MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics

Physics 8.01 Fall 2012

Problem Set 7 Collisions Solutions

Problem 1 Collision

A ping-pong ball and a bowling ball collide elastically on a frictionless surface. The magnitude of the initial velocity of the ping-pong ball is

vp,0 and the direction of the velocity is in the positive x -direction. The magnitude of the initial velocity of the bowling ball is

vb,0 and the direction of the velocity is in the negative x -direction. You may assume that the mass of the bowling ball is much greater than the mass of the ping-pong ball. After the collision what is the velocity of ping-pong? Solution: The collision is elastic and the relative velocity of the ping-pong ball and bowling ball has the same magnitude but opposite direction Because the bowling ball is much more massive than the ping-pong ball we can assume that the speed of the bowling ball does not change due to the collision. The initial relative velocity of the of the two objects is

!v rel ,0 =

!v p,0 !!vb,0 = vp,0 i ! vb,0 (! i) = (vp,0 + vb,0 )i .

The final relative velocity of the of the two objects is

!v rel , f =

!v p, f !!vb, f = vp, f (! i) ! vb,0 (! i) = (!vp, f + vb,0 )i

We know that

!v rel ,0 = !!v rel , f

Therefore

(vp,0 + vb,0 )i = !(!vp,0 + vb,0 )i .

So we can solve for the final speed of the ping-pong ball

vp, f = vp,0 + 2vb,0 .

Problem 2 Flyby

A satellite (with mass negligible compared to that of the Earth) is making a fly-by of the Earth. Let the velocity of the Earth be

!ve = ve i . The satellite’s orbit is symmetric about a line through the center of the Earth that is parallel to x -axis. Far from the Earth the magnitude of the satellite’s x -component of velocity is vi when approaching and vf when receding. For the case where vi = 4ve , (i) is the final speed vf of the satellite greater, equal to, or less than the initial speed vi of the satellite?, (ii) What is the final speed vf of the satellite in terms of the speed of the earth ve ? Answer We can think of the fly-by as an elastic collision because energy is constant. Hence the relative speed between the satellite and earth does not change. The relative speed as the satellite approaches earth is vi ! ve . The relative speed after the satellite is far away from earth is

v f ! ve . Because the relative speeds are equal vi ! ve = v f + ve ,

therefore v f = vi ! 2ve . Because vi = 4ve , the final speed of the satellite is

v f = 2ve .

Problem 3: A particle of mass m , moving in the x-direction, is acting on by a potential

U (x) = !U1

xx1

"

#$%

&'

3

!xx1

"

#$%

&'

2"

#$$

%

&''

, (1)

where U1 and x1 are positive constants and U (0) = 0 .

a) Sketch 1( ) /U x U as a function of 1/x x . b) Find the points where the force on the particle is zero. Classify them as stable or unstable.

Calculate the value of U (x) / U1 at these equilibrium points.

c) For energies E that lies in 0 < E < (4 / 27)U1 find an equation whose solution yields the turning points along the x-axis about which the particle will undergo periodic motion.

d) Suppose E = (4 / 27)U1 and that the particle starts at 0x = with speed 0v . Find v0 .

Solution: a) The figure below shows a graph of U (x) vs. x , with the choice of values

x1 = 1.5 m , U1 = 27 / 4 J , and E = 0.2 J .

b) The force on the particle is zero at the minimum of the potential which occurs at

Fx (x) = !

dUdx

(x) =U1

3x1

3

"

#$

%

&' x2 !

2x1

2

"

#$

%

&' x

"

#$

%

&' = 0 (2)

which becomes x

2 = (2x1 / 3)x . (3) We can solve Eq. (3) for the extrema. This has two solutions

x = (2x1 / 3) and x = 0 . (4) The second derivative is given by

d 2Udx2 (x) = !U1

6x1

3

"

#$

%

&' x !

2x1

2

"

#$

%

&'

"

#$

%

&' . (5)

Evaluating the second derivative at x = (2x1 / 3) yields a negative quantity

d 2Udx2 (x = (2x1 / 3)) = !U1

6x1

3

"

#$

%

&'

2x1

3!

2x1

2

"

#$

%

&'

"

#$

%

&' = !

2U1

x12 < 0 (6)

indicating the solution x = (2x1 / 3) represents a local maximum and hence is an unstable point. At x = (2x1 / 3) , the potential energy is given by the value U ((2x1 / 3)) = (4 / 27)U1 . Evaluating the second derivative at x = 0 yields a positive quantity

d 2Udx2 (x = 0) = !U1

6x1

3

"

#$

%

&' 0 !

2x1

2

"

#$

%

&'

"

#$

%

&' =

2U1

x12 > 0 (7)

indicating the solution x = 0 represents a local minimum and is a stable point. At the local minimum, x = 0 , the potential energy U (0) = 0 . c) Because the kinetic energy K(x) = E !U (x) > 0 must be always be positive, for energies in the range of

U (0) = 0 < E <U (2x1 / 3) =

4U1

27. (8)

the particle will undergo periodic motion, between the values xa < x < xb < 2x1 / 3, where xa and

xb are the turning points and are solutions to the equation

E =U (x) = !U1

xx1

"

#$%

&'

3

!xx1

"

#$%

&'

2"

#$$

%

&''

. (9)

For E >U (2x1 / 3) =

4U1

27, Eq. (9) has only one solution xa and for all values of x > xa

the kinetic energy K(x) = E !U (x) > 0 which means that the particle can “escape” to infinity but can never enter the region x < xa . For E <U (0) = 0 , the kinetic energy is negative for all values of x i.e.

K(x) = E !U (x) < 0; ! " < x < +" . All regions of space are forbidden.

d) If the particle has speed 0v at 0x = where the potential energy is zero U (0) = 0 , the energy of the particle is constant and equal to kinetic energy

E = K(0) =

12

mv02 . (10)

Therefore

(4 / 27)U1 =

12

mv02 (11)

which we can solve for the speed 0v , v0 = 8U1 / 27m . (12)

Problem 4 A block of mass mb sits at rest on a frictionless table; the block has a circular surface of radius R as shown in the figure. A small cube of mass mc and speed

vc,0 is incident upon the block; the cube moves frictionlessly on the table and the block.

a) What is the maximum height h above the table, attained by the cube (assume vc,0

is small enough that h < R )? b) The cube comes back down the block and when it leaves the block it is moving

along the table in the opposite direction(to the left in the figure above). What is the final speed of the block when the cube is no longer on it? (Hint: think about the energy and momentum conditions.)

Problem 5 One Dimensional Collision

A proton makes a head-on collision with an unknown particle at rest. The proton rebounds straight back with 4 / 9 of its initial kinetic energy. Find the ratio of the mass of the unknown particle to the mass of the proton, assuming that the collision is elastic. Solution: We choose as our system the proton and the unknown particle. Since we know that the collision is elastic the mechanical energy is constant. We also assume that there are no external forces acting so the momentum is constant. The collision is one dimensional so the constancy of momentum and energy provide two equations. In the statement of the problem the following quantities are not known: the mass of the proton 1m , the mass of the unknown particle 2m , the initial speed of the proton 1,0v , and the final speed of the unknown particle, 2, fv . Since we are told that the final kinetic energy of the proton is

1, 1,

2 21, 1 1,0 1

1 4 4 12 9 9 2f ifK m v K m v! "= = = # $% &

, (13)

we can determine the final speed of the proton

1, 1,23f iv v= . (14)

From the momentum condition we should be able to determine the final speed of the unknown particle, 2, fv in terms of the initial speed of the proton 1,0v . Then the speeds of all the objects can be expressed in terms of the initial speed of the proton 1,0v . So the initial speed of the proton 1,0v will cancel out in the energy equation which will then yield the desired ratio of the mass of the unknown particle to the mass of the proton, 2 1/m m .

In the figure below we depict the states of the system before and after the collision.

The x-component of the momentum is constant so 1 1,0 1 1, 2 2,f fm v m v m v= ! + . (15) Now substitute Eq. (14) into Eq. (15) yielding

1 1,0 1 1,0 2 2,23 fm v m v m v= ! + . (16)

We can solve Eq. (16) for the final speed of the unknowen particle in terms of the masses and the initial speed of the proton,

11, 1,0

2

53fmv vm

= . (17)

The kinetic energy is constant in the collision,

2 2 21 1,0 1 1, 2 2,

1 1 12 2 2f fm v m v m v= + . (18)

We first substitute Eq. (13) for the final kinetic energy of the proton into Eq. (18) yielding

2 2 21 1,0 1 1,0 2 2,

1 4 1 12 9 2 2 fm v m v m v= + . (19)

Eq. (19) simplifies to

2 21 1,0 2 2,

59 fm v m v= (20)

Now substitute our expression for the final speed of the unknown particle (Eq. (17)) into Eq. (20) yielding

2 2

2 21 11 1,0 2 1,0 1,0

2 2

55 259 3 9

m mm v m v vm m

! "= =# $

% &. (21)

We can solve Eq. (21) for the ratio of the mass of the unknown particle and the proton

2

1

5mm

= . (22)

Problem 6 Two Dimensional Collision A particle A of mass m is initially moving in the positive x-direction with a speed ,0Av and collides elastically with a second particle B of mass 2m , which is initially at rest.. After the collision the particle A moves with an unknown speed ,A fv , at an unknown angle ,A f! with respect to the positive x-direction. After the collision, particle B moves

with an unknown speed ,B fv , at an angle !B, f = 45o with respect to the positive x-

direction. Find ,A f! .

Solution: We choose as our system the two particles. Since we know that the collision is elastic the mechanical energy is constant. We also assume that there are no external forces acting so the momentum is constant. The collision is two dimensional so the change in each component of the momentum is zero due to the collision. The energy and momentum equations provide three independent equations. There are four unknown quantities, the mass m , and the final speeds of each particle, ,A fv and ,B fv , and the unknown angle

,A f! . Since particle B has twice the mass of particle A , the mass will cancel out of the momentum and energy equations so we should be able to solve for ,A f! . Initial state before the collision: The components of the momentum of the system 0 ,0 ,0A A B Bm m= +p v v! ! ! before the collision are given by

,0 ,0

,0 0.x A

y

p mv

p

=

= (1)

Final state after the collision: The components of the momentum , ,f A A f B B fm m= +p v v! ! ! after the collision are given by

, , , , ,

, , , , ,

cos 2 cos

2 sin sin .x f A f A f B f B f

y f B f B f A f A f

p mv mv

p mv mv

! !! !

= +

= " (2)

There are no any external forces acting on the system, so each component of the total momentum of the system remains constant during the collision, 0x , x , fp p= (3) 0y , y , fp p= . (4) These two equations become

,0 , , , ,

, , , ,

cos 2 cos

0 2 sin sin .A A f A f B f B f

B f B f A f A f

mv mv mvmv mv

! !! !

= +

= " (5)

The collision is elastic; the kinetic energy is the same before and after the collision, 0 fK K= , (6) or

2 2 2,0 , ,

1 1 1 22 2 2A A f B fmv mv mv= + . (7)

We have three equations, two momentum equations and one energy equation, with three unknown quantities, ,A fv , ,B fv and ,A f! since we are already given that and , 45oB f! = . We first rewrite the expressions in Equation (5), canceling the factors of m , as

, , ,0 , ,

, , , ,

cos 2 cos

sin 2 sinA f A f A B f B f

A f A f B f B f

v v vv v

! !! !

= "

=. (8)

We now square each expression in Equation (8) and then add the resulting expressions, yielding

vA, f

2 (cos2!A, f + sin2!A, f ) = (vA,0 " 2vB, f cos!B, f )2 + 4vB, f2 sin2!B, f . (9)

We can use the identities 2 2

, ,cos sin 1A f A f! !+ = and 2 2, ,cos sin 1B f B f! !+ = to simplify

Equation (9), yielding

vA, f

2 = vA,02 ! 4vA,0vB, f cos"B, f + 4vB, f

2 . (10) Substituting Equation (10) into Equation (7) yields

12

mvA,02 =

12

m(vA,02 ! 4vA,0vB, f cos"B, f + 4vB, f

2 ) +12

2mvB, f2 . (11)

Equation (11) simplifies to

0 = !4mvA,0vB, f cos"B, f + 6mvB, f

2 , (12) which may be solved for the final speed of object B ,

, ,0 ,2

cos3B f A B fv v != . (13)

Divide the second expression in Equation (8) by the first expression in Eq. (8), yielding

, , , ,

, , ,0 , ,

sin 2 sincos 2 cos

A f A f B f B f

A f A f A B f B f

v vv v v

! !! !

="

. (14)

Equation (14) simplifies to

, ,,

,0 , ,

2 sintan

2 cosB f B f

A fA B f B f

vv v

!!

!=

". (15)

We now substitute Eq. (13) into Eq. (15) yielding

,0 , ,

,

,0 ,0 , ,

22 cos sin3tan22 cos cos3

A B f B f

A f

A A B f B f

v

v v

! !!

! !

" #$ %& '=

" #( $ %& '

. (16)

Canceling the initial velocity of particle A in Eq. (16) yields

, ,

,

, ,

4 cos sin3tan41 cos cos3

B f B f

A f

B f B f

! !!

! !

" #$ %& '=" #($ %& '

. (17)

We now use the given information that , 45B f! = ! and hence sin 45 cos 45 2 / 2= =! ! in Eq. (17) which becomes

,2 / 3tan 21 2 / 3A f! = ="

. (18)

Hence the angle that particle A moves with the x-axis after the collision is given by 1

, tan 2 63.4A f! "= = ! . (19)

Problem 7: Comet Flyby of Jupiter A comet of mass mc is initially in the Oort Cloud which is a distance far enough away from the sun that you may approximate it as infinity. Due to random motions in the cloud, the comet picks up a negligibly small initial velocity but it now moves out of the cloud and heads directly towards the sun of mass ms . The planet Jupiter of mass

mj is directly in the way of the comet.

Assume Jupiter is in a nearly circular orbit around the sun with radius ,s jr and has tangential speed

v j , as measured in a reference frame at rest with the sun. When the comet reaches the orbit

of Jupiter, the comet has a speed vc,i = 2v j directed towards the sun. Jupiter’s gravity changes

the direction of the comet’s path by 90o until the comet is moving in a direction opposite of Jupiter. As seen by an observer moving with the planet Jupiter, the comet changes its direction

but not the magnitude of its velocity vc! .

Reference frame at rest with respect to sun:

before encounter after encounter

a) As seen in a reference frame moving with the planet Jupiter, what is the magnitude of

comet’s velocity vc! before the encounter? Express your answer in terms of jv .

b) As seen in a reference frame at rest with the sun, what is the comet’s speed

vc, f after it’s flyby of Jupiter? Did it speed up, stay the same speed, or slow down? Express your answer in terms of the initial speed of the comet

vc,i = 2v j .

Solution: We have two reference frames: the reference frame fixed to the sun and a reference frame fixed to Jupiter moving with velocity

!V = v j j with respect to the sun’s frame where we

have chosen unit vectors as shown in the figures below.

We know that from the Law of Addition of Velocities that the velocity of the comet in the reference frame moving with Jupiter is equal to

! !vc,i =

!vc,i "!V = vc,i i " v j j = 2v j i " v j j

So the magnitude of comet’s velocity !

vc,i before the encounter in Jupiter’s reference frame is

!vc,i =

! !vc,i = !vc,i "!V = vc,i i " v j j = ( 2v j )

2 + v j2 = 3v j .

In Jupiter’s reference frame the encounter is an elastic collision (we are ignoring the gravitational interaction between the comet-Jupiter system and the sun) and hence the relative speed remains unchanged. Since Jupiter is at rest, this means that after the encounter,

! !vc, f = ! !vc,i = 3v j

Therefore we can again use the Law of Addition of Velocities to find the velocity of the comet after the encounter in the sun’s reference frame

!vc, f =

! !vc,i +!V = " 3v j j+ v j j = v j (1" 3) j .

So the speed of the comet is

vc, f = ( 3 !1)v j < 2v j .

The comet has slowed down.

Problem 8 Challenge A thin target of lithium is bombarded by helium nuclei of energy 0E . The lithium nuclei are initially at rest in the target but are essentially unbound. When

a helium nucleus enters a lithium nucleus, a nuclear reaction can occur in which the compound nucleus splits apart into a boron nucleus and a neutron. The collision is inelastic, and the final kinetic energy is less than 0E by 2.8 MeV . ( 6 -131MeV=10 eV=1.6 10 J! ). The relative masses of the particles are: helium, mass 4 ; lithium, mass 7 ; boron mass 10 ; neutron, mass 1. The reaction can be symbolized

7 4 10 1Li + He B n 2.8 MeV! + " .

a) The minimum initial kinetic energy necessary for the reaction to take place is called the threshold energy, 0,thresholdE . What is 0,thresholdE for which neutrons can be produced? What is the energy of the neutrons at this threshold?

b) Show that if the incident kinetic energy falls in the range

0,threshold 0 0,threshold 0.27 MeVE E E< < + ,

the neutrons ejected in the same direction as the incoming helium (forward direction) do not all have the same energy but must have one or the other of two possible energies. By considering the reaction in a reference frame moving with the velocity of the center of mass of the system, explain why there must be two distinct energies.