lai (yl8859) – HW13 – Ross – (15200)
1
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printout
should
have
6
questions.
Multiplechoice questions may continue on
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before answering.
001
10.0 points
A 9 kg steel ball strikes a wall with a speed
of 11 m
/
s at an angle of 42
.
5
◦
with the normal
to the wall.
It bounces off with the same
speed and angle, as shown in the figure.
x
y
11 m
/
s
9 kg
11 m
/
s
9 kg
42
.
5
◦
42
.
5
◦
If the ball is in contact with the wall for
0
.
198 s, what is the magnitude of the average
force exerted on the ball by the wall?
Correct answer: 737
.
277 N.
Explanation:
Let :
M
= 9 kg
,
v
= 11 m
/
s
,
and
θ
= 42
.
5
◦
.
The
y
component of the momentum is un
changed. The
x
component of the momentum
is changed by
Δ
P
x
=
−
2
M v
cos
θ .
Therefore, using impulse formula,
F
=
Δ
P
Δ
t
=
−
2
M v
cos
θ
Δ
t
=
−
2 (9 kg) (11 m
/
s) cos 42
.
5
◦
0
.
198 s
bardbl
vector
F
bardbl
=
737
.
277 N
.
Note:
The direction of the force is in negative
x
direction, as indicated by the minus sign.
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 Winter '08
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 mechanics, Acceleration, Mass, General Relativity, Velocity, Lai

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