PHYSICS 1010
T#5
Questions and Solutions
Chapter 9, Problem 63
A body of mass 2.0 kg makes an elastic collision with another body at rest and
continues to move in the original direction but with 1/4 of its original speed.
(a)
What is the mass of the
other body?
(b)
What is the speed of the twobody center of mass if the initial speed of the 2.0 kg body
was 4 m/s?
63.
(a) Let
m
1
be the mass of the body that is originally moving,
v
1
i
be its velocity before the collision, and
v
1
f
be its velocity after the collision. Let
m
2
be the mass of the body that is originally at rest and
v
2
f
be its
velocity after the collision. Then, according to Eq. 967,
We solve for
m
2
to obtain
We combine this with
to obtain .
(b) The speed of the center of mass is
.
Chapter 9, Problem 65
In Figure particle 1 of mass
m
1
= 3.0 kg
slides rightward along an
x
axis on a frictionless floor with a speed of 2.0 m/s.When it reaches
x
= 0, it
undergoes a onedimensional elastic collision with stationary particle 2 of mass
m
2
= 0.4 kg. When
particle 2 then reaches a wall at
x
w
= 70 cm, it bounces from the wall with no loss of speed. At what
position (in cm) on the
x
axis does particle 2 then collide with particle 1?
65.
We use Eq 967 and 968 to find the velocities of the particles after their first collision (at
x
= 0 and
t
= 0):
v
1
f
= v
1
i
=
(2.0 m/s)
= m/s
v
2
f
= v
1
i
=
(2.0 m/s) =
m/s
≈
1.7 m/s.
At a rate of motion of 1.7 m/s, 2
x
w
= 140 cm (the distance to the wall and back to
x
= 0) will be traversed
by particle 2 in 0.82 s.
At
t
= 0.82 s, particle 1 is located at
x
= (–2/7)(0.82) = –23 cm,
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Documentand particle 2 is “gaining” at a rate of (10/7) m/s leftward; this is their relative velocity at that time.
Thus,
this “gap” of 23 cm between them will be closed after an additional time of (0.23 m)/(10/7 m/s) = 0.16 s
has passed.
At this time (
t
= 0.82 + 0.16 = 0.98 s) the two particles are at
x
= (–2/7)(0.98) = –28 cm.
CHAPTER 10 ROTATION
Linear position: x
Angular position: θ
Linear displacement: Δx
Angular displacement: Δθ
Linear speed: v
Angular speed: ω
Linear acceleration: a
Angular acceleration: α
Mass: m – is a measure of the tendency of an
object to resist changes in its linear motion.
Moment of inertia: I – is a measure of the resistance of an
object to changes in its rotational motion. I = Σ m
i
r
i
2
Force: F
Torque: τ – measures the tendency of a force
to rotate an object about an axis. τ = rFsin(,)
Linear momentum: p
Angular momentum: L
The net external force on a system of particles
equals the time rate of change of the total linear
momentum of the system:
ΣF
ext
= dp
total
/dt
The net external torque acting on a system about some axis
passing through an origin in an inertial frame equals the
time rate of change of the total angular momentum of the
system about that origin
Στ
ext
= dL
total
/dt
Law of conservation of linear momentum
: the
total linear momentum of a system of particles is
conserved if no net external force is acting on the
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '09
 Rupinder
 Physics, Mass

Click to edit the document details