Ap physics semester one final exam review 1 version b PDF

Preview

Summary

Download Ap physics semester one final exam review 1 version b PDF

Description

Review: 1 AP Physics Semester 1 Final exam 20% of Semester 1 Grade 10:30 am to 12:30 pm Tuesday, Dec 19th 2017 30 minutes - Free Response questions 90 minutes – Multiple-Choice questions 120 minutes total

AP Physics Questions and Grading Curve

AP Physics - Final Exam Semester 1 - Dec 19th 2017

1

Position

REVIEW

REVIEW

REVIEW

Draw Free Body Diagrams for each of the two blocks. Then, for each FBD, draw vector(s) for all forces acting on that block. Ignore the forces that cancel each other in the vertical dimension. Focus on the forces in the horizontal dimension. Start with ma = ? What is the force or forces acting on that block?

REVIEW

REVIEW

When you see ramps, think of sine of the angle for the component of gravity that works on the plane of the ramp. And thus the normal force must be the cosine of the angle. When there is friction (other than air friction), think of the normal force and the coefficient of friction.

REVIEW

REVIEW

When you see pulleys, draw FBDs and remember that the tension in continuous segments of string (or wire or cable..) is the same. Write out equations “ma = ….” for each FBD. Remember displacement equation to connect displacement with velocity, acceleration and time.

REVIEW

REVIEW

When you see pulleys, draw FBDs and remember that the tension in continuous segments of string (or wire or cable..) is the same. Write out equations “ma = ….” for each FBD. m1a = T1+ (- m1g) m2a = - T2 + (m2g) T1 = T2 Solve for a. Then plug in the value for a in either of the equations for tension.

T1 T2

REVIEW

REVIEW

Find net Force. Then equate that net Force to the total mass x acceleration of that total mass. Example: Fnet = (5 – 1)kg (9.8 m/s2) = 40 N 40 N = (5 + 1) kg (a) a = 40/6 = 6.67 m/s2

REVIEW

REVIEW

The question sounds confusing. Don’t worry. Just draw a diagram of the situation, label the vectors and write out the relevant equation. “Just barely freeing…” is a big hint that you can pretend that nothing is moving in this situation and thus there is no unbalanced force, so all forces must cancel each other. Since Linda creates a horizontal force, the horizontal components (sine) of the rope tension must cancel it.

REVIEW

See next page for detailed solution for part c.

REVIEW

2014 AP Physics 1 Test Question

REVIEW

2

3

2014 AP Physics 1 Test Question

REVIEW

2

3

2014 AP Physics 1 Test Question

REVIEW

4 5

21

2014 AP Physics 1 Test Question

REVIEW

4 5

1000N(2s)(0.5)+=+impulse+for+t+=+0+to+2 -500N(1s)+=+impulse+for+t+=+3+to+4 1000Ns-500Ns+=+net+impulse+=+500+Ns 500+Ns+=+m(Vf – Vi) 500+Ns+=+4000+kg+(Vf – Vi) (Vf – Vi)+=+500/4000+m/s+=+0.125+m/s

Compare+initial+and+ final+momenta+for+ the+two+balls.+ Rubber+ball+loses+ more+momentum+ than+clay+ball+to+the+ glider.+

22

2014 AP Physics 1 Test Question

REVIEW

6

7

23

2014 AP Physics 1 Test Question

REVIEW

6

7

24

2015 AP Physics 1 Test Question

REVIEW

8

(A) Neither graph alone is sufficient; both graph 1 and graph 2 are needed together. (B) Graph 1 is sufficient. (C) Neither graph alone is sufficient; additional information regarding the impulse on the cart is necessary. (D) Neither graph alone is sufficient; additional information regarding the cart’s acceleration is necessary.

25

2015 AP Physics 1 Test Question

REVIEW

8

Graph+1+shows+Force+applied+ over+a+time+interval.+This+is+F+ Δt,+which+is+impulse.+ F+Δt =+m+(Vf – Vi) Vf can+be+determined+ because+all+other+variables+ are+known+(m,+Vi)+ or+ will+be+measured+and+ recorded+on+Graph+1+(+F,++Δt )

(A) Neither graph alone is sufficient; both graph 1 and graph 2 are needed together. (B) Graph 1 is sufficient. (C) Neither graph alone is sufficient; additional information regarding the impulse on the cart is necessary. (D) Neither graph alone is sufficient; additional information regarding the cart’s acceleration is necessary.

26

2015 AP Physics 1 Test Question

REVIEW

9

27

2015 AP Physics 1 Test Question

REVIEW

9

28

2015 AP Physics 1 Test Question

REVIEW

10

29

2015 AP Physics 1 Test Question

REVIEW

10

30

2015 AP Physics 1 Test Question

REVIEW

11

12

31

2015 AP Physics 1 Test Question

REVIEW

11 Graph+shows+Force+applied+over+a+time+interval.+This+is+F+Δt,+whic is+impulse.+ F+Δt =+m+(Vf – Vi) 500N(.004s)(0.5)+++500N(.004s)(0.5)+=+0.05kg+(Vf - - 10+m/s) 2Ns+=+0.05kg+(Vf ++10+m/s)+ 40m/s+– 10m/s+=+Vf ;+++++30+m/s+=+Vf moving+to+the+right+side

2016 AP Physics 1 Test Question

12

Cons+of+Momentum: 1kg(2m/s)+++0.5kg(- 4m/s)+=+1.5kg(Vf) 2+– 2+=+1.5+(Vf) 0+=+1.5+(Vf)++ Vf =+0+m/s KEf =+½+(1.5+kg)+(0)2 =+0+J 32

VARIATIONS+OF+TEST+QUESTIONS+for+DAILY+GRADE+Dec+4+during+clas

REVIEW Variation-Problem:-2

10+N 5+N 0 - 5+N - 10+N

1+++++2++++3++++4++++5 ms Time++ A spacecraft of mass 10 kg is traveling in a straight line in the positive direction. Engines can be fired so that the force exerted on the spacecraft is in the positive or negative direction. The graph above shows data for the force during one interval. Which of the following is the best estimate of the net change in the speed of the spacecraft from time t = 0 to time t = 5 ms ?

(A) - 1.25 x 10-2 m/s (B) - 2.50 x 10-3 m/s (C) - 2.50 x 10-2 m/s (D) - 1.25 x 10-3 m/s

33

2014 AP Physics 1 Test Question

REVIEW

Variation-Problem:-2-Solution 10+N 5+N 0 - 5+N - 10+N

1+++++2++++3++++4++++5 ms Time++ A spacecraft of mass 10 kg is traveling in a straight line in the positive direction. Engines can be fired so that the force exerted on the spacecraft is in the positive or negative direction. The graph above shows data for the force during one interval. Which of the following is the best estimate of the net change in the speed of the spacecraft from time t = 0 to time t = 5 ms ?

(A) - 1.25 x 10-2 m/s (B) - 2.50 x 10-3 m/s (C) - 2.50 x 10-2 m/s (D) - 1.25 x 10-3 m/s - 5N(2ms) + - 5N(2ms)(0.5) + - 10N(2ms)(0.5) = impulse for t = 0 to 4 ms + 5N(1ms)(0.5) = impulse for t = 4 to 5 ms - 25 N ms + 2.5 N ms = net impulse = ( - 22.5 N ms ) = ( - 0.0225 Ns ) ( - 0.0225 Ns ) = m(Vf – Vi) ( - 0.0225 Ns ) = 10 kg (Vf – Vi) ( - 0.0225 Ns ) / 10 kg= (Vf – Vi) (Vf – Vi) = - 0.00225 m/s = (- 2.25 x 10-3 m/s) 34

Variation Problem:-3-Dec.-4,-2017

VARIATIONS+OF+TEST+QUESTIONS+for+DAILY+GRADE+Dec+4+during+clas

REVIEW 2015 AP Physics 1 Test Questio

2m

10m

2m

(A) - (3/4) Vo (B) - (1/5) Vo (C) - (2/3) Vo (D) - (2/5) Vo

10m

(9/4) Vo 3 Vo (1/3) Vo (5/2) Vo

35

Variation-Problem:-3-Solution

REVIEW 2015 AP Physics 1 Test Questio

2m

10m

2m

(A) - (3/4) Vo (B) - (1/5) Vo (C) - (2/3) Vo (D) - (2/5) Vo

10m

(9/4) Vo 3 Vo (1/3) Vo (5/2) Vo

36

REVIEW

AP Physics Monday Nov 27 Quiz (Daily Grade) AP Physics 1 – Practice Workbook – Book 1

2016+AP+Exam+Question

2

1) A car of mass m, traveling at speed v, stops in time t when maximum braking force is applied. If this same braking force is used to stop a car of mass 2m traveling at speed v, how much time would be required to stop this car? (A) 1⁄2 t (B) t (C) √2 t (D) 2t

3

3) Two objects, P and Q, have the same momentum. Q can have more kinetic energy than P if it has: (A) More mass than P (B) The same mass as P (C) More speed than P (D) The same speed at P

4

4) A spring is compressed between two objects with unequal masses, m and M, and held together. The objects are initially at rest on a horizontal frictionless surface. When released, which of the following is true? (A) The total final kinetic energy is zero. (B) The two objects have equal kinetic energy. (C) The speed of one object is equal to the speed of the other. (D) The total final momentum of the two objects is zero.

5

5) Two football players with mass 75 kg and 100 kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively. If they grab each other as they collide, the combined speed of the two players just after the collision would be: (A) 2 m/s. (B) 3.4 m/s (C) 4.6 m/s (D) 7.1 m/s

1

REVIEW

AP Physics Monday Nov 27 Quiz (Daily Grade) Solution AP Physics 1 – Practice Workbook – Book 1

2016+AP+Exam+Question

2

1) A car of mass m, traveling at speed v, stops in time t when maximum braking force is applied. If this same braking force is used to stop a car of mass 2m traveling at speed v, how much time would be required to stop this car? (D) 2t (A) 1⁄2 t (B) t (C) √2 t

3

3) Two objects, P and Q, have the same momentum. Q can have more kinetic energy than P if it has: (A) More mass than P (B) The same mass as P (C) More speed than P (D) The same speed at P

4

4) A spring is compressed between two objects with unequal masses, m and M, and held together. The objects are initially at rest on a horizontal frictionless surface. When released, which of the following is true? (A) The total final kinetic energy is zero. (B) The two objects have equal kinetic energy. (C) The speed of one object is equal to the speed of the other. (D) The total final momentum of the two objects is zero.

5

5) Two football players with mass 75 kg and 100 kg run directly toward each other with speeds of 6 m/s and 8 m/s respectively. If they grab each other as they collide, the combined speed of the two players just after the collision would be: (A) 2 m/s. (B) 3.4 m/s (C) 4.6 m/s (D) 7.1 m/s

1

mVi + FΔt = mVf 6[(12 - 8)/(2 - 0)] + FΔt = 6[(9 12)/(5 - 2)] 12 + FΔt = - 6 Momentum for first 2 seconds is 12 kg m/s Momentum for last 3 seconds is - 6 kg m/s So change in momentum is: 6 – 12 = - 18 kg m/s

REVIEW

AP Physics Monday Nov 27 Quiz (Daily Grade) AP Physics 1 – Practice Workbook – Book 1

8. Two carts are held together. Cart 1 is more massive than Cart 2. As they are forced apart by a compressed spring between them, which of the following will have the same magnitude for both carts. (A)change of velocity (B) force (C) speed (D) velocity

9. A ball with a mass of 0.50 kg and a speed of 6 m/s collides perpendicularly with a wall and bounces off with a speed of 4 m/s in the opposite direction. What is the magnitude of the impulse acting on the ball? (A)1 Ns (B) 5 Ns (C) 2 m/s (D) 10 m/s

10. A cart with mass 2m has a velocity v before it strikes another cart of mass 3m at rest. The two carts couple and move off together with a velocity of: (A) v/5 (B) 2v/5 (C) 2v/3. (D) (2/5)½ v

REVIEW

AP Physics Monday Nov 27 Quiz (Daily Grade) Solution AP Physics 1 – Practice Workbook – Book 1

8. Two carts are held together. Cart 1 is more massive than Cart 2. As they are forced apart by a compressed spring between them, which of the following will have the same magnitude for both carts. (A)change of velocity (B) force (C) speed (D) velocity

9. A ball with a mass of 0.50 kg and a speed of 6 m/s collides perpendicularly with a wall and bounces off with a speed of 4 m/s in the opposite direction. What is the magnitude of the impulse acting on the ball? (A)1 Ns (B) 5 Ns (C) 2 m/s (D) 10 m/s

10. A cart with mass 2m has a velocity v before it strikes another cart of mass 3m at rest. The two carts couple and move off together with a velocity of: (A) v/5 (B) 2v/5 (C) 2v/3. (D) (2/5)½ v

Actual AP Physics 1 Exam Questions - 2015

REVIEW 36. Assuming that the velocity versus time graphs are straight lines for both spaceships, which of the following graphs best represents the net force exerted on Ship 2?

REVIEW 36. Assuming that the velocity versus time graphs are straight lines for both spaceships, which of the following graphs best represents the net force exerted on Ship 2?

1

If the velocity versus time graphs are straight lines then the the acceleration of Ship 2 must be constant, thus a must be a constant number and thus net force, F = ma must be a constant product. C shows a constant value for this net Force.

Actual AP Physics 1 Exam Questions - 2015

REVIEW

2

At+highest+point,+the+stone’s+vertical+velocity+is+zero+ and+its+horizontal+velocity+is+constant+throughout+its+ flight+(gravity+acts+perpendicular+to+horizontal+flight).+ So+there’s+no+change+in+velocity+or+speed+at+that+ moment.+

D f C

3

4

(Vf - Vi)/+t+=+a,+so+Vf =+Vi+++at,+plug+in+values: Vf =+2+m/s+++(+- 0.5+m/s2)t When+Vf...