homework 08 – GAUTHIER, JOSEPH – Due: Oct 26 2007, 1:00 am
1
Question 1, chap 11, sect 4.
part 1 of 1
10 points
An open cart on a level surface rolls without
frictional loss through a downpour of rain.
The rain falls vertically downward as shown
below.
As the cart rolls, an appreciable amount of
rain water accumulates in the cart.
v
rain
rain water
cart
The speed of the cart will
1.
increase because of conservation of mo
mentum.
2.
decrease because of conservation of me
chanic energy.
3.
increase because of conservation of me
chanic energy.
4.
remain the same because the raindrops
are falling perpendicular to the direction of
cart’s motion.
5.
decrease because of conservation of mo
mentum.
correct
Explanation:
This is an inelastic collision in the direction
along which the cart is rolling.
Only mo
mentum
vectorp
along that direction is conserved.
Because the raindrops fall vertically, they do
not carry momentum horizontally.
Assume
Δ
m
of rain water accumulates on the cart:
p
i
=
p
f
m v
= (
m
+ Δ
m
)
v
′
.
Therefore
v
′
=
m
m
+ Δ
m
v
v
′
< v .
The speed of the cart will
decrease
because
of conservation of momentum.
Question 2, chap 11, sect 1.
part 1 of 2
10 points
A 52 kg pole vaulter falls from rest from a
height of 4.9 m onto a foam rubber pad. The
pole vaulter comes to rest 0.27 s after landing
on the pad.
a) Calculate the athlete’s velocity just be
fore reaching the pad.
Correct answer:
−
9
.
805
m
/
s (tolerance
±
1
%).
Explanation:
Let :
m
= 52 kg
,
Δ
y
=
−
4
.
9 m
,
t
= 0
.
27 s
,
and
a
=
−
9
.
81 m
/
s
2
.
Since
v
i
= 0 m/s,
v
2
f
= 2
a
Δ
y
v
f
=
±
radicalbig
2
a
Δ
y
=
±
radicalBig
2 (
−
9
.
81 m
/
s
2
) (
−
4
.
9 m)
=
−
9
.
805 m
/
s
,
which is 9
.
805 m
/
s directed downward.
Question 3, chap 11, sect 1.
part 2 of 2
10 points
b) Calculate the constant force exerted on
the pole vaulter due to the collision.
Correct answer:
1888
.
37
N (tolerance
±
1
%).
Explanation:
Since
v
f
= 0 m/s,
vector
F
Δ
t
=
m
Δ
vectorv
=
mvectorv
f
−
mvectorv
i
=
−
mvectorv
i
F
=
−
m v
i
Δ
t
=
−
(52 kg) (
−
9
.
805 m
/
s)
0
.
27 s
=
1888
.
37 N
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homework 08 – GAUTHIER, JOSEPH – Due: Oct 26 2007, 1:00 am
2
directed upward.
Question 4, chap 11, sect 1.
part 1 of 1
10 points
A 1 kg steel ball strikes a wall with a speed
of 4
.
05 m
/
s at an angle of 51
.
1
◦
with the
normal to the wall.
It bounces off with the
same speed and angle, as shown in the figure.
x
y
4
.
05 m
/
s
1 kg
4
.
05 m
/
s
1 kg
51
.
1
◦
51
.
1
◦
If the ball is in contact with the wall for
0
.
156 s, what is the magnitude of the average
force exerted on the ball by the wall?
Correct answer:
32
.
6058
N (tolerance
±
1
%).
Explanation:
Let :
M
= 1 kg
,
v
= 4
.
05 m
/
s
,
and
θ
= 51
.
1
◦
.
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 (1 kg) (4
.
05 m
/
s) cos 51
.
1
◦
0
.
156 s
bardbl
vector
F
bardbl
=
32
.
6058 N
.
Note:
The direction of the force is in negative
x
direction, as indicated by the minus sign.
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 Fall '07
 Gilbert
 Momentum, m/s, JOSEPH

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