Phys 1120, CAPA #2 solutions.
NOTE: Here, I will provide "generic" solution. Since the numbers and some details vary a little,
your problems may look a wee bit different, but this solution should tell you enough to solve
your problem) The
to your particular questions are available through CAPA (on the
login page, click on "view previous set". Just be sure to fill in the set number first, before
COMMENT: You should ALWAYS try to solve problems using symbols only, even
) on "number crunching " problems. There are lots of reasons - it's easier to check
your work, it's easier to see if the units are correct, it's easier for a TA to see if you've made a
mistake, it's easier to see if various limits make sense, it's easier to see the PHYSICS when you
have a formula instead of plugging numbers in right away. It's how any professional, in any field,
will solve real life problems, even numerical ones. Get in the habit of this! We will ask exam
questions where we won't give any numbers, you will have to work on a symbolic level.
1) Particle 1 has a given mass and acceleration, so Newton-II says
F(on 1) = m(1)*a(1) is given to us.
But Newton-III says F(on 2) is the same magnitude!
N-II says for object 2, F(on 2) = m(2)*a(2). The problem asks for m(2):
m(2) = F(on 2) / a(2) (I'm considering magnitude only) = F(on 1) / a(2) = m(1)*a(1)/a(2).
All the numbers on the right are given, just plug in.
2) Now use F = k Q1 Q2 / R^2. The problem says Q1=Q2, call it Q, and we know F
from the previous problem. R was given. (Don't forget to convert cm to m first!)
So Q^2 = F*R^2/ k, or Q = Sqrt[ F*R^2/k] = Sqrt[ m(1)*a(1) * R^2 / k]
Plug in the numbers now.
3) A force diagram is needed here. Let me describe it, I
have three force arrow on the charge:
F = qE straight to the LEFT (from electricity), F = mg
straight DOWN (from gravity),
and tension T along the string direction.
The question asks for how much charge is needed).
What I'm thinking is that the object is in equilbrium.
That means the NET FORCE on it is zero. So if I write
down Newton's law (F(net)=ma), with a=0 (that's what
equilbrium implies), then I will have TWO equations to
help me, the x- and y- components. There are TWO unknowns (the Tension ithe rope, and the
charge on the ball.) So 2 equations, 2 unknowns, we should be ok!
Components of Fnet=ma: Do x- and y- separately.
F_netx = -qE + Tsin(theta) = 0 (Do you see why I use sin(theta) for the x component! Look
carefully at the picture in CAPA, and which angle is named theta.)