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Momentum/impulse check & help needed

by JWDavid
Tags: check, momentum or impulse
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JWDavid
#1
Nov18-08, 03:28 PM
P: 23
1. The problem statement, all variables and given/known data
A 52.4 g ball is thrown from the ground into the air with an initial speed of 16.3 m/s at an angle 27.4o above the horizontal
a) What are the values of kinetic energy of the ball initially and just before it hits the ground
b) Find the corresponding values of the momentum (magnitude and direction) and the change in momentum.
c) Show that the change in momentum is equal to the weight of the ball multiplied by the time of flight, and thereby find the time of flight.


2. Relevant equations
KE = mv2/2
p = mv
J = p2 - p1 = F[tex]\Delta[/tex]t = m*a*t


3. The attempt at a solution
a) 0.0524*16.32/2 = 6.96 J both going up and coming down
b) 0.0524*16.3 = 0.854 So p1 0.854 @ 27.4o
p2 0.854 @ -27.4o
[tex]\Delta[/tex]p = p2 - p1 = 1.71
c) Please check a and b - because I just can't seem to get this to work

I've used hmax = V2SIN2[tex]\theta[/tex]/2g
and t = squareroot (2*hmax/g) to determine that t should equal 0.76s

but every time I try to use my change of momentum (b) to get this answer they don't agree.

Thanks in advance
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alphysicist
#2
Nov18-08, 03:34 PM
HW Helper
P: 2,249
Hi JWDavid,

Quote Quote by JWDavid View Post
1. The problem statement, all variables and given/known data
A 52.4 g ball is thrown from the ground into the air with an initial speed of 16.3 m/s at an angle 27.4o above the horizontal
a) What are the values of kinetic energy of the ball initially and just before it hits the ground
b) Find the corresponding values of the momentum (magnitude and direction) and the change in momentum.
c) Show that the change in momentum is equal to the weight of the ball multiplied by the time of flight, and thereby find the time of flight.


2. Relevant equations
KE = mv2/2
p = mv
J = p2 - p1 = F[tex]\Delta[/tex]t = m*a*t


3. The attempt at a solution
a) 0.0524*16.32/2 = 6.96 J both going up and coming down
b) 0.0524*16.3 = 0.854 So p1 0.854 @ 27.4o
p2 0.854 @ -27.4o
[tex]\Delta[/tex]p = p2 - p1 = 1.71
I don't believe this is correct. The momentum vectors are not parallel, so you need to use vector subtraction to find the change in momentum.
G01
#3
Nov18-08, 03:42 PM
HW Helper
G01's Avatar
P: 2,688
Remember that your change in momentum is a vector not a scalar. You are subtracting the momenta like scalars, not like vectors.

HINT:

The x components of the momenta must remain constant since there is no force in that direction. Thus:

[tex]\Delta \vec{p}=\Delta p_y \hat{y}[/tex]

jamesrc
#4
Nov18-08, 03:43 PM
Sci Advisor
PF Gold
P: 478
Momentum/impulse check & help needed

Momentum is a vector quantity so while you may have calculated the before and after momentum correctly, you cannot simply subtract them the way you did.

Also, your calculation for time of flight is only the time for the ball to fall from its max height, not the full time of flight.

Note that the only force acting on the ball is gravity, hence the impulse on the ball is that force (the weight of the ball) times the time during which that force is applied (the entire time of flight). This is what accounts for the change in y-velocity. The x-velocity of the ball remains unchanged.

Edit: I see I was too slow on the first part...
JWDavid
#5
Nov18-08, 04:42 PM
P: 23
Okay, so t should be 2*root(...) or 1.52
meaning that if it is possible to prove c) then 1.52s * 0.0524g (ball mass) should equal delta-p... but this means that delta-p only equals 0.0796 which can't be right?!?
Because if I vector calculate p2-p1 for p(x) = 0 for
p(y) = -8.54SIN27.4 - 8.54SIN27.4 = -3.93 - 3.93 = -6.86 Ns
which does not equal TOF when divided by mass ... soooooo... help...
jamesrc
#6
Nov18-08, 04:49 PM
Sci Advisor
PF Gold
P: 478
Look at your units. And your addition... you are closer than you think.

You've got your units for impulse and your units for momentum wrong.
alphysicist
#7
Nov18-08, 05:02 PM
HW Helper
P: 2,249
Quote Quote by JWDavid View Post
Okay, so t should be 2*root(...) or 1.52
meaning that if it is possible to prove c) then 1.52s * 0.0524g (ball mass) should equal delta-p... but this means that delta-p only equals 0.0796 which can't be right?!?
No, I believe you forgot to multiply by g.

Because if I vector calculate p2-p1 for p(x) = 0 for
p(y) = -8.54SIN27.4 - 8.54SIN27.4 = -3.93 - 3.93 = -6.86 Ns
This is closer to what I was talking about, but the momentum magnitude is 0.854, not 8.54, in your units.
JWDavid
#8
Nov18-08, 05:20 PM
P: 23
you're right I did forget g, and I did forget to transfer the error correction I noted from my paper of (DECIMAL) 854 not 8.54.

But I'm still not feeling it:

I now have the [tex]\Delta[/tex]p = -.854Sin27.4 - .854SIN27.4 = .686 Ns for the change in impulse.

I now have the time (calculated using 2* ( t = root(2h/g)) equal to 1.52 s (as a check)

But when I take the impulse and divide by the weight I get:
.686 / (.0524*9.8) = 1.34 s and this is as close as I've gotten.

and this is Ns / kg * m / s^2 --- N = kg * m / s^2 so for units we have Ns/N so seconds is right.

I'm just not seeing it, but I really do appreciate all the help so far.
alphysicist
#9
Nov18-08, 06:23 PM
HW Helper
P: 2,249
Quote Quote by JWDavid View Post
you're right I did forget g, and I did forget to transfer the error correction I noted from my paper of (DECIMAL) 854 not 8.54.

But I'm still not feeling it:

I now have the [tex]\Delta[/tex]p = -.854Sin27.4 - .854SIN27.4 = .686 Ns for the change in impulse.
I think you have a calculator error here. Can you check it again?
JWDavid
#10
Nov19-08, 12:45 AM
P: 23
Quote Quote by alphysicist View Post
I think you have a calculator error here. Can you check it again?
See NOW THIS is the problem with growing up not being allowed to use calculators!
I did this portion in my head - and got it way wrong!

Thanks got it now.


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