Question 1 Question 2 Question 3 Question 4 Question 5 Question 6 Question 7 Question 8 Question 9

You are the technical advisor for the David Letterman Show. Your task is to design a circus stunt in which Super Dave, who weighs 750 N, is shot out of a cannon that is 40o above the horizontal. The “cannon” is actually a 1m diameter tube that uses a stiff spring to launch Super Dave. The manual for the cannon states that the spring constant is 1800 N/m. The spring is compressed by a motor until its free end is level with the bottom of the cannon tube, which is 1.5m above the ground. A small seat is attached to the free end of the spring for Dave to sit on. When the spring is released, it extends 2.75m up the tube. Neither the seat nor the chair touch the sides of the 3.5m long tube, so there is no friction. After a drum roll, the spring is released and Super Dave will fly through the air. You have an airbag 1m thick for Super Dave to land on. You know that the airbag will exert an average retarding force of 3000 N in all directions. You need to determine if the airbag is thick enough to stop Super Dave safely – that is, he is slowed to a stop by the time he reaches ground level. Consider the seat and spring to have negligible mass. Ignore air resistance.

• Question 1• Question 2• Question 3• Question 4• Question 5• Question 6• Question 7• Question 8• Question 9• Question 10

• Question 11• Question 12• Question 13• Question 14• Question 15• Question 16• Question 17• Question 18• Question 19• Paired Problem 1

• A : i only• B : ii only• C : iii and iv only• D : all of the above

1. Which of the following physics principles are most suited to solve this problem? i) Kinematical considerations

ii) Linear momentum conservationiii) Mechanical energy conservationiv) Work-Energy Theorem

This is a tedious way because it involves vectors. Also, it only applies to the

“projectile motion” part of the problem.

Choice: A

Incorrect

There is only a collision at the airbag. We need to analyze other parts of the

problem. Considering the energy of the system at different points will be more

helpful.

Choice: B

Incorrect

Using these principles makes the problem easier to solve. The work-energy theorem is useful in situations

where you need to relate a body’s speed at two different points in its motion. The energy approach is useful

when a problem includes motion with varying forces along a curved path. However, conservation of total mechanical energy requires that only conservative

forces do work.

Choice: C

Correct

All of these principles are applicable. We are asked to find the easiest way. Since we are not dealing with time

explicitly, we should be applying the work-energy theorem and conservation of

mechanical energy.

Choice: D

Incorrect

2. Which form of energy does the spring-Dave-earth system possess just before the spring is released? Assume the reference height is the floor.

• A : i only• B : ii only• C : iii only• D : ii and iii only

i) Kinetic Energy (K)ii) Spring Potential Energyiii) Gravitational Potential Energy

There is no kinetic energy before the spring is released because

Super Dave has zero initial velocity (vo=0).

Choice: A

Incorrect

This is one type of potential energy (PE) associated with this system, but there is another type of PE associated with this system as well. Remember that Super Dave is displaced vertically before the

spring is released.

Choice: B

Incorrect

This is one of the forms of potential energy (PE) associated with this system, but there is another type of PE associated with this system as well. Remember that

the spring is initially compressed.

Choice: C

Incorrect

The system has gravitational and spring potential energy, because Super Dave

begins above the floor (where PE=0) and

the spring is initially compressed.

Choice: D

Correct

3. Which of the following conditions are required for the Law of Conservation of Mechanical Energy to hold for a system? -U is elastic potential energy. -The subscripts i and f stand for initial and final.                                      -W is work done by non-conservative forces                                     -E is total mechanical energy.

• A: The work done by non-conservative forces must be zero.• B: Energy is not created or destroyed, but can change forms. • C:

• D: Ef=Ei or (Uf+Kf=Ui+Ki)

€

ΔU+ΔK =0

The other choices are either statements of conservation of total mechanical

energy (not the requirements) or are incorrect.

Choice: A

Correct

This is how the Law can be expressed in words, but there are

more requirements.

Choice: B

Incorrect

Choice: C

Incorrect

This is the mathematical expression of the Law, but there

are more requirements.

This is the same as choice C, just in a different form. There are

more requirements.

Choice: D

Incorrect

4. Since the gravitational and elastic forces are conservative, and we ignore air resistance, we can apply the Law of Conservation of Mechanical Energy to this system. How do we (mathematically) express the initial mechanical energy of the system before the spring is released?

A: Ei=mgh

C:

D: Ei=0

B:

€

Ei =12 ⎛ ⎝ ⎜

⎞ ⎠ ⎟kΔx2

€

Ei =mgh+12 ⎛ ⎝ ⎜

⎞ ⎠ ⎟kΔx2

h=height

Δx=compression of the spring

k=spring constant

g=gravitational acceleration (9.8m/s2)

m= Dave’s mass

Think about the spring, which also adds to the initial

mechanical energy since it is compressed.

Choice: A

Incorrect

What about the gravitational potential energy? Recall that the system begins above the floor.

Choice: B

Incorrect

Gravity and the compressed spring contribute to total mechanical

energy. These forms of energy are expressed correctly.

Choice: C

Correct

Choice: D

Incorrect

Kinetic energy is the only type of mechanical energy that is initially

zero.

5. What happens to the energy that was stored in the spring right after the spring is released and Super Dave is launched?

• A: It transforms to gravitational PE only.• B: It transforms to K only.• C: Some of it transforms to gravitational PE and

some to K.• D: It is lost.

Choice: A

Incorrect

Dave gains kinetic energy as the spring is released, because he gains a non-zero velocity.

Recall : where v is Super Dave’s velocity.

€

K =12 ⎛ ⎝ ⎜

⎞ ⎠ ⎟mv2

Since Super Dave is shot at an angle above the horizontal, he gains

height which increases his gravitational PE (mgh).

Choice: B

Incorrect

Choice: C

Correct

The elastic PE transforms partly to gravitational PE and partly to kinetic energy, because Super Dave gains height and velocity.

Choice: D

Incorrect

Since we are ignoring air resistance, there is no non-conservative force

involved. Therefore, there is no energy loss.

6. Dave’s energy right at the instance of impact with the airbag consists of which form of energy?:

• A: only K• B: only gravitational PE• C: both gravitational PE and K• D: none of the above

Choice: A

Incorrect

The top of the airbag is 1m above the floor, so there should be some

gravitational potential energy at the point where Super Dave first makes contact

with the airbag.

Choice: B

Incorrect

If Super Dave had no kinetic energy, we wouldn’t have to worry about him getting hurt. Super Dave is traveling with a non-zero velocity. Therefore, he has potential

energy.

Choice: C

Correct

Dave’s energy at this point consists of both kinetic energy and gravitational PE, because he is traveling with a non-zero velocity and is still displaced vertically.

Choice: D

Incorrect

He has both kinetic energy and gravitational PE. Remember that he is

traveling with a non-zero velocity and is not quite at floor level at this point.

7. Can we still use the principle of mechanical energy conservation after Dave hits the air bag?

• A: yes• B: no• C: we don’t have enough information to decide

Since the airbag softens Dave’s landing, there is a retarding force, which is a non-conservative force. Therefore, we can not

use the principle of conservation of mechanical energy.

Choice: A

Incorrect

The retarding force (non-conservative force), which comes from the airbag resisting Dave’s motion, does work. Thus, we can not use this principle.

Choice: B

Correct

We do have enough information. The airbag causes a non-conservative

force to do work on Dave. This retarding force disallows the use of the principle of

mechanical energy conservation.

Choice: C

Incorrect

8. After the impact with the airbag, which physics principles should we use to solve the problem? • A: kinematics• B: work-energy theorem• C: Impulse-Momentum Theorem• D: all are possible ways

Kinematical considerations will not be useful here.

Choice: A

Incorrect

Choice: B

Correct

Using the work-energy theorem will be the best

method to solve the problem.

Choice: C

Incorrect

This method is not useful due to insufficient information.

Kinematics and the impulse-momentum theorem will not be helpful here. This problem can be solved by the work-

energy theorem.

Choice: D

Incorrect

9. Let’s put the concepts that we have considered in questions 1-8 into mathematical form.

Which of the following equations correctly describes our application of conservation of total mechanical energy from the point just before the release of the spring to the point just before impact?

A:

C:

D:

B:

€

(1/2)kΔx2 = (1/2)mv2

€

(1/2)kΔx2 + mgh i = (1/2)mv2

€

(1/2)kΔx2 + mgh i = (1/2)mv2 + mgh f

€

mghi =(1/2)mv2 + mgh f

There is gravitational PE at the points just before release and just before impact,

because Dave is above floor level (which we are defining as the zero of

gravitational potential energy), in both cases.

Choice: A

Incorrect

What about the gravitational PE just before impact when Dave is still at least 1 m above the floor?

Choice: B

Incorrect

This expression contains all of the forms of energy that are involved at

these two points.

Choice: C

Correct

Choice: D

Incorrect

In addition to the gravitational PE just before release, remember that the

compressed spring stores elastic potential energy which can change into other forms

of mechanical energy.

10. From the equation we found in question 9, which one of the following expressions is true for v2 after simplification?

• A :

• B :

• C :

€

v2 =kΔx2 +mghi −mghf

€

v2 =(kΔx2 + 2mg(hi −h f))

m

€

v2 =2g(hi −h f)

Check your algebra. Pay attention to the mass and the factor of 2.

Choice: A

Incorrect

Correct algebra.

Choice: B

Correct

Spring PE is set to zero in this expression, which is incorrect.

Choice: C

Incorrect

11. From the information given in the original problem statement, what is the value of the initial compression of the spring ?

A: 1.5 m

C: 1800 N/m

D: 2.75 m

B: 3.5 m

€

Δx

E: 1 m

This is the initial height of Dave in his seat atop the compressed spring.

Choice: A

Incorrect

This is the length of the tube and has nothing to do with the physics

of the problem.

Choice: B

Incorrect

This is the stiffness (spring constant k) of the spring.

Choice: C

Incorrect

This is the initial compression of the spring. (The amount the spring is compressed from

its equilibrium position.)

Choice: D

Correct

Choice: E

Incorrect

This is the height of the top surface of the airbag from the

floor.

12. Also from the information given in the original problem: What does the 750 N value represent?

• A : Dave’s mass (m)• B : Dave’s weight (W=mg)• C : the retarding force of the air• D : the spring constant

The problem states that Super Dave weights 750 N. Weight differs from mass by a factor of g (the value of

gravitational acceleration near Earth).

Choice: A

Incorrect

This value is given as Super Dave’s weight in the problem statement.

mg = W = weight of Dave

Choice: B

Correct

We can ignore the retarding force of air in this problem. It is stated in the problem that

Super Dave weighs 750 N.

W = 750 N = mg

Choice: C

Incorrect

The spring constant (k) is in units of N/m and is given elsewhere in the

original problem. Super Dave is said to

weigh 750 N.

Choice: D

Incorrect

13. Now that we have Identified the relevant values:

Calculate the numerical value of v.

Recall that v is Super Dave’s velocity

just before impact with the airbag.

Dave’s mass is 76.53 kg. from

€

v2 =(kΔx2 + 2mg(hi −h f))

m

v2 =1800 N/m( ) 2.75m( )

2+ 2 76.53kg( ) 9.8m /s 2

( ) 1.5m −1m( )

76 .53kg

v2 =187 .7m2 /s 2

v = 187 .7m2 /s 2 =13.7m /s

Reasoning:

€

m=Wg

Now that we found Dave’s speed just before impact, we can focus on the “airbag-Dave”

system.

Here we will assume that Super Dave hits the air bag at normal incidence (straight down

onto the air bag). This makes the calculation much simpler, but is only true as a limiting

case.

14. Here, we are only concerned with Dave’s motion in       the vertical direction.      Finish the following statement correctly.

The retarding force Fretarding from the airbag on Dave…

• A : is perpendicular to Dave’s direction of motion.

• B : is in the same direction as Dave’s direction of motion.

• C : is opposite to Dave’s direction of motion.• D : has no direction.

There could be no energy loss if this were the case and Dave would get hurt

regardless of the strength of the airbag’s force.

Choice: A

Incorrect

This is nonsense. Landing on an airbag would not provide a force

that would cause one to accelerate downward.

Choice: B

Incorrect

The retarding force of the airbag opposes the direction of motion. This

force does work on Dave. It causes him to slow down and hopefully helps him

to land safely.

Choice: C

Correct

It does have a specific direction. Force is a vector quantity.

Choice: D

Incorrect

15. What does the work-energy principle say?

• A : A system’s change in PE is equal to the work done on the system by the resultant force.

• B : the work done is equal to the energy dissipated in the form of heat

• C : the work done by the resultant force acting on the system is equal to the change in the system’s K

• D : there is no such principle

This statement is true if all of the work is done by

conservative forces. Even so, it is not the Work-Energy

Theorem.

Choice: A

Incorrect

This is not always true, and is not relevant to this problem.

Choice: B

Incorrect

This is the correct explanation of the work-energy principle.

In mathematical form:

Choice: C

Correct

€

K f −K i =Wtotal

12mvf

2 −12mvi

2 =Wtotal

There is such a principle.

Choice: D

Incorrect

16. In general, total work can be expressed as:

Where y represents change vertical distance and is the angle between the direction of motion and the direction of force, remember we are only considering the vertical component of the retarding force.

What is Wtotal in the Dave-airbag system?

A:

B:

C

€

Wtotal=Ftotal Δy cos(θ)( )

€

Wtotal=−(Fretarding )Δy

€

Wtotal=−Fretarding−mg( )Δy

€

Wtotal=mgΔy

Choice: AIncorrect

After Dave hits the airbag, the total force (we only care about the y-component) acting upon him is a

combination of gravitational force and the retarding force from the airbag. These forces are in opposite directions,

making the total force equal to:

€

Ftotal=(Fretarding −mg)

Choice: BCorrect

This is the correct expression for total work in this system, because gravitational force and the retarding force from the airbag act on Dave in opposite directions. Also, cos()=-1 because =180º.

Choice: CIncorrect

After Dave hits the airbag, the total force (we only care about the y-component) acting upon him is a

combination of gravitational force and the retarding force from the airbag. These forces are in opposite directions,

making the total force equal to:

€

Ftotal=(Fretarding −mg)

17. Assuming that Dave is going to come to a stop at some point after he hits the airbag and applying the work-energy principle, which one of the following expressions is correct for the Dave-airbag system? (vi is the Dave’s velocity just before impact. It’s the velocity v that we found earlier.)

• A :

• B :

• C :

€

12mvf

2 −12mvi

2 =0

€

−12mvi

2 =−Fretarding−mg( )Δy

€

12mvf

2 = Fretarding−mg( )Δy

There must be some work done if Dave is going to stop. The retarding force from the airbag does work on him.

Choice: A

Incorrect

Since Kf = 0, this is the correct expression.

Choice: B

Correct

The kinetic energy before impact with the airbag should be included. Since

Dave stops moving, vf=0.

Choice: C

Incorrect

18. For Dave to land safely, what must be true about y?

• A : y > 1m• B : y = 0• C : y < 1m• D : y = 1m

This wouldn’t be safe because he would hit the hard floor.

Choice: A

Incorrect

This implies that vi = 0

Choice: B

Incorrect

If y<1m, Super Dave would stop before the airbag is fully compressed

and would not hit the floor.

Choice: C

Correct

This is risky. Dave would hit the floor.

Choice: D

Incorrect

19. What is the numerical value of y?

Solution

Applying the principle of conservation of mechanical energy from the point just before the release of the spring to the point just before impact, we found:

Solution

In analysis after the impact, we applied the work-energy theorem. Finding:

Continue

€

vi2 =(kΔx2 + 2mg(hi −h f))

m

vi2 =187 .7m2 /s 2

vi = 187 .7m2 /s 2 =13.7m /s

€

12mvi

2 = Fretarding−mg( )Δy Rearranging to solve for y we get:

€

Δy=12

mvi2

(Fretarding −mg)

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟

Δy =1

2

76.5kg(187 .7m2 /s 2 )

3000 N− 750N

⎛

⎝ ⎜

⎞

⎠ ⎟

Δy = 3.2m

Since Δy is greater than 1m, Super Dave will not land safely. We need a thicker airbag, or one that has a stronger retarding force.

Paired Problem1. In the track shown below, section AB is a quadrant of a circle of 1.0m radius. A block is released at point A and slides without friction until it reaches point B. The horizontal part is not smooth. If the block comes to rest 3.0m after point B, what is the coefficient of kinetic friction mk on the horizontal surface?