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Ch 8 HW Day 5 Solus: 78, 82, 86, 87
*78 ••
Picture the Problem We can apply conservation of momentum and
the definition of an elastic collision to obtain equations relating the
initial and final velocities of the colliding objects that we can solve
for v1f and v2f.
Apply conservation
of momentum to the
elastic collision of
the particles to
obtain:
m1v1f  m2v2f  m1v1i  m2v2i
(1)
m1v12f  12 m2 v22f  12 m1v12i  12 m2 v22i
Relate the initial and
final kinetic energies
of the particles in an
elastic collision:
1
2
Rearrange this equation
and factor to obtain:
m2 v22f  v22i  m1 v12i  v12f




or
m2 v2f  v2i v2f  v2i 
 m1 v1i  v1f v1i  v1f 
Rearrange equation (1)
to obtain:
Divide equation (2)
by equation (3) to
obtain:
(2)
m2 v2f  v2i   m1 v1i  v1f 
v2f  v2i  v1i  v1f
(3)
Rearrange this
equation to obtain
equation (4):
v1f  v2f  v2i  v1i
(4)
Multiply equation
(4) by m2 and add it
to equation (1) to
obtain:
Solve for v1f to
obtain:
Multiply equation
(4) by m1 and
subtract it from
equation (1) to
obtain:
Solve for v2f to obtain:
m1  m2 v1f  m1  m2 v1i  2m2v2i
v1 f 
m1  m2
2m2
v1i 
v2 i
m1  m2
m1  m2
m1  m2 v2f  m2  m1 v2i  2m1v1i
2m1
m2  m1
v2 f 
v1i 
v 2i
m1  m2
m1  m2
Remarks: Note that the velocities satisfy the condition that
v2f  v1f  v2i  v1i  . This verifies that the speed of
recession equals the speed of approach.
82 •
Picture the Problem Let the mass of the bullet be m, that of the
wooden block M, the pre-collision velocity of the bullet v, and the
post-collision velocity of the block+bullet be V. We can use
conservation of momentum to find the velocity of the block with
the bullet imbedded in it just after their perfectly inelastic
collision. We can use Newton’s 2nd law to find the acceleration of
the sliding block and a constant-acceleration equation to find the
distance the block slides.
Using a constantacceleration equation,
relate the velocity of
the block+bullet just
after their collision to
their acceleration and
displacement before
stopping:
Solve for the distance
the block slides before
coming to rest:
Use conservation of
momentum to relate
the pre-collision
velocity of the bullet
to the post-collision
velocity of the
block+bullet:
0  V 2  2ax
because the final velocity of the
block+bullet is zero.
V2
x  
2a
mv  m  M V
(1)
Solve for V:
m
V 
v
mM
Substitute in equation
(1) to obtain:
1  m

x   
v
2a  m  M 

Apply  F  ma to the
block+bullet (see the
FBD in the diagram):
F
x
(3)
and
F
y
Use the definition of
the coefficient of
kinetic friction and
equation (4) to obtain:
2
(2)
  f k  m  M a
Fn  m  M g  0
f k  k Fn  k m  M g
Substitute in equation (3):
 k m  M g  m  M a
Solve for a to obtain:
a  k g
Substitute in equation (2)
to obtain:
1  m

x 
v

2 k g  m  M 
2
(4)
Substitute numerical values and evaluate x:

1
x 
20.22 9.81 m/s 2

2



0.0105 kg

750 m/s   0.130 m
 0.0105 kg  10.5 kg


86 ••
Picture the Problem This nuclear reaction is 5Li   + p + 3.15 
1013 J. To conserve momentum, the alpha particle and proton
must move in opposite directions. We’ll apply both conservation of
energy and conservation of momentum to find the speeds of the
proton and alpha particle.
Use conservation of
momentum in this process to
express the alpha particle’s
velocity in terms of the
proton’s:
pi  pf  0
and
0  mp vp  m v
mp
mp
Solve for v and substitute
for m to obtain:
v 
Letting E represent the
energy released in the
reaction, apply conservation
of energy to the process:
K p  K  E
Substitute for v:
m
vp 
4mp
vp  14 vp
or
1
2
mp vp2  12 m v2  E
1
2
mp vp2  12 m 14 vp   E
2
Mult by 32 (least common
denominator) to get rid of all
fractions!
Solve for vp and substitute for
m to obtain:
Substitute numerical values
and evaluate vp:
vp 
vp 
32 E
32 E

16mp  m
16mp  4mp



32 3.15  10 13 J
20 1.67  10  27 kg

 1.74  10 7 m/s
Use the relationship between
vp and v to obtain v:
87

v  14 vp  14 1.74 107 m/s

 4.34 106 m/s
•••
Picture the Problem The pictorial representation shows the
projectile at its maximum elevation and is moving horizontally.
It also shows the two fragments resulting from the explosion.
We chose the system to include the projectile and the earth so
that no external forces act to change the momentum of the
system during the explosion. With this choice of system we can
also use conservation of energy to determine the elevation of the
projectile when it explodes. We’ll also find it useful to use
constant-acceleration equations in our description of the motion
of the projectile and its fragments.
(a) Use
conservation of
momentum to
relate the velocity
of the projectile
before its explosion
to the velocities of
its two parts after
the explosion:
The only way this
equality can hold is
if:


pi  pf



m3 v 3  m1v1  m2 v 2
m3v3ˆi  m1vx1ˆi  m1v y1ˆj  m2v y 2ˆj
m3 v3  m1v x1
and
m1v y1  m2 v y 2
Express v3 in terms
of v0 and substitute
for the masses to
obtain:
vx1  3v3  3v0 cos
 3120 m/s cos30  312 m/s
and
v y1  2 v y 2
(1)
Using a constantacceleration
equation with the
downward
direction positive,
relate vy2 to the
time it takes the 2kg fragment to hit
the ground:
With Ug = 0 at the
launch site, apply
conservation of
energy to the climb
of the projectile to
its maximum
elevation:
Solve for y:
Substitute
numerical values
and evaluate y:
Substitute in
equation (2) and
evaluate vy2:
Substitute in
equation (1) and
y  v y 2 t  g t 
2
1
2
y  12 g t 

t
2
vy2
(2)
K  U  0
Because Kf = Ui = 0,
or
 Ki  U f  0
 12 m3v y20  m3 gy  0
y 
v y20
2g
2

v0 sin 30

2g


120 m/s sin30  2
y 

2 9.81m/s 2
vy2


 183.5 m

183.5 m  12 9.81m/s 2 3.6 s 

3.6 s
 33.3 m/s
v y1  233.3 m/s   66.6 m/s
2
evaluate vy1:
Express v in vector
form:
1

v1  vx1iˆ  v y1 ˆj

(b) Express the
total distance d
traveled by the 1kg fragment:
Relate x to v0 and
the time-toexplosion:
Using a constantacceleration
equation, express
texp:
Substitute
numerical values
and evaluate texp:
Substitute in
equation (4) and
evaluate x:
Relate the distance
traveled by the 1kg fragment after
the explosion to the
time it takes it to
reach the ground:
312 m/s  iˆ  66.6 m/s  ˆj
d  x  x'
(3)
x  v0 cos  t exp 
(4)
texp
v0 sin 


g
g
texp 
vy0
120 m/s sin30   6.12 s
9.81 m/s 2
x  120 m/s cos306.12 s 
 636.5 m
x'  vx1t'
Using a constantacceleration
equation, relate the
time t for the 1kg fragment to
reach the ground
to its initial speed
in the y direction
and the distance to
the ground:
y  v y1t'  g t' 
1
2
2
183.5 m = 66.6 m/s t – 4.905 t2
Use QF, Polysmelt, or graph
Substitute to
obtain the
quadratic
equation:
Solve the quadratic
equation to find
t:
t = 15.9 s
Substitute in
equation (3) and
evaluate d:
d  x  x'  x  vx1t'
 636.5 m  312 m/s 15.9 s 
 5.61km
(c) Express the
energy released in
the explosion:
Eexp  K  K f  K i
(5)
Find the kinetic
energy of the
fragments after the
explosion:
K f  K1  K 2  12 m1v12  12 m2v22
1kg  312 m/s 2  66.6 m/s 2 
2
 12 2 kg 33.3 m/s 

1
2
 52.0 kJ
Find the kinetic
energy of the
projectile before
the explosion:
K i  12 m3v32  12 m3 v0 cos  
Substitute in
equation (5) to
determine the
energy released in
the explosion:
Eexp  K f  Ki  52.0 kJ  16.2 kJ
2

1
2
3 kg 120 m/s  cos 302
 16.2 kJ
 35.8 kJ