
#1
Nov1808, 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 27.4^{o} 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 = mv^{2}/2 p = mv J = p_{2}  p_{1} = F[tex]\Delta[/tex]t = m*a*t 3. The attempt at a solution a) 0.0524*16.3^{2}/2 = 6.96 J both going up and coming down b) 0.0524*16.3 = 0.854 So p_{1} 0.854 @ 27.4^{o} p_{2} 0.854 @ 27.4^{o} [tex]\Delta[/tex]p = p_{2}  p_{1} = 1.71 c) Please check a and b  because I just can't seem to get this to work I've used h_{max} = V^{2}SIN^{2}[tex]\theta[/tex]/2g and t = squareroot (2*h_{max}/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 



#2
Nov1808, 03:34 PM

HW Helper
P: 2,250

Hi JWDavid,




#3
Nov1808, 03:42 PM

HW Helper
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] 



#4
Nov1808, 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 yvelocity. The xvelocity of the ball remains unchanged. Edit: I see I was too slow on the first part... 



#5
Nov1808, 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 deltap... but this means that deltap only equals 0.0796 which can't be right?!? Because if I vector calculate p2p1 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... 



#6
Nov1808, 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. 



#7
Nov1808, 05:02 PM

HW Helper
P: 2,250





#8
Nov1808, 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. 



#9
Nov1808, 06:23 PM

HW Helper
P: 2,250





#10
Nov1908, 12:45 AM

P: 23

I did this portion in my head  and got it way wrong! Thanks got it now. 


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