*This preview shows
pages
1–3. Sign up
to
view the full content.*

Chapter 11 Solutions
Problem
1. What is the impulse associated with a 650-N force acting for 80 ms?
Solution
The impulse for a constant force (from Equation 11-1) is
I
=
!
F
dt
=
(650 N)(0.08 s)
=
52 N
"
s
, in the direction of the
force.
Problem
6. A proton moving in the positive
x
direction at
4.3 Mm/s
collides with a nucleus. The collision lasts 0.12 fs, and the
average impulsive force is
42
î
+
17
ˆ
j
μ
N
.
(a) Find the velocity of the proton after the collision. (b) Through what angle
has the proton’s motion been deflected?
Solution
(a) Equation 11-2 gives
!
p
=
m
v
f
"
m
v
i
=
F
av
!
t
,
so
v
f
=
v
i
+
F
av
!
t
=
m
=
(
4
.
3 Mm
/
s
)
î
+ (
42
î
+
17
ˆ
j
)(
10
"
6
N
)
#
(
0
.
12
!
10
"
15
s
)
=
(
1
.
67
!
10
"
27
kg
)
=
(
7
.
32
î
+
1
.
22
ˆ
j
)
Mm
/
s.
(b)
!
f
=
tan
"
1
(1.22
=
7.32)
=
9.48
°
,
CCW from the
x
-axis, which
is the deflection. (Note:
v
f
=
7.42 Mm/s
.
Problem
11. (a) Estimate the impulse imparted by the force shown in Fig. 11-18. (b) What is the average impulsive force?
Solution
(a) The impulse is the area under the curve in Fig. 11-18. By counting boxes, each of which has “area”
(0.25 ms)(0.5 N)
=
1.25
!
10
"
4
N
#
s,
we find
I
'
(
55
.
5 boxes
)
!
(
1
.
25
!
10
"
4
N
#
s
/
box)
=
6
.
94
!
10
"
3
N
#
s
.
(The direction of
I
is in the
direction of
F
, assumed to be constant.) (b)
F
av
=
I
=
!
t
=
6
.
94
"
10
#
3
N
$
s
=
3
"
10
#
3
s
=
2
.
31 N
.
FIGURE 11-18 Problem 11 Solution.

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*