Physics 101 Quiz #4 Solution
Oct 8, 2004
Name
20 Minutes. Box your answers.
Problem 1
A block of mass m slides down an
inclined plane and then up a circular ramp
as shown in the diagram. Both the ramp
and the plane have height h. Please
express all your answers in terms of the
given variables and g, the acceleration
due to gravity.
h
θ
(a) [2 points] Assume all the surfaces are frictionless. If the block begins from rest at the
top of the inclined plane what will its speed be at the bottom of the plane?
Mechanical energy is conserved in this part, so the value of the mechanical energy of the
block is the same at the top and the bottom of the ramp
Etop = mgh
Etop = Ebottom
√
→
→
v = 2gh .
1 2
1 2
Ebottom =
mgh =
mv
mv
2
2
(b) [2 points] Now assume that the inclined plane has friction (µ k , µs > 0) but not enough
to prevent the block from sliding down the plane. What will the speed of the block be at
the bottom of the plane in this situation?
Mechanical energy is no longer conserved but the difference between mechanical energy
at the top and bottom of the the ramp is equal to the work done by the frictional force
Etop = mgh
Etop − Ebottom = Wfriction
1 2
1
Ebottom =
mv
→
mgh − mv 2 = µk mgh cot θ
2
2
Wfriction = (Ffriction ) ∗ (lenght of ramp)
!
"
#
h
v = 2gh (1 − µk cot θ) .
= (µk mg cos θ) ∗
sin θ
(c) [2 points] After exiting the inclined plane in part (b), how high will the block rise on
the circular ramp? The circular ramp continues to be frictionless.
In this case mechanical energy is again conserved. From part (a) it should rise to a height
2
of h = v2g .
hrise = h (1 − µk cot θ) .
Continued on the other side....
15 m/s
Problem 2
A stone is thrown horizontally with an
initial speed of 15 m/s from the top of
a 5 meter tower. The influence of air
resistance may be ignored throughout
this problem.
5m
(a) [2 points] What is the speed of the stone when it is halfway (height=2.5 meters) to the
ground?
Since
air
resistance
can
be
ignored,
1 2
Etop = mgh + mvtop
2
h 1 2
Ehalf = mg + mvhalf
2 2
→
vhalf =
$
mechanical
energy
is
conserved.
Etop = Ehalf
h 1 2
1 2
= mg + mvhalf
mgh + mvtop
2
2 2
2
vtop
+ gh = 17 m/s
(b) [2 points] When the stone hits the ground what angle does its velocity vector make with
the ground?
Since only the vertical component of velocity changes during the fall, conservation of
mechanical energy tells us :
Etop = Ebottom
1 2
1 2
mgh + mvtop
= 0 + mvbottom
2
2
1 2
1 2
1 2
1
mgh + mvx top + mvy top =
mvx bottom + mvy2 bottom .
2
2
2
2
This last line breaks the velocity vector into components. Since the horizontal velocity is
unchanged during the fall and the initial vertical velocity is zero,
1 2
mv
2 y bottom
#
=
2gh .
mgh =
vy bottom
Then, if we call θ the angle the velocity vector makes with the ground when it hits
tan θ =
vy
vx
bottom
bottom
=
√
2gh
vx top
= 0.66 → θ = 33◦ .
PRINT YOUR NAME:
QUIZ 4
Physics 101, Fall 2005
Friday, October 14th , 2005
SOLUTIONS
Problem 1. Three balls, A, B, and C are thrown from the top of a cliff of height h = 100 m
overlooking a flat land. The speed of the the three balls is identical: vA = vB = vC = 10 m/s.
The angle in the figure is φ=π/4.
VC
φ
φ
VA
VB
h
a. (2 pts) Which one of the three balls will have the largest speed at the moment it makes
contact with the ground? Box your answer and detail your reasoning on the side.
• Ball A
• Ball B
• Ball C
• They all reach ground with the same speed
The three balls have equal initial kinetic energy. They will also have equal final kinetic energy,
as they drop by the same height. So they all reach ground with the same speed.
b. (1 pts) Which of the three balls will reach the ground in the shortest time? Box your
answer and detail your reasoning on the side.
• Ball A
• Ball B
• Ball C
• They all reach ground at the same time
• Ball B and C reach ground at the same time, before ball A.
Ball B reaches the ground first. What matters here is only the motion in the y direction. Since
ball B is the only one with an initial velocity directed towards ground, it will reach the ground
first.
QUIZ 4
Friday, October 14th , 2005
Physics 101, Fall 2005
Problem 2. A block (mass m = 10 kg) starts sliding from the top of the two ramps with
initial null velocity. Along the first inclined, straight ramp, the block slides over the surface
with a coefficient of kinetic friction µk = 2/3. The angle of incline of the first ramp with
respect to the horizontal direction is φ. The motion along the second ramp - an arc of a
circumference with a radius r = 10m - is frictionless. The body leaves the curved ramp at
an angle θ0 with respect to the vertical.
a. (3 pts) What is the kinetic energy 12 mv12 of the body at the end of the first ramp (y = r)?
Express the kinetic energy as a function of r, g, m, and µk . Provide a numeric estimate for
the kinetic energy as well.
Let’s call with the index 0 quantities at the top of the first ramp and with the index 1 quantities
at the bottom of the first ramp. Energy conservation tells us that:
Wnc = ∆KE + ∆P E = KE1 − KE0 + P E1 − P E0 =
1
mv 2 + mgr − mg(2r)
2 1
(1)
The work of the non conservative forces is the work done by the kinetic friction. It is negative,
since kinetic friction opposes the motion. We can calculate the normal force from the free body
diagram: N = mg cos φ. Total displacement along the first inclined ramp is s = r/ sin φ.
Wnc = f~k · ~s = −fk s = −µk N
r
r
cos φ
= −µk mg cos φ
= −µk mgr
= −µk mgr
cos φ
sin φ
sin φ
(2)
where we made use of cos φ = sin φ since φ = π/4. Therefore:
1
10 kg × 9.8 m/s2 × 10 m
1
mv12 = mgr − µk mgr = mgr =
= 327 J
2
3
2
(3)
b. (2 pts) Express the kinetic energy on the second ramp as a function of θ, r, g, m, and
µk . (if you could not complete the first part of this problem, make a resonable guess for a
numeric value of 12 mv12 , and give your answer as function of this value as well).
No frictional forces over the second ramp. 0 = ∆KE + ∆P E = KE2 − KE1 + P E2 − P E1 .
Therefore:
1
1
1
4
2
2
mv = mv1 + mgr(1 − cos θ) = mgr + mgr(1 − cos θ) = mgr
− cos θ
(4)
2 2
2
3
3
c. (2 pts) What is the value of the take off angle θ0 ? Express the angle as a function of r,
g, m, and µs . Provide a numeric estimate.
The body takes off when the normal force goes to zero. That happens when the component of
the weight in the radial direction, mg cos θ, equals mass times centripetal acceleration.
mv22
1
1
= mgr cos θ → mv22 = mgr cos θ
r
2
2
Substituting the expression for the kinetic energy of the body, 12 mv22 = mgr
get:
1
4
3
4
8
mgr cos θ = mgr
− cos θ → cos θ = → θ = arccos
2
3
2
3
9
(5)
4
3
− cos θ , we
(6)