1

Physics 181

Assignment 2

Solutions 1. A mass m is projected straight up with initial speed

v

0. Find the height to which

it rises before coming back down: a. If there is no air resistance. b. If there is an air resistance force opposite to the velocity, with magnitude

mbv , where v is the instantaneous speed and b is a positive constant. c. Show that your answer to (b) reduces to your answer to (a) when b! 0 .

[Use the expansion ln(1 + x) = x !

12

x2+ ... for

x << 1 .]

a. Simplest is to use conservation of energy, with U = 0 at the launch point. Then we have

1

2mv

0

2= mgy

max, so

ymax = v0

2 /2g .

b. There are two approaches. One can solve for things as a funciton of time, or one can use energy principles. The latter is more direct in this case.

First the time approach. From the 2nd law we have m!!y = !mg ! mb!y , or

!!y + b!y = !g . We note that it can be written

d

dt( !y + by) = !g , so

!y + by = !gt + A . The initial conditions are

y(0) = 0 and

!y(0) = v0 ; thus

A = v

0 and we have

!y + by = v

0! gt .

For the particular solution we try yp = A + Bt , obtaining

B + b(A + Bt) = v0 ! gt , so B + bA = v

0 and bB = !g . Solving for the

constants we have

yp =

v0

b+

g

b2

!"#

$%&'

g

bt .

For the solution to the homogeneous equation we try yc = Ce!t and find

C(! + b)e!t= 0 , so ! = "b . The total solution is

y = Ce!bt

+v

0

b+

g

b2

"#$

%&'!

g

bt

Applying y(0) = 0 to evalutate C we have after rearranging:

y =g

b21 +

bv0

g

!

"#$

%&(1 ' e'bt ) '

g

bt .

2

The velocity is

!y =g

b1 +

bv0

g

!

"#$

%&e'bt '

g

b.

At the maximum height this is zero, which occurs at the time when

ebt

= 1 + bv0 / g , or t = (1/b)ln(1 + bv0 / g) . Substituting in the formula for y

we obtain after some rearrangement

ymax =

v0

b!

g

b2ln(1 + bv0 / g) .

Now the energy approach. The general principle is: The work done by non-conservative forces is equal to the change in E. (Here E = T +U .) In this case the non-conservaive force is air resistance: !mbv .

Applying this principle as the particle falls distance dy, we have

(!mbv) "dy = d 1

2mv2 ! mgy#

$%&= mv "dv ! mgdy .

Rearranging we have

dy = !

vdv

bv + g.

We integrate both sides, noting that when y = y

max, v = 0 .

dy

o

ymax

! = "vdv

bv + gv0

0

! .

Carrying out the integral on the right side gives the answer immediately. c. For small b we expand the log as suggested. The first term cancels with

v0 /b , while the second term gives

v0

2 /2g , as in (a). [If you have a graphing calculator, look at

y

max vs. b for

v

0= g = 2 , so that

v0

2 /2g = 1 .]

2. A mass m falls from rest under gravity, with air resistance force opposite to the

velocity and of magnitude mkv2 , where v is the instantaneous speed. When its

speed is v it has fallen a distance y. a. Show that for two different speeds we have

y2! y

1=

1

2kln

g ! kv1

2

g ! kv2

2

"

#$$

%

&''

.

b. Show that as k ! 0 (no air resistance) this formula gives

3

y2! y

1=

v2

2! v

1

2

2g.

[Use the same expansion as in Prob. 1]

a. We use the energy approach, as in 1(b) above:

(!mkv2 ) "dy = mv "dv ! mg "dy ,

so

dy =vdv

g ! kv2.

Integrating we have

dyy

1

y2

! =vdv

g " kv2v1

v2

! .

This gives the answer.

b. Write g ! kv2

= g(1 ! kv2 / g) , use ln(a/b) = ln a ! ln b , then apply the expansion.

3. A projectile is fired up a hill of slope angle 30° . The elevation angle of the

projectile above the horizontal is ! > 30° , and its initial speed is v

0. Let the

distance up the slope of the hill from where the projectile is launched be s. a. At what value of s does the projectile land

on the hill? b. For what value of ! is that value a

maximum? c. What is that maximum value of s? [Neglect air resistance.]

a. The equations in cartesian coordinates (with origin at the launch point) are

x(t) = v0 cos! " t, y(t) = v0 sin! " t # 1

2gt2 .

Solving the first for t and substituting in the second, we have:

y = x tan! "gx2

2v0

2cos

2!

.

On the slope x = scos30° = s ! ( 3 /2) and y = ssin 30° = s ! (1/2) , so

s = 3 ! s tan" #3g

4v0

2cos

2"

! s2 .

v

0

!

30°

4

The root s = 0 is the launch point. The one we want is

s =4v0

2

3g[ 3 !sin" cos" # cos2

"] .

b. Set ds/d! = 0 to find 3(cos2! " sin2

!) + 2sin! cos! = 0 . At this point it helps to note that cos2! = cos

2! " sin

2! and sin 2! = 2sin! cos! . We find

tan 2! = " 3 , so 2! = 120° , and ! = 60° .

c. Substituting into the formula for s we find

smax

=2v

0

2

3g. This is 2/3 of the

maximum range on a horizontal surface.

4. A superball of mass M is dropped from height h onto a floor, where it makes an

elastic collision and rebounds. A negligible distance above the top of the ball, a marble of mass m is dropped at the same time. Immediately after the ball rebounds the marble collides elastically with it. Neglect the sizes of the objects. a. Draw a picture of the ball and marble immediately after the ball rebounds

from the floor but before the marble collides with it. Indicate the velocities of the two objects at that instant. Do the same thing for the situation immediately after the marble collides with the ball.

b. Find the speeds of the two objects in the first drawing. c. Find the speeds of the two objects in the second drawing. d. Show that if m > M/3 the ball will strike the floor again soon after its

collision with the marble. e. If M >> m it is as though the marble hits a rigid but moving floor in its

collision. Use this, and consider the collision between the two objects as seen in a reference frame moving with the ball in the first picture in (a), to show that the marble will rebound to 9 times h. Check that your answer to (c) agrees with this.

a. Drawings as shown above.

b. By conservation of energy v

0= 2gh . The elastic collsion with the

massive floor reverses the direction of the velocity at the same speed. c. Take upward as positive. Then conservation of momentum gives

Mv

0! mv

0= Mv

1+ mv

2,

and conservation of energy gives

v

0

v

0

v

1

v

2

5

1

2Mv

0

2+

1

2mv

0

2=

1

2Mv

1

2+

1

2mv

2

2 .

Let ! = m/M . Then these two equations become:

(1 !" )v0 = v1 +"v2

(1 +" )v02= v1

2+"v2

2

Some algebra allows us to solve for the final velocities:

v2 =

3 !"

1 +"v0 , v1 =

1 ! 3"

1 +"v0 .

d. We see that if m > M/3 ( 3! > 1 ) then v

1< 0 , meaning that M is moving

toward the floor after the collision, so it will strike the floor again. e. In the frame moving (upward) with M before the collision with m, the

downward velocity of m is 2v

0. The elastic collision (for very large M)

merely reverses the direction, so after the collision m is moving upward with speed

2v

0 relative to M, therefore with speed

3v

0 relative to the

floor. In the solution in (c) as ! " 0 we see that v

2! 3v

0 as claimed.

5. It is observed that the speed v of a particle of mass m varies with its position x

according to the formula v = !/x , where ! is a constant. Find the force F(x) on the particle.

We have

!v =dv

dx

dx

dt= !

"

x2

v = !"

2

x3

. From the 2nd law F = m !v = !m"

2 /x3 .

6. A train is moving horizontally at constant speed V when a person sitting in a seat

on it throws a ball of mass m directly forward at speed v relative to the train. a. Find the kinetic energy given to the ball as measured on the train. b. Find the same quantity as measured in a frame at rest on the ground. [This

is the difference between its kinetic energy before and after it is thrown.] c. The person throwing the ball does the work to create the kinetic energy in

(a). Account for the difference between that and the energy in (b). Who or what is doing the extra work? [What force keeps the person making the throw at rest relative to the train?]

a. The person throwing the ball does not move relative to the train as the ball is thrown, so the kinetic energy provided is just

1

2mv

2 .

6

b. The ball’s speed relative to the ground after it is thrown is v +V , and before it is thrown it has speed V, so the change in its kinetic energy is

12

m (v +V )2 !V2"

#$%=

12

mv2+ mvV .

c. The force exerted by the seat keeps the person at rest relative to the train, preventing him from recoiling with momentum mv . Assume the time it takes to throw the ball is !t . Then the average force from the seat is F = mv/!t . During that time the train moves a distance !x =V!t . In the frame at rest on the ground, the force acts through this distance, so it does work W = F!x = mvV . This accounts for the extra kinetic energy.

7. The force on a particle in one dimension is F(x) = !4x + 4x3 .

a. Find U(x) , taking U(0) = 0 .

b. Make a plot of U(x) in the region where U is not negative. (A graphing calculator is useful for this.)

c. Describe the motion for total energy: (1) E > 1 ; (2) 0 < E < 1 and x starting from zero; (3) 0 < E < 1 and x starting from ! .

d. Locate the equilibrium points, specifying whether they are stable or unstable.

a. We have U(x) = ! F dx

0

x

" = 2x2 ! x

4 .

b. Graph below.

c. (1) The maximum value of U is 1, so for E > 1 the kinetic energy is always

positive and there are no turning points. The motion is unbound.

7

(2) There is a minimum in U at x = 0 , so in this region for 0 < E < 1 there are two turning points. The motion is oscillation between these points.

(3) If the particle comes in from ! with E < 1 , there will be a turning point, and the motion will reverse and go back to ! .

d. Setting F = 0 we find x = x3 , which has three solutions: x = 0 (a

minimum, so stable equilibrium); x = ±1 (maxima, so unstable).

8. The two members of a double star have masses m

1 and

m

2. Under the influence

of Newtonian gravity (attractive force F = Gm1m2 /r

2 , where r is the distance between the objects) they both orbit around their center of mass in circles. Prove that the period T of their orbits obeys Kepler’s 3rd law: T2

! r3 .

The CM is at distance

r1=

m2

m1+ m

2

r from m

1, and distance

r2=

m1

m1+ m

2

r from

m

2. Since they orbit in circles, the 2nd law gives

F = m

1r1!

2 , and similarly for

m

2. Thus we have

Gm

1m

2

r2

= m1

m2

m1+ m

2

r!2 , or

r

3!

2= G(m1 + m2 ) . Since

! = 2" /T , we have Kepler’s 3rd law. [It also holds when the orbits are ellipses.]

9. Consider the following force laws (in the usual xyz coordinate system): (1)

F = (ayz + bx)i + (axz + bz)j+ (axy + by)k ;

(2) F = ar /r2 , where

r = xi + yj+ zk .

In these, a, b, and c are constants. a. Prove that each of these is a conservative force. b. For each, find the potential energy function (up to an additive constant).

The test is whether ! " F = 0 . This means !xFy = !yFx , etc. If this is satisfied, then

we have U(x, y,z) = ! F "dr# = ! (Fx dx + Fy dy + Fz dz)#

(1) Direct calculation shows that ! " F = 0 . We will choose U(0,0,0) = 0 . Then

U(x, y,z) = ! Fx(y = z = 0)dx

0

x

" ! Fy(x = x,z = 0)dy0

y

" ! Fz(x = x, y = y)dz0

z

" .

This gives U(x, y,z) = !axyz ! byz ! 1

2bx2 .

8

(2) All central forces are conservative. To prove it for this one, we note that

!

xr = x/r , so that

!

xr

n= nr

n"1!

xr = nr

n"2x , etc. For the integral for U(r) we

choose a path such that dr is parallel to r so that r !dr = r dr . We find

U(r) = !

a

r2

r dr = !a

rdr = !a ln r + C"" .