PHYS2014
Fall 2010
Benton
OSU Physics Dept.
1
Physics 2014: General Physics I
Solutions to Homework Assignment 2
Due:
Friday, 10 September 2010 at the beginning of your Recitation session.
Problems:
1. A pair of ice skaters are practicing a figure skating move on the ice. The two skaters are
separated by a distance of 15 m. If the first skater skates at a constant velocity of 5.0 m/s along a
line initially making a 40
°
angle with respect to the line connecting the two skaters, with what
velocity will the second skater need to skate at so that she meets the first skater at a right angle?
Solution: We are asked to find the velocity,
v
2
, for skater 2 such that she covers a distance
s
2
while skater 1 covers a distance
s
1
skating at a velocity of
v
1
. Skater 1 will take time
t
to skate a
distance
s
1
at velocity
v
1
. Skater 2’s velocity,
v
2
, must be such that she skates a distance
s
2
in that
same time
t
. Let
s
1
be the distance that skater 1 must travel and
s
2
be the distance skater 2 must
travel as shown in the diagram below.
From this diagram, it is clear that:
1
2
15 m cos40
11.49 m
15 m sin40
9.64 m
s
s
=
⋅
=
=
⋅
=
r
r
Time t is then:
1
m
5.0
s
v
=
a
θ = 40
ο
1
5
m
2
?
v
=
a
skater 1
skater 1
1
m
5.0
s
v
=
a
θ = 40
ο
2
?
v
=
a
skater 1
skater 1
1
s
a
2
s
a
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Fall 2010
Benton
OSU Physics Dept.
2
1
1
11.49 m
3.3 s
m
5.0
s
t
v
s
=
=
=
And the velocity
v
2
is:
2
2
9.64 m
m
4.2
s
3.3 s
s
v
t
=
=
=
2. I set up my inclined plane apparatus at an angle
θ
. Then I roll my billiard ball up the track with
an initial velocity of 3 m/s as shown in the diagram above. It takes the ball a total of 1.2 s to roll
up the track, come to a stop, and then roll back down into my hand. Neglecting frictional effects,
find the angle
of the track?
Solution: The problem is symmetrical in that it will take exactly the same amount of time for the
ball to roll up the ramp and come to a stop as it will for the ball to roll down the ramp starting
from rest and reaching a final speed of 3 m/s. Thus we only need to consider half the problem,
for which the time is 0.6 s. Let just consider that part of the problem where the ball starts at
velocity
v
0
= 3 m/s, rolls up the ramp while it is being slowed down by the acceleration due to
gravity, then comes to a stop after 0.6s.
We can rotate the axis by
θ°
so that the velocity vector is parallel to the xaxis. In that case, the
ball will be accelerated by that component of gravity parallel to the xaxis.
sin
x
a
g
= −
We can use our first kinematic equation
f
v
0
0
1
1
0
2
sin
m
3
s
sin
sin
30.7
m
9.80
0.6 s
s
v
at
v
g
t
v
gt
−
−
=
+
−
= −
−
−
=
=
=
−
−
r
3. You are on the target range preparing to shoot a new rifle when the range safety officer comes
up to you and asks you what the muzzle velocity is of your rifle. (Muzzle velocity is the velocity
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 Fall '08
 Nandi
 Acceleration, Work, Velocity, Physics Dept, OSU Physics Dept

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