### Present Remotely

Send the link below via email or IM

CopyPresent to your audience

Start remote presentation- Invited audience members
**will follow you**as you navigate and present - People invited to a presentation
**do not need a Prezi account** - This link expires
**10 minutes**after you close the presentation - A maximum of
**30 users**can follow your presentation - Learn more about this feature in our knowledge base article

# AP Physics B Free Response 2010 B

No description

by

Tweet## Rechel Geiger

on 18 April 2013#### Transcript of AP Physics B Free Response 2010 B

A small block of mass 0.15 kg is placed at point A at a height 2.0 m above the bottom of a track, as shown in the figure above, and is released from rest. It slides with negligible friction down the track, around the inside of the loop of radius 0.60 m, and leaves the track at point C at a height 0.50 m above the bottom of the track.

(worth a total of 10 points) 2010 AP Physics B Free Response (Form B) A Calculate the speed of the block when it leaves the track at point C. (two points) What the AP readers are looking for:

Your ability to use understand conservation of energy while answering this questions. Now we have to GUESS. Givens:

mass = 0.15 kg

gravity = 9.8 m/s/s

original height = 2.0 m

final height = 0.50 m

Your Unknown?

Final velocity. Now choose an Equation.

mgh + (1/2)mv^2 = mgh + (1/2)mv^2 will do.

But since we know initial velocity is 0, simplify:

mgh = mgh + (1/2)mv^2 Substitute...

(0.15 kg)(9.8 m/s/s)(2.0 m) = (0.15 kg)(9.8 m/s/s)(0.5 m) + (1/2)(0.15 kg)(v^2).

And Solve!

v = 5.4 m/s If answered correctly, the reader will have awarded you one point for correctly using conservation of energy and one point for the correct answer. B On the figure below, draw and label the forces (not components) that act on the block when it is at the top of the loop at point B. (three points) What the AP readers are looking for:

Your ability to correctly draw and label all forces WITHOUT any extraneous forces. Analyze what forces are acting of the block when at the top of the loop.

- Gravity is pulling down.

- The normal force of the track is also pushing perpendicularly to the block. Draw the forces you have identified.

mg N If answered correctly, you will be awarded one point for CORRECTLY drawing AND labeling the force of gravity, one point for correctly drawing and labeling the normal force, and one point for not having any extraneous forces. C Calculate the minimum speed the block can have at point B without losing contact with the track. (two points) What the reader is looking for:

-Your ability to recognize that force is equal to the centripetal force

-Knowledge that for minimum speed, normal force must equal 0. Givens:

m = 0.15 kg

r = 0.60 m

g = 9.8 m/s/s

normal force = 0 N Unknown:

velocity Equation:

net force = force of gravity + normal force = centripetal force

mg = mv^2/r Substitute:

(0.15 kg) (9.8 m/s/s) = (0.15 kg)(v^2)/(0.60 m) And solve.

v = 2.4 m/s If answered properly, you will earn one point for correctly equating the forces and one point for your correct answer. D Calculate the minimum height h about the bottom of the track at which the block can be released and still go around the loop without losing contact with the track. (three points) What the AP readers want to see:

Your ability to analyze the conservation of energy in this situation while recognizing the application of a previous answer. Givens:

m = 0.15 kg

g = 9.8 m/s/s

v = 2.4 m/s (given in answer to part c)

height at top of loop = 2r = 1.2 m Unknown:

initial height Equation:

Use the conservation of energy equation as in part A.

mgh = mgh + (1/2)mv^2

Substitute:

(0.15 kg)(9.8 m/s/s)(h) = (0.15 kg)(9.8 m/s/s)(1.2 m) + (1/2)(0.15 kg)(2.4 m/s)^2 Solve:

h = 1.5 m If you answered this question correctly, the reader will have awarded you one point for correctly using the conservation of energy equation, one point for appropriately substituting the velocity found in part C, and one point for finding the correct answer. Rechel Geiger

Full transcript(worth a total of 10 points) 2010 AP Physics B Free Response (Form B) A Calculate the speed of the block when it leaves the track at point C. (two points) What the AP readers are looking for:

Your ability to use understand conservation of energy while answering this questions. Now we have to GUESS. Givens:

mass = 0.15 kg

gravity = 9.8 m/s/s

original height = 2.0 m

final height = 0.50 m

Your Unknown?

Final velocity. Now choose an Equation.

mgh + (1/2)mv^2 = mgh + (1/2)mv^2 will do.

But since we know initial velocity is 0, simplify:

mgh = mgh + (1/2)mv^2 Substitute...

(0.15 kg)(9.8 m/s/s)(2.0 m) = (0.15 kg)(9.8 m/s/s)(0.5 m) + (1/2)(0.15 kg)(v^2).

And Solve!

v = 5.4 m/s If answered correctly, the reader will have awarded you one point for correctly using conservation of energy and one point for the correct answer. B On the figure below, draw and label the forces (not components) that act on the block when it is at the top of the loop at point B. (three points) What the AP readers are looking for:

Your ability to correctly draw and label all forces WITHOUT any extraneous forces. Analyze what forces are acting of the block when at the top of the loop.

- Gravity is pulling down.

- The normal force of the track is also pushing perpendicularly to the block. Draw the forces you have identified.

mg N If answered correctly, you will be awarded one point for CORRECTLY drawing AND labeling the force of gravity, one point for correctly drawing and labeling the normal force, and one point for not having any extraneous forces. C Calculate the minimum speed the block can have at point B without losing contact with the track. (two points) What the reader is looking for:

-Your ability to recognize that force is equal to the centripetal force

-Knowledge that for minimum speed, normal force must equal 0. Givens:

m = 0.15 kg

r = 0.60 m

g = 9.8 m/s/s

normal force = 0 N Unknown:

velocity Equation:

net force = force of gravity + normal force = centripetal force

mg = mv^2/r Substitute:

(0.15 kg) (9.8 m/s/s) = (0.15 kg)(v^2)/(0.60 m) And solve.

v = 2.4 m/s If answered properly, you will earn one point for correctly equating the forces and one point for your correct answer. D Calculate the minimum height h about the bottom of the track at which the block can be released and still go around the loop without losing contact with the track. (three points) What the AP readers want to see:

Your ability to analyze the conservation of energy in this situation while recognizing the application of a previous answer. Givens:

m = 0.15 kg

g = 9.8 m/s/s

v = 2.4 m/s (given in answer to part c)

height at top of loop = 2r = 1.2 m Unknown:

initial height Equation:

Use the conservation of energy equation as in part A.

mgh = mgh + (1/2)mv^2

Substitute:

(0.15 kg)(9.8 m/s/s)(h) = (0.15 kg)(9.8 m/s/s)(1.2 m) + (1/2)(0.15 kg)(2.4 m/s)^2 Solve:

h = 1.5 m If you answered this question correctly, the reader will have awarded you one point for correctly using the conservation of energy equation, one point for appropriately substituting the velocity found in part C, and one point for finding the correct answer. Rechel Geiger