
f=μN=μmg
for now?
it looks much prettier and is easier to see what we are talking about and identify the parts

Answer 23

lol ok

Answer 24

also, it is not uncommon for things to cancel algebraically, so unless you want to keep doing the same redundant calculations in you calculator again and again
(times mass, divided by mass, then times mass again for example) I advise you to do it the cool way

now back to the matter at hand, so what do we need to make this solveable?
we need to think about the definition of work, which is what?

Answer 25

W=Fd

Answer 26

yes, that is one
and we know the force acting on the doll, and the distance over which it acts
and what has this got to do with the kinetic energy of the object?

Answer 27

W=0.717(11.9)
W=8.5J...Doesn't that mean that KE=8.5J? But this is the work that friction does on the object. correct?

Answer 28

ah numbers!
I don't want to reach for my calculator...
oh wait, I do read you say "Doesn't that mean that KE=... But this is the work that friction does on the object. correct?"

the answer to that is a big "yes"
the other definition for work besides the one you gave a simple version of is that work equal the change in energy of an object over a distance

Answer 29

so that is the important part, that\[W=fd=\Delta K\]in general\[W=\Delta K+\Delta U\]but here there is no change in potential energy, so \(\Delta U=0\)

Answer 30

so now you should be able to deduce the initial velocity of the sponge-pencil object

Answer 31

So how does that give us kinetic energy? Because we still don't know v...

Do I substitute Fd=mv^2? I'm confused as to how to proceed...

Answer 32

you know m, d, and F, solve for v

Answer 33

0.085J=mv^2? but it's change in v right

Answer 34

well yes, that is a good observation, that is not necessarily really \(v_0\) but is actually \(\Delta v\)
however in this case what is the final velocity?

Answer 35

0. but that would give us 0.085=0

Answer 36

oh no, no it wouldn't. sry.

Answer 37

0.085=0.225(v^2)
vo=0.615m/s

Answer 38

ok now I will check you numbers, let me get a calculator
in the meantime, you realize what happened exactly here?

Answer 39

Yup. I got it, thanks so far

Answer 40

we have that\[fd=\Delta K=\frac12m(v_f-v_o)^2\]but since vf=0 we get\[fd=\frac12mv_0^2\]

Answer 41

I got 0.860 for vo, i forgot the 1/2.

Answer 42

that's why I'm gonna show you why it's better to keep numbers until the end
I also should mention that it helps with sigfigs too....

Answer 43

so that's\[
W=\Delta K\]\[fd=m\Delta v\]\[\mu N=\frac12m(\cancel{v_f}^0-v_i)^2=\frac12mv_i^2\]\[\mu \cancel mg=\frac12\cancel mv^2\]\[\mu g=v^2\implies v=\sqrt{\mu g}\]so all that using of the mass in our calculations is totally an unnecessary waste of time

Answer 44

Hey Turing I need to leave for soccer practice, so I'll get back in 2 hours or so to complete this question...sorry :/

Answer 45

that's cool, while in soccer meditate on the fact that m didn't need to be there
and yeah, realize the problem is not done; that's not the velocity we actually want

Answer 46

typo I forgot the 2 now lol\[
W=\Delta K\]\[fd=m\Delta v\]\[\mu N=\frac12m(\cancel{v_f}^0-v_i)^2=\frac12mv_i^2\]\[\mu \cancel mg=\frac12\cancel mv^2\]\[2\mu g=v^2\implies v=\sqrt{2\mu g}\]

Answer 47

it magically.... oops :/

Answer 48

I put it in the wrong place!

Answer 49

\[W=\Delta K\]\[fd=m\Delta v\]\[\mu Nd=\frac12m(\cancel{v_f}^0-v_i)^2=\frac12mv_i^2\]\[\mu \cancel mgd=\frac12\cancel mv^2\]\[2\mu gd=v^2\implies v=\sqrt{2\mu gd}\]what mistake? I don't know what you mean.
(...:P)

Answer 50

they keep creeping in!

Answer 51

but now I think we're good
from here it is a momentum problem

Answer 52

Interesting discussion here on this problem!

Answer 53

indeed, do you know how to proceed from here?
if not I will be back in 10 min

Answer 54

oh you are not the asker lol

Answer 55

correct.

Answer 56

well that proves my level of absent-mindedness
I'm glad you enjoyed me acting the fool, but even fools must eat
later!