Calculating Momentum of a Thrown Ball of Dough - Step-by-Step Guide

  • Thread starter Thread starter mortho
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
To calculate the momentum of a 0.14 kg ball of dough thrown upward at 24 m/s, the momentum formula P=mv is used. At maximum height, the velocity is zero, resulting in zero momentum. Halfway to maximum height, the velocity can be determined using kinematic equations, but the user is struggling to find the final velocity due to a lack of information on force or acceleration. The discussion highlights the challenges of solving physics problems without complete data and the importance of patience in seeking help from the forum community. Overall, the thread emphasizes the collaborative nature of academic support in solving homework problems.
mortho
Messages
99
Reaction score
0
Dough"s Momentum

Homework Statement


A 0.14 kg ball of dough is thrown straight up into the air with an initial velocity of 24 m/s.

a)Find the momentum of the of dough at its maximum height.
b)Find the momentum of the ball of dough halfway to its maximum height on the way up.



Homework Equations



P=mv

The Attempt at a Solution



Okay this problem sounds easy so i straight plugged in (0.14*24) but that wasn't right. so i tried figuring out Vf and use that but in order to find Vfinal i don't have Force for Vf=ft/m so i tried squareroot 2ax but i don't have a either. I am so lost . Please help. THanks
 
Physics news on Phys.org
sorri no one seemed to be responding when i was posting this and i really needed to get this solved ..so i changed the title .
 
mortho said:
sorri no one seemed to be responding when i was posting this and i really needed to get this solved ..so i changed the title .

Please keep in mind that this is not a service rendered by a paid staff. The people helping out in the Homework Forum are volunteering their time and a lot of them are students or other people in academic settings. That will mean right now that there are relatively fewer of them online since they are still on winter break. (I myself am not a student, but the university I'm at won't start classes again for two more weeks...) Waiting a few hours for a reply under these circumstances is not unusual, although this place has gotten noticeably livelier since Tuesday.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top