What is the magnitude of the total momentum

  • Thread starter lollypop
  • Start date
  • #1
33
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hi, i did the first part of this problem but on
the second i'm stuck.

A golf ball with mass 4.5×10^-2 kg is moving
in the +x-direction with a speed of 8.65 m/s ,
and a baseball with mass 0.145 kg is moving in
the -y-direction with a speed of 7.2m/s .
--What is the magnitude of the total momentum
of the system that consists of the two balls?
(i got for this one 1.06 kg*m/s using p=mv for
each ball and adding them )
--What is the direction of the total momentum
of the system that consists of the two balls?
Express your answer as an angle measured from
below the x-axis (for this part i tried doing
arctangent but the answer is not right)

any help is good, thanks.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Write the two momentum vectors in "component" notation:

"A golf ball with mass 4.5×10^-2 kg is moving
in the +x-direction with a speed of 8.65 m/s " so it's momentum is 4.5x10^-2*8.65= 0.38925 in the positive x direction:
0.38925i+ 0j
(i is the unit vector in the positive x-direction and j is the unit vector in the positive y-direction.)

" baseball with mass 0.145 kg is moving in
the -y-direction with a speed of 7.2m/s" so its momentum is 0.145*7.2=
1.044 in the negative y direction:
0i- 1.044j

The total momentum vector for the system is the sum of those two vectors: 0.38925i- 1.044j

You can NOT just add the raw values- they are not in the same direction! (and I don't see how you could have gotten "1.06" in any case.)

The magnitude of the momentum is the "length" of that vector which is, by the Pythagorean theorem, √(0.389252+ 1.0442)= 1.114 kg m/s approximately.

Yes, you should be able to find the angle using arctan:
The vector diagram should give you a right triangle with legs of length .38925 and 1.044 (and, of course, hypotenuse of length 1.114).
The tan(θ)= .38925/1.044= .3728 so θ= 20.5 degrees (make sure your calculator is in "degree mode" if you want angles in degrees).

Now, check your diagram to see where that angle is! I have intentionally done the "wrong" angle so you will need to determine what the angle is measured from "below the x-axis".
 
  • #3
33
0
thans you, i did get the right angle after ur explanation, it was 70 degrees, thanks again! :)
 

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