AP Physics I
Study Guide: Impulse, Momentum and Energy – Answer Key
Chapter 7 Linear Momentum
•
Momentum is defined as _____ mass times velocity _____________
•
Impulse is defined as force x time and is equal to the __ change in
momentum.
•
In all interactions between isolated objects, momentum is conserved
(constant).
This means that the total momentum of the system __before__ the collision is
equal to the total momentum of the system __after___ the collision.
1. There are three types of collisions that can occur
i.
ii.
iii.
__elastic___collisions where objects collide and there is no
energy lost
_completely inelastic_ collisions where objects stick together
and move together as one object after the collision.
_inelastic_collisions – where objects collide, there is some
energy lost and the objects do not stick together. Most collisions
are of this type.
•
Momentum __is__ conserved but kinetic energy _is not__ conserved in an
inelastic collision.
•
In an inelastic collision, _kinetic__ energy is converted to internal elastic
_potential__ energy when the objects deform. Some kinetic energy is also
converted to sound energy and internal energy (heat).
•
In an _elastic_ collision, two objects return to their original shapes and move
away from the collision separately.
•
Both _momentum and _kinetic energy__ are conserved in an elastic collision.
•
In an elastic collision, the speed of approach is equal to the speed of
separation or v1 – v2 = v2’ – v1’
Review questions:
1. A car with a mass of 1000 kg is traveling at 30 m/s when the driver applies
the brakes. The car eventually comes to a stop. What is the impulse
applied to the car?
ans. - 30000 Ns
2. If the car in question 1 takes 0.8 seconds to come to a stop, what was the
average braking force?
ans. – 37500 N
3. A ball moving with a momentum of 5 kg m/s has an elastic collision with a
stationary ball. After the collision the first ball is at rest and the second ball
which had been stationary is moving. What is the momentum of the
second ball?
ans. 5 kg m/s
4. A 65.0 kg ice skater moving to the right at 2.5 m/s throws a 0.15 kg
snowball to the right with a velocity of 32.0 m/s with respect to the ground.
a. What is the velocity of the ice skater after throwing the snowball?
ans. 2.4 m/s
b. A second skater initially at rest with a mass of 60 kg catches the snowball.
What is the velocity of the second skater after catching the snowball in a
completely inelastic collision?
ans. 0.08 m/s
5. If two objects collide and one is initially at rest, is it possible for both to be
at rest after the collision? Explain your answer.
No! Initial momentum not zero, final can’t be zero!
6. Two carts of masses 5.0 kg and 3.0 kg move toward each other on a
frictionless track with speeds of 5.0 m/s and 3.0 m/s respectively. The
carts stick together after colliding head on.
a. Find the final speed.
ans. 2.0 m/s
b. What is the loss (if any) of kinetic energy?
60 J
7. A 25.0 g marble sliding to the right at 20 cm/s catches up to and collides
with a 10.0 g marble moving in the same direction at 15.0 cm/s. After the
collision, the 10.0 g marble moves to the right at 22.1 cm/s.
a. Find the velocity of the 25.0 g marble.
ans. 0.17 m/s
b. Is the collision elastic?
Yes. KEf = KEi
8. A 15.0-kg object moving in the +x direction at 5.5 m/s collides head-on
with a 10.0-kg object moving in the -x direction at 4.0 m/s. Find the final
velocity of each mass if: (a) the objects stick together; (b) the collision is
elastic; (c) the 15.0-kg object is at rest after the collision; (d.) the 10.0-kg
object is at rest after the collision; (e) the 15.0-kg object has a velocity of
4.0 m/s in the -x direction after the collision. Are the results in (c), (d), and
(e) "reasonable"? Explain.
a) 1.7 m/s
b) -2.1 m/s, 7.4 m/s
c) 4.3 m/s; reasonable
d) 2.8 m/s; not reasonable (15 kg object passes through
10 kg object).
e) -4.0 m/s, 10.3 m/s; not reasonable (KE gained).
9. A 1500-kg car traveling at 30 m/s east collides with a 2000-kg car traveling
at 25.0 m/s south. The two cars stick together after the collision. What is
the velocity of the cars immediately after the collision?
v’ = 19.2 m/s at 48 degrees south of east.
Chapter 6 Work and Energy
Key Concepts
Use these words to fill in the blanks. All words are used; some may be used
more than once.
conserved
time
destroyed
position
created
kinetic
kinetic energy
displacement
area
mass
potential
force
work
speed
Work is done only when there is a __ displacement ___ in the direction of a
component of the net force.
Work can be calculated from the ___ area ___ under the ___ force ___ vs.
displacement curve.
Power is the __ work _ done per unit __time___.
Objects in motion have kinetic energy because of their __mass_ and _speed_.
The net work done on or by an object is equal to the change in the __kinetic
energy_______ of the object.
Potential energy is energy associated with an object’s _position_______. It
represents the objects potential to have _kinetic______ energy or do
_work______.
Energy can be transformed but can never be _created__ or _destroyed__.
Mechanical energy is the sum of the __potential__ and __kinetic____ energy in
a system.
In the absence of friction, mechanical energy is __conserved__ (remains
constant).
Equations:
Wnet = Fnet*d*cosφ
P = Work/time (and also Force x velocity)
KE = 1/2 mv2
Wnet = ΔKE
PEgravitational = mgh
PEelastic = 1/2kx2 (where x is the displacement from the unstretched position).
Mechanical Energy ME = PE + KE (Where PE can be either PEgravitational or
PEelastic or both).
PEi + KEi = PEf + KEf
(Conservation of Mechanical Energy)
Chapter 6 Review Problems:
1. How much work is done when a lift raises a 2000 kg car to a height of 2
meters?
ans. 39200 J
2. What is the change in potential energy of the car in question 1?
ans. 39200 J
3. If the car in problem 1 is lifted in 10 seconds, how much power is being used?
ans. 3920 W
4. How much work is done when a 1000N force, applied to the right, moves an
object 100 m to the right on a frictionless surface.
ans. 105 J
5. A car engine generates 100000 W of power when traveling at 20 m/s. What
net force is being applied to the car?
ans. 5000 N
6. How much work is done on a vacuum cleaner pulled 10 m by a force of 60 N
at an angle of 45 degrees to the horizontal?
ans. 424 J
7. A 1000 kg car is traveling at 35 m/s when the driver steps on the brakes and
slows down to 15 m/s over a distance of 100 m. What force was applied by the
brakes?
ans. -5000 N
8. A force of 100 N is applied horizontally to a 10.0-kg mass on a level surface.
The coefficient of kinetic friction between the mass and the surface is 0.25. If the
mass is moved a distance of 50 m, what is the change in its kinetic energy?
ans. 3775 J
9. A spring with a force constant of 500 N/m is stretched 5 cm. How much energy
is stored in the spring?
ans. 0.625 J
10. A 1500-kg car moving at 25 m/s hits an initially uncompressed horizontal
spring with spring constant of 2.0 × 106 N/m. What is the maximum compression
of the spring? (Neglect the mass of the spring.)
ans. 0.68 m
11. When a skier with a mass of 50 kg starts from rest on a hill she has 100000 J
of PEg relative to the base. Assuming no friction, how fast is she traveling when
she reaches the bottom of the hill? How fast was the skier going when she was
half way down the hill?
Half way : ans. 44 m/s
Bottom : ans. 63 m/s
Multiple Choice Answers:
1. B
2. C
3. B
4. A
5. D
6. D
7. A
8. D
9. A
10. B and C
Free response solutions:
11. a) Energy conservation with position B set as h=0. Ua = Kb
vb = 2 gL
b) Forces at B, Ft pointing up and mg pointing down. Apply Fnet(c)
2
Fnet(C) = m1vb / r
Ft – m1g = m1(2gL) / L
Ft = 3m1g
c) Apply momentum conservation perfect inelastic
m1v1i= (m1+m2) vf
vf =
pbefore = pafter
m1
2 gL
(m1 + m2 )
d) Projectile. First find time to travel from B to D using the y direction equations
2
2
dy = viyt + ½ g t
L = 0 + gt / 2
t=
2L
g
Then use vx = dx / t
dx =
m1
m1
2L
2 gL
=
2L
(m1 + m2 )
g
(m1 + m2 )
The dx found is measured from the edge of the second lower cliff so the total horizontal distance
would have to include the initial x displacement (L) starting from the first cliff.
è
⎡ 2m1
⎤
m1
2 L + L = L ⎢
+ 1⎥
(m1 + m2 )
⎣ (m1 + m2 ) ⎦
12. a) (a) Apply energy conservation
Utop = Kbottom
2
2
mgh = ½ m v
Mgh = ½ (M) (3.5vo)
2
h = 6.125 vo / g
(b) momentum conservation
M(3.5vo) = Mv + (1.5M)(2vo)
v = 0.5 vo
(c) WNC = ∆K (Kf-Ki) Kf = 0
2
– fk d = 0 – ½ (1.5M)(2vo)
2
µk (1.5 M) g (d) = 3Mvo
2
µk = 2vo / gD
d) – The coliision is inelastic because the kinetic energy changes
Justification – show kinetic energy difference
2
2
2
2
Kf- Ki – [0.5M(0.5 vo) + 0.5(1.5M)(2vo) ] – 0.5M(3.5 vo) = -3Mvo