15–39.
SOLUTION
A ballistic pendulum consists of a 4-kg wooden block
originally at rest, u=0°. When a 2-g bullet strikes and
becomes embedded in it, it is observed that the block swings
upward to a maximum angle of u=6°. Estimate the speed
of the bullet. 1.25 m
θ
1.25 m
θ
SOLUTION
(
d
+
)
Σm(v1)=Σm(v2)
0 +0=–
80
32.2
vA+
60
32.2
(vb)x
vA=0.75(vb)x
v
b
=v
A
+v
b
>
A
(
d
+
)
(vb)x=–vA+4
a12
13 b
vA=1.58 ft>s S
Ans.
(
d
+
)
Σm(v1)=Σm(v2)
60
32.2
(2.110) =
a
80
32.2
+60
32.2 b
v
v=0.904 ft>s
Ans.
*15–40.
The boy jumps off the flat car at A with a velocity of
v
=4 ft>s
relative to the car as shown. If he lands on the
second flat car B, determine the final speed of both cars
after the motion. Each car has a weight of 80 lb. The boy’s
weight is 60 lb. Both cars are originally at rest. Neglect the
mass of the car’s wheels.
v 4 ft/s
13
12
5
AB
15–41.
A 0.03-lb bullet traveling at strikes the 10-lb
wooden block and exits the other side at as shown.
Determine the speed of the block just after the bullet exits
the block, and also determine how far the block slides
before it stops.The coefficient of kinetic friction between
the block and the surface is mk=0.5.
50 ft>s
1300 ft>s
SOLUTION
a0.03
32.2 b113002a12
13 b+0=a10
32.2 bnB+a0.03
32.2 b1502a4
5b
A
:
+
B
©m1n1=©m2n2
5
12
3
4
5
13
1300 ft/s
50 ft/s
15–42.
SOLUTION
Ans.
Ans.
Ans.t=0.216 s
a10
32.2 b13.482–51t2=0
A
:
+
B
mv1+©LFdt=mv2
N=504 lb
–a0.03
32.2 b113002a5
13 b–10112
A
10–3
B
+N112
A
10–3
B
=a0.03
32.2 b1502a3
5b
A
+c
B
mv1+©
LFdt=mv2
vB=3.48 ft>s
a0.03
32.2 b113002a12
13 b+0=a10
32.2 bvB+a0.03
32.2 b1502a4
5b
A
:
+
B
©m1v1=©m2v2
A
0.03-lb bullet traveling at strikes the 10-lb
wooden
block and exits the other side at as shown.
Determine
the speed of the block just after the bullet exits
the
block. Also, determine the average normal force on the
block
if the bullet passes through it in 1 ms, and the time the
block
slides before it stops.The coefficient of kinetic
friction between the block and the surface is
mk=0.5.
50 ft>s
1300 ft
>
s
5
12
3
4
5
13
1300 ft/s50 ft/s
SOLUTION
15–43.
The 20-g bullet is traveling at 400 m
>
s when it becomes
embedded in the 2-kg stationary block. Determine the
distance the block will slide before it stops. The coefficient
of kinetic friction between the block and the plane is
m
k=0.2.
400 m/s
*15–44.
A toboggan having a mass of 10 kg starts from rest at Aand
carries a girl and boy having a mass of 40 kg and 45 kg,
respectively.When the toboggan reaches the bottom of the
slope at B, the boy is pushed off from the back with a
horizontal velocity of ,measured relative to
the toboggan. Determine the velocity of the toboggan
afterwards. Neglect friction in the calculation.
vb>t=2m>s
SOLUTION
we have
Relative Velocity: The relative velocity of the falling boy with respect to the
toboggan is .Thus, the velocity of the boy falling off the toboggan is
[1]
Conservation of Linear Momentum: If we consider the tobbogan and the riders as
a system, then the impulsive force caused by the push is internal to the system.
Therefore, it will cancel out. As the result, the linear momentum is conserved along
the xaxis.
[2]
Solving Eqs. [1] and [2] yields
Ans.
yb=6.619 m s
yt=8.62 m>s
A
;
+
B
(10 +40 +45)(7.672) =45yb+(10 +40) yt
mTyB=mbyb+
A
mt+mg
B
yt
A
;
+
B
yb=yt–2
yb=yt+yb/t
yb/t=2m>s
yB=7.672 m>s
0+2795.85 =1
2(10 +40 +45) y2
B+0
T1+V1=T2+V2
v
b/t
v
t
B
A
3m
15–45.
1
v1
x
h
y
u
z
The block of mass mis traveling at in the direction
shown at the top of the smooth slope. Determine its speed
and its direction when it reaches the bottom.u2
v2
u1
v
1
SOLUTION
15–46.
The two blocks A and B each have a mass of 5 kg and are
suspended from parallel cords. A spring, having a stiffness
of k=60 N>m, is attached to B and is compressed 0.3 m
against A and B as shown. Determine the maximum angles
u and f of the cords when the blocks are released from rest
and the spring becomes unstretched.
AB
2 m2 m
θφ
15–47.
BA
10 km/h20 km/h
k 3 MN/m
The 30-Mg freight car Aand 15-Mg freight car Bare moving
towards each other with the velocities shown. Determine the
maximum compression of the spring mounted on car A.
Neglect rolling resistance.
SOLUTION
*15–48.
Blocks Aand Bhave masses of 40 kg and 60 kg,
respectively.They are placed on a smooth surface and the
spring connected between them is stretched 2 m. If they are
released from rest, determine the speeds of both blocks the
instant the spring becomes unstretched.
SOLUTION
Ans.
Ans.vB=2.19 m s
vA=3.29 m>s
0+1
2118021222=1
214021nA22+1
216021nB22
T
1+V
1=T
2+V
2
0+0=40 nA–60 nB
(:
+)©mn1=©mn2
k 180 N/m
AB
15–49.
A boy A having a weight of 80 lb and a girl B having a weight
of 65 lb stand motionless at the ends of the toboggan, which
has a weight of 20 lb. If they exchange positions, A going to B
and then B going to A’s original position, determine the final
position of the toboggan just after the motion. Neglect
friction between the toboggan and the snow.
SOLUTION
4 ft
AB
SOLUTION
15–50.
A boy A having a weight of 80 lb and a girl B having a weight
of 65 lb stand motionless at the ends of the toboggan, which
has a weight of 20 lb. If A walks to B and stops, and both walk
back together to the original position of A, determine the final
position of the toboggan just after the motion stops. Neglect
friction between the toboggan and the snow. 4 ft
AB
15–51.
The 10-Mg barge B supports a 2-Mg automobile A. If
someone drives the automobile to the other side of the
barge, determine how far the barge moves. Neglect the
resistance of the water.
40 m
A
B
SOLUTION
*15–52.
3.5 m
A
B30
SOLUTION
evah ew,12–41.qE gniylppA. si
(1)
Relative Velocity:The velocity of the crate is given by
(2)
The magnitude of vCis
(3)
Conservation of Linear Momentum: If we consider the crate and the ramp as a
system, from the FBD, one realizes that the normal reaction NC(impulsive force)is
internal to the system and will cancel each other.As the result, the linear momentum
is conserved along the xaxis.
(4)
Solving Eqs. (1), (3), and (4) yields
Ans.
From Eq. (2)
Thus, the directional angle of is
vC
f
vC=30.866016.3562–1.1014i–0.516.3562j=54.403i–3.178j6m>s
vC>R=6.356 m>s
vR=1.101 m>s=1.10 m>svC=5.43 m>s
0=8.660 vC>R–50 vR
(:
+)0=1010.8660 vC>R–vR2+401–vR2
0=mC1vC2x+mRvR
=2v2
C>R+v2
R–1.732 vRvC>R
vC=2(0.8660 vC>R–vR22+1–0.5 vC>R22
=10.8660 vC>R–vR2i–0.5 vC>Rj
=-vRi+1vC>Rcos 30°i–vC>Rsin 30°j2
vC=vR+vC>R
171.675 =5v2
C+20 v2
R
0+171.675 =1
21102v2
C+1
21402v2
R
T
1+V
1=T
2+V
2
1019.81211.752=171.675 N #m
The free-rolling ramp has a mass of 40 kg.A 10-kg crate is
released from rest at Aand slides down 3.5 m to point B.If
the surface of the ramp is smooth, determine the ramp’s
speed when the crate reaches B.Also, what is the velocity of
the crate?
7
529
sB=71.4 mm S
Ans:
sB=71.4 mm S
15–55.
SOLUTION
(1)
(2)
Solving Eqs. (1) and (2),
Ans.
nb=-4.44 m>s=4.44 m>s;
nc=5.04 m>s;
–nb=nc–0.6
(;
+)nb=nc+nb>c
(3 +0.5)(4.952) =(3)nc–(0.5)nb
(;
+)©mn1=©mn2
nB=4.952 m>s
0+(3 +0.5)(9.81)(1.25) =1
2(3 +0.5)(nB)2
2+0
T
h
e cart
h
as a mass of 3
k
g an
d
ro
ll
s free
l
y
d
own t
h
e s
l
ope.
When it reaches the bottom, a spring loaded gun fires a
0.5-kg ball out the back with a horizontal velocity of
measured relative to the cart. Determine the
final velocity of the cart.
vb>c=0.6 m>s,
v
b/c
v
c
B
A
1.25 m
*15–56.
Two boxes Aand B, each having a weight of 160 lb, sit on
the 500-lb conveyor which is free to roll on the ground. If
the belt starts from rest and begins to run with a speed of
determine the final speed of the conveyor if (a) the
boxes are not stacked and Afalls off then Bfalls off, and (b)
Ais stacked on top of Band both fall off together.
3ft>s,
Thus,
When a box falls off, it exerts no impulse on the conveyor, and so does not alter the
momentum of the conveyor.Thus,
a) Ans.
b) Ans.vc=1.17 ft>s;
vc=1.17 ft>s;
vb=1.83 ft>s:vc=1.17 ft>s;
vb=-vc+3
a:
+bvb=vc+vb>c
0=a320
32.2 b(vb)–a500
32.2 b(vc)
a:
A
B
15–57.
The 10-kg block is held at rest on the smooth inclined plane
by the stop block at A.If the 10-g bullet is traveling at
when it becomes embedded in the 10-kg block,
determine the distance the block will slide up along the
plane before momentarily stopping.
300 m>s
SOLUTION
Conservation of Energy:The datum is set at the blocks initial position. When the
block and the embedded bullet is at their highest point they are habove the datum.
gniylppA. si ygrene laitnetop lanoitativarg riehT
Eq. 14–21, we have
Ans.d
=
3.43 >sin 30°
=
6.87 mm
h=0.003433 m =3.43 mm
0+1
2(10 +0.01)
A
0.25952
B
=0+98.1981h
T
1+V
1=T
2+V
2
(10 +0.01)(9.81)h=98.1981h
v=0.2595 m>s
0.01(300 cos 30°) =(0.01 +10) v
mb(vb)x¿=(mb+mB)vxœ
30
A
300 m/s
533
Ans:
( vA)2=0.353 m>s
( vB)2=2.35 m>s