Physics 140  Discussion 10  Solutions
8.13.
I
DENTIFY
:
The force is constant during the 1.0 ms interval that it acts, so
.
t
=
∆
J
F
r
r
2
1
2
1
(
).
m
=
 =

J
p
p
v
v
r
r
r
r
r
S
ET
U
P
:
Let
x
+
be to the right, so
1
5 00 m/s.
x
v
= + .
Only the
x
component of
J
r
is nonzero, and
2
1
(
).
x
x
x
J
m v
v
=

E
XECUTE
:
(a)
The magnitude of the impulse is
3
3
(2 50 10 N)(1 00 10 s)
2 50 N s.
J
F t
=
∆ =
.
×
.
×
= .
⋅
–
The
direction of the impulse is the direction of the force.
(b)
(i)
2
1
.
x
x
x
J
v
v
m
=
+
2 50 N s.
x
J
= + .
⋅
2
2 50 N s
5 00 m/s
6 25 m/s.
2 00 kg
x
v
+ .
⋅
=
+ .
= .
.
The stone’s velocity has
magnitude 6.25 m/s and is directed to the right. (ii) Now
2 50 N s
x
J
=  .
⋅
and
2
2 50 N s
5 00 m/s
3 75 m/s.
2 00 kg
x
v
 .
⋅
=
+ .
= .
.
The stone’s velocity has magnitude 3.75 m/s and is directed to
the right.
E
VALUATE
:
When the force and initial velocity are in the same direction the speed increases
and when they are in opposite directions the speed decreases.
__________________________________________________________________________________
8.29.
I
DENTIFY
:
The horizontal component of the momentum of the system of the rain and freight
car is conserved.
S
ET
U
P
:
Let
x
+
be the direction the car is moving initially. Before it lands in the car the rain
has no momentum along the
x
axis.
E
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 Winter '10
 Uher
 Force, Mass, Velocity, m/s, Ax vBx

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