Exam 3 !
Name!
!
Question 1!
A 2,000 kg car in neutral rolls down on a 14.8° incline a distance of 50 meters. It start our rolling
at 15 m/s and finishes at 10 m/s. During this process, assumes there is no friction and air
resistance produces a constant 1,000 N of force and that the brakes are applied producing
another constant force on the car.!
a.! How much work is done by the braking force?!
15 m/s
b.! What is the magnitude of the braking force?!
c.! What is the magnitude of the total impulse applied on
the car? !
10 m/s
50 m
d.! How much time does this process take?!
14.8°
e.! What is the stopping power of the brakes?!
Solution!
a.! How much work is done by the braking force?!
For this I will use the work-energy theorem. The forces on
the car are shown.!
Wnc = ΔK + ΔU !
normal
braking
air
resistance
The non-conservative work is done by the air resistance
and braking. The total non-conservative work is !
gravity•sin(14.8°)
gravity
Wnc = Wair +Wbrake = −(1000)(50) +Wbrake = −5 ×104 +Wbrake !
The change in kinetic energy is!
1
1
1
ΔK = mv f2 − mvi2 = (2000)(102 − 152 ) = −1.25 ×105 J !
2
2
2
The change in potential energy is!
ΔU = −mgh = −(2000)(9.8)(50 sin14.8°) = −2.5 ×105 J !
Together,!
−5 ×104 +Wbrake = −1.25 ×105 − 2.5 ×105 !
Wbrake = −3.25 ×105 J !
b.! What is the magnitude of the braking force?!
The magnitude of the braking force is!
Wbrake = − Fbrake Δx
⇒
W
−3.25 ×105
Fbrake = − brake = −
= 6,500 N !
Δx
50
Fbrake = 6,500 N page 1
c.! What is the magnitude of the total impulse applied to the car? !
For this I will use the impulse-momentum theorem. Let the down-incline direction be the positive
direction.!
!
!
I total = Δp !
The change in the momentum of the car is!
!
!
!
Δp = mv f − mvi = (2000)(10)x̂ − (2000)(15)x̂ = −10, 000 kg ⋅ m/s x̂ !
!
I total = 10, 000 kg ⋅ m/s !
d.! How much time does this process take?!
To find the elapsed time, we just need the net force. This is due to gravity, air resistance, and
braking.!
!
!
!
!
ΣF = Fgravity + Fair + Fbrake = +(2000)(9.8)sin14.8° − 1, 000 − 6500 = −2,500 N !
Therefore, !
!
!
I total = ΣFΔt
⇒
− 10, 000 = (−2500)Δt !
Δt = 4 s !
e.! What is the stopping power of the brakes?!
The stopping power is just the work done by the brakes over the time it acted.!
Pavg =
!
Wbrake
3.25 ×105
=
!
Δt
4
Pavg = 8.13 ×104 W !
page 2
Question 2!
A ball at the end of a 1 meter long pendulum is released from a certain
angle. At the bottom of the swing, it strikes directly at a second
identical ball that is at rest on a frictionless surface. The collision is
elastic and the second ball exits the collision with a speed of 2 m/s. At
what angle was the ball on the pendulum released?!
θ
Solution!
Here is the fast way. Since the masses are equal and the collision is
elastic and the second ball starts out at rest and the momentum is conserved, the
of the first ball at the bottom must also have been 2 m/s.!
speed
Here is the longer way. For the collision, the momentum is conserved. The second ball starts out
at rest. The masses are equal.!
p1,i + p2,i = p1,f + p2,f !
v1,i = v1,f + 2 !
The kinetic energy is also conserved.!
2
2
v1,i
= v1,f
+4!
Together, !
2
2
v1,i
= (v1,f + 2)2 = v1,f
+4!
2
2
v1,f
+ 4v1,f + 4 = v1,f
+4
⇒
4v1,f = 0
⇒ v1,f = 0 !
v1,i = v1,f + 2 ⇒ v1,i = 2 !
Since the mechanical energy of a pendulum is conserved, the height from which the ball was
released must be!
K f = Ui !
1 2
mv = mgh
2 f
⇒ h=
v f2
2g
=
22
= 0.20408 m !
2(9.8)
At this height, the angle would have been!
h = L(1 − cos θ) ⇒
!
cos θ = 1 −
h
L
⎛
⎛
h⎞
0.20408 ⎞⎟
⎟⎟ !
⇒ θ = cos−1 ⎜⎜ 1 − ⎟⎟⎟ = cos−1 ⎜⎜ 1 −
⎜⎝
⎜
⎟
⎟⎠
L⎠
1
⎝
θ = 37.3° !
page 3
Question 3!
A ball starts out at rest on the ground at point A. It is kicked
and obtains a velocity as indicated at point B immediately after
being kicked just above the ground. It reaches its maximum
height at point C. It falls back down to point D right before
hitting the ground at the same height as point B. At point E, it
comes to a complete stop when the ground compressed it by
some amount exactly like a spring.!
Assume that there is no other non-conservative work done on
the ball other than the kick. Answer the following questions for
the ball.!
A
B
C
D
E
a.! At which point(s) is the kinetic energy minimum?!
!
b.! At which point(s) is the kinetic energy maximum?!
!
c.! At which point(s) is the total potential energy minimum?!
!
d.! At which point(s) is the total potential energy maximum?!
!
e.! At which point(s) is the momentum minimum?!
!
f.! At which point(s) is the momentum maximum?!
!
Solution!
a.! At which point(s) is the kinetic energy minimum?!
A and E since the ball is at rest.!
b.! At which point(s) is the kinetic energy maximum?!
B and D since the ball is moving the fastest there.!
c.! At which point(s) is the total potential energy minimum?!
A since the ball has no potential energy there.!
d.! At which point(s) is the total potential energy maximum?!
E since all of the kinetic energy is not potential energy.!
e.! At which point(s) is the momentum minimum?!
A and E since the ball is at rest.!
f.! At which point(s) is the momentum maximum?!
B and D since the ball is moving the fastest there.!
!
page 4
Question 4!
Consider a ball and the Earth as one system.!
a.! Circle all of the following processes through which the total mechanical energy of the system is
conserved.!
!
i.! The ball falls through a certain distance without air resistance.!
!
ii.! The ball rolls through a certain distance on a flat surface without friction.!
!
iii.! The ball swings across a certain angle at the end of a pendulum.!
!
iv.! The ball sinks through a certain depth of water at constant speed.!
!
!
!
!
b.! Circle all of the following processes through which the total momentum of the system is
conserved.!
!
i.! The ball falls through a certain distance without air resistance.!
!
ii.! The ball rolls through a certain distance on a flat surface without friction.!
!
iii.! The ball swings across a certain angle at the end of a pendulum.!
!
iv.! The ball sinks through a certain depth of water at constant speed.!
!
!
!
!
Solution!
a.! Circle all of the following processes through which the total mechanical energy of the system is
conserved.!
!
i.! The ball falls through a certain distance without air resistance.!
!
ii.! The ball rolls through a certain distance on a flat surface without friction.!
!
iii.! The ball swings across a certain angle at the end of a pendulum.!
What you are look for here is whether any non-conservative work is done. The water in iv does.!
b.! Circle all of the following processes through which the total momentum of the system is
conserved.!
!
i.! The ball falls through a certain distance without air resistance.!
!
ii.! The ball rolls through a certain distance on a flat surface without friction.!
What you are look for here is whether any external impulse is applied to the system. The tension
in iii and the force from the water in iv do.!
page 5