Question

A 1.50 x 10^3 kg car leaves a parking lot. Thirty seconds later it is moving along a highway at 72km/hr.

a. What is the car's change in momentum
b. What average force does the motor produce to bring about this change in momentum.

Answer 1

whats p? and what problem?

Answer 2

so isn't it the change in momentum we are looking for?

Answer 3

Isn't it \(p=mv\) ?

Answer 4

isn't it change in p = mvf-mvi?

Answer 5

oh yea, let theEric help us. I am not good at Physics

Answer 6

@jacobsmith222
Yeah, the change would be like
\(p_\text{72km/h}-p_\text{parked}=\Delta p\)

Answer 7

then what is p=mv?

Answer 8

That is the definition of momentum. Thanks @Loser66! You got kinetic energy there, hehe... :)

Answer 9

So \(p\) is momentum.
And \(m\) is the mass.
And \(v\) is the velocity.

Answer 10

okay that makes sense. Would it be initial or final velocity for v in the definition

Answer 11

So you see that it's a simple equation! And not hard to remember! If you ever got hit by something, you want it to have less momentum.
Lots of momentum: football player, 200 pounds, running full speed.
Not much momentum: a fly buzzing around, lazily.

\(p=mv\)

Answer 12

okay so how would we solve this problem?

Answer 13

When you look at the change, it's always \(\text{final - initial}\).

And here's a trick for substituting. When you look at \(p_\text{parked}\) for example, you replaced it with \(m_\text{parked}v_\text{parked}\). Of course, \(m\) is always the same, so you can just say \(mv_\text{parked}\).

Answer 14

Okay so is it going to be 1.5x10^3(72)-1.5x10^3(0)

Answer 15

also for the problem yesterday about finding the magnitude of the impulse, is it NS and change of p=FxT

Answer 16

Yeah, that's the calculation for part (a)!

Answer 17

and final units is kg/km/hr

Answer 18

actually, kg * km / hr since its mass * velocity. :)

Answer 19

okay and also for the problem yesterday about finding the magnitude of the impulse, is it NS and change of p=FxT

Answer 20

Newton seconds for impulse? I think so!

Answer 21

okay and how would we achive b?

Answer 22

Was I in a problem with impulse yesterday? Hehe... I don't remember...

Answer 23

It was A force of 30 N is applied to a hockey puck for .8s. Find the magnitude of the impulse

Answer 24

So, impulse if often \(J\) - is that what you're using?

Answer 25

p

Answer 26

Ah, see, I use \(p\) for momentum and \(J\) for impulse.
If you use \(p\) for impulse, what do you use for momentum?
I would do \(J=\bar F \times t\).

Answer 27

But I think there is a more helpful equation.

Answer 28

p=fxt

Answer 29

\(J=\dfrac{\Delta p}{\Delta t}\) by definition, right?

Answer 30

Scratch that - I want to use your variables.

Answer 31

wait before we do this. How do I find b?

Answer 32

Haha, we can combine our two impulse equations... We can find impulse with my last equation, and we can solve for \(\bar F\) in your first equation. The bar just means "average."

What variable do you use for momentum?

Answer 33

p

Answer 34

And what do you use for impulse?

Answer 35

idk. All we have is change in p =fxt

Answer 36

I made a mistake before, \(\bar F=\dfrac{\Delta p}{\Delta t}\), sorry! But that's how you would do part (b)... THAT's the definition that will help us!

Answer 37

And that, really, is the same as your equation. No impulse involved there! :)

Answer 38

so what exactly would i plug in

Answer 39

\(\bar F=\dfrac{\Delta p}{\Delta t}\Longleftrightarrow \Delta p=\bar F\ \Delta t\)

Answer 40

could i just do p=ma and find a and plug it in?

Answer 41

\(F=ma\), not \(p=ma\)!
And it will take some work to find \(a\), and you don't have to! But that is one way.

But give my equation a shot. Force is the change in momentum over a change in time. You know both the change in momentum (you calculated it) and the change in time (it's given in the problem).

Answer 42

108000/30

Answer 43

3600

Answer 44

Mind the units! :) Sorry for my delay...

Answer 45

Notice the momentum is in kg km/hr and your time, 30, is in seconds.

So your 3600 is 3600 kg*km/hr/s, which isn't quite Newtons! It actually isn't a unit of force, without the time unit being the same, I think.

Answer 46

I got 3600kgm/s^2

Answer 47

Oh, haha, sorry! Let me check. Those would be the right units, sorry! :)

Answer 48

also cna you help me out with

1. A force of 8.0 N acts on a 2.0 kg mass for 5.0s
a. what is the change in momentum of the mass
b. what is the change in the velocity of the mass

2. A force of 6.00 N acts on a 3.00 kg object for 10.0 s
a. What is the objects change in momentum
b. What is its change in velocity

Answer 49

I suggest you ask that in another post! That way others can see and help too.
I'll look at it in between my distractions!

Answer 50

thanks!!

Answer 51

Will they be different since its the objects change in momentum and velocity and not the change in velocity and momentum of the mass?

Answer 52

Change in momentum is 108,000kg*km/hr.
Change in time is 30s.
So average force is \(\dfrac{108,000\text{ [kg*km/hr]}}{30\text{ [s]}}=3,600\text{ kg*km/hr/s}\)

And since...
\(\mathrm{ 1km=1000m}\)
\(\mathrm{1hr=3600s}\)

we see by substitution that...

\(\mathrm{3,600 kg*km/hr/s=3,600kg*1,000m/3600s/s=1,000 kg*m/s^2=1,000N}\)

I think those are the right units.

Answer 53

I apologize but where did 108000 come from

Answer 54

That is the change in momentum! You actually calculated it above! (see the attachment)
\(\mathrm{1.50\times10^3kg\quad\times\quad72km/hr\quad-\quad1.50\times10^3kg\quad\times\quad0\\=108,000\ kg*km/hr}\)

Answer 55

oh sorry I meant for my new problem

Answer 56

Post a new one! There will be much less clutter that way..

Answer 57

2. A force of 6.00 N acts on a 3.00 kg object for 10.0 s
a. What is the objects change in momentum
b. What is its change in velocity