Momentum & Energy Conceptuals! Answers not given, sorry. --- SOLVED QUESTIONS. 32) How many joules of potential energy does a 1 kg book gain when it is tossed across the room at a speed of 2 m/s? How much energy is imparted to the wall it accidentally encounters? 33) Which has the greater kinetic energy - a car traveling at 30 km/h or a half-as-heavy car traveling at 60 km/h? --- UNSOLVED. @AllTehMaffs, help please (below)! T-T 40) If a moving object doubles its speed, how much more momentum does it have? How much more energy? 58) Consider the swinging balls apparatus. If two balls are lifted and released, momentum is conserved as two balls pop out the other side with the same speed. Can you explain why this never happens (Is this a typo? I Googled it and it said something about one ball popping out).
What do ya think 32 a) is?
A) would be... uh 2 joules?
What kind of energy is it asking for?
Joules? c: I believe it's kinetic...
Like, directly, what does the question ask about the energy? Does it ask for the potential energy of the book or the kinetic energy of the book?
Nope. Doesn't say
32) "How many joules of potential energy does....."
OH. that. Well, yes; but I thought that was a typo for "kinetic"/
That's a pretty big typo.... :P
You can never be faulted for answering a question as written - you can always be faulted for incorrectly inferring something not there.
My answer is 2 joules of kinetic due to conservation of momenta \[PE_0 + \cancel{KE_0}^0 = KE_f + \cancel{PE_f}^0\]Well, it's happened before (e.g. km/s when it should have been m/s)...
First, that's conservation of energy. Second, it doesn't apply here because it doesn't say any which way the books is being thrown. If it's being thrown horizontally, then there is no change in height, therefore no change in potential energy - ergo, what can you absolutely know about the potential energy of the book with the given info? (it's a trick question)
Sorry, I meant energy LOL o.o what?
What can you absolutely know about the potential energy of the book with the given information?
It's 0?
For the book one, there is zero change in potential energy, correct. ^^
Part b is not zero, however.
\[PE_0=0,~KE_f=?\] Eh... how do I solve o_o;;
You do know K though, you wrote it before.
It's what you originally thought the energy of the book was.
... I forgot...
The thread didn't forget :/
2 Joules?
Maybe - prove it to me. How did you get that before?
aka - What's the formula for Kinetic Energy?
I got 2 joules from conservation of energy...
\[KE_{i \ book}=\frac{1}{2}mv^2 = \frac{1}{2}(1kg)(4m^2/s^2)=2J\] So if that's all the energy that the book has to start, and when it hits the wall it has zero velocity \[KE_{f \ book} = \frac{1}{2}mv^2=\frac{1}{2}(1kg)(0m^s/s^2) = 0J\] Then CoE \[E_i=E_f\] \[KE_{i \ book} = KE_{f \ book} + W_{NC \ wall}\] where WNC is the Non-Conservative Work done by the wall to stop the book - aka, the amount of energy that is imparted to the wall. Therefor, how much energy is imparted to the wall when the book hits it?
?????
Isn't it 2m/s?
\[ v=2m/s\] \[v^2 = 4 m^2/s^2\] and energy is in joules, so the energy imparted to the wall is also in units of joules
Ooookay.... got that
And so you know that the book has 2 Joules of energy via kinetic energy when thrown. When it hits the wall, what happens to the book's velocity?
2 J transferred to the wall
Bingo ^^
The book's velocity goes to zero, and the energy that the book loses is imparted to the wall.
Okay. Is #33 "both have the same KE"?
Plug em both into \[K_1 = \frac{1}{2}m_1v_1^2\] \[K_2=\frac{1}{2}m_2v_2^2\] \[K_1 \overset{?}{=} K_2\]
How did you do that last equals sign? Also, I did do that and ended up getting the same KE.
Then you made an error, unfortunately, and it's magic!! ^^
1. what? 2. -_-
1) you did it wrong. try it again. 2) \overset{?}{=}
1. okay 2. thanks --- car 1\[\frac{1}{2}mv^2=\frac{1}{2}m\times \frac{30km\times 10^3m\times 1h}{1h\times 1km\times 3600s}\approx 4.17\times m\]
you didn't actually square that velocity
oh I forgot to type that in.
... WAIT
car 1: 69.9m car 2: 278m
What's this m business?
you're not measuring distance
It's mass. Also, can momenta cancel (and can energy cancel)?
you're not using momentum, so I don't understand that, nor do I understand can "energy cancel" :/
Also, if you're just not going to actually put the mass in, the coefficient needs actual units. Otherwise that's in units of mass
Oh, no it's another question: "Can momenta cancel? Can energies cancel?" Well, I don't have a given mass...
you have a relationship between them (and sorry, I thought that they were given - the other things still need units though!) \[m_1=m\] \[m_2=m/2\] You can also do that with the velocities \[v_1 = v = 30km/h\] \[v_2 = 2v = 60km/h\] \[K_1 =\frac{m_1v_1^2 }{2}=\frac{mv^2}{2}\] \[K_2 = \frac{m_2v_2^2}{2} = \frac{(m/2)(2v)^2}{2} = mv^2\] \[K_2 = 2K_1\]
Also, can you think of an example of momenta canceling? How about energies canceling?
Nope. o.o
What happens if two balls of equal mass travel towards each other at equal velocities and inelastically collide, leaving them stuck together and still?
zero energy
and momentum
p1=-p2 \[p_1 + p_2 = -p_2+p_2 = 0\] The energies don't specifically cancel though - the energy just changes form. there's sound and heat and light and whatnot from the collision, so only the momentum cancel each other.
Ah, right... so energies do NOT cancel.
@AllTehMaffs I solved for the kinetic energies and they're still the same? #32
I just showed that they're not equal - Car 2 has twice as much kinetic energy as car 1. It's the long post like 10 comments up.
and you mean 33 right?
oops yeah Also... if a moving object doubles its speed, how much IMPULSE does it have? In the previous question (asking for momentum and energy) I got double the momentum and 4x the energy.
@AllTehMaffs I looked and couldn't find the long post...
It's the one that ends with \[K_2=2K_2\] - you didn't get that one? :/ :(
derp, \[K_2 = 2K_1\]
I got it but I can't turn the information in my work into that...
So you should use either \[m_1=m \\ m_2 = m/2\] or \[m_1 = 2m \\ m_2 = m\] Then for the v's \[v_1 = v = 30km/h = \frac{25}{3}m/s\] \[ v_2 = 2v = 2(30 km/h) = 2\left(\frac{25}{3} m/s\right)\] Then plug the values into each expression \[K_1 = \frac{m_1v_1^2}{2}\] \[K_2 = \frac{m_2v_2^2}{2}\] and compare them ^_^ Remember that \[(ab)^2 = a^2b^2\] ie \[(2a)^2 = 4(a^2)\]
okeys :p
40) Just like before, you can define \[v_1=v \\ v_2 = 2v\] Then compare \[ p_1 \ \text{vs} \ p_2\] and \[K_1 \ \text{vs} \ K_2\] 58) Needs a picture or more detail. What *is the "swinging balls apparatus?" - cannot infer anything based on given info....
Also, waaaay up above, you mentioned that you found the energy of the book with conservation of energy, and I said "that doesn't apply here." I meant that the way you had written the equation was not the correct method for the problem. \[E_i = E_f\] was true, but there wasn't any potential energy to speak of, so saying that the initial and final potential energies were different was incorrect. Conservation of energy still applied, however, it was just that the wall ate up the energy that was imparted to the book by the thrower. ^_^
okey
ԅ(≖ ω ≖ԅ)
o.o?! @♪Chibiterasu
How goes energy?
It goes fine (as soon as I have enough time to finish the rest). @AllTehMaffs The "swinging ball apparatus" is a Newton's cradle.
Join our real-time social learning platform and learn together with your friends!