Download Physics

AP Physics 4: Linear Momentum and Energy
A.
Momentum and Energy formulas
1. impulse changes object's momentum: Ft = mv = p
2. quantity of motion (vector quantity): p = mv (kg•m/s)
3. work changes object's energy: W = F||d (J)
a. measured in joules (1 J = 1 N•m)
a. Only component of F parallel to d does work
F

b.
d
1. F|| = Fcos  W = (Fcos)d
2. include sign W > 0 when F  d 
3. F  when d  (orbit)
variable force—stretching a spring
1. graph spring force (Fs) vs. position (x)
Fs = kx
slope = k
Area = W
x
2. Fs = kx  slope = Fs/x = k
3. W = Fsx  area = ½x(kx) = ½kx2 = W
c. power is the rate work is done: P = W/t (W)
1. measured in watts (1 W = 1 J/s)
2. P = W/t = F(d/t) = Fvav (v is average)
3. graphing
a. P = W/t  slope of W vs. t graph
b. P = Fvav  area under F vs. v graph
4. kilowatt-hour, 1KWh = 3.6 x 106 J
4. mechanical energy
a. work-energy theorem: work done to an object
increases mechanical energy; work done by an
object decreases mechanical energy
b. scalar quantity, like work
c. kinetic energy—energy of motion
1. positive only
2. K = ½mv2 = p2/2m
Steps
Algebra
v2 = v02 + 2ad
start with
assume Ko = 0  vo = 0 v2 = 2ad
ad = ½v2
solve for ad
K = W = Fd
start with
substitute ma for F
K = (ma)d = m(ad)
substitute ½v2 for ad
K = m(½v2)
K = ½mv2
rearrange
mv = p
start with
m2v2 = p2
square both sides
divide both sides by 2m m2v2/2m = p2/2m
substitute K for ½mv2
K = p2/2m
d. potential energy—energy of relative position
1. gravitational potential energy
a. based on arbitrary zero
(usually closest or farthest apart)
b. Ug = mgh (near the Earth's surface)
Steps
Algebra
Ug = W = Fd
start with
substitute mg for F
Ug = (mg)d
substitute h for d
Ug = mgh
c. Ug = -GMm/r (orbiting system)
1. G = 6.67 x 10-11 N•m2/kg2
2. r = distance from center to center
3. Ug = 0 when r is Ug < 0 for all
values of r because positive work is
needed reach Ug = 0
2. spring (elastic) potential energy, Us = ½kx2
a. Us = W to stretch the spring
b. see work by a variable force above
Name __________________________
B.
Solving Work-Energy Problems
1. work done on object A by a "nonconservative" force
(push or pull, friction) results in the change in amount
of mechanical energy
2. work done on object A by a "conservative" force (gravity,
spring) results in the change in form of mechanical
energy (U  K) for object A, but no loss in energy
a. conservative forces (Fg and Fs)
1. Fg  d : Ug  K, Fg  d : K  Ug
2. Fs  d : Us  K, Fs  d : K  Us
b. process isn't 100 % efficient
1. friction (W = Ffd) reduces mechanical energy
2. mechanical energy is converted into random
kinetic energy of the object's atoms and the
temperature increases = heat energy—Q
3. total energy is still conserved
3. examples
Process
Energy
Work done to pull a pendulum bob off center W  Ug
1 Transformation release pendulum
Ug  K
Work done by hit a stationary object
KW
Work done to throw a ball into the air
WK
K  Ug
2 Transformation ball rises and falls
Work done by falling ball dents ground
KW
Work done to load a projectile in spring-gun W  Us
3 Transformation release projectile
Us  K
Work done by projectile penetrates target
KW
4. general solution for work-energy problems

determine initial energy of the object, Eo
o if elevated h distance: Ug = mgh
o if accelerated to v velocity: K = ½mv2
o if spring compressed x distance: Us = ½kx2

determine energy added/subtracted due to an external
push or pull: Wp = ±F||d

determine energy removed from the object by friction:
Wf = Ffd = (mgcosd
o d is the distance traveled
o  is the angle of incline (0o for horizontal)

determine resulting energy, E' = Eo ± Wp – Wf

determine d, h, x or v
o if slides a distance d: 0 = Eo ± Wp – mgcosd'
o if elevates a height h: E' = mgh'
o If compresses a spring x: Us: E' = ½kx'2
o if accelerated to velocity v: E' = ½mv'2

general equation (not all terms apply for each problem)
K + Ug + Us ± Wp – Wf = K' + Ug'+ Us'
½mv2 + mgh + ½kx2 ± Fpd – Ffd = ½mv'2 + mgh' + ½kx'2
C. Solving Collision Problems
1. collision between particles doesn't change the total
amount of momentum because the impulse on A equals
the impulse on B, but in the opposite directions
mAvA + mBvB = mAvA’ + mBvB’
Steps
Algebra
FA = -FB
start with Newton's Law
multiply both side by t
FAt = -FBt
substitute mv for Ft
mAvA = -mBvB
mA(vA’ – vA) = -mB(vB’ – vB)
substitute v' – v for v
-mAvA – mBvB = -mAvA’ – mBvB’
collect like v terms
mAvA + mBvB = mAvA’ + mBvB’
multiply by -1
b. two particles collide and stick together
1. inelastic collisions (vA' = vB')
2. mAvA + mBvB = (mA + mB)v’
Steps
Algebra
mAvA + mBvB = mAvA’ + mBvB’
start with
substitute v' for vA' and vB' mAvA + mBvB = mAv’ + mBv’
mAvA + mBvB = (mA + mB)v’
simplify
c. two particles collide and bounce off
1. elastic collisions (vA'  vB')
2. both energy and momentum are conserved
a. two unknowns  need two equations
b. mAvA + mBvB = mAvA’ + mBvB’
c. ½mAvA2 + ½mBvB2 = ½mAvA’2 + ½mBvB’2
d. simpler equation: vA + vA' = vB + vB’
mAvA + mBvB =
½mAvA2 + ½mBvB2 =
mAvA’ + mBvB’
½mAvA’2 + ½mBvB’2
mA(vA – vA') = mB(vB' – vB)
mA(vA2 – vA'2) = mB(vB'2 – vB2)
mA(vA2 – vA'2) = mB(vB'2 – vB2)
mA(vA – vA') = mB(vB' – vB)
(vA – vA')(vA + vA') = (vB – vB')(vB + vB')
(vA – vA') = (vB' – vB)
3. solving two equations and two unknowns

fill in vA and vB into vA + vA' = vB + vB'

write expression for vA’ in terms of vB'

substitute vA' expression in equation:
mAvA + mBvB = mAvA’ + mBvB’

solve for vB’

solve for vA’ using the expression for vA' above
4. collisions in two dimensions
a. px is conserved independently of py
b. elastic collision
1. mAvAx + mBvBx = mAvAx’ + mBvBx’
2. mAvAy + mBvBy = mAvAy’ + mBvBy’
3. solve two equations & two unknowns
c. inelastic collision
1. mAvAx + mBvBx = (mA + mB)vx'
2. mAvAy + mBvBy = (mA + mB)vy'
3. solve two equations & two unknowns
d. object explodes into two pieces mA and mB
a.
2.


1. (mA + mB)v = mAvA’ + mBvB’
2. opposite inelastic collision equation
solve ballistics problems
M

(vM = 0)
m
vm
bullet collides inelastically with block: mvm = (M + m)v'
block swings or slides (conservation of energy)
o block swings like a pendulum to height h
o K = Ug  ½(M + m)v'2 = (M + m)gh  h = v'2/2g
o block slides a distance d along a rough surface
o K = Wf  ½(M + m)v'2 = (M + m)gd  d = v'2/2g
A.
Momentum and Energy Formulas
Human Power Lab
Run up a flight of stairs at the football stadium and calculate
the power that you generated.
a. Collect the following data.
weight Fg-lbs
Number of steps N
time t
b.
dy
Questions 4-6 Use the following options
(A) W > 0
(B) W = 0
(C) W < 0
4. Work done by a person holding a 10-kg box.
B—without displacement, work = 0 (W = Fd)
5.
height of step y
width of step x
Calculate the following from the data.
Formula
Calculation
Work done by a person pulling a 10-kg box across a rough
floor at constant speed.
A—force and distance are in opposite directions
6.
dy = Ny
Work done by friction when a 10-kg box is pulled across a
rough floor at constant speed.
C—force and distance are in the opposite directions
dx
dx = Nx
d
d = (dx2 + dy2)½
7.
A box is being pulled up a rough incline by a rope. How
many forces are doing work on the box?
(A) 1
(B) 2
(C) 3
(D) 4
C—friction, gravity and tension
v
v = d/t
Fg Fg = Fg-lbs x 4.45
Questions 8-9 Alice applied 10 N of force over 3 m in 10 s. Bill
applies the same force over the same distance in 20 s.
8. Who did more work?
(A) Alice
(B) Bill
(C) They did the same
C—W = fd, which is the same for Alice and Bill
m
Fg = mg
9.
K
K = ½mv2
Ug Ug = Fgdy
P
c.
P = (Ug + K)/t
Calculate the power in horse power (1 hp = 746 W).
Who produced the greater power?
(A) Alice
(B) Bill
(C) They did the same
A—P = W/t, since Alice did the work in less time she
produced greater power
10. Car A has twice the mass of car B, but they both have the
same kinetic energy. How do their speeds compare?
(A) 2vA = vB
(B) 2vA = vB
(C) 4vA = vB
B—½mAvA2 = ½mBvB2
2mBvA2 = mBvB2  2vA2 = vB2  2vA = vB
11. A system of particles has total kinetic energy of zero. What
can you say about the total momentum of the system?
(A) p > 0
(B) p = 0
(C) p < 0
Questions 1-24 Briefly explain your answer.
1. An open cart rolls along a frictionless track while it is
raining. As it rolls, what happens to the speed of the cart
as the rain collects in it?
(A) increase
(B) the same
(C) decrease
C—rain increases mass, but momentum is conserved
and p = mv  increase mass = decrease v
Questions 2-3 Two boxes, one heavier than the other, are
initially at rest on a horizontal frictionless surface. The
same constant force F acts on each for 1 s.
2. Which box has more momentum after the force acts?
(A) light
(B) heavy
(C) tie
C—impulses are equal  change in momentum is equal
3.
Which box has the greater velocity after the force acts?
(A) light
(B) heavy
(C) tie
A—equal change in momentum: p = mv), but smaller
mass would produce greater change in v
B—kinetic energy is zero only when v = 0  p = mv = 0
12. A system of particles has total momentum of zero. Does it
necessarily follow that the total kinetic energy is zero?
(A) yes
(B) no
B—momentum is a vector quantity, a system of
particles could be moving, but in opposite directions
13. Two objects have the same momentum. Do these two
objects also have the same kinetic energy?
(A) yes
(B) no
B—K = ½mv2, where p = mv; since v is squared in K, but
not p, then differences in v will not have the same effect
Questions 14-15 Stone A has twice the mass as stone B. They
are dropped from a cliff and reach a point just above ground.
14. What is the speed of stone A compared to stone B?
(A) vA = vB
(B) vA = ½vB
(C) vA = 2vB
(D) vA = 4vB
A—acceleration due to gravity is equal and they fall the
same distance  velocities must be the same
15. What is the kinetic energy of stone A compared to stone B?
(A) KA = KB
(B) KA = ½KB
(C) KA = 2KB
(D) KA = 4KB
C—stone A has twice the mass as B  it has twice the K
B—Us = ½k(x)2: since x is positive Us is positive
Ug = mgh: since h is negative Ug is negative
25. What is the momentum of a 0.5-kg ball traveling at 18 m/s?
p = mv = (0.5 kg)(18 m/s) p = 9.0 kg•m/s
Questions 16-17 Car A and Car B are identical, but car A has
twice the velocity of car B.
16. What is the kinetic energy of car A compared to car B?
(A) KA = KB
(B) KA = ½KB
(C) KA = 2KB
(D) KA = 4KB
D—K = ½mv2  car A has 4 times the kinetic energy
17. Car A takes d meters to stop. What is the stopping distance
for car B?
(A) 2d
(B) 4d
(C) ½d
(D) ¼d
D—K = Wf = mgd; car B has ¼ the kinetic energy  it
takes ¼ the distance
18. The work Wo accelerates a car for 0 to v. How much work
is needed to accelerate the car from v to 3v?
(A) 2Wo
(B) 3Wo
(C) 8Wo
(D) 9Wo
C—W = K: K 0  v  v2 - 02 = v2
K from v to 3v  (3v)2 – v2 = 8v2
19. A golfer making a putt gives the ball an initial velocity of vo,
but he has misjudged the putt, and the ball only travels
one-quarter of the distance to the hole. What speed should
he have given the ball?
(A) 2vo
(B) 3vo
(C) 4vo
(D) 8vo
A—he needs 4 x the kinetic energy, which requires 2 x v
(K = ½mv2)
20. Which can never be negative?
(A) W
(B) K
(C) U
(D) p
B—mass can't be negative and v is squared  K = ½mv2
is always positive
21. You and your friend both solve a problem involving a skier
going down a slope, starting from rest. The two of you
have chosen different levels for h = 0 in this problem.
Which of the quantities will you and your friend agree on?
I. Ug
II. Ug
III. K
(A) I only
(B) II only (C) III only (D) II and III
D—your absolute value of Ug is different, but the change
is the same and K depends on v not h
22. Two paths lead to the top of a big hill. A is steep and direct,
while the B is twice as long but less steep. The change in
potential energy on path A compared to path B is
(A) UA < UB
(B) UA = UB
(C) UA > UB
B—the beginning and end points are the same  h is
the same
23. How does the work required to stretch a spring 2 cm
compare with the work required to stretch it 1 cm?
(A) W2 = W1
(B) W2 = 2W1
(C) W2 = 4W1
(D) W2 = 8W1
C—W = Us = ½k(x)2, since x is 2 x then W = 22 x
24. A mass attached to a vertical spring causes the spring to
stretch and the mass to move downward. Which is true
about the sign for Us and Ug?
(A) +Us, +Ug
(B) +Us, –Ug
(C) –Us, +Ug
(D) –Us, –Ug
26. What force is generated by a racket, which strikes a 0.06-kg
tennis ball that reaches a speed of 65 m/s in 0.03 s?
Ft = mv
F(0.03 s) = (0.06 kg)(65 m/s)F = 130 N
27. How much work is done to move a 50-kg crate horizontally
10 m against a 150-N force of friction?
W = F||d = (150 N)(10 m) = 1500 J
28. How much work is done to pull a 100-kg crate horizontally
10 m using a force of 100 N at 30o?
W = F||d = (100 N)(cos30)(10 m) = 866 J
29. How much work is done to carry a 100-kg crate 10 m up a
30o ramp?
W = F||d = (1000 N)(sin30)(10 m) = 5000 J
30. Why is work not needed to keep the earth orbiting the sun?
Work is not needed to keep the Earth orbiting the sun
because there is no force in the direction of motion.
31. How much power is needed to change the speed of a
1500-kg car from 10 m/s to 20 m/s in 5 s?
P = W/t = ½m(v2 – vo2)/t
P = ½(1500)(202 – 102)/5 = 45,000 W
32. How much power does a 75-kg person generate when
climbing 50 steps (rise of 25 cm per step) in 12 s?
a. in Watts
P = W/t = U/t = mgh/t
P = (75 kg)(10 m/s2)(50 x 0.25 m)/12 s = 780 W
b.
in horse power (1hp = 746 W)
780 W x 1 hp/746 W = 1.05 hp
33. How much power is needed to maintain a speed of 25 m/s
against a total friction force of 200 N?
P = Fvav = (200 N)(25 m/s) = 5000 W
34. How long will it take a 1750-W motor to lift a 285-kg piano
to a sixth-story window 16 m above?
P = W/t = mgh/t
1750 W = (285 kg)(10 m/s2)(16 m)/t  t = 26 s
35. A 1000-kg car travels at 30 m/s against 600-N of friction.
a. How much power does the car engine deliver?
P = Fvav = (600 N)(30 m/s) = 18,000 W
b.
The car accelerates at 2 m/s2. How much power does
the car engine deliver now?
P = Fvav = (ma + Ff)vav
P = [(1000 kg)(2 m/s2) + 600 N](30 m/s) = 78,000 W
The car goes up a 10o incline at 30 m/s. How much
power does the car engine deliver now?
c.
P = Fvav = (mgsin + Ff)vav
P = [(1000 kg)(10 m/s2)(sin10) + 600 N](30 m/s) = 70,000 W
36. What is the kinetic energy of a 2-kg block moving at 9 m/s?
K=
½mv2 =
½(2 kg)(9
m/s)2
42. A box sliding on a frictionless flat surface runs into a fixed
spring, which compresses a distance x to stop the box. If
the initial speed of the box is doubled, how much would the
spring compress?
(A) ½x
(B) x
(C) 2x
(D) 4x
C—Us = K  ½kx2 = ½mv2  x  v (double v = double x)
= 81 J
37. What is the gravitational potential energy of a 2-kg block
that is 6 m above zero potential energy?
Questions 43-44 Alice and Bill start from rest at the same
elevation on frictionless water slides with different shapes.
 Bill
 Alice
Ug = mgh = (2 kg)(10 m/s2)(6 m) = 120 J
38. What is the gravitational potential energy of the earth-moon
system? (MEarth = 5.97 x 1024 kg, Mmoon = 7.35 x 1022 kg,
distance between Earth and moon, r = 3.84 x 108 m)
Ug = -GMm/r
Ug = -(6.67 x 10-11)(5.97 x 1024)(7.35 x 1022)/3.84 x 108
Ug = -7.6 x 1028 J
39. Consider the following spring (k = 100 N/m).
a. Calculate the force (F = kx) needed to stretch a spring
from 0.0 m to 0.5 m.
x (m)
0.0
0.1
0.2
0.3
0.4
0.5
0N
10 N
20 N
30 N
40 N
50 N
F (N)
b. Graph the data
40 N
0N
0.2 m
Calculate the area under the graph.
0.4 m
A = ½BH = ½(0.50)(50) = 12.5 N•m
d.
44. Who makes it to the bottom in the least amount of time?
(A) Alice
(B) Bill
(C) both are the same
A—Alice is faster at all times except the very end (she
has fallen a greater h at each instance)
45. A cart starting from rest rolls down a hill and at the bottom
has a speed of 4 m/s. If the cart were given a push, so its
initial speed at the top of the hill was 3 m/s, what would be
its speed at the bottom?
(A) 4 m/s
(B) 5 m/s
(C) 6 m/s
(D) 7 m/s
Us = ½kx2 = ½(100)(0.50)2 = 12.5 J
B. Solving Work-Energy Problems
Questions 40-47 Briefly explain your answer.
40. Three balls of equal mass start from rest and roll down
different ramps. All ramps have the same height. Which
ball has the greatest speed at the bottom of the ramp?
 
B
46. You see a leaf falling to the ground with constant speed.
When you first notice it, the leaf has initial total mechanical
energy Ei. You watch the leaf until just before it hits the
ground, at which point it has final total mechanical energy
Ef. How do these total energies compare?
(A) Ei < Ef
(B) Ei = Ef
(C) Ei > Ef
C—energy is lost due to air resistance
How much potential energy is stored in the stretched
spring?
A
C—Ug = K  mgh = ½mv2  v = (2gh)½, since h is the
same v is the same
B—K' = Ko + Ug
½mv2 = ½m(3)2 + ½m(4)2  v2 = 32 + 42 = 25  v = 5 m/s
20 N
c.
43. At the bottom whose velocity is greater?
(A) Alice
(B) Bill
(C) tie
C
47. You throw a ball straight up into the air. In addition to
gravity, the ball feels a force due to air resistance.
Compared to the time it takes the ball to go up, the time to
come back down is
(A) less
(B) equal
(C) greater
C—at each elevation, the ball is traveling faster on the
way up compared to the way down
48. A rock is dropped from 20 m. What is the final velocity?
a. Use kinematics to solve this problem.
v2 = vo2 + 2ad
v2 = 2(-10 m/s2)(-20 m) = ±20 m/s
b.
(A) A
(B) B
(C) C
(D) All the same
D—the same amount of Ug is converted to K  same
speed
41. A stationary block slides down a frictionless ramp and
attains a speed of 2 m/s. To achieve a speed of 4 m/s, how
many times higher must the block start from?
(A) 2 times
(B) 4 times
(C) 8 times
B—a speed of 4 m/s requires 4 x as much K  the ball
must lose 4 x Ug = 4h
Use energy to solve this problem.
Ug-0 + K0 = Ug-t + Kt  mgh = ½mv2
(10 m/s2)(20 m) = ½v2  v = ±20 m/s
49. A pendulum bob reaches a maximum height of 0.6 m above
the lowest point in the swing, what is its fastest speed?
Ug-0 + K0 = Ug-t + Kt  mgh = ½mv2
(10 m/s2)(0.6 m) = ½v2  v = 3.5 m/s
50. How far must a 1 kg ball fall in order to compress a spring
0.1 m? (k = 1000 N/m)
Ug-0 + Us-0 = Ug-0 + Us-0  mgh = ½kx2
(1)(10 m/s2)h = ½(1000 kg)(0.1 m/s)2  h = 0.50 m
51. A 10-kg box is initially at the top of a 5-m long ramp set at
53o. The box slides down to the bottom of the ramp. The
force of friction is 31 N. Determine the
a. potential energy at the top of the ramp.
Ug = mgh = mgLsin
Ug = (10 kg)(10 m/s2)(5 m)sin53 = 400 J
b.
work done by friction during the slide.
Wf = Ffd = (-31 N)(5 m) = -155 J
c.
velocity of the box at the bottom of the ramp.
½mv2
Ug + Wf = Kt =
400 J – 155 J = ½(10 kg)v2 v = 7 m/s
52. A spring (k = 500 N/m) is attached to the wall. A 5-kg
block on a horizontal surface ( = 0.25) is pushed against
the spring so that the spring is compressed 0.2 m. The
block is released and propelled across the surface.
a. Determine the potential energy of the spring.
Us = ½kx2 = ½(500 N/m)(0.2 m)2 = 10 J
b.
Determine the distance that the block travels.
Us = Wf = Ffd = mgd
10 J = (0.25)(5 kg)(10 m/s2)d d = 0.8 m
C.
Solving Collision Problems
Collision Lab
Observe an elastic collision between a swinging 200-g
weight and a golf ball, and compare the actual post-collision
velocities with the theoretical velocities.
a. Collect the following data.
200-g Weight (A)
Golf Ball (B)
string length LA
mass mB
initial angle A
25o
final angle A'
b.
table height dyB
distance dxB'
Calculate the following from the data.
Formula
Calculation
dyA dyA = LA(1 – cosA)
vA vA2 = 2gdyA
dyA' dyA' = LA(1 – cosA')
vA' vA'2 = 2gdyA'
tB'
dyB = ½gtB'2
vB' vB' = dxB'/tB'
c.
Calculate the following theoretical values.
Formula
Calculation
vA = -(vA' – vB')
vA'
mAvA = mAvA' + mBvB'
vB' vB' = vA + vA'
d.
Calculate the percent differences.
Formula
Calculation
vA'
% = 100|v|/vtheory
vB'
Questions 53-59 Briefly explain your answer.
53. A small car and a large truck collide head-on and stick
together. Which one has larger momentum change?
(A) car
(B) truck
(C) tie
C—impulses are equal (Newton's third law)  change in
momentums are equal
54. A small beanbag and a bouncy rubber ball are dropped
from the same height above the floor. The both have the
same mass. Which would hurt more if it hit you on the
head?
(A) beanbag
(B) rubber ball (C) doesn't matter
B—the rubber ball would bounce back which would
generate a greater p and impulse force
55. A box slides with initial velocity 10 m/s on a frictionless
surface and collides inelastically with a stationary identical
box. What is the final velocity of the combined boxes?
(A) 0 m/s
(B) 5 m/s
(C) 10 m/s (D) 20 m/s
B—mAvA + mBvB = (mA + mB)v'
mA(10 m/s) + 0 = 2mAv'  v' = 5 m/s
Questions 56-57 A uranium nucleus (at rest) undergoes fission
and splits into two fragments, one heavy and the other light.
56. Which fragment has the greater momentum?
(A) heavier one (B) lighter one (C) tie
C—initial momentum = final momentum = zero  equal
momenta in opposite directions
57. Which fragment has the greater speed?
(A) heavier one (B) lighter one (C) tie
B—momentums are equal, p = mv, but with less mass,
the lighter one has greater speed
58. Alice (50 kg) and Bill (75 kg) are standing on slippery ice
and push off of each other. If Alice slides at 6 m/s, what
speed does Bill have?
(A) 2 m/s
(B) 3 m/s
(C) 4 m/s
(D) 6 m/s
C—0 = (mAvA' + mBvB'
0 = (50 kg)(6 m/s) + (75 kg )vB'  vB' = -4 m/s
59. A cannon sits on a stationary railroad flatcar with a total
mass of 1000 kg. When a 10-kg cannon ball is fired at a
speed of 50 m/s, what is the recoil speed of the flatcar?
(A) 0 m/s
(B) 0.5 m/s (C) 10 m/s (D) 50 m/s
B—0 = (mA + mB)v'
0 = (10)(50 m/s) + (1000)vB'  vB' = -0.5 m/s
60. A 25-kg child in a stationary 55-kg boat with a 5-kg
package throws the package out horizontally at 8 m/s.
What is the boat and child's resultant velocity?
0 = mAvA' + mBvB'
0 = (25 kg + 55 kg)v1' + (5 kg)(8 m/s)v = -0.5 m/s
61. An 85-kg safety running at 5 m/s tackles a 95-kg fullback
traveling at 4 m/s from behind. What is their mutual speed
just after the tackle?
mAvA + mBvB = (mA + mB)v'
(85)(5) + (95)(4) = (85+ 95)v'  v' = 4.47 m/s
62. A 0.45-kg ice puck, moving east with a speed of 3.0 m/s,
has a head-on elastic collision with a 0.9-kg puck initially at
rest. What are the resulting speeds and directions?
vA + vA' = vB + vB'
3.0 + vA' = 0 + vB'  vB' = vA' + 3.0
mAvA + mBvB = mAvA' + mBvB'
(.45 kg)(3.0 m/s) = (45 kg)(vA') + (.9 kg)(vA' + 3.0)
vA' = -1 m/s (west)
vB' = vA' + 3.0 = -1 m/s + 3 = 2 m/s (east)
63. A 1-kg block traveling at 5 m/s in the direction of 30o south
of east collides and sticks with a 2-kg block traveling north
at 3 m/s. Determine
a. The x-component of the resulting velocity, vx'.
px: mAvAcosA + mBvBcosB = (mA + mA)vx'
(1)(5)cos-30 + 0 = (1 + 2)vx'  vx' = 1.4 m/s
b.
The resultant speed.
v' = (vx'2 + vy'2)½
v' = [(1.4 m/s)2 + (1.2 m/s)2]½ = 1.9 m/s
d.
The resultant direction.
tan = vy'/vx'
tan = 1.2 m/s/1.4 m/s = 0.81   = 54o
64. A 0.50-kg softball is traveling at 40 m/s. A bat makes
contact with the ball for 0.025 s, after which, the ball's
velocity is 35 m/s in the opposite direction (v = -75 m/s).
a. Determine the change in the ball's momentum.
p = mv = (0.5 kg)(-75 m/s)
p = -37.5 kg•m/s
b.
67. Determine the velocities after the following elastic
collisions between mA (1 kg) and mB (3 kg).
vA = + 2 m/s
vB = 0 m/s
A = 0o
B = 0o
vA + vA' = vB + vB'
2 + vA' = 0 + vB'  vB' = vA' + 2
mAvA + mBvB = mAvA' + mBvB'
(1 kg)(2 m/s) = (1 kg)(vA') + (3 kg)(vA' + 2)
vA' = -1 m/s (west)
vB' = vA' + 2 = -1 m/s + 2 = 1 m/s (east)
68. A 15-g bullet penetrates a 1.1-kg block of wood. As a
result, the block slides along a surface ( = 0.85) for 9.5 m.
a. How much work is done by friction?
W = F||d = Ffd = mgd
W = (0.85)(1.1 kg)(10 m/s2)(9.5) = 89 J
The y-component of the resulting velocity, vy'.
py: mAvAsinA + mBvBsinB = (mA + mA)vy'
(1.0)(5)sin-30 + (2)(3)sin90 = (1 + 2)vy'  vy' = 1.2 m/s
c.
(mA + mB)v = mAvA' + mBvB'
0 = (1 kg)(-12 m/s) + (3 kg)vB'
vB' = 4 m/s (east)
b.
What is the velocity of the system just after the impact?
W = K = ½mv2
89 J = ½(1.1 kg)v2  v = 13 m/s
c.
What is the velocity of the bullet just before the impact?
mAvA + mBvB = (mA + mB)v'
(0.015 kg)vA + 0 = (1.1 kg)(13 m/s) vA = 950 m/s
69. An 18-g bullet traveling at 230 m/s buries itself in a 3.6-kg
pendulum hanging on a 2.8-m long string.
a. Determine for the bullet/pendulum just after impact.
(1) velocity
mAvA + mBvB = (mA + mB)v'
(0.018 kg)(230 m/s) + 0 = (3.6 kg)v'  v' = 1.1 m/s
(2) kinetic energy
K = ½mv2 = ½(3.6 kg)(1.1 m/s)2 = 2.2 J
Determine the average force exerted by the bat.
Ft = mv
F(0.025 s) = -37.5 kg•m/s  F = -1500 N
65. Determine the velocity after the following inelastic
collisions between mA (1 kg) and mB (3 kg).
vA = + 2 m/s
vB = -1 m/s
A = 0o
B = 0o
mAvA + mBvB = (mA + mB)v'
(1 kg)(2 m/s) + (3 kg)(-1 m/s) = (1 kg + 3 kg)v'
v' = -0.25 m/s (west)
vA = + 5 m/s
vB = 3 m/s
A = 53o
B = 90o
px: mAvAcosA + mBvBcosB = (mA + mB)vx'
(1)(5cos53) + (3)(3cos90) = (1 + 3)vx'  vx' = 0.75 m/s
py: mAvAsinA + mBvBsinB = (mA + mB)vy'
(1)(5sin53) + (3)(3sin90) = (1 + 3)vy'  vy' = 3.25 m/s
v' = (vx'2 + vy'2)½ = [(.75 m/s)2 + (3.25 m/s)2]½
v' = 3.33 m/s
tan = vy'/vx' = 3.25 m/s/0.75 m/s = 4.33   = 77o
66. Stationary mA (1 kg) and mB (3 kg) are separated by a
compressed spring, which is then released. What is the
resulting velocity of mB, when vA is -12 m/s?
b.
How high does the pendulum rise?
K = Ug = mgh
2.2 J = (3.6 kg)(9.8 m/s2)h  h = 0.062 m
70. Andre hits a 0.06-kg tennis ball straight up into the air with
a 300-N force. The ball remains on the racket for 0.25 m.
a. Using dynamics and kinematics, determine
(1) The acceleration.
Fnet = ma
300 N = (0.06 kg)a a = 5000 m/s2
(2) The initial velocity.
v2
v2
= vo2 + 2ad
= vo2 + 2ad = 2(5000 m/s2)(0.25 m)v = 50 m/s
(3) The maximum height reached by the tennis ball.
v2
= vo2 + 2ad
0 = (50 m/s)2 + 2(-10 m/s2)d d = 125 m
b.
Use energy to determine the maximum height reached
by the tennis ball.
F||d = mgh
(300 N)(0.25 m) = (0.06 kg)(10 m/s2)h  h = 125 m
71. How much power is used to lift 100 kg a distance of 2 m in
4 s?
P = W/t = mgh/t
P = (100 kg)(10 m/s2)(2 m)/(4 s) = 500 W
72. A 1,000-kg car maintains a constant speed of 30 m/s against
a combined friction and air resistance force of 550 N.
a. How much power is needed to cruise at 30 m/s?
Pf = Ffvav
Pf = (550 N)(30 m/s) = 16,500 W
b.
How steep an incline can the car climb if the engine
can generate 50,000 W of power?
P = Pclimb + Pf = Fgv + Pf = mgsin + Pf
50,000 W = (10,000sinN)(30 m/s) + 16,500 W  = 6o
73. A 1-kg block is pushed down against a spring (k = 500 N/m),
which is compressed 0.1 m. The block is released and
propelled vertically.
a. Determine the potential energy of the spring.
U = ½kx2
U = ½(500 N/m)(0.1 m)2 = 2.5 J
b.
74. A 10-kg box is initially at the top of a 5-m long ramp set at
30o. The box slides down to the bottom of the ramp. The
force of friction is 26 N. Determine the
a. potential energy at the top of the ramp.
Ug = mgh = mgLsin
Ug = (10 kg)(10 m/s2)(5 m)sin30 = 250 J
height at , h
b. Calculate the following from the data.
Formula
Calculation
Pendulum
h
h = h – ho
Ug Ug = (mA + mB)gh
K
K = Ug
v'
K = ½(mA + mB)v'2
vA mAvA + 0 = (mA + mB)v'
Kinematics
dy = ½gt2
t
vA vA = dx/t
Velocimeter
vA vA = (vkm/hr)/3.6
c.
Calculate the percent differences with the velocimeter.
Formula
Calculation
P
% = 100|v|/vvelocimeter
K
velocity of the box at the bottom of the ramp.
Ug – Wf = K'  Ug – Wf = ½mv2
250 J – 130 J = ½(10 kg)v2 v = 4.9 m/s
75. A 0.050-kg bullet traveling at 1,000 m/s penetrates a 10-kg
block of wood.
a. What is the velocity of the block after impact?
mAvA + mBvB = (mA + mB)v'
(0.050 kg)(1,000 m/s) + 0 = (10.05 kg)v' v' = 5 m/s
b.
lowest height, ho
work done by friction during the slide.
Wf = Ffd
W = (26 N)(5 m) = 130 J
c.
Velocimeter
Determine the maximum height reached by the block.
Us = Ug = mgh
2.5 J = (1 kg)(10 m/s2)h  h = 0.25 m
b.
pendulum angle, 
Practice Multiple Choice (No calculator)
Briefly explain why the answer is correct in the space provided.
1
2
3
4
5
6
7
8
9
10 11 12
C
A
D
A
A
D
C
D
A
B
A
D
13 14 15 16 17 18 19 20 21 22 23 24
C
A
B
C
D
C
C
D
C
C
B
D
25 26 27 28 29 30 31 32 33 34 35 36
C
B
B
B
A
C
A
D
A
C
A
D
1. The force on an object versus time is graphed below.
How far does the block travel along a rough surface
( = 0.25) before stopping?
K = W = F||d = Ffrd  ½mv2 = mgd
½(5 m/s)2 = (0.25)(10 m/s2)d  d = 5 m
c.
How high does the block rise if it were suspended
from a long string?
K = Ug  ½mv2 = mgh
½(5 m/s)2 = (10 m/s2)h  h = 1.25 m
Ballistic Pendulum Lab
Measure the velocity of a projectile using pendulum data,
kinematic data and the velocimeter, and comparing the
three values.
a. Collect the following data.
Pendulum
Kinematics
bearing mass, mA
height, dy
pendulum mass, mB
distance, dx
The object's change in momentum, in kg•m/s, from 0 to 4 s is
(A) 40
(B) 20
(C) 0
(D) -20
Ft above the graph equals Ft below the graph
 +J – J = 0 and no change in momentum.
2.
A 3-kg block, initially at rest, is pulled along a frictionless,
horizontal surface with a force shown as a function of time
t by the graph.
8.
An object of mass m is moving with speed vo to the right on
a horizontal frictionless surface when it explodes into two
pieces. Subsequently, one piece of mass 2/5m moves with
a speed ½vo to the left. The speed of the other piece of
the object is
(A) vo/2
(B) vo/3
(C) 7vo/5
(D) 2 vo
mv = m1v1' + m2v2'  mvo = (2/5m)(-1/2vo) + (3/5m)v2'
vo = -2/10vo + 3/5v2'  v2' = 2vo
9.
The speed of the block at t = 3 s is
(A) 3 m/s
(B) 4 m/s
(C) 6 m/s
(D) 8 m/s
J = Ft = area under the graph = ½(6 N)(3 s) = 9 N•s
J = mv  9 N•s = (3 kg)v  v = 3 m/s
3.
Two pucks, where mI = 3mII, are attached by a stretched
spring and are initially held at rest on a frictionless surface.
The pucks are released simultaneously. Which is the
same for both pucks as they move toward each other?
(A) Speed
(B) Velocity
(C) Acceleration
(D) Magnitude of momentum
mIvI = mIIvII and mI = 3mII, then vII = 3 mI  speed,
velocity and acceleration will be different.
4.
A 2,000-kg railroad car rolls to the right at 10 m/s and
collides and stick to a 3,000-kg car that is rolling to the left at
5 m/s. What is their speed after the collision?
(A) 1 m/s
(B) 2.5 m/s (C) 5 m/s
(D) 7 m/s
mAvA + mBvB = (mA + mB)v'
(2,000)(10) + (3,000)(-5) = (5,000)v'  v' = 1 m/s
5.
A 5-kg block with momentum = 30 kg•m/s, sliding east
across a horizontal, frictionless surface, strikes an obstacle.
The obstacle exerts all impulse of 10 N•s to the west on the
block. The speed of the block after the collision is
(A) 4 m/s
(B) 8 m/s
(C) 10 m/s (D) 20 m/s
po + J = pt 30 kg•m/s + (- 10 N•s) = 20 kg•m/s
pt = mv  v = pt/m = (20 kg•m/s)/(5 kg) = 4 m/s
6.
In the diagram, a block of mass M initially at rest on a
frictionless horizontal surface is struck by a bullet of mass
m moving with horizontal velocity v.
Two objects of mass 0.2 kg and 0.1 kg, respectively, move
parallel to the x-axis. The 0.2 kg object overtakes and
collides with the 0.1 kg object. Immediately after the
collision, the y-component of the velocity of the 0.2 kg
object is 1 m/s upward.
What is the y-component of the velocity of the 0.1 kg object
immediately after the collision?
(A) 2 m/s downward
(B) 0.5 m/s downward
(C) 0 m/s
(D) 0.5 m/s upward
mAvAy + mBvBy = mAvAy' + mBvBy'
0 = (0.2 kg)(1 m/s) + (0.1 kg)vBy'  vBy' = -2 m/s
10. Two particles of equal mass mo
moving with equal speeds vo
along paths inclined at 60° to the
x-axis, collide and stick together.
Their velocity after the collision
has magnitude
(A) vo/4
(B) vo/2
(C) vo/2
(D) 3vo/2
vy' = 0, movocos60 + movocos60 = (mo + mo)vx'
movo(½) + movo(½) = 2movx'  vx' = vo/2
11. Student A lifts a 50-N box to a height of 0.4 m in 2.0 s.
Student B lifts a 40-N box to a height of 0.50 m in 1.0 s.
Compared to student A, student B does
(A) the same work but develops more power
(B) the same work but develops less power
(C) more work but develops less power
(D) less work but develops more power
WA = F||d = (50 N)(.4 m) = 20 J = WB = (40 N)(.50 m) = 20 J
PA = W/t = 20 J/2.0 s = 10 W < PB = 20 J/1.0 s = 20 W 
12. What is the change in gravitational potential energy for a
50-kg snowboarder raised a vertical distance of 400 m?
(A) 50 J
(B) 200 J
(C) 20,000 J (D) 200,000 J
Ug = mgh = (50 kg)(10 m/s2)(400 m) = 200,000 J
What is the velocity of the bullet-block system after the
bullet embeds itself in the block?
(A) (M + v)m/M
(B) (m + v)m/M
(C) (m + M)v/M
(D) mv/(m + M)
mAvA + mBvB = (mA + mB)v'
mv + 0 = (m + M)v'  v' = mv/(m + M)
7.
A disc of mass m is moving to the right with speed v when
it collides and sticks to a second disc of mass 2m. The
second disc was moving to the right with speed v/2.
13. How high is a 50-N object moved if 250 J of work is done
against the force of gravity?
(A) 2.5 m (B) 10 m
(C) 5 m
(D) 25 m
W = F||d
250 J = (50 N)d  d = 5 m
14. What is the spring potential energy when a spring (k = 80
N/m) is stretched 0.3 m from its equilibrium length?
(A) 3.6 J
(B) 12 J
(C) 7.2 J
(D) 24 J
Us = ½kx2 = ½(80 N/m)(0.3 m)2 = 3.6 J
The speed of the composite body after the collision is
(A) v/3
(B) v/2
(C) 2v/3
(D) 3v/2
mAvA + mBvB = (mA + mB)v'
mv + (2m)(v/2) = (m + 2m)v'  v + v = 3v'  v' = 2v/3
15. What is the kinetic energy of a 5-kg block that slides down
an incline at 6 m/s?
(A) 20 J
(B) 90 J
(C) 120 J
(D) 240 J
K = ½mv2 = ½(5 kg)(6 m/s)2 = 90 J
Questions 16-17 A weight lifter lifts a mass m at constant speed
to a height h in time t.
16. How much work is done by the weight lifter?
(A) mg
(B) mh
(C) mgh
(D) mght
W = F||d = mgh
17. What is the average power output of the weight lifter?
(A) mg
(B) mh
(C) mgh
(D) mgh/t
P = W/t = mgh/t
18. The graphs show the position d versus time t of three
objects that move along a straight, level path.
Initial energy, Ug = mgh = (40 N)(8 m) = 320 J
Ug – Wf = K  320 J – 50 J = 270 J
24. A 50-kg diver falls freely from a diving platform that is 3 m
above the surface of the water. What is her kinetic energy
at 1 m above the water?
(A) 0
(B) 500 J
(C) 750 J
(D) 1000 J
mgh3 = K + mgh1
(50)(10)(3) = K + (50)(10)(1)  K = 1000 J
25. A 1000 W electric motor lifts a 100 kg safe at constant
velocity. The vertical distance through which the motor
can raise the safe in 10 s is most nearly
(A) 1 m
(B) 3 m
(C) 10 m
(D) 32 m
P = W/t = mgh/t
1000 W = (100 kg)(10 m/s2)h/10 s  h = 10 m
Questions 26-28 The vertical height versus gravitational potential
energy for an object near earth's surface is graphed below.
Ug (J)
Which has no change in kinetic energy?
(A) II only (B) III only (C) I and II (D) I and III
80
K = 0  constant velocity (no acceleration). In I
velocity is constant and II is stationary.
60
19. Which is a scalar quantity that is always positive or zero?
(A) Power
(B) Work
(C) Kinetic energy
(D) Potential Energy
All energy/power values are scalar, kinetic energy can
only be positive (v2).
Questions 20-21 A constant force of 900 N pushes a 100 kg
mass up the inclined plane at a uniform speed of 4 m/s.
40
20
0
h (m)
0.5
1.5
2.5
3.5
26. What is Ug when the object is 2.25 m above the surface?
(A) 50 J
(B) 45 J
(C) 60 J
(D) 55 J
Ug at 2.25 m is 45 J
27. What is the mass of the object?
(A) 1.5 kg (B) 2.0 kg (C) 2.5 kg
20. The power developed by the 900-N force is
(A) 400 W (B) 800 W (C) 900 W (D) 3600 W
P = Fv = (900 N)(4 m/s) = 3600 W
21. The gain in potential energy when the mass goes from the
bottom of the ramp to the top.
(A) 100 J
(B) 500 J
(C) 1000 J (D) 2000 J
Ug = mgh = (100 kg)(10 m/s2)(1 m) = 1000 J
22. What is the maximum height that a 0.1-kg stone rises if 40
J of work is used to shoot it straight up in the air?
(A) 0.4 m (B) 4 m
(C) 40 m
(D) 400 m
W = Ug = mgh
40 J = (0.1 kg)(10 m/s2)h  h = 40 m
23. A 40-N block is released from rest on an incline 8 m above
the horizontal.
What is the kinetic energy of the block at the bottom of the
incline if 50 J of energy is lost due to friction?
(A) 50 J
(B) 270 J
(C) 320 J
(D) 3100 J
(D) 3.0 kg
data point (1.0, 20) and Ug = mgh.
m = Ug/gh = (20 J)/(10 m/s2)( 1.0 m) = 2 kg
28. What does the slope of the graph represent?
(A) mass of the object
(B) gravitational force on the object
(C) kinetic energy of the object
(D) potential energy of the object
slope = y/x = Ug/h
Ug = mgh  Ug/h = mg, which is gravitational force
29. If an object with greater mass was graphed instead of the
object graphed above, how would the slope of the graph
differ from the above graph?
(A) more positive
(B) less positive
(C) equal but negative
(D) be the same
Greater mass = greater gravitational force and slope =
gravitational force  the slope would be greater.
30. Which is the graph of the spring potential energy of a
spring versus elongation from equilibrium?
(A)
(B)
(C)
(D)
Us = ½kx2  Us increases with the square of the
elongation, which forms a curve that slopes upward.
31. Which is the graph of the gravitational potential energy of
an object versus height? (Assume height << earth's radius)
(A)
(B)
(C)
(D)
a.
How much potential energy does the spring have just
before it turns back?
Us = ½kx2 = ½(180 N/m)(0.023 m)2 = 0.048 J
c.
How much work is done by friction as it slides 7.3 cm?
Us5 – Wf = Us2.3
Wf = Us5 – Us2.3 = 0.225 J – 0.048 J = 0.177 J
Ug = mgh  Ug increases in proportion to h, which
forms a straight diagonal line.
32. An object falls freely near earth's surface. Which graph
best represents the relationship between the object's
kinetic energy and its time of fall?
(A)
(B)
(C)
(D)
d.
What is the coefficient of friction?
Wf = mgd7.3
0.177 J = (0.62)(10)(0.073)   = 0.39
2.
A larger block with mass M slides down and strikes a
smaller block with mass m, where M = 3m. The blocks
stick and fall to the floor.
K increases with time so B and C are incorrect. K =
½mv2 and v = vo + at  K  t2, which forms a parabola.
33. A system consists of two masses m1 and m2
(m1 < m2). The objects are connected by a string,
hung over a pulley and then released. When the
object of mass m2 has descended a distance h,
the change in potential energy of the system is
(A) (m1 – m2)gh
(B) m2gh
(C) (m1 + m2)gh
(D) ½(m1 + m2)gh
a.
Mgh = ½MvM2  (10)(0.30) = ½vM2  vM = 2.45 m/s
Ug
Ug-1 + Ug-2
Ug = m1g(+h) + m2g(-h) = (m1 – m2)gh
b.
34. From the top of a 70-m-high building, a 1-kg ball is thrown
directly downward with an initial speed of 10 m/s. If the
ball reaches the ground with a speed of 30 m/s, the energy
lost to friction is most nearly
(A) 50 J
(B) 250 J
(C) 300 J
(D) 450 J
c.
How far horizontally from the table's edge do the
blocks land?
dy = ½gt2  0.90 = ½(10)t2  t = 0.42 s
½mv'2
35. A block of mass M is released from rest at the top of an
inclined plane of height h. If the plane is frictionless, what
is the speed of the block at the bottom of the incline?
(A) (2gh)½ (B) 2Mgh
(C) 2MghR2 (D) 5gh
Ug = K (no rolling because of zero friction)
mgh = ½mv2 v = 2gh = (2gh)½
36. Block of mass m slides on a horizontal frictionless table
with an initial speed vo. It then compresses a spring of
force constant k and is brought to rest. How much is the
spring compressed from its natural length?
(A) vo2/2g (B) mgvo/k (C) mvo/k
(D) vo(m/k)½
What is the velocity of the combined blocks after the
collision?
MvM = (M + m)v'  3m(2.45) = 4mv' v' = 1.84 m/s
Ug + K – Wf = K  mgh +
– Wf =
(1)(10)(70) +½(1)(4)2-Wf = ½(1)(30)2  Wf = 300 J
½mv2
What is the velocity of the larger block just before it
collides with the smaller block?
4.
dx = v't = (1.84)(0.42) = 0.77 m
Students are to calculate the spring constant k of a spring
that initially rests on a table. When the spring is
compressed a distance x from its uncompressed length Lo
and then released, the top rises to a maximum height h
above the point of maximum compression. The students
repeat the experiment, measuring h with various masses m
taped to the top of the spring.
K = Us
½mvo2 = ½kx2  x = vo(m/k)½
1.
Practice Free Response
A 0.62-kg block is attached to the spring (k = 180 N/m).
When the system is compressed 5.0 cm and released, it
slides a total of 7.3 cm before turning back.
a.
Us = Ug 
½kx2 = mgh  h = kx2/2mg
b.
a.
How much potential energy does the spring have
when compressed 5.0 cm?
Us = ½kx2 = ½(180 N/m)(0.05 m)2 = 0.225 J
Derive an expression for the height h in terms of m, x,
k, and fundamental constants.
Given that the experimental variables are mass and
height, what quantities should be graphed so that the
slope of a best-fit straight line can be used to calculate
the spring constant k?
1/m and h