ABRHS PHYSICS (H) NAME: ____________________ Unit 5: Momentum

2015-16

Text: Chapter 9 Problems (p. 229-240) #1: 18, 20, 27, 31, 37 (momentum & impulse) #2: 40, 42, 45, 46, 100 (conservation of momentum) #3: 49, 103, 123, 129 (collisions) Vocabulary: momentum, impulse, center of mass, conservation of momentum, elastic collision, inelastic collision Math: definitions:

p = m v r =

mixi∑M

= 1M

xdm∫

derived formulas:

J =

Fdt∫ = (

Favet)

v1 f = m1−m2

m2 +m1( ) v1i +2m2

m2 +m1( ) v2i v2 f = 2m1

m2+m1( ) v1i +m2−m1( )m2+m1( ) v2i

skills: no new math skills Objectives: You should be able to do the following: • Explain what is meant by the center of mass and why it is important - especially in regards to

Newton's Laws and momentum. • calculate the center of mass for a collection of masses. • Explain the concept of conservation of momentum. Use examples to support your explanation. • Explain what is meant by the term “impulse” and how it relates to momentum. Use examples to

support your explanation. • Compare and contrast the phrases “elastic collision” and “inelastic collision.” Use examples to

support your explanation. • Calculate impulse (or change in momentum) given an appropriate graph and information. • Do problems involving the following ideas:

♦ an object (or objects stuck together), perhaps moving, explodes into two or more pieces with different masses and velocities.

♦ an object bouncing off a wall experiences an impulse (and change in momentum) ♦ two objects collide (either elastically or inelastically) in one dimension and have new

velocities after the collision. ♦ two objects collide (either elastically or inelastically) in two dimensions, and have new

velocities after the collision.

ABRHS PHYSICS (H) NAME: ________________

Center of Mass

side 1

Your book defines and shows the center of mass equations in the very beginning of chapter 9, but I would like to develop the ideas based off of our lab experience, and hopefully it will make a little more conceptual sense. Imagine two different masses separated by some distance. Somewhere in between is their "center of mass." Let's pragmatically define it as the point where they would be balanced - like we did with the introduction to torques lab. The diagram below shows two masses - and the "x" in the middle is the center of mass.

m1 m2

d2d1 If the "x" is the center of mass, then the following would be true:

m1d1 = m2d2 That is the most basic way to think about the center of mass. It's just the weighted center of the masses. If you suspend an object from its center of mass, it will be balanced. (In a uniform gravitational field the center of mass is also the center of gravity.) To show how this is the same as the book definition, let's just draw the same thing, but shifted along the horizontal axis. In the diagram below, r is the vector that points from the origin to the center of mass, x1 and x2 are the distances to the two masses.

m1 m2

d2d1

r

x1

x2

x=0

Looking at the diagram, let's replace the d1 with r - x1 and d2 with x2 - r. Then we get

m1(r − x1) = m2 (x2 − r) Which we rearrange to get

(m1 +m2 )r = m2x2 +m1x1

r = m2x2 +m1x1m1 +m2

Your textbook just gives the above as the basic definition for the center of mass between two particles. This can then be generalized for lots of masses as (using M for the total mass)

r =mixi∑M

We generalize that a little more into three dimension by simply doing the above three different times - one for each dimension. If we wanted to talk about an infinite number of infinitely small masses spread around, the above summation would be treated as an integral:

ABRHS PHYSICS (H) NAME: ________________

Center of Mass

side 2

r = 1M

xdm∫

In three dimensions we would just treat each dimension independently - so we would have three different summations (or integrals), one for each dimension. Why is the center of mass important? Basically, when we do Newton's Laws involving large objects, the Second Law applies to the center of mass. Add up all the external forces acting on an object as if they were acting on the center of mass. This is what we did earlier in the year - and it will still apply in the future when we look at rotation. Questions: 1. Four point masses are arranged as shown in the diagram. Notice that the origin passes through

one of the masses labeled "2m." Where is the center of mass? (Worded another way, what is the center of mass vector?)

a

b

2m

m

3m

2b

2m x

y

2. Three identical, uniform masses each of mass m and length a are arranged as

three sides of a square as shown in the diagram. Where is the center of mass? 3. Imagine that there are a bunch of different masses arranged on the x-axis, on either side of the

origin. After doing r =mixi∑M

for all the masses, you get r = 0. What does that mean?

m, a

ABRHS PHYSICS (H) NAME________________

Lab 9-1: Impulse

Purpose: 1. To compare the impulse exerted on a cart and compare it to the change in momentum of the cart.

Equipment: 1 cart w/plunger 2 500 gram bars motion detector, force probe Procedure: l. Arrange a cart, force probe and motion detector on a level track as shown. Open up Logger Pro.

force probe(held firmly)

motion detectorcart (with plunger)

2. Hold the force probe firmly in place and zero the probe. Start collecting data – when you hear

the motion detector clicking, give the cart a quick shove so that it ends up colliding and bouncing off the force probe.

3. From the velocity vs time graph, record the velocity of the just before and after the collision by clicking on the “Examine” button.

4. From the force vs. time graph, determine the total impulse acting on the cart and record this in the data table below. (To determine the area under a curve in Logger Pro, click on the integrate button or under Analyze, choose Integral.)

5. Repeat the above, but give the cart a bigger initial speed. (You may have to switch the force probe to read 50 N - in which case you need to restart Logger Pro.)

6. Repeat the above two more times, each time adding one of the 500 gram black bars. Calculations: 1. Calculate the change in momentum of the cart for each trial. Show your work here, and record

your results in the table. 2. How did you determine the impulse acting on the cart by the spring? Data and Results:

trial mass (kg)

vi (m/s)

vf (m/s)

Impulse (N•s)

∆p (kg•m/s)

no bar 0.5 no bar 0.5 1 bar 1.0 2 bars 1.5

Conclusions: 1. In general, how did the impulse of the spring compare to the change in momentum of the cart? 2. For each trial, what happened to the kinetic energy of the cart?

ABRHS PHYSICS (H) NAME: _________________

Momentum & Impulse

side 1

1. A 5.0 kg object with a speed of 30 m/s strikes a steel plate at an angle of 45º and rebounds with the same speed and angle. What is the change (magnitude and direction) of the linear momentum of the ball (relative to the steel plate.)

2. A 3 kg ball with an initial velocity of 15i m/s experiences an impulse of –20i + 30j Ns over a time

interval of 0.04 seconds. What is the final velocity of the ball? 3. A super ball (mass = 0.035 kg) is thrown with a velocity of 15 m/s into a wall. It bounces back

with a speed of 10 m/s. The ball and wall were in contact for only 0.02 seconds. What was the average force of the wall on the ball?

4. A 1500 kg car traveling at 10 m/s somehow

experiences a net force as shown in the diagram. (The force is in the direction that the car was moving.) What is the final speed of the car?

5. A tennis ball of mass 0.1 kg bounces off a wall. It had an original speed of 30 m/s and bounces

straight backwards with a speed of 20 m/s. The force verses time graph showing the impulse of the wall on the tennis ball is shown. What was the maximum force on the tennis ball?

t (ms)10 20 30

Fmax

(N)

Answers: 1) 212 kg•m/s, ⊥ and away from surface 2) 8.3i + 10j m/s 3) (-) 43.75 N 4) 20 m/s 5) (-) 333 N

t (s)4 8

F (N)

3000

2000

1000

ABRHS PHYSICS (H) NAME: ___________________

Lab 9-2: Explosions

side 1

Purpose: To determine if momentum and kinetic energy are conserved in an explosion. Materials: 1 track 1 cart with plunger 1 cart with no plunger 2 500 gram bars 2 motion detectors Procedure:

A B

1. As always, make sure the track is level on the lab table. 2. Start up Logger Pro and open the file "18 Momentum Energy Coll." The motion detector

plugged into “Dig 1” is Position 1 (the red line) and the motion detector plugged into “Dig 2” is Position 2 (the blue line).

3. On the appropriate cart, push in the plunger so that it locks in place. Push it in and lift it up a little so that the notch in the plunger hooks on the cart. (Ask your teacher for help if needed. This is surprisingly difficult.)

4. Place both carts on the track. Start recording data by clicking on “Collect”. Using a meter stick, tap the little knob on the cart to release the spring plunger, pushing the carts apart. Record the velocities for both carts after the spring is released. Include any negative signs!

5. Place a 500 gram mass on the non-spring cart, and repeat. 6. Place the second 500 gram mass on the same cart and repeat. 7. Calculate the momentums and kinetic energies in the rest of the table. Humor me and write

the equations for momentum and kinetic energy below:

Data and Results:

Mass (kg)

Initial Velocity

(m/s)

Final Velocity

(m/s)

Initial Momentum

(kg•m/s)

Final Momentum

(kg•m/s)

Initial Kinetic Energy

(J)

Final Kinetic Energy

(J)

Cart A 0.5 0

Cart B 0.5 0

Totals --- --- --- Cart A 0.5 0

Cart B 1.0 0

Totals --- --- --- Cart A 0.5 0

Cart B 1.5 0

Totals --- --- ---

ABRHS PHYSICS (H) NAME: ___________________

Lab 9-2: Explosions

side 2

Conclusions: 1. In terms of Newton’s Third Law, why do the two carts go in opposite directions? 2. What was true about the initial momomentum of the two carts before the explosion? Why does

this make total sense? 3. In general, compare the momentum of cart A to the momentum of cart B after each explosion. 4. What was true about the total momentum of the carts after the explosion? 5. What is meant by the phrase "conservation of momentum?" How can this apply when nothing

was moving, and then both carts were moving after the explosion? 6. Was kinetic energy conserved in the explosion? How about energy in general? Explain. 7. The momentums of each cart were (hopefully) always equal and opposite. What was true about

the kinetic energies of the two carts? 8. What was true about the total kinetic energies of the carts each trial? Also explain why your

answer makes sense. 9. Imagine there is an explosion between two carts, A and B. A has more mass than B.

a. Which will experience a greater change in momentum? b. Which will experience a greater impulse? c. Which will experience a greater change in velocity? d. Which will experience a greater force? e. Which will experience a greater acceleration?

ABRHS PHYSICS (H) NAME: ______________ Momentum & Recoil

Answers: 1. 1.7 m/s 2. recoil=250 m/s! 3. A:B = 3:2 4. 1.46 m/s 5. 7.7 m/s 6) 37,500 m

1. A 1.5 kg gun can fire a 0.005 kg bullet with a speed of 500 m/s. What is the “recoil speed” of the gun?

2. Imagine someone invented a super-light gun that has a mass of only 0.010 kg, yet could still fire

a 0.005 kg bullet with a speed of 500 m/s. Why would this gun be really dangerous for the person firing the gun?

3. Two astronauts are floating in space. One of them pushes the other, sending astronaut A to the

left at 0.5 m/s and the astronaut B to the right at 0.75 m/s. What is the ratio of their masses? 4. A 70 kg person is at rest in a 40 kg canoe by a dock. In a horrible attempt to get out of the canoe,

the person tries to jump forward, sending the canoe backward. If the relative speed of the person with respect to the canoe is 4 m/s, what is the speed of the person with respect to the dock?

5. A 7 kg object is sliding along a frictionless surface with an initial speed of 5 m/s. For some

reason, it explodes into 2 pieces. A 4 kg piece of it is now only sliding with a speed of 3 m/s. How fast is the other piece traveling?

6. A projectile is fired from a gun at an angle of 45º with the horizontal and with a muzzle velocity

of 500 m/sec. At the highest point in its flight, the projectile explodes into two fragments of equal mass. One fragment, whose speed immediately after the explosion is zero, falls vertically down. How far from the gun does the other fragment land, assuming a level terrain and ignoring air resistance?

ABRHS PHYSICS (H) NAME: ___________________

Lab 9-3: Inelastic Collisions

side 1

Purpose: To determine if momentum and kinetic energy are conserved in a collision in which two objects stick to each other after the collision.

Materials: 1 track 2 carts 1 500 gram bar Procedure:

velcro sides

this car at rest

BA

1. Arrange the empty carts so that the velcro ends face each other as shown in the diagram.

Start the motion detector and give the cart a little push. Make sure the resulting graph shows the velocity both before and after the collision.

2. Record the velocity of cart A alone, and the velocity of the carts combined. (Probably easiest to do with the "Examine" tool and inspecting the velocity graph just before and after the collision.)

3. Place a 500 gram bar in cart B (the target cart) and repeat. 4. Place the 500 gram bar in cart A instead and repeat. 5. Calculate the momentums and kinetic energies in the rest of the table:

Data and Results:

Mass (kg)

Initial Velocity

(m/s)

Final Velocity

(m/s)

Initial Momentum

(kg•m/s)

Final Momentum

(kg•m/s)

Initial Kinetic Energy

(J)

Final Kinetic Energy

(J)

Cart A 0.5

Cart B 0.5 0

Totals --- --- --- Cart A 0.5

Cart B 1.0 0

Totals --- --- --- Cart A 1.0

Cart B 0.5 0

Totals --- --- --- Conclusions: 1. Was momentum conserved in these collisions? Explain. 2. Was kinetic energy conserved in these collisions? Explain.

ABRHS PHYSICS (H) NAME: ___________________

Lab 9-3: Inelastic Collisions

side 2

Follow-up Questions: 1. A 2 kg cart traveling at 4 m/s crashes and sticks to a 1 kg cart initially at rest. How fast are the

two carts going when they are stuck together? 2. A 2 kg cart traveling at 2.5 m/s crashes and sticks to a 3 kg cart initially at rest. How fast are

the two carts going when they are stuck together? 3. A 2 kg cart crashes and sticks to a 1 kg cart initially at rest. If they are going at 2 m/s when they

are stuck together, how fast was the first cart going by itself? 4. A 2 kg cart crashes and sticks to a 3 kg cart initially at rest. If they are going at 1.75 m/s when

they are stuck together, how fast was the first cart going by itself? 5. A cart traveling at 3.2 m/s collides and sticks to a 1.2 kg cart that is initially at rest. The two

carts have a speed of 2.5 m/s right after the collision. What was the mass of the first cart? 6. It turns out that any time two objects collide and stick, kinetic energy is lost. Why? How can

energy be conserved then? 7. Imagine a poor moth is floating in the air on the highway. A car comes along and collides with

the moth, sadly resulting in the moth being smeared on the windshield of the still moving car. a. Who experienced a greater change in velocity? b. Who experienced a greater acceleration? c. Who experienced a greater force? d. Who experienced a greater impulse? e. Who experienced a greater change in momentum?

ABRHS PHYSICS (H) NAME________________

Lab 9-4: Elastic Collisions

side 1

Purpose: 1. To examine the instantaneous momentum and kinetic energy during an elastic collision between two carts.

2. To determine if momentum / kinetic energy are conserved in an elastic collision. Materials: 1 track 2 carts (one of each) two 500 gram bars 2 motion detectors Procedure:

this cart at rest

A B

magnets

1. Make sure the track is level and place one of the carts in the middle of the track. Place the

second cart at one end of the track. Place the two motion detectors at either end of the track. (see diagram above.) Make sure that the magnets in the carts are facing each other, so that the carts bounce without touching.

2. Start Logger Pro and open the file "18 Momentum Energy Coll." 3. Start the motion detectors and give cart A a small push so that it “crashes” into cart B.

Make sure the carts don’t physically touch – the collision should all happen through the magnetic fields. Record the velocities of the carts both before and after the collision. Keep the signs!

4. Place two 500 gram bars in cart B (the target cart) and repeat. 5. Place two 500 gram bars in cart A instead and repeat. 6. For the last trial, give both carts a push towards each other. You can do what you want with

the masses – just make sure that if the masses are the same you don’t push them with the same speed.

7. By creating new calculated columns as needed, make the following two graphs for the last trial. Check with your teacher before printing.

I) pA, pB and ptotal vs time. and II) KA, KB and Ktotal vs time.

Data and Results:

Mass (kg)

Initial Velocity

(m/s)

Final Velocity

(m/s)

Initial Momentum

(kg•m/s)

Final Momentum

(kg•m/s)

Initial Kinetic

Energy (J)

Final Kinetic

Energy (J)

Cart A 0.5

Cart B 0.5 0

Totals --- --- ---

Cart A 0.5

Cart B 1.5 0

Totals --- --- ---

Cart A 1.5

Cart B 0.5 0

Totals --- --- ---

ABRHS PHYSICS (H) NAME________________

Lab 9-4: Elastic Collisions

side 2

Cart A

Cart B

Totals --- --- --- Conclusions: 1. In general, how did the total momentum before the carts crashed and bounced compare to the

total momentum after the carts crashed? 2. In general, how did the total kinetic energy before the carts crashed and bounced compare to the

total kinetic energy after the carts crashed? 3. What happened to momentum during the collision in the last trial? (Look at the graph you

printed in step 7.) Can you say that it was conserved? 4. What happened to kinetic energy during the collision in the last trial? (Look at the graph you

made in step 7.) Can you say that it was conserved? 5. Compare and contrast an inelastic collision with an elastic collision.

ABRHS PHYSICS (H) NAME: ______________ Momentum & Collisions

side 1

1. A 1500 kg car rear-ends a 2000 kg at a stop sign. Immediately after the collision, the two cars have a speed of 13 m/s. (They then skid to a stop, but that is not part of this problem.) How fast was the 1500 kg car moving just before the collision?

2. Maya and Miguel are on the bumper cars and have a head-on collision. Maya (total mass 300 kg)

is traveling to the right at 5 m/s. Miguel (250 kg) is traveling to the left at 3 m/s. After the collision, Maya is only traveling at 1 m/s, but still to the right. What is the final velocity of Miguel?

3. A stream of 40 gram bullets, fired horizontally with a speed of 1000 meters per second, strikes a

10 kg wooden block that is free to move on a horizontal frictionless tabletop. What is the speed of the block after it has absorbed 15 bullets?

4. A 0.5 kg ball is traveling with a speed of 7 m/s when it collides with a 0.4 kg ball traveling in the

opposite direction with a speed of 3 m/s. After the collision, the first ball (0.5 kg) is traveling with a speed of 1 m/s, in the same direction that it was before. a. What is the speed and direction of the second ball after the collision? b. Which ball experienced the greater change in momentum? change in velocity? impulse? c. If the collision lasted for 0.25 seconds, what was the force on the second ball?

5. Two cars approach a 90º intersection. Neither driver is paying attention to what they are doing,

and they collide. After the collision, the cars stick together, and skid to a stop in 14.8 meters at an angle as shown. The two cars have masses of 1300 kg and 1600 kg, as shown. You happen to know that the coefficient of friction between the tires and the road was µ = 0.3. How fast were the drivers going just prior to the collision?

m = 1300 kg

m = 1600 kg28 degrees

the two cars skid to stop along this line, distance of 14.8 meters

(initially traveling straight up)

(initially traveling to the left)

ABRHS PHYSICS (H) NAME: ______________ Momentum & Collisions

side 2

6. A ball of mass 0.25 kg has a one-dimensional elastic collision with a ball of mass 0.35 kg. After the collision the smaller ball has a velocity of 6.5 m/s and the larger ball has a velocity of 9 m/s. What was the velocity of each ball before the collision?

The 7 to 9 all use the same diagram (but the numbers are different for each problem.)

m1 m2

v1

v2

1

2

vipre-collision post-collision

7. A ball of mass 0.2 kg is traveling horizontally with a speed of 3.5 m/s. It collides with a ball of

0.4 kg, initially at rest. After the collision, the first ball is traveling with a velocity of 3 m/s at an angle of 30º above the original direction of motion. What is the velocity of the other ball?

8. m1 = 0.25 kg and m2 = 0.15 kg. The speeds after the collision are v 1 = 4 m/s and v2 = 6 m/s. If θ1

= 30º. a. Was the collision elastic? (Hint: Find the initial speed first.)

b. What was the impulse on m1? 9. Both masses are 0.25 kg. The initial speed of m1 was 5 m/s and the collision was elastic. If θ2 =

30º, how fast were the masses going after the collision? Answers: 1. a) 30.3 m/s 2) 1.8 m/s 3) 56.6 m/s 4. a) 4.5 m/s b) same, 0.4 kg, same c) (-) 12 N 5) 1600 kg: 15 m/s; 1300 kg: 9.9 m/s 6) 9.5 & 6.9 m/s 7) 0.45i – 0.75j m/s 8. a) 6.5 m/s; not elastic b) –0.76i + 0.5j Ns 9) v1=2.5 m/s; v2=4.3 m/s

ABRHS PHYSICS NAME: ________________

Elastic Collisions

side 1

Energy and momentum are always conserved in a collision, no matter what happens. Momentum iseasy to deal with because there is only “one form” of momentum, (p=mv), but you do have toremember that momentum is a vector. Energy is tricky because it has many forms, the mosttroublesome being heat, but also sound and light. If kinetic energy is conserved in a collision, it iscalled an elastic collision. In an elastic collision, the total kinetic energy is conserved because theobjects in question “bounce perfectly” like an ideal elastic. An inelastic collision is one where someof the of the total kinetic energy is transformed into other forms of energy, such as sound and heat.Any collision in which the shapes of the objects are permanently altered, some kinetic energy isalways lost to this deformation, and the collision is not elastic. It is common to refer to a“completely inelastic” collision whenever the two objects remain stuck together, but this does notmean that all the kinetic energy is lost; if the objects are still moving, they will still have somekinetic energy.

General Equation Derivation: Elastic Collision in One Dimension

Given two objects, m1 and m2, with initial velocities of v1i and v2i, respectively, how fast will they begoing after they undergo a completely elastic collision? We can derive some expressions for v1f and v2f

by using the conservation of kinetic energy and the conservation of momentum, and a lot of algebra.

Begin by making the following conservation statements:

Conservation of Kinetic Energy: 1

2m1v1i

2+ 1

2m2v2i

2= 1

2m1v1f

2+ 1

2m2v2 f

2

Conservation of Momentum: m1v1i +m2v2i = m1v1 f +m2v2 f

To solve for v1f and v2f (which is really two equations in two unknowns), we need some algebratricks to simplify the substitutions. Take both equations and group them according to the masses:put all the m1’s on one side of the equation and all the m2’s on the other. We’ll also cancel out allthe 1/2’s at this point.

Conservation of Kinetic Energy becomes: m1v1i

2−m1v1f

2= m2v2 f

2−m2v2i

2

which can be simplified as m1 v1i

2− v1f

2( ) = m2 v2 f2− v2 i

2( ) eqn. 1

Conservation of Momentum becomes: m1v1i −m1v1f = m2v2 f −m2v2i

which can be simplified as m1 v1i − v1f( ) = m2 v2 f − v2 i( ) eqn. 2

Now comes the algebra fun. Divide equation 1 by equation 2.

m1 v1i2−v1 f

2( )m1 v1i −v1 f( )

=m2 v2 f

2−v2i

2( )m2 v2 f −v2i( )

After all the cancellations, we are left with: v1i + v1f = v2 f + v2i eqn. 3

Solving for v1f we get: v1 f = v2 f + v2i − v1i eqn. 4

Now we take equation 4 and substitute back into one of our original equations to solve for v2f. Sincethe momentum equation is easier, lets use that.

Conservation of Momentum becomes: m1v1i +m2v2i = m1(v2 f + v2i − v1i) + m2v2 f

ABRHS PHYSICS NAME: ________________

Elastic Collisions

side 2

Now do some algebra... m2v2 f +m1v2 f = m1v1i +m2v2i −m1 v2i − v1i( )

m2 +m1( )v2 f = m1v1i +m1v1i + m2v2i −m1v2i

v2 f =

m1+m1( )m2+m1( )

v1i +m2−m1( )m2+m1( )

v2i

Until we get: v2 f =

2m1

m2+m1( )v1i +

m2−m1( )m2+m1( )

v2i

Now we substitute this result back into equation 4 do some algebra to solve for v1f.

Equation 4 becomes: v1 f =

2m1

m2 +m1( )v1i +

m2 −m1( )m2 +m1( )

v2i

+ v2i − v1i

Now do some algebra.... v1 f =

2m1

m2 +m1( )− 1

v1i +

m2 −m1( )m2 +m1( )

+ 1

v2i

v1 f =

2m1− m2 +m1( )m2 +m1( )

v1i +

m2−m1( )+ m2 +m1( )m2 +m1( )

v2i

Until we get: v1 f =

m1−m2

m2 +m1( )v1i +

2m2

m2 +m1( )v2i

Questions:

1. What would happen if an object were to have a completely elastic collision with an identicalobject initially at rest?

2. What would happen if an object were to have a completely elastic collision with an identicalobject not initially at rest?

3. What would happen if a really small object were to collide with a really massive object initiallyat rest? (i.e. m2 >> m1)

4. What would happen if a really massive object were to collide with a really small object atinitially at rest? (i.e. m2 << m1)

5. Imagine holding a really light object on top of a really massive object, and then dropping both ofthem at the same time onto the ground. If all the collisions are elastic, and the objects aredropped from a height of h, how high will the little object bounce?

6. In the previous question, what should be the ratio of the masses if the bottom mass had novelocity after all the collisions? How high would the little mass bounce?