Momentum Help: Solving a Ball of Dough's Halfway Point Velocity Problem

[hshphyss]

Momentum Help! please

Can anyone help me with this problem?

A 0.19 kg ball of dough is thrown straight up into the air with an initial velocity of 13 m/s. What is the momentum of the ball of dough halfway to its maximum height on the way up.

I tried (.19)(-13) and (.19)(13) but that was the wrong answer. Thank-you

 

[Pengwuino]

Well your momentum is your velocity * mass. While the mass does stay the same, the velocity will not. You are looking for the momentum half way through the trip so what do you think the momentum will be halfway through the trip up if its original momentum is 13m/s?

 

[hshphyss]

Oh okay so it will just be half?

 

[saschouch]

initial velocity is 13 m/s. mass is .19 kg. acceleration of gravity is -9.81 m/s^2. use those to find the velocity at a given time when it is halfway up. when you get the velocity there, use p=mv and you're done. so no, it's not half.

 

[hshphyss]

so it would be the v=(vi)(m)(g) and then plug that into the momentum formula. I got v=(-13)(.19)(-9.8)=24.21 and then 24.21(.19)=4.6 m/2 but that answer didn't work

 

[Tubs]

You have to use kinematics to determine what the velocity is at half the height, THEN you can use the momentum formula. HINT: the system has constant acceleration!

 

[Pengwuino]

so it would be the v=(vi)(m)(g) and then plug that into the momentum formula. I got v=(-13)(.19)(-9.8)=24.21 and then 24.21(.19)=4.6 m/2 but that answer didn't work

No it wouldn't.

You have to use the kinematic equation for velocity to determine how long it will take to get to the top. Using this number and the equation for position, determine the maximum height. Then use this equation again to determine what at what time value was the ball at 1/2 of it's maximum height. Using that time value, you can determine what the velocity was at that particular time.

 

[Jameson]

Yes, but to make it even shorter, you don't need to solve for the toal time. Just solve for the max height with the kinematic equation: v_{f}^2=v_{0}^2+2ad.

 

[Pengwuino]

You can't simplify it like that. its velocity 1/2 the way up isn't the same as its velocity when the velocity is 1/2 the initial.

 

[daniel_i_l]

You can use v_{f}^2=v_{0}^2+2ad
and find the final hight d where v_{f} = 0
then you can find the speed you want by useing the same equation were for
d you put half of the final hight.

 

[hshphyss]

once i find the final height (8.6)what equation would I use before I solve for momentum, the same 1? if so would it be 13^2= v^2 + 2(9.8)(8.6/2)?

 

[daniel_i_l]

You can use the same equation, but it is the final speed (vf) that you are looking for, not the initial one. You know that the initial speed (v0)is 13m/s. Now you want to find the final speed, half way up. so the 13^2 has to be on the other side.

 

[hshphyss]

I keep getting the wrong answer I'm not sure where I'm making my mistake.

So it would be 0^2=13^2+2(9.8)x and x=8.6 after i divide that by half i would do v^2=13^2+2(9.8)4.3 and I would get v^2=253.3 and i would get 15.9 but that is the wrong answer. This is an easy problem in comparison to the others I did with this assignment and I feel dumb because I can't seem to get it.

 

[daniel_i_l]

the acceleation in this case is negative because the mass is going up against gravity. That would be: v^2=13^2-2(9.8)4.3

 

[BerryBoy]

Remember!
The vector force of weight (due to gravity) is opposite the direction of initial velocity.. Recalculate your equation... (you are missing a minus sign).

Sam

 

[BerryBoy]

Thats the second time today I've replied at the same time as someone - I'm getting lucky

Sam

 

[hshphyss]

even when I do the work with the negative sign it still comes up as incorrect I don't know why

 

[daniel_i_l]

the speed should be right, maybe the problem is in the next part.

 

[BerryBoy]

Have you considered that the answer your trying to match may be wrong?

(It happens)! ;-)