pc1431 assignment 4 answers

Assignment 4: Linear Momentum

Due: 2:00am on Friday, March 1, 2013

Note: To understand how points are awarded, read your instructor's Grading Policy.

A Game of Frictionless Catch

Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, , is identical to the

combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest.

Chuck then picks up a ball of mass and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground

(ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is . The speed of the thrown ball relative to the ground is .

Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is .

When answering the questions in this problem, keep the following in mind:

1. The original mass of Chuck and his cart does not include the mass of the ball.

2. The speed of an object is the magnitude of its velocity. An object's speed will always be a nonnegative quantity.

Part A

Find the relative speed between Chuck and the ball after Chuck has thrown the ball.

Express the speed in terms of and .

Hint 1. How to approach the problem

All this question is asking is: "How fast are Chuck and the ball moving away from each other?" If two objects are moving at the same speed(with respect to the ground) in the same direction, their relative speed is zero. If they are moving at the same speed, , in opposite

directions, their relative speed is . In this problem, you are given variables for the speed of Chuck and the ball with respect to the ground,

and you know that Chuck and the ball are moving directly away from each other.

ANSWER:

Correct

Make sure you understand this result; the concept of "relative speed" is important. In general, if two objects are moving in opposite directions(either toward each other or away from each other), the relative speed between them is equal to the sum of their speeds with respect to theground. If two objects are moving in the same direction, then the relative speed between them is the absolute value of the difference of the theirtwo speeds with respect to the ground.

Part B

What is the speed of the ball (relative to the ground) while it is in the air?

Express your answer in terms of , , and .


Apply conservation of momentum. Equate the initial (before the ball is thrown) and final (after the ball is thrown) momenta of the systemconsisting of Chuck, his cart, and the ball. Use the result from Part A to eliminate from this equation and solve for .

Hint 2. Initial momentum of Chuck, his cart, and the ball

Before the ball is thrown, Chuck, his cart, and the ball are all at rest. Therefore, their total initial momentum is zero.

Hint 3. Find the final momentum of Chuck, his cart, and the thrown ball

What is the total momentum of Chuck, his cart, and the ball after the ball is thrown?

Express your answer in terms of , , , and .

Remember that and are speeds, not velocities, and thus are positive scalars.

ANSWER:

PC1431AY1213SEM2

Assignment 4: Linear Momentum Resources

=

Signed in as Mikael Lemanza Help Close

http://redirect.prod.mastering.pearsoncmg.com/redirect/helpLink?env=prod&productID=MasteringPhysics&role=student&type=standalone&topicID=

Correct

Since , you can use this equation to write in terms of . Then use the equation for in the above part to find a

relation between and .

ANSWER:

Correct

Part C

What is Chuck's speed (relative to the ground) after he throws the ball?



Use the answer to Part B to eliminate from the equation derived in Part A. Then solve for .

ANSWER:

Correct

Part D

Find Jackie's speed (relative to the ground) after she catches the ball, in terms of .

Express in terms of , , and .


Apply conservation of momentum. Equate the initial (before Jackie catches the ball) and final (after the ball is caught) momenta of thesystem consisting of Jackie, her cart, and the ball, and solve for .

Hint 2. Initial momentum

Just before Jackie catches the ball, the momentum of the system consisting of Jackie, her cart, and the ball is equal to the momentum ofthe ball as it flies through the air: .

Hint 3. Find the final momentum

What is the final momentum of the system after Jackie catches the ball?


ANSWER:

Correct

ANSWER:

Correct

=

=

=

=

=

Part E

Find Jackie's speed (relative to the ground) after she catches the ball, in terms of .

Express in terms of , , and .


In Part B, you found an expression for in terms of . You can substitute this expression for into the equation you found in Part D,

which will give you an expression for in terms of the desired quantities.

ANSWER:

Correct

A Girl on a Trampoline

A girl of mass kilograms springs from a trampoline with an initial upward velocity of meters per second. At height meters

above the trampoline, the girl grabs a box of mass kilograms.

For this problem, use meters per second per second for the magnitude of the

acceleration due to gravity.

Part A

What is the speed of the girl immediately before she grabs the box?

Express your answer numerically in meters per second.


Use conservation of energy. Find the initial kinetic energy of the girl as she leaves the trampoline. Then find her gravitational potential

energy just before she grabs the box (define her initial potential energy to be zero). According to the principle of conservation of

energy, . Once you have , use the definition of translational kinetic energy to find the girl's speed .

Hint 2. Initial kinetic energy

What is the girl's initial kinetic energy as she leaves the trampoline?

Express your answer numerically in joules.

ANSWER:

Correct

Hint 3. Potential energy at height

What is the girl's gravitational potential energy immediately before she grabs the box?


ANSWER:

=

= 1920

Correct

ANSWER:

Correct

Part B

What is the speed of the girl immediately after she grabs the box?

Express your answer numerically in meters per second.


Think of the process of grabbing the box as a collision. Though the girl and the box don't collide as such, any interaction between twoobjects that takes place extremely fast can be thought of as a collision. To find the velocity at a later time, which of the following principlescould you use?

ANSWER:

Correct

If the girl merely "grabs" the box, there are no external forces other than gravity, and in the limit that the "collision" takes placeinstantaneously, gravity does not change the momentum of the girl/box system.

Hint 2. Total initial momentum

What is the total momentum before the collision?

Answer in kilogram meters per second.

ANSWER:

Correct

The girl and the box travel with the same speed after she grabs it.

ANSWER:

Correct

Part C

Is this "collision" elastic or inelastic?

Hint 1. Definition of an inelastic collision

If two objects move together with the same velocity after a collision, the collision is said to be inelastic.

ANSWER:

= 1180

= 4.98

conservation of momentum alone

conservation of energy alone

both conservation of momentum and conservation of energy

Newton's second law

= 299

= 3.98

Correct

In inelastic collisions, some of the system's kinetic energy is lost. In this case the kinetic energy lost is converted to heat energy in the girl'smuscles as she grabs the box, and sound energy.

Part D

What is the maximum height that the girl (with box) reaches? Measure with respect to the top of the trampoline.

Express your answer numerically in meters.


Use conservation of energy. From Part B you know the velocity of the girl/box system just after the girl grabs the box. Therefore, you cancompute the kinetic energy of the girl/box system just after the collision. You can also compute the gravitational potential energy

of the girl/box system at this point. The sum of these two quantities must equal the gravitational potential energy of the girl/box

system at the height (where their velocity, and therefore kinetic energy, will be zero).

Hint 2. Finding

What is the girl/box system's gravitational potential energy immediately after she grabs the box?


ANSWER:

Correct

Hint 3. Finding

What is the girl/box system's kinetic energy immediately after she grabs the box?


ANSWER:

Correct

ANSWER:

Correct

Filling the Boat

A boat of mass 250 is coasting, with its engine in neutral, through the water at speed 1.00 when it starts to rain with incredible intensity. The

rain is falling vertically, and it accumulates in the boat at the rate of 100 .

Part A

What is the speed of the boat after time 0.500 has passed? Assume that the water resistance is negligible.

Express your answer in meters per second.


elastic

inelastic

= 1470

= 594

= 2.81

Since the rain originally has no momentum in the direction of motion of the boat and the water resistance is negligible, there are no externalforces acting on the boat. Thus, the horizontal component of momentum of the boat is conserved. As rain falls into the boat, then, the massof the boat increases, and the boat must slow down, as required by conservation of momentum.

Hint 2. Find the momentum of the boat before it starts to rain

What is the momentum of the boat before it starts to rain?

Express your answer in kilogram-meters per second.

Hint 1. Momentum defined

A particle of mass and velocity has momentum

.

ANSWER:

Correct

Since momentum is conserved along the direction of motion, the component of momentum parallel to the direction of motion of the boatmust be the same before and after it has started to rain. Express mathematically this equality and solve for the speed of the boat afterthe given time interval has elapsed.

Hint 3. Find the mass of the boat after it has started to rain

What is the mass of the boat after it has been raining for time 0.500 ?

Express your answer in kilograms.

Hint 1. The mass of water in the boat

After time , the boat has collected 100 kilograms of water from rain.

ANSWER:

Correct

Now calculate the component of momentum of the boat parallel to the direction of motion at the given time interval and set it equal tothe same component calculated before it started to rain.

ANSWER:

Correct

Part B

Now assume that the boat is subject to a drag force due to water resistance. Is the component of the total momentum of the system parallel to

the direction of motion still conserved?

ANSWER:

Correct

The boat is subject to an external force, the drag force due to water resistance, and therefore its momentum is not conserved.

Part C

250

300

0.833

yes

no

The drag is proportional to the square of the speed of the boat, in the form where . What is the acceleration of the boat

just after the rain starts? Take the positive axis along the direction of motion.

Express your answer in meters per second per second.


From Newton's second law of motion, you know that the net force acting on a system equals the time rate of change of momentum of thesystem. In this case, the net force acting on the boat is simply the drag force. To solve the problem, then, write Newton's second law ofmotion in terms of momentum and solve for the acceleration of the boat. Note that the drag force is opposite the direction of motion, whilethe momentum of the boat is parallel to the direction of motion. Also, keep in mind that the mass of the boat is not constant, and take thatinto account when you compute the rate of change of momentum with time.

Hint 2. Find the time rate of change of momentum of the boat

In the following expressions, denotes the mass of the boat before it starts to rain, is the velocity of the boat as a function of time ,

and is the rate at which rain is collected into the boat. Which one of the following expressions represents the time rate of change of

momentum of the boat?

Hint 1. The derivative of the product of functions

Since the momentum of a particle is the product of the mass of the particle and its velocity, when both the mass and the velocity ofthe particle vary with time, to calculate the time rate of change of momentum of the particle you need to use the product rule ofdifferentiation. Let and be, respectively, the mass and the velocity of the particle as functions of time. The time rate of

change of momentum of the particle is then

.

Hint 2. Find the mass of the boat

Rain accumulates in the boat at rate , and so the mass of the boat increases with time. Which of the following expressions

represents the mass of the boat at time ? Let represent the mass of the boat before it starts to rain.

ANSWER:

Correct

ANSWER:

Correct

Now set the rate of change of momentum equal to the drag force . Keep in mind that the drag force is opposite the direction of

motion, and take care to include the correct sign when you write Newton's law. The equation that you will obtain is a differentialequation. Solve it for . Then, substitute in and the appropriate values, such as 1.00 , to find the acceleration

just as the rain starts.

ANSWER:

Correct

Rocket Car

A rocket car is developed to break the land speed record along a salt flat in Utah. However, the safety of the driver must be considered, so theacceleration of the car must not exceed (or five times the acceleration of gravity) during the test. Using the latest materials and technology, the total

mass of the car (including the fuel) is 6000 kilograms, and the mass of the fuel is one-third of the total mass of the car (i.e., 2000 killograms). The car ismoved to the starting line (and left at rest), at which time the rocket is ignited. The rocket fuel is expelled at a constant speed of 900 meters per secondrelative to the car, and is burned at a constant rate until used up, which takes only 15 seconds. Ignore all effects of friction in this problem.

Part A

Find the acceleration of the car just after the rocket is ignited.

Express your answer to two significant figures.


The equation for the acceleration due to rocket propulsion is , where is the exhaust speed. To use this equation, first find

an expression for the rate of mass loss of the car.

Hint 2. find the rate of mass change

Find the rate that the rocket car's mass is changing.

Express your answer to three significant figures.

ANSWER:

Correct

ANSWER:

Correct

The driver of this car is experiencing just over , or two times the acceleration one normally feels due to gravity, at the start of the trip. This is

not much different from the acceleration typically experienced by thrill seekers on a roller coaster, so the driver is in no danger on this score.

Part B

Find the final acceleration of the car as the rocket is just about to use up its fuel supply.


Hint 1. What has changed?

What has changed from the time of the initial ignition of the rocket to the moment when the fuel is used up?

ANSWER:

Correct

Hint 2. Find the final mass

−2.11×10−3

= -133

= 20

the exhaust speed of the rocket relative to the car

the total mass of the car (including the fuel)

the rate of mass change of the car

Find the final mass of the car (including the fuel) after all the fuel has been used up.


ANSWER:

Correct

ANSWER:

Correct

The driver of this car is experiencing just over , or three times the acceleration one normally feels due to gravity, by the end of the trip. This

is the maximum acceleration achieved during the trip, and it is still very safe for the driver, who can easily withstand over with training.

Part C

Find the final velocity of the car just as the rocket is about to use up its fuel supply.


Hint 1. Find the change in speed

Write an expression for the change in speed of the car from start to finish: . You will need to make use of the differential equation

for rocket motion

,

if you don't know the equation for velocity of a rocket.

Express your answer in terms of the exhaust speed , the initial mass of the car (plus fuel) , and the final mass of the car

.

Hint 1. How to solve the differential equation

The differential equation for rocket motion is an example of a separable differential equation. It can be rewritten as

,

where the fact that has been used on the left hand side. Thus, integrating the left side gives the change in velocity. Integrate

the right side to get an expression for the change in velocity in terms of the initial and final masses.

ANSWER:

Correct

ANSWER:

Correct

At the end of the trip, the driver is going a bit over Mach 1, or one times the speed of sound. This problem was based loosely on the breakingof the sound barrier by the ThrustSSC team in October 1997.

= 4000

= 30

=

= 360

Pucks on Ice

Two hockey players, Aaron and Brunnhilde, are pushing two pucks on a frictionless ice rink. The pucks are initially at rest on the starting line.Brunnhilde is pushing puck B, which has a mass three times as great as that of puck A,which Aaron is pushing. The players exert equal constant forces of magnitude on their

pucks, directed horizontally, towards the finish line. They start pushing at the same time, andeach player pushes his or her puck until it crosses the finish line, a distance away.

Part A

Which puck reaches the finish line first?

Hint 1. Compute the relative acceleration of the pucks

If and are the magnitudes of the accelerations of pucks A and B, respectively, what is the value of the ratio ?

ANSWER:

Correct

ANSWER:

Correct

Part B

Let be the magnitude of the kinetic energy of puck A at the instant it reaches the finish line. Similarly, is the magnitude of the kinetic

energy of puck B at the (possibly different) instant it reaches the finish line. Which of the following statements is true?

Hint 1. Determine the simplest way to answer this question

There are several possible approaches to this problem. Which is the simplest?

Choose the best option.

ANSWER:

= 3

Both pucks reach the finish line at the same time.

Puck A reaches the finish line first.

Puck B reaches the finish line first.

More information is needed to answer this question.

Use (force equals mass times acceleration) to find the acceleration of each puck.

Use (relating distance traveled to acceleration and time) to find the time to the finish line.

Use the work-energy theorem.

Apply conservation of momentum and energy.

Correct

The work-energy theorem relates the initial and final kinetic energies of an object to the work done on that object:

, where is the final kinetic energy of the object, is the initial kinetic energy of the object, and is the work done

on the object. Since each player applies a constant force, the work done on each puck is easily computed.

Hint 2. Work done on puck A

Find , the work done on puck A over the distance .

ANSWER:

Correct

Hint 3. Work done on puck B

Find , the work done on puck B over the distance .

ANSWER:

Correct

ANSWER:

Correct

Part C

Let be the magnitude of the momentum of puck A at the instant it reaches the finish line. Similarly, is the magnitude of the momentum of

puck B at the (possibly different) instant it reaches the finish line. Which of the following statements is true?

Choose the best option.

Hint 1. Method 1: Compute the ratio of the pucks' velocities

The momentum of an object is the product of its mass and velocity. From the problem introduction, you know that . Find ,

the ratio of the velocity of puck A at the instant it reaches the finish line to the velocity of puck B at the (possibly different) instant it reachesthe finish line.

Hint 1. How to find the final velocities

You can easily compute the ratio using the (already determined) fact that the final kinetic energy of both pucks is the same. Write

the kinetic energy of each puck in terms of its velocity (for example, ). Set these expressions equal, and use the

known ratio of the masses.

ANSWER:

Correct

=

=

You need more information to decide.

= 1.73

Hint 2. Method 2: Use the impulse-momentum theorem

The impulse-momentum theorem states that.

You are given that both forces are the same, and you have compared the times in an earlier part.

ANSWER:

Correct

Three-Block Inelastic Collision

A block of mass moving with speed undergoes a completely inelastic collision with a stationary block of mass . The blocks then move, stuck

together, at speed . After a short time, the two-block system collides inelastically with a third block, of mass , which is initially stationary. The

three blocks then move, stuck together, with speed . All three blocks have nonzero mass.

Assume that the blocks slide without friction.

Part A

Find , the ratio of the velocity of the two-block system after the first collision to the velocity of the block of mass before the collision.

Express your answer in terms of , , and/or .

Hint 1. What physical principle to use

Apply the principle of conservation of linear momentum, noting that the mass of the two-block system is .

ANSWER:

Correct

Intuition and experience with the momentum equations lead to the following conclusions:

1. The blocks will slow down after collision ( , or ).

2. The greater the mass of block 1 for a fixed mass of block 2, the less the blocks will slow down after the collision ( increases

as the mass of block 1 increases with respect to the mass of block 2, but the ratio will still, of course, be less than 1).

The simplest equation that satisfies these criteria is . Try to use similar reasoning for the rest of this problem.

Part B

You need more information to decide.

=

Find , the ratio of the kinetic energy of the two-block system after the first collision to the kinetic energy of the block of mass before

the collision.


Hint 1. Formula for kinetic energy

The translational kinetic energy of an object of mass with speed is .

ANSWER:

Correct

Part C

Find , the ratio of the velocity of the three-block system after the second collision to the velocity of the block of mass before the

collisions.


Hint 1. Total mass of the blocks

After the second collision, the mass of the system is .

ANSWER:

Correct

The time between collisions is irrelevant to the velocities and energies of this problem. If you consider the second collision to take place 0units of time after the first collision, then the problem can be seen to be equivalent to the first problem with replaced by .

Part D

Find , the ratio of the kinetic energy of the three-block system after the second collision to the initial kinetic energy of the block of mass

before the collisions.


ANSWER:

Correct

Part E

Suppose a fourth block, of mass , is included in the series, so that the three-block system with speed collides with the fourth, stationary,

block. Find , the ratio of the kinetic energy of all the blocks after the final collision to the initial kinetic energy of the block of mass

before any of the collisions.

Express your answer in terms of , , , and/or .

Hint 1. How to approach the question

You can find the speed of the four blocks using the method of the previous parts. Alternatively, you may notice a pattern in your

=

=

=

expressions for and and use this pattern to find .

ANSWER:

Correct

You may proceed in this way to find the kinetic energy of an -block system after inelastic collisions. The result is

.

Conservation of Momentum in Two Dimensions Ranking Task

Part A

The figures below show bird's-eye views of six automobile crashes an instant before they occur. The automobiles have different masses andincoming velocities as shown. After impact, the automobiles remain joined together and skid to rest in the direction shown by . Rank these

crashes according to the angle , measured counterclockwise as shown, at which the wreckage initially skids.

Rank from largest to smallest. To rank items as equivalent, overlap them.

Hint 1. Conservation of momentum in two dimensions

Since momentum is a vector quantity, the x component of momentum and the y component of momentum must be individually conserved inany collision. Thus, the total x momentum before the collision must be equal to the total x momentum of the sliding wreckage after thecollision. The same is true for the total y momentum.

Hint 2. Determining the angle

Once the x and y momenta of the wreckage are determined, the exact angle through which the wreckage skids can be determined bytrigonometry. Determining the exact angle of this final momentum vector is accomplished the same way you would find the angle of anyvector, typically by finding the inverse tangent of the y component over the x component. (You can also determine the ranking withoutcalculating the exact angle at which the wreckage skids.)

ANSWER:

Correct

Surprising Exploding Firework

=

A mortar fires a shell of mass at speed . The shell explodes at the top of its trajectory (shown by a star in the figure) as designed. However, rather

than creating a shower of colored flares, it breaks into just two pieces, a smaller piece of mass and a larger piece of mass . Both pieces land

at exactly the same time. The smaller piece lands perilously close to the mortar (at a distance of zero from the mortar). The larger piece lands adistance from the mortar. If there had been no explosion, the shell would have landed a distance from the mortar. Assume that air resistance and

the mass of the shell's explosive charge are negligible.

Part A

Find the distance from the mortar at which the larger piece of the shell lands.

Express in terms of .

Hint 1. Find the position of the center of mass in terms of

The two exploded pieces of the shell land at the same time. At the moment of landing, what is the distance from the mortar to the

center of mass of the exploded pieces?

Express your answer in terms of .

Hint 1. Key idea

The explosion only exerts internal forces on the particles. The only external force acting on the two-piece system is gravity, so thecenter of mass will continue along the original trajectory of the shell.

ANSWER:

Correct

Hint 2. Find the position of the center of mass in terms of

The larger piece of the shell lands a distance from the mortar, and the smaller piece lands a distance zero from the mortar. What is ,

the final distance of the shell's center of mass from the mortar?

Express your answer in terms of .

Hint 1. A helpful figure

Here is a figure to help you visualize the situation.

=

ANSWER:

Correct

ANSWER:

Correct

± A Rocket in Deep Space

A rocket is fired in deep space, where gravity is negligible. In the first second it ejects of its mass as exhaust gas and has an acceleration of

15.4 .

Part A

What is the speed of the exhaust gas relative to the rocket?

Express your answer numerically in kilometers per second.


In deep space gravity is negligible and there is no air resistance; thus no external forces act on the rocket and the exhaust gas, and thetotal momentum of the system (rocket plus exhaust gas) is conserved. By applying conservation of momentum, one can derive a formula forthe acceleration of the rocket given the speed of the exhaust gas relative to the rocket, the original mass of the rocket, and the rate ofchange of the rocket's mass with time.

Hint 2. The acceleration of the rocket

By applying conservation of momentum to the system that comprises the rocket and the exhaust gas, one can derive a formula for theacceleration of the rocket in terms of the speed of the exhaust gas relative to the rocket and the mass of the rocket . In symbols,

,

where represents time. The quantity is the time rate of change of the mass of the rocket, and it is a negative quantity (the mass of

the rocket decreases continuously with time as it burns fuel). Since is a positive quantity (it is the speed of the exhaust gas), the

acceleration of the rocket is also positive.

(Another way to derive the formula above, is to apply the relationship to the exhaust gas and note that the force on the rocket is

equal and opposite to the force on the exhaust.)

Hint 3. Find the change in mass of the rocket

What is the rate of change of the mass of the rocket if is its original mass before the launch?

ANSWER:

=

=

Correct

Now substitute the expression for the rate of change of mass that you have just determined into the formula for the acceleration of therocket and solve for the speed of the exhaust gas.

ANSWER:

Correct

A Relation Between Momentum and Kinetic Energy

Part A

A cardinal (Richmondena cardinalis) of mass 4.50×10−2 and a baseball of mass 0.149 have the same kinetic energy. What is the ratio of the

cardinal's magnitude of momentum to the magnitude of the baseball's momentum?


Recall that the kinetic energy of an object (of mass and speed ) is given by , and the magnitude of the momentum by

. Combining these equations into a single expression can then be used to eliminate , giving an expression of the kinetic energy in

terms of the momentum instead of the velocity. We can then use this relation, along with the assumptions, to find the ratio of the momenta in terms of the masses.

Hint 2. Find a relationship between kinetic energy and momentum

Select the general expression for the kinetic energy of an object with mass and momentum .

ANSWER:

Correct

Use the fact that the kinetic energies of the cardinal and the baseball are the same to find an equation for the ratio .

ANSWER:

Correct

Part B

A man weighing 750 and a woman weighing 420 have the same momentum. What is the ratio of the man's kinetic energy to that of the

woman ?


= 2.46

= 0.550

As in the previous part, an expression for the momentum must be found in terms of the kinetic energy. Then the ratio of the kinetic energies must be found in terms of the weights, instead of the masses.

Hint 2. Find a relationship between momentum and kinetic energy

Select the general expression for the momentum of an object with mass and kinetic energy .

ANSWER:

Correct

Use the fact that the momenta of the man and woman are the same to find an equation for the ratio . Use the relationship

between mass and weight to express this ratio in terms of weights rather than masses.

ANSWER:

Correct

Collision at an Angle

Two cars, both of mass , collide and stick together. Prior to the collision, one car had been traveling north at speed , while the second was

traveling at speed at an angle south of east (as indicated in the figure). After the collision, the two-car system travels at speed at an angle

east of north.

Part A

Find the speed of the joined cars after the collision.

Express your answer in terms of and .

Hint 1. Determine the conserved quantities

Which of the following statements is true for the collision described?

ANSWER:

= 0.560

Correct

Apply conservation of momentum:.

Find both components (north and east) of the initial momentum using the information in the problem introduction. The magnitude

of is equal to the magnitude of the momentum vector for the two-car system after the collision: .

Hint 2. The component of the final velocity in the east-west direction

Find the component of in the east-west direction.


Hint 1. Find the east-west component of the initial momentum

What is , the magnitude of the total momentum of the two cars in the east-west direction? (Take eastward to be positive,

westward negative.)


ANSWER:

Correct

Now use the conservation of momentum equation to find .

ANSWER:

Correct

Hint 3. Find the north-south component of the final momentum

Find the component of in the north-south direction.


Hint 1. Find the north-south component of the initial momentum

What is the magnitude of the total momentum of the two cars in the north-south direction? (Take northward to be positive,

southward negative).


ANSWER:

Correct

ANSWER:

Momentum is conserved but kinetic energy is not conserved.

Kinetic energy is conserved but momentum is not conserved.

Both kinetic energy and momentum are conserved.

Neither kinetic energy nor momentum is conserved.

=

(east-west) =

=

(north-south) =

Correct

Hint 4. Math help

Let be the east-west component of , and the north-south component. Then

.

You will also need to use the following trignometric identity when you evaluate the right-hand side of the above equation in terms of and :

.

ANSWER:

Correct

Part B

What is the angle with respect to north made by the velocity vector of the two cars after the collision?

Express your answer in terms of . Your answer should contain an inverse trigonometric function using the notation asin, atan etc.

and not arcsin, arctan etc.

Hint 1. A formula for

Let be the east-west component of , and the north-south component. Then

,

since the angle asked for is the angle east of north.

ANSWER:

Correct

Two Worlds on a String

Two balls, A and B, with masses and are connected by a taut, massless string, and are moving along a horizontal frictionless plane. The

distance between the centers of the two balls is . At a certain instant, the velocity of ball B has magnitude and is directed perpendicular to the

string and parallel to the horizontal plane, and the velocity of ball A is zero.

Part A

Find , the tension in the string.

=

=

Express in terms of , , , and .

Hint 1. Descibe the nature of the motion

Describe the ongoing (not the instantaneous) motion of the system qualitatively.

ANSWER:

Correct

Hint 2. The key idea

For a ball to move in a circle, it must be subject to a centripetal force. It is the tension in the string that provides this force. Therefore,

, where is the linear speed of the rotational motion (relative to the point about which the ball rotates) and is the radius of the

motion. As the answer to the first hint suggests, and are not the same as and .

Hint 3. Find the velocity of the center of mass

Find , the translational speed of the system's center of mass.


Hint 1. How to compute the velocity of the center of mass

You can calculate the velocity of the center of mass of a system by computing the total momentum of the system and dividing

by the total mass of the system.

ANSWER:

Correct

Hint 4. Find the rotational speed

As the system slides across the horizontal plane, it will rotate about its center of mass. Find , the linear speed of ball B's rotational

motion relative to the center of mass.


Hint 1. How to compute ball B's rotational speed

To find the speed with which ball B rotates about the system's center of mass, you must subtract the translational speed of thecenter of mass from the ball's total speed :

.

ANSWER:

Correct

Hint 5. Find the radius of rotation

Find , the radius of rotational motion of ball B.


Hint 1. How to approach the question

Ball B moves in a circle around ball A while ball A remains at rest.

Both balls move in a circle about the midpoint of the string while sliding along the plane translationally.

Both balls move in a circle about the center of mass of the system while sliding along the plane translationally.

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Since ball B rotates about the center of mass of the system, will be the distance from the ball to the system's center of mass. In

other words, to find , calculate the distance between the center of mass and ball B.

Hint 2. Position of the center of mass

Recall that the position of the center of mass of a system is equal to the weighted average position of all the individual objectsconstituting the system (in this case, the two balls). The weighting factor for each object is its mass.

ANSWER:

Correct

Hint 6. Acceleration of ball B

Once you know ball B's radius of rotation and rotational speed , you can compute its acceleration using the centripetal acceleration

formula: . Find , the magnitude of the acceleration of ball B.

Express your answer in terms of , , , and .

ANSWER:

Correct

The the only force acting on the ball of mass is the tension in the string. Now that you know the acceleration, you can compute the

tension in the string using Newton's second law.

ANSWER:

Correct

Note that your answer is "symmetric" between the parameters and . This is as it should be: The tension should be the same

regardless of whether or initially moves. Only their relative velocity matters.

Score Summary:

Your score on this assignment is 99.9%.You received 49.94 out of a possible total of 50 points.

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pc1431 assignment 4 answers

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