homework 21 – BAUTISTA, ALDO – Due: Mar 22 2006, 4:00 am
1
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Question 1
Part 1 of 1.
10 points.
Two identical balls are on a frictionless hor
izontal tabletop. Ball
X
initially moves at 10
meters per second, as shown in Fgure on the
lefthand side. It then collides elastically with
ball
Y
, which is initially at rest. After the col
lision, ball
X
moves at 6 meters per second
along a path at 53
◦
to its original direction,
as shown in on the righthand side.
10 m/s
X
before
Y
after
6 m/s
X
0 m/s
53
◦
Which of the following diagrams best repre
sents the motion of ball
Y
after the collision?
1.
after
8 m/s
Y
37
◦
correct
2.
4 m/s
Y
after
3.
after
6 m/s
Y
53
◦
4.
after
4 m/s
Y
37
◦
5.
after
6 m/s
Y
37
◦
6.
0 m/s
Y
after
7.
6 m/s
Y
after
8.
after
4 m/s
Y
53
◦
9.
after
8 m/s
Y
53
◦
10.
8 m/s
Y
after
Explanation:
±rom conservation of momentum,
m
1
~v
1
i
+
m
2
~v
2
i
=
m
1
~v
1
f
+
m
2
~v
2
f
m~v
=
m~v
1
+
m~v
2
~v
=
~v
1
+
~v
2
.
This relation leads to the vector diagram be
low, where it is used that when two equal
mass have an elastic collision, and one of the
masses was originally at rest, they move per
pendicular to each other after the collision.
~v
2
θ
~v
1
φ
~v
Now 90
◦
=
φ
+
θ
, and with
φ
= 53
◦
θ
= 90
◦

53
◦
= 37
◦
Then

~v

cos
θ
=

~v
2


~v
2

= (10 m
/
s) cos 37
◦
= 8 m
/
s
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View Full Documenthomework 21 – BAUTISTA, ALDO – Due: Mar 22 2006, 4:00 am
2
Question 2
Part 1 of 3.
10 points.
A hockey puck of mass 0.6 kg with an ini
tial speed of 2
.
76 m
/
s collides elastically with
another equally massive puck initially at rest.
After the collision, the two hockey pucks are
observed to have the same speed, each recoil
ing symmetrically at equal magnitude but op
positely directed angles from the initial puck’s
direction.
v
0
v
Φ
Before
After
What is the angle Φ between the two recoil
ing pucks?
Correct answer: 90
◦
(tolerance
±
1 %).
Explanation:
x
y
Φ
θ
1
θ
2
v
0
By momentum conservation
x
:
m v
0
=
m v
cos
θ
1
+
m v
cos
θ
2
y
:
0 =
m v
sin
θ
1

m v
sin
θ
2
and so, from the second equation,
θ
1
=
θ
2
. By
energy conservation in an elastic collision:
1
2
m v
2
0
=
1
2
m v
2
+
1
2
m v
2
,
so
v
2
=
v
2
0
2
,
or
v
=
v
0
√
2
.
Inserting these results into the equation for
the
x
component of momentum, we get
v
0
= 2
v
cos
θ
1
=
√
2
v
0
cos
θ
1
and then cos
θ
1
=
1
√
2
,
which means that
θ
1
= 45
◦
.
Using that Φ +
θ
1
+
θ
2
= 180
◦
, we
get
Φ = 90
◦
.
Question 3
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 Spring '08
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 Energy, Friction, Kinetic Energy, Momentum, Work, m/s, – BAUTISTA

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