Phisic's 12 Help

  • Thread starter God64bit
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Problem 1
A 12.0 kg shopping cart rolls due south at 1.50 m/s . After striking the bumper of a car, it travels at 0.80 m/s, 30° E of S. What is the magnitude of the change in momentum sustained by the shopping cart?

a drew a free body diagram used the formula
Pyo + Py1 = P'yo + P'y1 and Pxo + Px1 = P'xo + P'x1
Py1 is some mithical object or the car which i am assumeing isnt moveing im takeing the y direction of its momentom gained + the X direction of its momentom gained adding them together yields
Sqare root of ((12(1.5)-12sin(60)(.8))^2 + (12sin(30)0.8)^2 = 11.8
now heres the problem 11.8 is not an answer what did i mess up? it doesnt seem like a calculator error wheres my thinking wrong?

Problem 2
Diagram = http://quizmebc.ca/images/exams/4010113.gif

x= the object labeled x
C for center of the beam
o for the object on the right
Assume clockwise negative
Sigma t = FxLx + FcLc - FoLo = 0
Rearranged
Lx=FcLc - FoLo / -Fx = 0
(.122)(9.8)(.2) - (.5)(9.8)(.25) / - (.2)(9.8) = Lx
Lx= 1.753 .... hmm considering the length of the entire bar is .9m i think something is wrong...

I'll probally end up have 6 or 7 questions today... i am allowed to ask so many ... dumb questions? (been going through and every 10-15 questions things just dont work right for some very odd reason.)
(big test and no one around for miles that knows theirs a diffrence between mass and wieght...)
 
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Answers and Replies

  • #2
dx
Homework Helper
Gold Member
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after hitting the bumper, the south component of the momentum is 12x0.8 cos 30 (you can see this from the diagram you drew). Find the difference in the south component before and after. Since the initial east component is zero, the change in east component is simply 12x0.8 sin30.
 

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