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HW#3b
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2D motion: Rule of thumb: treat horizontal directions motion separately from vertical motion.
Initial velocity v0 may have two components:
in x direction and y direction, v
0x
or v
0y
. Use
the given angle and sin or cos to calculate v
0x
or v
0y
.
A few questions have only x direction initial velocity, zero launch angle: v
0x
=v
0
, v
0y
=0.
Horizontal: X direction, constant velocity. no acceleration. set x=0 at the initial point. so that
x=v
0x
*t. (v0x is initial velocity's in horizontal direction)
If you can calculate the time t, you know the corresponding x position easily.
Vertical: Y direction, constant acceleration due to gravity, (y direction motion is the same as
the question we asked in the first quiz. )
set ground to be y=0, y
0
is equal to the y value at t=0 (platform, cliff, etc.
..), set upward to be
the positive direction and acceleration in y direction a = g.
If you know y direction v
0
, y
0
, a, maybe also final y position or final y velocity, you should be
able to solve y direction.
Rule of thumb: treat horizontal directions motion separately from vertical motion. Write 2
separate motion equations in x or y directions.
What limits the travel time? y direction motion equation.
What limits the maximum height? y direction motion equation.
What determines the horizontal distance? initial velocity in x direction and the total time.
Another good advice is to form study groups and discuss with your peers.
After the HW is due, you can request webassign to release the keys. Within a few days,
detailed solution steps will be put on my website
. You should still work hard on the HW
questions after the due date and understand them all. Otherwise, you will fail in Quiz
or Exams.
Problem 1 (Walker Ch. 4 P. 25).
A cork shoots out of a champagne bottle at an angle of [34.0]
degrees above the horizontal.
If the cork travels a horizontal distance of 1.30 m in [1.35] s, what was
its initial speed?
Solution.
This is a pretty darn easy problem, if we assume that it lands at the same vertical
position as it starts out, and assuming we remember the range equation,
()
2
0
sin 2
v
R
g
θ
=
We can just solve this for the initial speed
v
0
:
0
2
sin 2
9.8 m/s
1.30 m
sin 68
3.71 m/s
gR
v
=
=
°
=
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OK.
This is not the right answer.
Because it was wrong to assume that the cork had landed.
And we did not use the time that was given. It is risky to use these kind of equations without
carefully examine the conditions.
We should not forget the basic rules to treat 2D motions.
So . . . the information given tells us that the horizontal
component of the cork's velocity, which is constant, is
equal to
1.30 m
0.970 m/s
1.34 s
x
x
v
t
∆
==
=
∆
Now, referring to the diagram, the relation for the
x
component of
v
0
gives
0
0
cos
0.970 m/s
1.170 m/s
cos
cos34
x
x
vv
v
v
θ
=
⇒=
=
=
°
And that's right.
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 Spring '09
 JamesM.Lockhart
 Physics

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