© Peter Dourmashkin 2012

MASSACHUSETTS INSTITUTE OF TECHNOLOGY

Department of Physics 8.01 Physics Fall Term 2012

Exam 1: Practice Problems Solutions

One-Dimensional Kinematics: ,d ddt dt

= =r vv a! !!!

vx (t) ! vx,0 = ax ( "t ) d "t

"t =0

"t =t

# , x(t) ! x0 = vx ( "t ) d "t"t =0

"t =t

#

Initial conditions:

x0 and vx ,0 are the values at t = 0 . Newton’s Second Law: Force, Mass, Acceleration:

!F = m !a

Newton’s Third Law: 1,2 2,1= !F F

! !

Problem 1: Elevator

A person of mass mp stands on a scale in an elevator of mass mp . The scale reads the magnitude of the force F exerted on it from above in a downward direction. Starting at rest at t = 0 the elevator moves upward, coming to rest again at time t = t0 . The downward acceleration of gravity is g . The acceleration of the elevator during this period is shown graphically above and is given analytically by

ay (t) = ! "

2!t0

t . (1)

a) Find the maximum speed of the elevator. b) Find the total distance traveled by the elevator. Solutions: Part a) The velocity is the integral of the acceleration. Inspection of the graph shows the integral increases until t = t0 = 2 where the velocity reaches its maximum value. That value is given by the area of the triangle above the axis.

vy ,max = (1 / 2)!(t0 / 2) =

!4

t0 . (2)

Part b)

vy (t) ! vy0 = ay (t)dt

0

t

" = # !2#t0

t$

%&'

()dt

0

t

" = #t ! #t0

t2 . (3)

Because

vy0 = 0 , the y-component of the velocity is

vy (t) = !t " !

t0

t2 . (4)

Integrating again to find the position

y(t) ! y0 = vy (t)dt

0

t

" = #t ! #t0

t2$

%&'

()dt

0

t

" =#2

t2 !#3t0

t3 . (5)

Because we choose y0 = 0 ,

y(t) = !

2t2 "

!3t0

t3 . (6)

The total distance traveled by the elevator is the position at time t = t0

y(t0 ) =

!2

t02 "

!3t0

t03 =

!6

t02 .

Problem 2: Traffic Violation A car is driving through a green light at t = 0 located at x = 0 with an initial speed

vc,0 = 12 m ! s-1 . The acceleration of the car as a function of time is given by

ac =

0; 0 < t < t1 = 1s

!(6 m " s-3)(t ! t1); 1 s < t < t2

#$%

&%.

a) Find the speed and position of the car as a function of time. b) Graph the speed and position of the car as a function of time. c) A bicycle rider is riding at a constant speed of

vb,0 and at t = 0 is 17 m behind the car. The bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle.

Solution: a) We need to integrate the acceleration for both intervals. The first interval is easy, the speed is constant. For the second integral we need to be careful about the endpoints of the integral and the fact that the integral is the change in speed so we must subtract vc (t1) = vc0

vc (t) =

vc0; 0 < t < t1 = 1 s

vc (t1) + !(6 m " s-3)(t ! t1)t1

t

# ; 1 s < t < t2

$

%&

'&

.

After integrating we get

vc (t) =vc0; 0 < t < t1 = 1 s

vc0 ! (3 m " s-3)(t ! t1)2

t1

t; 1 s < t < t2

#$%

&%.

Now substitute the endpoint so the integral to finally yield

vc (t) =

vc0 = 12 m ! s-1; 0 < t < t1 = 1 s

12 m ! s-1 " (3 m ! s-3)(t " t1)2; 1 s < t < t2

#$%

&%.

For this one dimensional motion the change in position is the integral of the speed so

xc (t) =xc (0) + (12 m ! s-1)dt

0

t1

" ; 0 < t < t1 = 1s

xc (t1) + 12 m ! s-1 # (3m ! s-3)(t # t1)2( )t1

t

" dt; 1 s < t < t2

$

%

&&

'

&&

.

Upon integration we have

xc (t) =xc (0) + (12 m ! s-1)t; 0 < t < t1 = 1s

xc (t1) + (12 m ! s-1)(t " t1) " (1 m ! s-3)(t " t1)3( )t1

t; 1 s < t < t2

#$%

&%.

We choose our coordinate system such that xc (0) = 0 ,

therefore xc (t1) = (12 m ! s-1)(1 s)=12 m . So after substituting in the endpoints of the integration interval we have that

xc (t) =

(12 m ! s-1)t; 0 < t < t1 = 1s

12 m+(12 m ! s-1)(t " t1) " (1 m ! s-3)(t " t1)3; 1 s < t < t2

#$%

&%.

b) Graph the speed and position of the car as a function of time.

Solution: The graphs of the speed and position are shown below.

c) A bicycle rider is riding at a constant speed of vb,0 and at t = 0 is 17 m behind

the car. The bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle.

Solution: we are looking for the instant that t2 the car has come to rest. So we use our expression for the speed for the interval 1s < t < t2 ,

0 = vc (t2 ) = 12 m ! s-1 " (3 m ! s-3)(t2 " t1)2 . We can solve this for t2 :

(t2 ! t1)2 = 4 s2 . We have two solutions: (t2 ! t1) = 2 s or (t2 ! t1) = !2 s . The second solution

t2 = t1 ! 2 s = 1 s ! 2 s = !1s does not apply to our time interval and so

t2 = t1 + 2 s = 1 s + 2 s = 3 s . During the position of the car at t2 is then given by

xc (t2 ) = 12 m+(12 m ! s-1)(t2 " t1) " (1 m ! s-3)(t2 " t1)3

= 12 m+(12 m ! s-1)(2 s) " (1 m ! s-3)(2 s)3 = 28 m.

Since the bicycle is traveling at a constant speed with an initial position xb0 = !17 m , the position of the bicycle is given by

xb(t) = !17 m + vbt . The bicycle and car intersect at instant t2 = 3 s :

xb(t2 ) = xc (t2 ) . Therefore

!17 m + vb(3 s) = 28 m . So the speed of the bicycle is

vb =

(28 m + 17 m)(3 s)

= 15 m ! s-1 .

Problem 3 Catching a bus At the instant a traffic light turns green, a car starts from rest with a given constant acceleration, 5.0 !10"1 m # s-2 . Just as the light turns green, a bus, traveling with a given constant speed, 1.6 !101 m " s-1 , passes the car. The car speeds up and passes the bus some time later. How far down the road has the car traveled, when the car passes the bus? Solution: I. Understand – get a conceptual grasp of the problem Think about the problem. How many objects are involved in this problem? Two, the bus and the car. How many different stages of motion are there for each object? Each object has one stage of motion. For each object, how many independent directions are needed to describe the motion of that object? One, only. What information can you infer from the problem? The acceleration of the car, the velocity of the bus, and that the position of the car and the bus are identical when the bus just passes the car. Sketch qualitatively the position of the car and bus as a function of time (Figure 4.12).

Figure 4.12 Position vs. time of the car and bus. What choice of coordinate system best suits the problem? Cartesian coordinates with a choice of coordinate system in which the car and bus begin at the origin and travel along the positive x-axis (Figure 4.13). Draw arrows for the position coordinate function for the car and bus.

Figure 4.13 A coordinate system for car and bus. II. Devise a Plan - set up a procedure to obtain the desired solution

Write down the complete set of equations for the position and velocity functions. There are two objects, the car and the bus. Choose a coordinate system with the origin at the traffic light with the car and bus traveling in the positive x-direction. Call the position function of the car, x1(t) , and the position function for the bus, x2 (t) . In general the position and velocity functions of the car are given by

x1(t) = x1,0 + vx10t +

12

ax ,1t2

vx ,1(t) = vx10 + ax ,1t

In this example, using both the information from the problem and our choice of coordinate system, the initial position and initial velocity of the car are both zero,

x1,0 = 0

and vx10 = 0 , and the acceleration of the car is non-zero ax ,1 ! 0 . So the position and

velocity of the car is given by

x1(t) =

12

ax ,1t2 ,

vx ,1(t) = ax ,1t .

The initial position of the bus is zero,

x2,0 = 0 , the initial velocity of the bus is non-zero,

x2 (ta ) = vx ,20ta

vx ,20 ! 0 , and the acceleration of the bus is zero, ax ,2 = 0 . So the position

function for the bus is

x2 (t) = vx ,20t .

The velocity is constant,

vx ,2 (t) = vx ,20 .

Identify any specified quantities. The problem states: “The car speeds up and passes the bus some time later.” What analytic condition best expresses this condition? Let t = ta correspond to the time that the car passes the bus. Then at that time, the position functions of the bus and car are equal,

x1(ta ) = x2 (ta ) . How many quantities need to be specified in order to find a solution? There are three independent equations at time t = ta : the equations for position and velocity of the car

x1(ta ) =

12

ax ,1ta2

vx ,1(ta ) = ax ,1ta ,

and the equation for the position of the bus,

. There is one ‘constraint condition’

x1(ta ) = x2 (ta ) . The six quantities that are as yet unspecified are

x1(ta ) , x2 (ta ) , vx ,1(ta ) ,

vx ,20 , ax , ta So you need to be given at least two numerical values in order to completely specify all the quantities; for example the distance where the car and bus meet. The problem specifying the initial velocity of the bus,

vx ,20 , and the acceleration, ax , of the car with given values. III. Carry our your plan – solve the problem! The number of independent equations is equal to the number of unknowns so you can design a strategy for solving the system of equations for the distance the car has traveled in terms of the velocity of the bus

vx ,20 and the acceleration of the car ax ,1 , when the car

passes the bus. Let’s use the constraint condition to solve for the time t = ta where the car and bus meet. Then we can use either of the position functions to find out where this occurs. Thus the constraint condition, x1(ta ) = x2 (ta ) becomes

12

ax ,1ta2 = vx ,20ta .

We can solve for this time,

ta = 2

vx ,20

ax ,1

.

Therefore the position of the car at the meeting point is

x1(ta ) =

12

ax ,1ta2 =

12

ax ,1 2vx ,20

ax ,1

!

"#

$

%& = 2

vx ,202

ax ,1

.

IV. Look Back – check your solution and method of solution

Check your algebra. Do your units agree? The units look good since in the answer the two sides agree in units,

m!" #$ =

m2% s-2

m% s-2

!

"&

#

$'

and the algebra checks. Substitute in numbers. Suppose ax = 5.0 !10"1 m # s-2 and

vx ,20 = 1.6 !101 m " s-1 , Introduce

your numerical values for vx ,20 and ax , and solve numerically for the distance the car has

traveled when the bus just passes the car. Then

ta = 2

vx ,20

ax ,1

= 21.6 !101 m " s-1( )5.0 !10#1 m " s-2( ) = 6.4 !101s

x1(ta ) = 2

vx ,202

ax ,1

= 2( ) 1.6 !101 m " s-1( )2

5.0 !10#1 m " s-2( ) = 1.0 !103m

Check your results. Once you have an answer, think about whether it agrees with your estimate of what it should be. Some very careless errors can be caught at this point. Is it possible that when the car just passes the bus, the car and bus have the same velocity? Then there would be an additional constraint condition at time t = ta , that the velocities are equal,

vx ,1(ta ) = vx ,20 .

Thus

vx ,1(ta ) = ax ,1ta = vx ,20

implies that

ta =

vx ,20

ax ,1

.

From our other result for the time of intersection

ta =

2vx ,20

ax ,1

.

But these two results contradict each other, so it is not possible.

Problem 4 Softball Catch A softball is hit over a third baseman’s head. The third baseman, as soon as the ball is hit, turns around and runs straight backwards at a constant speed of 7.0m ! s"1 for a time interval 2.0s and catches the ball at the same height it left the bat. The third baseman was initially 18 m from home plate. What is the initial speed and angle of the softball when it left the bat? Ignore air resistance. Solution: It’s probably best to choose the origin at the point where the ball leaves the bat at time t = 0 . This way, both the initial and final vertical components of the ball’s position are zero.

As suggested by the diagram, use v0 for the ball’s initial speed, and !0 for the angle of the ball’s initial velocity with respect to the horizontal. Use the horizontal direction, from the batter to the fielder, as the ˆ+i direction, and the upward vertical as the ˆ+j direction. The equation for the horizontal component of the position of the ball is given by

xb(t) = vx ,0t (1) The third baseman is running at a constant velocity

vt ,0 = v1 . Note: We are using “ t ” in the subscript to denote the third baseman and “ b” in the subscript to denote the ball. The equation for the position of the third baseman is

xt (t) = xt ,0 + vt ,0t (2)

So at the end of the time interval catcht t= ! is the third baseman is a horizontal distance

xt (tcatch ) = xt ,0 + vt ,0tcatch = xt ,0 + v1 !t = 1.8 "101 m + (7.0m # s$1)(2.0s) = 32m (3)

from the point where the ball hits the bat. At the instant tcatch the third baseman catches the ball, we have that Eq. (1) becomes xb(tcatch ) = xt (tcatch ) (4) Setting t = tcatch into Eq. (1) and Eq. (2) and then substituting into the left hand side and right hand side of Eq. (4) yields

vx ,0tcatch = xt ,0 + vt ,0tcatch (5) Solve Eq. (5) for the horizontal component of the initial velocity of the ball:

vx0 =

xt ,0

t+ vt ,0 = 18m / 2s + 7.0m ! s"1 = 16m ! s"1 (6)

The equation for the vertical component of the velocity is given by 0( )y yv t v g t= ! (7) At the top of the fight, ( ) 0y topv t = (8) Substituting Eq. (8) into Eq. (7) yields 00 y topv gt= ! (9) The ball reached its highest point at a time

ttop = !t / 2 = 1.0s after it was hit.

We use this value for the instant the ball reaches the highest point of its trajectory and solve Eq. (9) for the initial vertical component of the velocity that is given by ( ) ( )2 1

0 / 2 9.8m s (1.0s) 9.8m sy topv gt g t ! != = " = # = # (10) . The initial speed and angle are then

( ) ( )2 22 2 1 1 1

0 0 0

101 1

0 10

16m s 9.8m s 18.8m s

9.8m stan tan 31.5 .16m s

x y

y

x

v v v

vv

!

" " "

"" "

"

= + = # + # = #

$ % $ %#= =& ' & '#( )( )!

(11)

Review: The range that the ball traveled and the initial angle and speed of the ball are consistent with reasonable values that you may encounter either playing or watching a softball game.

Problem 5: Jumping off a cliff A person, standing on a vertical cliff a height h above a lake, wants to jump into the lake but notices a rock just at the surface level with its furthest edge a distance s from the shore. The person realizes that with a running start it will be possible to just clear the rock, so the person steps back from the edge a distance d and starting from rest, runs at an acceleration that varies in time according to 1xa b t= and then leaves the cliff horizontally. The person just clears the rock. Find s in terms of the given quantities d , 1b , h , and the gravitational constant g . You may neglect all air resistance.

Solution: While the person is running, since the acceleration varies in time, in order to find the x-component of the velocity, we integrate the x-component of the acceleration with respect to time

( )0 1

0 021

12

( ) ( )t t

x x xv t v a t dt b t dt

b t

! = =

=

" " . (1)

Since the runner started from rest vx0 = 0 and therefore 21

12( )xv t b t= . (2) We now integrate the x-component of the velocity with respect to time to find the displacement

21

0 120 0

3116

( ) ( ) ( )t t

xx t x v t dt b t dt

b t

! = =

=

" " . (3)

Choose an origin from the point the runner began, so x0 = 0 , and the position function is 31

16( )x t b t= . (4)

Let t = t1 denote the instant the runner reaches the cliff, then 31

1 1 16( )x t d b t= = . (5) Hence

1/3

11

6dtb

! "= # $% &

. (6)

The x-component of the velocity of the runner at t = t1 is then

2/3

21 11 1 1 12 2

1

6( )xd

v t b t bb

! "= = # $

% &. (7)

Let’s restart our clock at 0t = when the runner leaves the cliff. Let’s pick our origin at the point of departure. Since the person is in free fall when in the air and we can neglect air resistance, the runner has acceleration ay = !g . So the y-component of the position of the runner is given by

201( )2

y t y g t= ! . (8)

At 0t = , 0 0y = , so

21( )2

y t h g t= ! . (9)

Let 2t t= denote the instant the runner hits the water, then

22 2

1( )2

y t h g t= ! = ! . (10)

So we can solve for the time of flight,

1/ 2

22htg

! "= # $% &

. (11)

The horizontal position of the runner from the edge of the cliff is given by 0( ) xx t v t= . (12)

Since we know the runner leaves the cliff with an x-component of the velocity equal to the x-component of the velocity of the runner at t = t1 , we have that

2/3

112

1

6( ) dx t b t

b! "

= # $% &

. (13)

At 2t t= , the horizontal distance to the rock is then

2/3 2 /3 1/ 2

1 12 1 2 12 2

1 1

6 6 2( ) d d hx t s b t b

b b g! " ! " ! "

= = =# $ # $ # $% &% & % &

. (14)

or

2/3 1/ 2

112

1

6 2d hs bb g

! " ! "= # $ # $

% &% &. (15)

Problem 6 Accelerating Wedge A 45o wedge is pushed along a table with constant acceleration

!A (see figure for choice

of axes and positive directions) according to an observer at rest with respect to the table. A block of mass m slides without friction down the wedge. Find its acceleration with respect to an observer at rest with respect to the table. Write down a plan for finding the magnitude of the acceleration of the block. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities. Also clearly state any assumptions you make. Be sure you include any free body force diagrams or sketches that you plan to use.

Solution: Choose a coordinate system for the block and wedge as shown in the figure below. Then

!A = Ax ,w i where

Ax ,w is the x-component of the acceleration of the wedge.

We shall apply Newton’s Second Law to the block sliding down the wedge. Since the wedge is accelerating, there is a constraint relation between the x - and y - components of the acceleration of the block. In order to find that constraint we choose a coordinate system for the wedge and block sliding down the wedge shown in the figure below. We shall find the constraint relationship between the components of the accelerations of the block and wedge by a geometric argument. From the figure above, we see that

tan! =

yb

l " (xb " xw ). (1)

Therefore

yb = (l ! (xb ! xw )) tan" . (2) If we differentiate Eq. (2) twice with respect to time noting that

2

2 0d ldt

= (3)

we have that

d 2 yb

dt2 = !d 2xb

dt2 !d 2xw

dt2

"

#$

%

&' tan( . (4)

Therefore

ab, y = !(ab,x ! Ax ,w ) tan" (5) recalling that

Ax ,w =

d 2xw

dt2 . (6)

We now draw a free body force diagram for the block (figure below)

Newton’s Second Law in the i - direction becomes

N sin! = mab,x . (7)

and the j -direction becomes

N cos! " mg = mab, y (8)

We can solve for the normal force from Eq. (7):

N =

mab,x

sin! (9)

We now substitute Eq. (5) and Eq. (9) into Eq. (8) yielding

mab,x

sin!cos! " mg = m("(ab,x " Aw,x ) tan!) . (10)

We now clean this up yielding

mab,x (cotan ! + tan!) = m(g + Aw,x tan!) (11)

Thus the x-component of the acceleration is then

ab,x =

g + Aw,x tan!cotan ! + tan!

(12)

From Eq. (5), the y -component of the acceleration is then

ab, y = !(ab,x ! Aw,x ) tan" = !

g + Aw,x tan"cotan " + tan"

! Aw,x

#

$%

&

'( tan" . (13)

This simplifies to

ab, y =

Aw,x ! g tan"cotan " + tan"

(14)

When 45! = ! , cotan 45 tan 45 1= =! ! , and so Eq. (12) becomes

ab,x =

g + Aw,x

2 (15)

and Eq. (14) becomes

ab, y =

A! g2

. (16)

The magnitude of the acceleration is then

a = ab,a

2 + ab, y22 =

g + Aw,x

2

!

"#

$

%&

2

+Aw,x ' g

2

!

"#

$

%&

2

(17)

a =

g 2 + Aw,x2

2

!

"#

$

%& .

Problem 7 Pulleys and Ropes Constraint Conditions Consider the arrangement of pulleys and blocks shown in the figure. The pulleys are assumed massless and frictionless and the connecting strings are massless and unstretchable. Denote the respective masses of the blocks as 1m , 2m and 3m . The upper pulley in the figure is free to rotate but its center of mass does not move. Both pulleys have the same radius R .

a) How are the accelerations of the objects related? b) Draw force diagrams on each moving object.

c) Solve for the accelerations of the objects and the tensions in the ropes.

Solution: Choose an origin at the center of the upper pulley. Introduce coordinate functions for the three moving blocks, 1y , 2y and 3y . Introduce a coordinate function Py for the moving pulley (the pulley on the lower right in the figure). Choose downward for positive direction; the coordinate system is shown in the figure below then.

a) The length of string A is given by

1A Pl y y R!= + + (1) where R! is the arc length of the rope that is in contact with the pulley. This length is constant, and so the second derivative with respect to time is zero,

2 2 2

1,1 ,2 2 20 A Py y P

d l d y d y a adt dt dt

= = + = + . (2)

Thus block 1 and the moving pulley’s components of acceleration are equal in magnitude but opposite in sign, , ,1y P ya a= ! . (3) The length of string B is given by 3 2 3 2( ) ( ) 2B P P Pl y y y y R y y y R! != " + " + = + " + (4) where R! is the arc length of the rope that is in contact with the pulley. This length is also constant so the second derivative with respect to time is zero,

2 2 2 2

2 3,2 ,3 ,2 2 2 20 2 2B Py y y P

d l d y d y d y a a adt dt dt dt

= = + ! = + ! . (5)

We can substitute Equation (3) for the pulley acceleration into Equation (5) yielding the constraint relation between the components of the acceleration of the three blocks, ,2 ,3 ,10 2y y ya a a= + + . (6) b) Free Body Force diagrams: The forces acting on block 1 are: the gravitational force 1m g

! and the pulling force 1,rT!

of the string acting on the block 1. Since the string is assumed to be massless and the pulley is assumed to be massless and frictionless, the tension AT in the string is uniform and equal in magnitude to the pulling force of the string on the block. The free body diagram is shown below.

Newton’s Second Law applied to block 1 is then

1 1 ,1

ˆ : A ym g T m a! =j . (7) The forces on the block 2 are the gravitational force 2m g

! and the string holding the

block, 2,rT!

, with magnitude BT . The free body diagram for the forces acting on block 2 are shown below.

Newton’s second Law applied to block 2 is 2 2 ,2

ˆ : B ym g T m a! =j . (8) The forces on the block 3 are the gravitational force 3m g

! and the string holding the

block, 3,rT!

, with magnitude BT . The free body diagram for the forces acting on block 3 are shown below.

Newton’s second Law applied to block 3 is 3 3 ,3

ˆ : B ym g T m a! =j . (9)

The forces on the moving pulley are the gravitational force Pm =g 0!! (the pulley is

assumed massless); string B pulls down on the pulley on each side with a force, ,P BT!

,

which has magnitude BT . String A holds the pulley up with a force ,P AT!

with the magnitude AT equal to the tension in string A . The free body diagram for the forces acting on the moving pulley are shown below.

Newton’s second Law applied to the pulley is ,

ˆ : 2 0B A P y PT T m a! = =j . (10) Since the pulley is massless we can use this last equation to determine the condition that the tension in the two strings must satisfy, 2 B AT T= (11) We are now in position to determine the accelerations of the blocks and the tension in the two strings. We record the relevant equations as a summary. ,2 ,3 ,10 2y y ya a a= + + (12) 1 1 ,1A ym g T m a! = (13) 2 2 ,2B ym g T m a! = (14) 3 3 ,3B ym g T m a! = (15) 2 B AT T= . (16) There are five equations with five unknowns, so we can solve this system. We shall first use Equation (16) to eliminate the tension AT in Equation (13), yielding 1 1 ,12 B ym g T m a! = . (17) We now solve Equations (14), (15) and (17) for the accelerations,

,22

By

Ta gm

= ! (18)

,33

By

Ta gm

= ! (19)

,11

2 By

Ta gm

= ! . (20)

We now substitute these results for the accelerations into the constraint equation, Equation (12),

2 3 1 2 3 1

4 1 1 40 2 4B B BB

T T Tg g g g Tm m m m m m

! "= # + # + # = # + +$ %

& '. (21)

We can now solve this last equation for the tension in string B ,

1 2 3

1 3 1 2 2 3

2 3 1

4 441 1 4B

g g m m mTm m m m m m

m m m

= =! " + +

+ +# $% &

. (22)

From Equation (16), the tension in string A is

1 2 3

1 3 1 2 2 3

824A B

g m m mT Tm m m m m m

= =+ +

. (23)

We find the acceleration of block 1 from Equation (20), using Equation (22) for the tension in string B,

2 3 1 3 1 2 2 3,1

1 1 3 1 2 2 3 1 3 1 2 2 3

2 8 44 4

By

T g m m m m m m m ma g g gm m m m m m m m m m m m m

+ != ! = ! =+ + + +

. (24)

We find the acceleration of block 2 from Equation (18), using Equation (22) for the tension in string B,

1 3 1 3 1 2 2 3,2

2 1 3 1 2 2 3 1 3 1 2 2 3

4 3 44 4

By

T g m m m m m m m ma g g gm m m m m m m m m m m m m

! + += ! = ! =+ + + +

. (25)

Similarly, we find the acceleration of block 3 from Equation (19), using Equation (22) for the tension in string B,

1 3 1 2 2 31 2,3

3 1 3 1 2 2 3 1 3 1 2 2 3

3 444 4

By

m m m m m mT gm ma g g gm m m m m m m m m m m m m

! += ! = ! =

+ + + +. (26)

As a check on our algebra we note that

1, 2, 3,

1 3 1 2 2 3 1 3 1 2 2 3 1 3 1 2 2 3

1 3 1 2 2 3 1 3 1 2 2 3 1 3 1 2 2 3

2

4 3 4 3 424 4 4

0.

y y ya a am m m m m m m m m m m m m m m m m mg g gm m m m m m m m m m m m m m m m m m

+ + =

+ ! ! + + ! ++ ++ + + + + +

=

Problem 8 Block and Pulley

A block of mass m1 rests motionless on an inclined plane that makes an angle ! with the horizontal. The coefficient of static friction between the block and plane is µs . A massless inextensible string is attached to one end of the block, passes over a fixed pulley, around a second freely suspended pulley, and is finally attached to a fixed support. The left hand part of the string is parallel to the plane. The sections of the string coming off the suspended pulley are vertical. The pulleys are massless, but an object of mass m2 is hung from the suspended pulley. Gravity acts downward. Part 1: Assume that if there were no friction the block would slide up the inclined plane. In what follows, express your answers in terms of m1 , m2 , ! , µs and the acceleration of gravity g (or some subset of these parameters).

a) Draw separate free body diagrams for the system consisting of the object hanging from the suspended pulley plus the suspended pulley, and the block on the inclined plane, showing and labeling carefully, all the forces that act on the objects.

b) Find the magnitude of the static friction force.

c) What is the condition on the masses of the blocks such that the block on the

incline plane just slides upward. Part 2: Now assume that the block on the incline plane is sliding upward. The coefficient of kinetic friction between the block and plane is µk . In what follows, express your answers in terms of m1 , m2 , ! , µk and the acceleration of gravity g (or some subset of these parameters).

d) Draw separate free body diagrams for the second (freely suspended) pulley, and the block on the inclined plane, showing and labeling carefully, all the forces that

act on the objects. e) Find the magnitude of the acceleration of the block on the incline plane.

Solution: We shall first assume that if there were no friction the block would slide up the incline plane. Therefore the static friction force points down the inclined plane. There is no acceleration so applying Newton’s Second Law for the block on the incline plane yields N ! m1g cos" = 0 (1) . T ! fs ! m1g sin" = 0 (2) Applying Newton’s Second Law to the system consisting of the object hanging from the suspended pulley plus the suspended pulley yields m2g ! 2T = 0 (3) We can solve for the magnitude of the static friction. Solve Eq. (3) for the tension T , and substitute that result into Eq. (2), and solve for magnitude of the static friction:

fs =

12

m2g ! m1g sin" (4)

Note that the maximum value of static friction is

fs,max = µsm1g cos! (5)

We can find the condition on the masses such that the block just slips up the inclined plane m2 > 2m1(µs cos! + sin!) (6) Now suppose the block is sliding up the incline plane T ! fk ! m1g sin" = m1ax1 (7) . The kinetic friction is given by fk = µk N = µk m1g cos! (8)

Thus T ! µk m1g cos" ! m1g sin" = m1ax1 (9) The force equation on block 2 plus the attached pulley is

m2g ! 2T = m2ay2 (10) Constraint: The length, call it L , of the string is fixed. Let R be the pulley radius. From our coordinate system we have that L = (D1 ! x) + "R / 2 + y + "R + (D2 + y) (11) After two derivatives with respect to time gives

0 = !ax1 + 2ay2 (12) Thus

ax1 = 2ay2 (13) Solve Eq. (9) for T and substitute the result into Eq. (10),

m2g ! 2(µk m1g cos" + m1g sin" + m1ax1) = m2ay2 (14)

Then use Eq. (13) yielding

ax1 = 2

m2 ! 2m1(µk cos" + sin")4m1 + m2

#

$%&

'(g (15)

When m1 = 0 block 2 is in free fall with an acceleration g , which makes sense. This result is positive as long as m2 > 2m1(µk cos! + sin!) . When m2 = 2m1(µk cos! + sin!) the tension in the string just balances the friction force agreeing with our result in Eq. (6) provided we assume that µs = µk . Because µs > µk , when the block just slips, it will immediately slide because the kinetic friction force is smaller than the static friction force. If we want to find the acceleration when the block on the incline plane slides downward then we must reverse the direction of the kinetic friction. Thus Eq. (9) becomes T + µk m1g ! m1g sin" = m1ax1 (16)

Otherwise the equations are the same hence

ay2 =

m2 ! 2m1(!µk cos" + sin")4m1 + m2

g (17)

ax1 = 2

m2 ! 2m1(!µk cos" + sin")4m1 + m2

#

$%&

'(g (18)

When m2 = 2m1(!µk cos" + sin") the gravitational force just balances the tension in the string and the friction force. For m2 < 2m1(!µk cos" + sin") , ax1 < 0 and the block slides down the inclined plane.

Problem 9: Winter Sports Calvin and Hobbes are riding on a sled. They are trying to jump the gap between two symmetrical ramps of snow separated by a distance W as shown above. Each ramp makes an angle ! with the horizontal. They launch off the first ramp with a speed Lv . Calvin, Hobbs and the sled have a total mass m .

a) What value of the initial launch speed Lv will result in the sled landing exactly at the peak of the second ramp? Express your answer in terms of some (or all) of the parameters m , ! , W , and the acceleration of gravity g . Include in your answer a brief description of the strategy that you used and any diagrams or graphs that you have chosen for solving this problem. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities.

Answer: The sled is undergoing projectile motion with an initial launch speed Lv in the direction given by a positive ! with respect to the horizontal. We choose a coordinate system as shown in the figure below:

Then the components of the acceleration are given by 0xa = and ya g= ! . We are given the range W , so we can use the equations for the y-components of position to find the time of flight and then use the x-component of position to find the initially launch speed Lv .

The y-component of the position is given by

21( ) sin2Ly t v t g t!= " (1)

Since the sled returns to it’s initial starting point at the instant ft t= , Eq. (1) becomes

21( ) 0 sin2f L f fy t t v t g t!= = = " (2)

We can solve Eq. (2) for the time of flight

2 sinLf

vtg

!= (3)

The x-component of the position is given by ( ) cosLx t v t!= (4) At the end of its flight. Eq. (4) becomes ( ) cosf L fx t W v t!= = (5) Substitute Eq. (3) into Eq. (5) to find that

2 22 cos sin sin 2L Lv vWg g! ! != = (6)

We can now solve Eq. (6) for the launch speed

sin 2LgWv

!= . (7)

b) Hobbes was clutching a bag of tiger food when they left the first ramp. He got so excited while in the air that he let go of the bag at the top of the flight, lightening the total mass attached to the sled by 10%. Explain qualitatively how this will affect the results you found in a). Answer: When we apply Newton’s Second Law to the sled, noting that the gravitational force is proportional to the mass, the mass cancels out and the acceleration is independent of the mass. Therefore when Hobbes releases the bag, the acceleration does not change, hence the trajectory remains the same. (Note that if Hobbes threw the bag in any direction with some speed then the trajectory would change do to recoil; a process will study shortly.)

c) Explain qualitatively what happens to the released bag of tiger food. Answer: Since the bag of food was released from rest, it’s trajectory is the same as the sled.

d) There is friction between the sled and the snow which can be modeled by

coefficient of kinetic friction kµ . What is the minimum value of kµ that will assure that the sled comes to rest somewhere on the second ramp? Assume an infinitely long ramp. Include in your answer a description of the plan that you will use to solve this part of the problem and any relevant diagrams or graphs that you find useful. Make sure you clearly state which concepts you plan to use to calculate any relevant physical quantities.

Answer: In order for the sled to eventually come to a stop (on an infinitely long hill) the acceleration must point uphill. We draw a free body force diagram with directions chosen as shown in the figure below. Since the friction force points uphill, we can apply Newton’s Second Law to find a condition such that the acceleration up hill is less than zero.

The forces on the sled (and riders) are gravity m

!g = mg( icos! " jsin!) (8) and the contact force between the ground and the sled k k

ˆ ˆN f= + = !C N f j i! !!

. (9) The components of the vectors in Newton’s Second Law, m=F a

! ! , are ksin xmg f ma! " = (10) cos .yN mg ma!" = (11) The condition that the sled slide down the ramp is expressed by

ay = 0 and the model for kinetic friction is k kf Nµ= . (12) The condition

ay = 0 in Eq. (11) gives

cosN mg != , (13) and so Eq. (12) becomes k k cosf mgµ != . (14) Substitute Eq. (14) into Eq. (10) to solve for the x -component of acceleration ( )ksin cosxa g ! µ != " . (15) In order for the sled to come to rest, the x -component of acceleration must be less than zero ( )ksin cos 0xa g ! µ != " < . (16) Hence we arrive at the condition that ktan! µ< . (17)

Problem 10 Bicycle and the Falling iPod A bicycle rider is traveling at a constant speed along a straight road and then gradually applies the brakes during a time interval

0 < t < t f until the bicycle comes to a stop. Assume that the magnitude of the braking force increases linearly in time according to

F = bt ,

0 < t < t f

where 0b > is a constant. The cyclist and bicycle have a total mass m . At the instant the person applies the brakes, a horizontal distance d from the rider, the wind blows and snaps an iPod off the branch of the tree with a horizontal speed pov in the direction shown in the figure below. The ipod was initially a height h above the ground. The cyclist catches the iPod at the instant the cyclist has come to a stop. You may assume that the cyclist catches it at a height s above the ground.

a) At what time 1t did the cyclist catch the ipod? Express your answer in terms of the quantities, b , m , h , s , pov , d , and g as needed.

b) What was the initial speed of the cyclist? You may leave your answer in terms of 1t from part a) and any other quantities as needed.

Solution: There are two objects moving, the bicyclist and the iPod. We shall choose a coordinate system for the bicyclist and the iPod as shown in the figure below.

We can use Newton’s Second Law to calculate the acceleration of the bicyclist. We can then integrate the acceleration to find the x-component of the velocity of the bicyclist. We then use the equations for free fall to determine the time it takes the iPod to fall a distance h s! . The x-component of the velocity of the bicyclist is zero at this instant so we can solve for the initial x-component of the velocity of the bicyclist.

a) At what time 1t did the cyclist catch the ipod? Express your answer in terms of the quantities, b , m , h , s , pov , d , and g as needed.

From Newton’s Second Law, m=F a! ! , are in the x-direction:

ˆ : xbt ma! =i . (1) Thus the x-component of the acceleration is

xb

a tm

= ! (2)

We can now integrate the x-component of the acceleration of the bicyclist to get the x-component of the velocity of the bicyclist

20

0 0

( )2

t t

bx bx xb b

v t v a dt t dt tm m

! "# = = # = #$ %& '( ( . (3)

So the x-component of the velocity of the cyclist as a function of time is given by

vbx (t) = vbx0 !

b2m

t2 (4)

The y-component of the position of the iPod is given by

2( ) ( )2gy t h s t= ! ! (5)

The bicyclist catches the iPod at the instant when

21 1( ) 0 ( )

2gy t t h s t= = = ! ! (6)

We can solve eq. (6) for the flight time:

12( )h stg!= (7)

b) What was the initial speed of the cyclist? You may leave your answer in terms of

1t from part a) and any other quantities as needed.

The bicyclist comes to a full stop at the instant t1 so we can substitute Eq. (7) into Eq. (4) and solve for the initial x-component of the velocity of the bicyclist

20 1

( )2bxb b h s

v tm mg

!= = (8)

Alternatively we can integrate Eq. (4) to find that the x-component of the position the bicyclist when it just came to a stop

1 1

2 31 0 0 0 1 1

0 0

( )2 6

t t

b b bx bx bxb b

x t x v dt v t dt v t tm m

! "# = = # = #$ %& '( ( . (9)

We chose 0 0x = and so substituting 0 0x = and Eq. (7) into Eq. (9) yields

31 0 1 1( )

6b bxb

x t v t tm

= ! (10)

During the time the iPod fell, it traveled a horizontal distance 1 1 1( )p po pox t v t v t= = (11) So 1 1( ) ( )b px t d x t= + (12) Substituting the results above yields

30 1 1 16bx po

bv t t d v t

m! = + (13)

We can now solve for bxov :

20 1

16bx pob d

v t vm t

= + + . (14)

Note: The parameters b , d , and pov are not independent. They are related as follows. We can integrate Eq. (4) using the value for 0bxv that we found in Eq. (8) to find that the x-component of the position the bicyclist when it just came to a stop

2 2 30 0

0 0 0

( ) ( )( )2 2 6

f ft tt

b f b bx bx f fb b h s b b h s b

x t x v dt v t dt t dt t tm mg m mg m

! "# #! "# = = # = # = #$ %$ %& ' & '( ( ( .(15)

We chose 0 0x = and so substituting 0 0x = and Eq. (7) into Eq. (15)yields

3

2( )( )3b fb h s

x tm g! "#= $ %$ %& '

(16)

During the time the iPod fell, it travelled a horizontal distance

2( )( )p f po f poh s

x t v t vg!= = (17)

So ( ) ( )b f p fx t d x t= + (18) Substituting the results above yields

3

2( ) 2( )3 pob h s h s

d vm g g! "# #= +$ %$ %& '

(19)

We can now solve for the constant b in terms of the other constants and pov :

3

2( ) 2( )3 poh s h sb m d vg g

! " ! "# #= +$ % $ %& ' & '

. (20)

Problem 11 Two Block Pull Consider two blocks that are resting one on top of the other. The lower block has mass

2M and is resting on a nearly frictionless surface. The upper block has mass 1 2M M< . Suppose the coefficient of static friction between the blocks is sµ .

a) What is the maximum force with which the upper block can be pulled horizontally so that the two blocks move together without slipping? Draw as many free body force diagrams as necessary. Identify all action-reaction pairs of forces in this problem.

b) What is the maximum force with which the lower block can be pulled

horizontally so that the two blocks move together without slipping? Draw as many free body force diagrams as necessary. Identify all action-reaction pairs of forces in this problem.

Solution: a) Suppose we take as our system both blocks. Then the free body diagram is shown below

Applying Newton’s Second Law to the system yields

i : F1 = ( M1+ M2 )a

j : NG2 ! ( M1+ M2 ) g = 0 (21)

Thus a = F1 / ( M1+ M2 ) (22)

However we have no knowledge of the internal forces between the surfaces of block 1 and block 2, so we cannot determine whether or not the force

!F1 will cause them to slip

relative to each other. In order to determine some condition

!F1 due to the fact that static friction cannot exceed

some maximum value we must draw free-body force diagrams for the respective blocks as shown below. Note that the static friction force is the force that accelerates the lower block.

The action-reaction pairs are:

i. The gravitational force (magnitude 1M g ) that the earth exerts on the upper block and the gravitational force (not shown) that the block exerts on the earth.

ii. The gravitational force (magnitude 2M g ) that the earth exerts on the lower block

and the gravitational force (not shown) that the block exerts on the earth. iii. The normal contact force

!N21 that the lower block 2 exerts on the upper block 1

and the normal contact force !N12 that the upper block 1 exerts on the lower block

2. iv. The static friction force

!f21 , directed to the left in the figure, that the lower block

2 exerts on the upper block 1 and the static friction force !f12 , directed to the right,

that the upper block 1 exerts on the lower block 2.

v. The upward contact force of magnitude NG2 that the surface (ground, table, or other) exerts on the lower block and the downward contact force that the lower block exerts on the surface

vi. Whatever or whomever is pushing the block will have a reaction force of

magnitude 1F directed to the left.

Applying Newton’s Second Law to the upper block yields

i : F1 ! f21 = M1 a1

j : N21 ! M1 g = 0 (23)

where 1a is the acceleration of the upper block. Applying Newton’s Second Law to the lower block yields

i : f12 = M2 a2

j : NG ! N12 ! M2 g = 0 (24)

where 2a is the acceleration of the lower block. The just-slipping conditions are that the accelerations are equal, 1 2a a a! = , (25) and that the static friction has its maximum value, f21 = ( f21)max = µs N21 . (26) We can now solve the above equations for the acceleration of the blocks. The first equation in (23) implies N21 = M1 g (27) and so Equation (26) becomes f21 = µs M1g . (28) Substituting Equations (28) and (25) into the first equation in (24), s 1 2M g M aµ = (29) and the acceleration of the blocks is

s 1 11 2 s

2 2

M g Ma a a gM M

µ µ= = = = . (30)

The pushing force is then found by substituting Equation (30) into the first equation in (23),

s 1 11 s 1 1 s 1

2 2

1M g MF M g M M gM M

µµ µ! "

= + = +# $% &

. (31)

Equivalently, we could have substituted our result for the acceleration from Eq. (22) into Eq. (30) resulting in

a = µsg

M1

M2

= F1 / ( M1 + M2 ) (32)

which yields the same result as Eq. (31) for the magnitude of the maximum possible pushing force on the upper block before the blocks slip. Check this result in the limits that one block is much heavier than the other. If

1 2M M! , the lower block won’t move at all, and 1 s 1F M gµ! , the maximum static friction force. If 1 2M M! , the lower block acts more or less as a lubricant, and the maximum force becomes very large.

b) The free-body force diagrams are shown below when we push the lower block,

The static friction force is now the only force that makes the upper block accelerate. The action-reaction pairs are the same as in part a) (except for (vi), the reaction force of magnitude 2F on whatever or whomever is doing the pushing). We again apply Newton’s Second Law to each block. Applying Newton’s Second Law to the upper block yields

i : f21 = M1 a1

j : N21 ! M1g = 0. (33)

Applying Newton’s Second Law to the lower block yields

i : F2 ! f12 = M2 a2

j : NG ! N12 ! M2g = 0. (34)

The just-slipping conditions are identical to Equations (25) and (26). Substitution of these into the first equation in (33) yield a = µsg (35)

After substituting Eq. (35) and Equation (25) in the first equation in (34), we find that the maximum pushing force is 2 s 1 2 s s 1 2( )F M g M g g M Mµ µ µ= + = + . (36) Note that if we draw a free-body diagram on the compound system, the free-body force diagram looks like

Newton’s Second Law in the horizontal direction then becomes

i : F2 = M1 + M2( )a (37) which is easily solved for the acceleration

a =

F2

M1 + M2

. (38)

Then substituting Eq. (38) into Eq. (35) yields

a =

F2

M1 + M2

= µsg (39)

which we can solve for the maximum pushing force as given by Eq. (36). We note by comparing Equation (30) to Equation (35) that the acceleration of the blocks is greater when you push the lower block than when you push the upper block (recall that the problem statement specified 1 2M M< ). Recall that the static friction force is the only force accelerating the block that you are not pushing so when you push the lower block ( fs )max ! M1a ! M1F2 (push lower book) . (40) When you push the upper block ( fs )max ! M2a ! M2 F1 (push upper book) . (41) The maximum value of static friction between the blocks has the same magnitude in both cases. Therefore the product

M2 F1 = M1 F2 (42) or

F1

F2

=M1

M2

(43)

So M1 / M2 < 1 , then F1 / F2 < 1 and you can push the lower block with a greater force

F2 > F1 .