This preview shows pages 1–3. Sign up to view the full content.
Week 6 homework
IMPORTANT NOTE ABOUT WEBASSIGN:
In the WebAssign versions of these problems, various details have been changed, so that
the answers will come out differently.
The method to find the solution is the same, but
you will need to repeat part of the calculation to find out what your answer should have
been.
WebAssign Problem 1:
Consult
Interactive Solution 7.9
at
www.wiley.com/college/cutnell
for a review of problemsolving skills that are involved
in this problem. A stream of water strikes a stationary turbine blade horizontally, as the
drawing illustrates. The incident water stream has a velocity of + 16.0 m/s, while the
exiting water stream has a velocity of – 16.0 m/s. The mass of water per second that
strikes the blade is 30.0 kg/s. Find the magnitude of the average force exerted on the
water by the blade.
REASONING
During the time interval
∆
t
, a mass
m
of water strikes the turbine
blade. The incoming water has a momentum
m
v
0
and that of the outgoing water is
m
v
f
. In order to change the momentum of the water, an impulse
(
29
t
Σ
∆
F
is applied
to it by the stationary turbine blade. Now
(
29
t
Σ
∆
=
F
t
∆
F
, since only the force of the
blade is assumed to act on the water in the horizontal direction. These variables are
related by the impulsemomentum theorem,
t
∆
F
=
m
v
f

m
v
0
, which can be solved
to find the average force
F
exerted on the water by the blade.
SOLUTION
Solving the impulsemomentum theorem for the average force gives
(
29
=
m
m
m
t
t

=

∆
∆
f
0
f
0
v
v
F
v
v
The ratio
(
29
/
m
t
∆
is the mass of water per second that strikes the blade, or 30.0 kg/s,
so the average force is
(
29
(
29
(
29
(
29
30.0 kg/s
16.0 m/s
16.0 m/s
960 N
m
t
=

=


+
=

∆
f
0
F
v
v
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThe magnitude of the average force is
960 N .
WebAssign Problem 2:
A 55kg swimmer is standing on a stationary 210kg .oating
raft. The swimmer then runs off the raft horizontally with a velocity of +4.6 m/s relative
to the shore. Find the recoil velocity that the raft would have if there were no friction and
resistance due to the water.
REASONING
The sum of the external forces acting on the swimmer/raft system is zero,
because the weight of the swimmer and raft is balanced by a corresponding normal
force and friction is negligible. The swimmer and raft constitute an isolated system,
so the principle of conservation of linear momentum applies. We will use this
principle to find the recoil velocity of the raft.
SOLUTION
As the swimmer runs off the raft, the total linear momentum of the
swimmer/raft system is conserved:
{
s s
r
r
Total momentum
Total momentum
before swimmer
after swimmer
starts running
runs off raft
0
m v
m v
+
=
14243
where
m
s
and
v
s
are the mass and final velocity of the swimmer, and
m
r
and
v
r
are
the mass and final velocity of the raft. Solving for
v
r
gives
(
29
(
29
s s
r
r
55 kg
4.6 m/s
1.2 m/s
210 kg
m v
v
m
+
=

=

=

WebAssign Problem 3:
The lead female character in the movie
Diamonds Are Forever
This is the end of the preview. Sign up
to
access the rest of the document.
 Fall '08
 RICHARDSON, B
 Physics, Work

Click to edit the document details