Version 001 – Review - Mechanics – tubman – (20131)
This print-out should have 45 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Distance Time Graph 01
001 (part 1 of 6) 10.0 points
Consider the following graph of motion.
1
The slope of the graph is the same everywhere, so the graph indicates constant positive velocity.
003 (part 3 of 6) 10.0 points
What is the speed from 2 s to 4 s?
1. 15 m/s
Distance (m)
50
2. 10 m/s correct
40
30
3. 0 m/s
20
10
0
4. 5 m/s
0
1
2 3 4
Time (sec)
5
How far did the object travel between 2 s
and 4 s?
1. 30 m
2. 20 m correct
5. 20 m/s
Explanation:
v=
∆d
40 m − 20 m
=
= 10 m/s .
∆t
2s
004 (part 4 of 6) 10.0 points
Consider the following graph of motion.
3. 10 m
400
350
4. 50 m
Distance (m)
300
5. 40 m
Explanation:
The particle moved from 40 m to 20 m, so
250
200
150
100
50
0
∆d = 40 m − 20 m = 20 m .
002 (part 2 of 6) 10.0 points
The graph indicates
1
2
3
4 5 6
Time (sec)
7
8
9
How far did the object travel between 3 s
and 9 s?
1. 150 m
1. constant position.
2. 400 m
2. constant velocity. correct
3. 100 m
3. no motion.
4. 50 m
4. increasing velocity.
5. 200 m
5. decreasing velocity.
6. 500 m
Explanation:
0
7. 300 m
Version 001 – Review - Mechanics – tubman – (20131)
2
Explanation:
8. 450 m
10. 250 m
Explanation:
The particle moved from 50 m to 400 m, so
the distance
∆d = 400 m − 50 m = 350 m .
005 (part 5 of 6) 10.0 points
The graph indicates
1. constant position.
v=
4. no motion.
5. decreasing velocity.
Explanation:
The slopes are steeper as time goes on, so
the velocities are increasing.
3. 60 m/s
4. 25 m/s
5. 20 m/s
6. 58 m/s correct
7. 40 m/s
8. 47 m/s
9. 50 m/s
4
3
2
1
−1
b
4 5 6 7
time (s)
What is the position at 2 seconds?
1
2
3
8
b
9
Correct answer: 15 m.
Explanation:
The initial position given in the problem is
10 m.
bb
5
b
4
velocity (m/s)
2. 30 m/s
b
0 b
006 (part 6 of 6) 10.0 points
What is the average speed from 3 s to 9 s?
1. 36 m/s
b
5
2. constant velocity.
3. increasing velocity. correct
∆d
400 m − 50 m
=
= 58 m/s .
∆t
6s
Velocity vs Time 05
007 (part 1 of 3) 10.0 points
Consider the plot below describing motion
along a straight line with an initial position of
x0 = 10 m.
velocity (m/s)
9. 350 m correct
3
2
1
0 bb
−1
b
1
2
3
4 5
time (s)
b
6 7
8
9b
The position at 2 seconds is 10 meters plus
the area of the triangle (shaded in the above
plot)
1
(2 s − 0 s)
2
× (5 m/s − 0 m/s)
x = 10 m +
= 15 m ;
Version 001 – Review - Mechanics – tubman – (20131)
Basket Time
010 (part 1 of 3) 10.0 points
Neglect: Air friction.
Your teacher tosses a basketball. The ball
gets through the hoop (lucky shot).
b
b
b
b b b b b
b
008 (part 2 of 3) 10.0 points
What is the position at 6 seconds?
/s
18 m
b
b
b
b
b
b
◦
69
b
b
2.517 m
Explanation:
The position is 15 m plus the area of the
trapezoid from 2 s to 6 s
1
x = 15 m + (6 s − 2 s)
2
× (5 m/s + 5 m/s)
b
b
b
Correct answer: 35 m.
b
3.048 m
however, it can also be calculated:
1
x = xi + vi (tf − ti ) + (tf − ti )2
2
= (10 m) + (0 m/s) (2 s − 0 s)
1
+ (2.5 m/s2 ) (2 s − 0 s)2
2
= 15 m .
3
= 35 m ;
however, it can also be calculated:
1
x = xi + vi (tf − ti ) + (tf − ti )2
2
= (15 m) + (5 m/s) (6 s − 2 s)
1
+ (0 m/s2 ) (6 s − 2 s)2
2
= 35 m .
ℓ
Figure: Not drawn to scale.
How long does it take the ball to reach its
maximum height?
009 (part 3 of 3) 10.0 points
What is the position at 8 seconds?
Explanation:
Correct answer: 1.71474 s.
Correct answer: 34.3333 m.
Explanation:
The position is 35 m minus the area of the
triangle from 6 s to 8 s
1
x = (35 m) + (8 s − 6 s)
2
× (−0.666667 m/s − 0 m/s)
= 34.3333 m ;
however, it can also be calculated
1
x = xi + vi (tf − ti ) + (tf − ti )2
2
= 35 m + (0 m/s) (8 s − 6 s)
1
+ (−0.333333 m/s2 ) (8 s − 6 s)2
2
= 34.3333 m .
Let : α = 69◦ ,
v0 = 18 m/s ,
h1 = 2.517 m ,
h2 = 3.048 m , and
ytop = maximum height of
ball′ s trajectory .
Note: The horizontal distance to the basket
ℓ is superfluous for Parts 1 and 2. This problem has two distinct parts, vertical motion in
Parts 1 and 2, then horizontal motion in Part
3.
Version 001 – Review - Mechanics – tubman – (20131)
Basic Concepts:
v = v0 sin α − g t
Or symbolically, we have
(1)
v02 sin2 α
ytop − h1 =
(2)
2g
1
(3)
ytop − h1 = v0 sin α t − g t2
2
Solution: Using Eq. 1, the time t1 to reach
the maximum height, (i.e., vytop = 0, the velocity at the top) is
v0
t1 =
sin(69◦ )
g
(18 m/s)
=
sin(69◦ )
(9.8 m/s2 )
= 1.71474 s .
1 2
gt
2 1
1
= (9.8 m/s2 ) (1.71474 s)2
2
= 14.4076 m , or
v02 sin2 α
=
2g
(18 m/s)2 sin2 (69◦ )
=
2 (9.8 m/s2 )
= 14.4076 m , so
= y1 + h1
= (14.4076 m) + (2.517 m)
= 16.9246 m .
y1 =
ytop
4
011 (part 2 of 3) 10.0 points
How long does it take the ball to reach the
hoop?
Correct answer: 3.39758 s.
Explanation:
The time t2 for the ball to decend from the
top into the basket is
1/2
2 [ytop − h2 ]
t2 =
g
1/2
2 [(16.9246 m) − (3.048 m)]
=
(9.8 m/s2 )
= 1.68284 s .
1/2
2 [ytop − h2 ]
t2 =
g
1/2
2 2
v0 sin α 2 [h2 − h1 ]
−
=
g2
g
2
(18 m/s)2 sin (69◦ )
=
(9.8 m/s2 )2
2 [(3.048 m) − (2.517 m)]
−
(9.8 m/s2 )
= 1.68284 s .
1/2
The total time of the ball’s trajectory is
t = t1 + t2
= (1.71474 s) + (1.68284 s)
= 3.39758 s .
012 (part 3 of 3) 10.0 points
What is the horizontal length ℓ of the shot?
Correct answer: 21.9165 m.
Explanation:
The horizontal length of the shot ℓ is
ℓ = v0 cos α t
= (18 m/s) cos(69◦ ) (3.39758 s)
= 21.9165 m .
Ball on a String
013 10.0 points
A ball on the end of a string is whirled around
in a horizontal circle of radius 0.38 m. The
plane of the circle is 1.2 m above the ground.
The string breaks and the ball lands 2.66 m
away from the point on the ground directly
beneath the ball’s location when the string
breaks.
The acceleration of gravity is 9.8 m/s2 .
Find the centripetal acceleration of the ball
during its circular motion.
Correct answer: 76.0317 m/s2 .
Version 001 – Review - Mechanics – tubman – (20131)
Explanation:
In order to find the centripetal acceleration
of the ball, we need to find the initial velocity
of the ball. Let y be the distance above the
ground. After the string breaks, the ball has
no initial velocity in the vertical direction, so
the time spent in the air may be deduced from
the kinematic equation,
y=
1 2
gt .
2
Solving for t,
⇒t=
r
2y
.
g
Let d be the distance traveled by the ball.
Then
d
d
vx = = r
.
t
2y
g
Hence, the centripetal acceleration of the ball
during its circular motion is
vx2
ac =
r
d2 g
=
2yr
= 76.0317 m/s2 .
Accelerating Elevator
014 10.0 points
An elevator starts from rest with a constant
upward acceleration and moves 1 m in the first
1.6 s. A passenger in the elevator is holding a
4.3 kg bundle at the end of a vertical cord.
What is the tension in the cord as the elevator accelerates? The acceleration of gravity
is 9.8 m/s2 .
5
T
g
aelevator
mg
Let h be the distance traveled and a the
acceleration of the elevator. Since the initial
velocity is zero,
h = v0 t +
a=
1 2 1 2
at = at
2
2
2h
.
t2
The equation describing the forces acting on
the bundle is
Fnet = m a = T − m g
2h
T = m (g + a) = m g + 2
t
2 (1 m)
2
= (4.3 kg) 9.8 m/s +
(1.6 s)2
= 45.4994 N .
Force and Motion 01
015 (part 1 of 7) 10.0 points
A sled on ice moves in the ways described in
the questions below. Friction is so small that
it can be ignored. A person wearing spiked
shoes standing on the ice can apply a force to
the sled and push it along the ice.
Direction of Force
Correct answer: 45.4994 N.
Explanation:
Direction of Force
h = 1 m,
t = 1.6 s ,
m = 4.3 kg ,
andg = 9.8 m/s2 .
Which force would keep the sled moving
toward the right and speeding up at a steady
rate (constant acceleration)?
Version 001 – Review - Mechanics – tubman – (20131)
1. A force acting toward the left and of
constant strength (magnitude).
2. A force acting toward the left and decreasing in strength (magnitude).
6
7. No applied force is needed. correct
Explanation:
Since the sled moves at a constant velocity,
there is no acceleration, and no applied force.
4. A force acting toward the right and of
constant strength (magnitude). correct
017 (part 3 of 7) 10.0 points
The sled is moving toward the right.
Which force would slow it down at a steady
rate (constant acceleration)?
5. A force acting toward the left and increasing in strength (magnitude).
1. A force acting toward the right and decreasing in strength (magnitude).
6. A force acting toward the right and increasing in strength (magnitude).
2. A force acting toward the right and increasing in strength (magnitude).
7. A force acting toward the right and decreasing in strength (magnitude).
3. A force acting toward the left and increasing in strength (magnitude).
Explanation:
According to Newton’s second law, force
is proportional to acceleration and they are
in the same direction. Since the sled speeds
up with constant acceleration to the right,
the force must be of constant magnitude and
toward the right.
4. A force acting toward the left and of
constant strength (magnitude). correct
016 (part 2 of 7) 10.0 points
Which force would keep the sled moving toward the right at a steady (constant) velocity?
7. A force acting toward the left and decreasing in strength (magnitude).
1. A force acting toward the right and decreasing in strength (magnitude).
Explanation:
In order to slow down the sled at a steady
rate, the force must be in the opposite direction (left) and of constant strength.
3. No applied force is needed.
2. A force acting toward the right and increasing in strength (magnitude).
3. A force acting toward the right and of
constant strength (magnitude).
4. A force acting toward the left and decreasing in strength (magnitude).
5. A force acting toward the left and of
constant strength (magnitude).
6. A
force acting toward the left and
increasing in strength (magnitude).
5. No applied force is needed.
6. A force acting toward the right and of
constant strength (magnitude).
018 (part 4 of 7) 10.0 points
Which force would keep the sled moving toward the left and speeding up at a steady rate
(constant acceleration)?
1. No applied force is needed.
2. A force acting toward the right and of
constant strength (magnitude).
3. A force acting toward the left and of
constant strength (magnitude). correct
Version 001 – Review - Mechanics – tubman – (20131)
7
4. A force acting toward the left and decreasing in strength (magnitude).
1. A force acting toward the right and increasing in strength (magnitude).
5. A force acting toward the left and increasing in strength (magnitude).
2. A force acting toward the right and of
constant strength (magnitude). correct
6. A force acting toward the right and increasing in strength (magnitude).
3. A force acting toward the left and decreasing in strength (magnitude).
7. A force acting toward the right and decreasing in strength (magnitude).
4. A force acting toward the left and increasing in strength (magnitude).
Explanation:
The same reason as Part 1.
5. A force acting toward the left and of
constant strength (magnitude).
019 (part 5 of 7) 10.0 points
The sled started from rest and was pushed
until it reached a steady (constant) velocity
toward the right.
Which force would keep the sled moving at
this velocity?
1. A force acting toward the left and decreasing in strength (magnitude).
2. A force acting toward the left and increasing in strength (magnitude).
3. No applied force is needed. correct
6. No applied force is needed.
7. A force acting toward the right and decreasing in strength (magnitude).
Explanation:
The same reason as Part 3.
021 (part 7 of 7) 10.0 points
The sled is moving toward the left.
Which force would slow it down at a steady
rate (constant acceleration)?
1. No applied force is needed.
4. A force acting toward the left and of
constant strength (magnitude).
2. A force acting toward the right and of
constant strength (magnitude). correct
5. A force acting toward the right and increasing in strength (magnitude).
3. A force acting toward the right and decreasing in strength (magnitude).
6. A force acting toward the right and decreasing in strength (magnitude).
4. A force acting toward the left and decreasing in strength (magnitude).
7. A force acting toward the right and of
constant strength (magnitude).
5. A force acting toward the right and increasing in strength (magnitude).
Explanation:
The same reason as Part 2.
6. A force acting toward the left and of
constant strength (magnitude).
020 (part 6 of 7) 10.0 points
The sled is slowing down at a steady rate and
has an acceleration to the right.
Which force would account for this motion?
7. A force acting toward the left and increasing in strength (magnitude).
Explanation:
The same reason as Part 3.
Version 001 – Review - Mechanics – tubman – (20131)
Elevator Free Body Diagram
022 10.0 points
A man stands in an elevator in the university’s
administration building and is accelerating
upwards. (During peak hours, this does not
happen very often.)
Elevator
Cable
8
5.
Fman, elevator Felevator, cable
6.
Felevator, cable
Fman, earth
Choose the correct free body diagram for
the man, where Fi,j is the force on the object
i, from the object j.
1.
Fman, cable
2.
Fman, floor
Fman, earth
Explanation:
Only the forces acting directly on the man
are to be in the free body diagram. Therefore,
the force from the cable should be omitted,
while those from gravity and from the floor’s
normal force should be included.
Force vs Time
023 10.0 points
A 0.32 kg mass is initially at rest and is free
to move with negligible friction along the xaxis. The figure below shows the value of an
applied force as a function of time.
b
b b
b
b b
5
4 b
3.
Felevator, cable Fman, floor
Force ( N )
correct
b
3
b
2
4.
Fman, acceleration
b
b
1
b
0 b
b
−1
Fman, earth
b
−2
b
b
b
b
b
b
b
0
1 2 3 4 5 6 7 8 9 10
time ( s )
∆t
Calculate the kinetic energy K of the mass
when it reaches 8 s.
Correct answer: 400 J.
Explanation:
The kinetic energy is one-half times the
Version 001 – Review - Mechanics – tubman – (20131)
1
mass times the velocity squared K = m v 2 .
2
The velocity is the initial velocity plus the
acceleration times the time v = v0 +a t , where
the acceleration is the force divided by the
F
mass a =
.
m
Since the velocity is constant over the time
intervals (∆t = 1 second) and the initial velocity v0 = 0 meters per second, we have
Let :
W = F d = (48 N) (10 m) = 480 J .
∆vi
i=1
8
X
ai ∆t
i=1
8
X
Fi
m
i=1
(∆t)2
=
2m
(∆t)2 8
X
!2
Fi
i=1
!2
!2
Block Jump Ramp 01
025 (part 1 of 3) 10.0 points
A block starts at rest and slides down a frictionless track. It leaves the track horizontally,
flies through the air, and subsequently strikes
the ground.
(∆t)2
!2
F01 + F12 + F23 + F34
2m
2
+ F45 + F56 + F67 + F78
(1 s)2 h
(4 N) + (2 N)
=
2 (0.32 kg)
+ (3 N) + (−1 N) + (5 N)
i2
+ (0 N) + (5 N) + (−2 N)
2
(1 s)2 16 N
=
0.64 kg
483 g
v
b
b
b
b
b
What is the speed of the ball when it leaves
the track? The acceleration of gravity is
9.81 m/s2 .
Correct answer: 6.71759 m/s.
Explanation:
Let : g = −9.81 m/s2 ,
m = 483 g , and
h1 = 2.3 m .
The kinetic energy at 8 seconds is 400 J.
This problem has a different plot for each
version.
Correct answer: 480 J.
b
x
= 400 J .
m
h1
v
b b b b
b b
h
Work Done on a Block
024 10.0 points
Lee pushes horizontally with a force of 48 N
on a 29 kg mass for 10 m across a floor.
Calculate the amount of work Lee did.
b b b b
b b
2.2 m
1
= m
2
v0 +
8
X
h2
1
= m
2
F = 48 N and
d = 10 m .
The mass of the object is ignored since there
is no friction present, so
1
m vf2
2
1
= m
2
=
Explanation:
4.5 m
K=
9
x
b
b
b
b
b
b
Version 001 – Review - Mechanics – tubman – (20131)
Choose the point where the block leaves the
track as the origin of the coordinate system.
While on the ramp,
10
Now choose ground level as the origin. Energy conservation gives us
K f = Ui
K b = Ut
1
m vx2 = −m g h1
2
vx2 = −2 g h1
p
vx = −2 g h1
q
= −2 (−9.81 m/s2 ) (2.3 m)
= 6.71759 m/s .
026 (part 2 of 3) 10.0 points
What horizontal distance does the block
travel in the air?
Correct answer: 4.49889 m.
Explanation:
1
m vf2 = −m g h
2
p
vf = −2 g h
q
= −2 (−9.81 m/s2 ) (4.5 m)
= 9.39628 m/s .
Alternate Solution:
p
vy = −2 g h2
q
= −2 (−9.81 m/s2 ) (2.2 m)
= 6.56993 m/s , so
q
vf = vx2 + vy2
q
= (6.71759 m/s)2 + (6.56993 m/s)2
= 9.39628 m/s .
Let :
h2 = −2.2 m .
With the point of launch as the origin,
h2 =
t=
1 2
gt
2
s
2 h2
.
g
Thus
x = vx t = vx
s
2 h2
g
s
= (6.71759 m/s)
Holt SF 06Rev 25
028 10.0 points
A tennis player places a 64 kg ball machine
on a frictionless surface. The machine fires a
0.059 kg tennis ball horizontally with a velocity of 38 m/s toward the north.
What is the final velocity of the machine?
Correct answer: −0.0350312 m/s.
Explanation:
Let north be positive.
2 (−2.2 m)
−9.81 m/s2
= 4.49889 m .
Let : m1 = 64 kg ,
m2 = 0.059 kg ,
v2,f = 38 m/s .
and
The initial momentum is zero, so
027 (part 3 of 3) 10.0 points
What is the speed of the block when it hits
the ground?
Correct answer: 9.39628 m/s.
Explanation:
Let :
h = 4.5 m .
0 = m1 ~v1,f + m2 ~v2,f
m2 v2,f
m1
(0.059 kg) (38 m/s)
=−
64 kg
v1,f = −
= −0.0350312 m/s ,
Version 001 – Review - Mechanics – tubman – (20131)
which is 0.0350312 m/s toward the south.
Net Forces 02
029 (part 1 of 3) 10.0 points
A 52.4 N object is in free fall.
What is the magnitude of the net force
which acts on the object?
Correct answer: 52.4 N.
Explanation:
During free fall, the net acceleration is
down, so
~ net =
F
X
~ down −
F
X
~ up .
F
During free fall with no air resistance, the only
force acting is the weight, so the net force is
its weight.
030 (part 2 of 3) 10.0 points
What is the magnitude of the net force when
the object encounters 19.3 N of air resistance?
Correct answer: 33.1 N.
Explanation:
The weight acts down and the air resistance acts up with the net acceleration acting
downward, so
~ net = W
~ −F
~ = 52.4 N − 19.3 N = 33.1 N .
F
031 (part 3 of 3) 10.0 points
What is the magnitude of the net force when it
falls fast enough to encounter an air resistance
of 52.4 N?
Correct answer: 0 N.
Explanation:
The net force is now zero, since the air
resistance is the same as the weight.
Arcade Game
032 10.0 points
In an arcade game a 0.08 kg disk is shot across
a frictionless horizontal surface by compressing it against a spring and releasing it.
11
If the spring has a spring constant of
153 N/m and is compressed from its equilibrium position by 7 cm, find the speed with
which the disk slides across the surface.
Correct answer: 3.06125 m/s.
Explanation:
From conservation of mechanical energy, we
have:
1
1
m vf2 = k x2i ,
2
2
s
k x2i
m
vf =
=
s
(153 N/m)(0.07 m)2
0.08 kg
= 3.06125 m/s .
Holt SF 05Rev 57
033 (part 1 of 2) 10.0 points
A 76 kg man jumps from a window 1.8 m
above a sidewalk.
The acceleration of gravity is 9.81 m/s2 .
a) What is his speed just before his feet
strike the pavement?
Correct answer: 5.94273 m/s.
Explanation:
Basic Concepts: Conservation of Mechanical Energy
Ui = K f
since vi = 0 m/s and hf = 0 m.
1
K = mv 2
2
Ug = mgh
Given:
m = 76 kg
hi = 1.8 m
g = 9.81 m/s2
Solution:
1
mghi = mvf 2
2
p
vf = 2ghi
q
= 2(9.81 m/s2 )(1.8 m)
= 5.94273 m/s
Version 001 – Review - Mechanics – tubman – (20131)
034 (part 2 of 2) 10.0 points
b) If the man jumps with his knees and ankles locked, the only cushion for his fall is
approximately 0.44 cm in the pads of his feet.
Calculate the magnitude of the average
force exerted on him by the ground in this
situation.
5
Correct answer: 3.05002 × 10 N.
Explanation:
Basic Concepts:
Wnet = ∆K = Kf − Ki = −Ki
since Kf = 0 J.
Wnet = Fnet d cos θ = −Fnet d
since θ = 180◦ ⇒ cos θ = −1.
The initial kinetic energy when the man
strikes the pavement equals the initial gravitational potential energy.
Ki = Ug = mgh
1
K0 = m v 2
2
1
159 g
=
(20 m/s)2
2 1000 g/kg
= 31.8 J
036 (part 2 of 3) 10.0 points
Find the potential energy at its highest position.
Correct answer: 30.2418 J.
Explanation:
U1 = m g h
159 g
=
(9.8 m/s2 )(19.4082 m)
1000 g/kg
= 30.2418 J
037 (part 3 of 3) 10.0 points
Find the magnitude of the energy lost due to
air resistance.
Correct answer: 1.5582 J.
Given:
d = 0.44 cm
Solution:
−Fnet d = −mgh
mgh
Fnet =
d
(76 kg)(9.81 m/s2 )(1.8 m)
=
0.0044 m
= 3.05002 × 105 N
Arrow Shot Straight Up
035 (part 1 of 3) 10.0 points
A 159 g arrow is shot straight up into the air
with a speed of 20 m/s. It reaches a maximum
height of 19.4082 m.
Find the initial kinetic energy. The acceleration of gravity is 9.8 m/s2 .
Correct answer: 31.8 J.
Explanation:
12
Explanation:
Elost = E0 − E1 = K0 − U1
= 31.8 J − 30.2418 J = 1.5582 J
Bouncing a Superball
038 10.0 points
A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is
from 23 m/s downward to 14 m/s upward.
1
If the contact time with the sidewalk is
800
s, what is the magnitude of the average force
exerted on the superball by the sidewalk?
Correct answer: 1776 N.
Explanation:
Let : m = 60 g = 0.06 kg ,
vu = 14 m/s ,
vd = 23 m/s , and
∆t = 0.00125 s
Version 001 – Review - Mechanics – tubman – (20131)
Choose the upward direction as positive.
The impulse is
I = F ∆t = ∆P
= m vu − m (−vd )
= m (vu + vd )
m (vu + vd )
F =
∆t
(60 g) (14 m/s + 23 m/s)
=
0.00125 s
= 1776 N .
Ballistic Block
039 (part 1 of 3) 10.0 points
Assume: The bullet penetrates into the block
and stops due to its friction with the block.
The compound system of the block plus the
bullet rises to a height of 9 cm along a circular
arc with a 16 cm radius.
Assume: The entire track is frictionless.
A bullet with a m1 = 30 g mass is fired
horizontally into a block of wood with m2 =
4.76 kg mass.
The acceleration of gravity is 9.8 m/s2 .
16 cm
13
The mechanical energy is conserved after
collision. Choose the position when the system stops at height h, where the kinetic energy is 0 and the potential energy is given
by
(mbullet + mblock ) g h = 4.22478 J ,
which is the total energy after collision.
040 (part 2 of 3) 10.0 points
Taking the same parameter values as those in
Part 1, determine the initial velocity of the
bullet.
Correct answer: 212.062 m/s.
Explanation:
During the rising process the total energy
is conserved
1
(mbullet + mblock ) vf2 and
2
Ef = (mbullet + mblock ) g h , so
p
vf = 2 g h
q
= 2 (9.8 m/s2 ) (0.09 m)
Ei =
= 1.32816 m/s .
The linear momentum is conserved in a collision.
9 cm
vbullet
30 g
4.76 kg
pi = mbullet vi
pf = (mbullet + mblock ) vf .
Therefore
mbullet + mblock
vf
mbullet
(0.03 kg) + (4.76 kg)
=
(0.03 kg)
× (1.32816 m/s)
= 212.062 m/s .
vi =
Calculate the total energy of the composite
system at any time after the collision.
Correct answer: 4.22478 J.
Explanation:
Let : r = 16 cm = 0.16 m ,
h = 9 cm = 0.09 m ,
mblock = 4.76 kg , and
mbullet = 30 g = 0.03 kg .
041 (part 3 of 3) 10.0 points
Denote vbullet to be the initial velocity, find
the momentum of the compound system immediately after the collision.
p
1. pf = mbullet g h
Version 001 – Review - Mechanics – tubman – (20131)
p
1
2. pf = (mbullet + mblock ) g h
2
3. pf = mbullet vbullet correct
4. pf = mblock vbullet
p
5. pf = mblock g h
7. pf =
√
Before the collision vb = v0 and vw = 0 (the
wooden block is not moving); after the colli′
sion vb′ = vw
≡ v ′ (the bullet is stuck). Therefore, immediately after the collision the block
and the bullet move up at velocity
v′ =
6. pf = (mbullet + mblock ) vbullet
mbullet + mblock vbullet
1
(mbullet + mblock ) vbullet
2
√
9. pf = mbullet + mblock g h
p
10. pf = (mbullet + mblock ) g h
8. pf =
Lifting a Wood Block
042 10.0 points
A 8.81 g bullet is fired vertically into a 2.2 kg
block of wood.
v0
The bullet gets stuck in the block, and the
impact lifts the block 0.07 m up. (That is, the
block — with the bullet stuck in it — rises
0.07 m up above its initial position, and then
falls back down.)
Given g = 9.8 m/s2 . What was the initial
velocity of the bullet?
Correct answer: 293.67 m/s.
Explanation:
Basic Concepts: Momentum conservation during the collision, and energy conservation afterwards.
During the collision, the net momentum of
the block and the bullet is conserved,
′
mb vb + mw vw = Pnet = mb vb′ + mw vw
.
mb vb
.
mb + mw
Note that the mechanichal energy is not conserved during the collision; indeed, most of
the bullet’s kinetic energy is dissipated by the
friction between the bullet and the block.
After the collison, the momentum of the
bullet+block system is no longer conserved
because of external gravitational forces. However, the mechanical energy is conserved after
the collision because there is no kinetic friction at work. Thus
Explanation:
As in part 2, due to conservation of linear
momentum,
pf = pi = mbullet vbullet .
14
Mnet v ′2
+ Mnet gh = const,
2
hence given initial velocity v1 = v ′ (just after
the collision) and initial height h1 = 0, it
follows that the maximal height of the block
after the collision is
Hmax
v12
=
=
2g
mb vb
mb + mw
2 .
2g.
Consequently, given the maximum height
Hmax and the masses of the bullet and the
block, we find that the original velocity of the
bullet was
vb =
mb + mw p
× 2gHmax = 293.67 m/s.
mb
Holt SF 06G 01
043 (part 1 of 3) 10.0 points
A 0.013 kg marble sliding to the right at 20.2
cm/s on a frictionless surface makes an elastic head-on collision with a 0.013 kg marble
moving to the left at 18.9 cm/s. After the
collision, the first marble moves to the left at
18.9 cm/s.
Find the velocity of the second marble after
the collision.
Version 001 – Review - Mechanics – tubman – (20131)
Explanation:
Correct answer: 0.202 m/s.
Explanation:
Let :
m1
v1,i
v2,i
v1,f
1
1
2
2
m1 v1,f
+ m2 v2,f
2
2
1
= (0.013 kg)(−0.189 m/s)2
2
1
+ (0.013 kg)(0.202 m/s)2
2
= 0.000497413 J .
EK f =
= m2 = 0.013 kg ,
= +20.2 cm/s ,
= −18.9 cm/s , and
= −18.9 cm/s .
where to the right is positive.
m1 ~v1,i + m2 ~v2,i = m1 ~v1,f + m2 ~v2,f
m1 v1,i + m2 v2,i − m1 v1,f
m2
= v1,i + v2,i − v1,f
= 0.202 m/s + (−0.189 m/s)
− (−0.189 m/s)
v2,f =
= 0.202 m/s
to the right.
044 (part 2 of 3) 10.0 points
What is the total kinetic energy before the
collision?
Correct answer: 0.000497413 J.
Explanation:
1
1
2
2
m1 v1,i
+ m2 v2,i
2
2
1
= (0.013 kg)(0.202 m/s)2
2
1
+ (0.013 kg) (−0.189 m/s)2
2
= 0.000497413 J .
EK i =
045 (part 3 of 3) 10.0 points
What is the total kinetic energy after the
collision?
Correct answer: 0.000497413 J.
15