Download kg g 75 600 50 m/s

75g 600 m/s
50 kg
Figure 1: Problem 1
1. A 75-g projectile traveling at 600 m/s strikes and becomes embedded in
the 50-kg block, which is initially stationary. Compute the energy lost
during the impact. Express your answer as an absolute value | ∆E | and
as a percentage n of the original system energy E.
Solution. Since the force of impact is internal to the system composed of
the projectile and block and since there are no other external forces acting
on the system in the horizontal line of motion, it follows that the linear
momentum of the system is conserved.
m1 v1 i = (m1 + m2 )vi
(1)
where m1 is the mass of the projectile, m2 is the mass of the block, and v
is the velocity after the impact. Thus
75(10−3 )(600) = 75(10−3) + 50 v
45
v =
= 0.900 m/s
(2)
50.075
The original energy E is
E=
1
(75)(10−3 )(6002 ) = 13500 J
2
(3)
The energy after the impact is
E′ =
1
75(10−3 ) + 50 (0.9002) = 20 J
2
(4)
The energy lost is
∆E = E − E ′ = 13480 J
n is
n=
∆E
13480
=
(100) = 99.9%
E
13500
(5)
(6)
2. Determine the magnitude HO of the angular momentum of the 2-kg sphere
about point O (a) by using the vector definition of angular momentum and
(b) by using an equivalent scalar approach. The center of the sphere lies
in the x − y plane.
Solution.
(a) The position vector from O to the sphere is
r = 12i + 5j m
1
(7)
y
7 m/s
2 kg 45
5m
12 m
x
O
Figure 2: Problem 2
The velocity of it is
7 cos 45◦ i + 7 sin 45◦ j
4.95i + 4.95j m/s
=
=
v
(8)
The angular momentum is
HO
= r × mv
= (12i + 5j) × 2(4.95i + 4.95j)
= 69.3k kgm2 /s
(b) The distance between the sphere and O is
p
r = 122 + 52 = 13 m
(9)
(10)
The angle between the velocity vector and the line from O to the
sphere is
5
−1
= 22.62◦
(11)
θ = tan
12
The equivalent scalar of angular momentum then is
rmv sin (45◦ − 22.62◦) = 69.3 kgm2 /s
(12)
3. At a certain instant, the particle of mass m has the position and velocity
shown in the figure, and it is acted upon by the force F . Determine its
angular momentum about point O and the time rate of change this angular
momentum.
Solution. The position and velocity vectors are
r
v
=
=
ai + bj + ck
vj
2
(13)
(14)
z
F
m
v
O
a
b
c
y
x
Figure 3: Problem 3
The angular momentum H O can be calculated as
H
=
=
r × mv
(ai + bj + ck) × mvj
=
mv(−ci + ak)
(15)
When F = F k, the time rate of change of H O is
Ḣ O
= r×F
= F (bi − aj)
3
(16)