Physics 101 Quiz #4 SolutionOct 8, 2004 Name

20 Minutes. Box your answers.

Problem 1A block of mass m slides down aninclined plane and then up a circular rampas shown in the diagram. Both the rampand the plane have height h. Pleaseexpress all your answers in terms of thegiven variables and g, the accelerationdue to gravity.

!

h

(a) [2 points] Assume all the surfaces are frictionless. If the block begins from rest at thetop 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 theblock is the same at the top and the bottom of the ramp

Etop = mgh

Ebottom =1

2mv2

! Etop = Ebottom

mgh =1

2mv2

!v =

"2gh .

(b) [2 points] Now assume that the inclined plane has friction (µk, µs > 0) but not enoughto prevent the block from sliding down the plane. What will the speed of the block be atthe bottom of the plane in this situation?Mechanical energy is no longer conserved but the difference between mechanical energyat the top and bottom of the the ramp is equal to the work done by the frictional force

Etop = mgh

Ebottom =1

2mv2

Wfriction = (Ffriction) # (lenght of ramp)

= (µkmg cos !) #!

h

sin !

"

!

Etop $ Ebottom = Wfriction

mgh $ 1

2mv2 = µkmgh cot !

v =#

2gh (1 $ µk cot !) .

(c) [2 points] After exiting the inclined plane in part (b), how high will the block rise onthe 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 heightof h = v2

2g .

hrise = h (1 $ µk cot !) .

Continued on the other side....

Problem 2A stone is thrown horizontally with aninitial speed of 15 m/s from the top ofa 5 meter tower. The influence of airresistance may be ignored throughoutthis problem.

5 m

15 m/s

(a) [2 points] What is the speed of the stone when it is halfway (height=2.5 meters) to theground?Since air resistance can be ignored, mechanical energy is conserved.

Etop = mgh +1

2mv2

top

Ehalf = mgh

2+

1

2mv2

half

! Etop = Ehalf

mgh +1

2mv2

top = mgh

2+

1

2mv2

half

vhalf =$

v2top + gh = 17 m/s

(b) [2 points] When the stone hits the ground what angle does its velocity vector make withthe ground?Since only the vertical component of velocity changes during the fall, conservation ofmechanical energy tells us :

Etop = Ebottom

mgh +1

2mv2

top = 0 +1

2mv2

bottom

mgh +1

2mv2

x top +1

2mv2

y top =1

2mv2

x bottom +1

2mv2

y bottom .

This last line breaks the velocity vector into components. Since the horizontal velocity isunchanged during the fall and the initial vertical velocity is zero,

mgh =1

2mv2

y bottom

vy bottom =#

2gh .

Then, if we call ! the angle the velocity vector makes with the ground when it hits

tan ! = vy bottom

vx bottom=

!2gh

vx top= 0.66 ! ! = 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 = 100moverlooking a flat land. The speed of the the three balls is identical: vA = vB = vC = 10m/s.The angle in the figure is φ=π/4.

VB

φ

h

VC

VAφ

a. (2 pts) Which one of the three balls will have the largest speed at the moment it makescontact 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 youranswer 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. Sinceball B is the only one with an initial velocity directed towards ground, it will reach the groundfirst.

QUIZ 4 Physics 101, Fall 2005 Friday, October 14th, 2005

Problem 2. A block (mass m = 10 kg) starts sliding from the top of the two ramps withinitial null velocity. Along the first inclined, straight ramp, the block slides over the surfacewith a coefficient of kinetic friction µk = 2/3. The angle of incline of the first ramp withrespect to the horizontal direction is φ. The motion along the second ramp - an arc of acircumference with a radius r = 10m - is frictionless. The body leaves the curved ramp atan angle θ0 with respect to the vertical.

a. (3 pts) What is the kinetic energy 12mv2

1 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 forthe kinetic energy as well.

Let’s call with the index 0 quantities at the top of the first ramp and with the index 1 quantitiesat the bottom of the first ramp. Energy conservation tells us that:

Wnc = ∆KE + ∆PE = KE1 −KE0 + PE1 − PE0 =12mv2

1 + mgr −mg(2r) (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 bodydiagram: N = mg cos φ. Total displacement along the first inclined ramp is s = r/ sinφ.

Wnc = ~fk · ~s = −fks = −µkNr

cos φ= −µkmg cos φ

r

sinφ= −µkmgr

cos φ

sinφ= −µkmgr (2)

where we made use of cos φ = sinφ since φ = π/4. Therefore:

12mv2

1 = mgr − µkmgr =13mgr =

10 kg × 9.8 m/s2 × 10 m2

= 327 J (3)

b. (2 pts) Express the kinetic energy on the second ramp as a function of θ, r, g, m, andµk. (if you could not complete the first part of this problem, make a resonable guess for anumeric value of 1

2mv2

1, and give your answer as function of this value as well).

No frictional forces over the second ramp. 0 = ∆KE + ∆PE = KE2 − KE1 + PE2 − PE1.Therefore:

12mv2

2 =12mv2

1 + mgr(1− cos θ) =13mgr + mgr(1− cos θ) = mgr

(43− cos θ

)(4)

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 ofthe weight in the radial direction, mg cos θ, equals mass times centripetal acceleration.

mv22

r= mgr cos θ → 1

2mv2

2 =12mgr cos θ (5)

Substituting the expression for the kinetic energy of the body, 12mv2

2 = mgr(

43 − cos θ

), we

get:12mgr cos θ = mgr

(43− cos θ

)→ 3

2cos θ =

43→ θ = arccos

89

(6)