School District #23 Science 10 INDIVIDUAL FEEDBACK … · INDIVIDUAL FEEDBACK Report 200047085 ......

School District #23 Science 10Science Dept.

INDIVIDUAL FEEDBACK Report 200047085

54.55

1. An 810 kg dragster is being decelerated by a parachute at 2.5 m/s2 as shown in the diagram.

What is the tension in the cord at this moment?

7.9 x 103 N

5.9 x 103 N

2.0 x 103 N

0 N

A.

B.

C.

D.

Solution:

The tension in the cord is the only force supplying the net force, so:

T = ma

T = 810 x 2.5

T = 2025 N

5. The system of blocks on a frictionless surface in the diagram below is accelerating at 2.0 m/s2.

What is the tension in the cord at X?

16 N

2.0 N

6.0 N

8.0 N

A.

B. C.

D.

MTERM2 Version 1

5. Solution:

Examining the FBD for each object and starting with the 3 kg mass:

Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N

7. What is the impulse on an object of mass 3.5 kg if a constant unbalanced force of 2.9 N acts on

it for 4.7 s?

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

14 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s

9. A 75 kg boy is riding on a 30 kg cart travelling at a speed of 2.0 m/s relative to the ground. The

boy jumps off in such a way as to land on the ground with zero velocity. What is the change in

speed of the cart?

3.0 m/s

0 m/s

2.0 m/s

5.0 m/s

A.

B. C.

D.

MTERM2 Version 1

9. Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s

10. A stationary object explodes into two fragments. A 4.0 kg fragment moves westwards at

3.0 m/s. What are the speed and kinetic energy of the remaining 2.0 kg fragment?

SPEED KINETIC ENERGY

6.0 m/s 36 J

6.0 m/s 18 J

4.2 m/s 36 J

4.2 m/s 18 J

A.

B. C.

D.

Solution:

Speed: use conservation of momentum

change in momentum = 0

pi = p

f , since the object was stationary the initial momentum equals zero.

pf = m

4v

4 - m

2v

2 , where p

f = 0

v2 = m

4v

4/m

2

v2 = 4 x 3 / 2

v2 = 6 m/s

Kinetic Energy: Ek = 0.5 mv2

Ek = 0.5 x 2 x 62

Ek = 36 J

12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?

4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.

Solution:

Conservation of energy

Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J

MTERM2 Version 1

The following information is to be used to answer questions 14 & 15.

A pendulum with a bob of mass 0.750 kg is initially displaced to the right as shown in the

diagram. Upon release, the bob rises to a maximum height of 0.240 m on the other side.

14. If there were no friction, what would the speed of the bob be as it first passed through its lowest

point?

3.60 m/s

2.17 m/s

2.21 m/s

3.57 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s

15. How much heat energy is produced because of friction?

0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J

19. An object of mass 250 kg accelerated from a velocity of 10 m/s to 45 m/s in 12 seconds. What's

the work done by the object?

2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J

MTERM2 Version 1

21. A 10 kg block initially at rest is pulled 13 m across a floor by a 50 N force.

If friction does 380 J of work over this distance, what is the block's final velocity?

8.7 m/s

7.3 m/s

14 m/s

11 m/s

A.

B. C.

D.

Solution:

Work by force = F•d = 650 J

Energy lost due to friction = 380 J

Net work on box = 270 J

0.5mv2 = 270 v = 7.3 m/s

MTERM2 Version 1



40.91

3. A cable joins the two masses shown below and passes over the frictionless peg 'P'. What is the

acceleration of the system?

9.8 m/s2

14.9 m/s2

3.1 m/s2

6.4 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2

4. A cart on a frictionless surface is attached to a hanging mass of 8. 2 kg.

If this system accelerates at 3.5 m/s2, what is the mass m of the cart?

23 kg

31 kg

6.0 kg

15 kg

A.

B. C.

D.

Solution:

The force causing the acceleration comes from the force of gravity acting on the 8.2 kg mass,

and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg

MTERM2 Page 1 Version 2



8.0 N

16 N

2.0 N

6.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N

6. When the speed of a truck is doubled its momentum changes by a factor of:

2

4

1/4

1/2

A.

B. C.

D.

Solution:

momentum is directly proportional to velocity, so doubling v doubles p

8. A ball strikes a wall perpendicularly with an initial speed of 4 m/s, bouncing off the wall at

4 m/s in the opposite direction. Which of the following statements correctly compares the ball's

momentum and kinetic energy before and after the collision?

Its momentum is different but its kinetic energy is the same

Its kinetic energy is different but its momentum is the same

Its momentum is different and its kinetic energy is different

Its momentum is the same and its kinetic energy is the same

A.

B. C.

D.


8. Solution:

Momentum is a vector, so the change in direction affects momentum. Energy is a scalar which

is not affected by the change in direction.



speed of the cart?

5.0 m/s

3.0 m/s

0 m/s

2.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s




6.0 m/s 18 J

6.0 m/s 36 J

4.2 m/s 18 J

4.2 m/s 36 J

A.

B. C.

D.

Solution:



pi = p


pf = m

4v

4 - m

2v

2 , where p

f = 0

v2 = m

4v

4/m

2

v2 = 4 x 3 / 2

v2 = 6 m/s


Ek = 0.5 x 2 x 62

Ek = 36 J


11. A 2.0 kg ball moving at 15 m/s south hits a stationary 5.2 kg ball and bounces back at

4.5 m/s north. Calculate the velocity of the 5.2 kg ball immediately after impact.

7.5 m/s north

7.5 m/s south

4.0 m/s north

4.0 m/s south

A.

B. C.

D.

Solution:

∆p=0

pf = p

i (South is positive)

m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2

= 7.5 m/s in the south direction


6.13 x 105 J

4.47 x 105 J

1.66 x 105 J

7.79 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J






point?

3.57 m/s

3.60 m/s

2.17 m/s

2.21 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s


1.76 J

1.84 J

0.074 J

0.081 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J



2.4 x 105 J

2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J




11 m/s

14 m/s

7.3 m/s

8.7 m/s

A.

B. C.

D.

Solution:




0.5mv2 = 270 v = 7.3 m/s




90.91


1/2

1/4

4

2

A.

B. C.

D.

Solution:




8.7 m/s

7.3 m/s

14 m/s

11 m/s

A.

B. C.

D.

Solution:




0.5mv2 = 270 v = 7.3 m/s




72.73



3.1 m/s2

6.4 m/s2

9.8 m/s2

14.9 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2








A.

B. C.

D.

Solution:





4.0 m/s north

4.0 m/s south

7.5 m/s north

7.5 m/s south

A.

B. C.

D.


11. Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2



1.66 x 105 J

4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J



4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

2.5 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J


23. A vehicle accelerates in three stages. Each stage takes the same time. The change in velocity

for each stage is shown below.

STAGE CHANGE IN VELOCITY

I from rest to speed v

II from speed v to speed 2v

III from speed 2v to speed 3v

Assuming no friction, which statement best describes the power developed by the engine in the

three intervals?

The power is highest in Interval I.

The power is highest in Interval II.

The power is highest in Interval III.

The power is the same in each of the three intervals.

A.

B. C.

D.

Solution:

Since the change in speed takes the same time, then the increase in kinetic energy will be

greatest in Interval III according to the equation Ek = 0.5mv2

P = E/t, so the largest energy increase in the same time will cause the largest power developed.




40.91



5.9 x 103 N

7.9 x 103 N

0 N

2.0 x 103 N

A.

B.

C.

D.

Solution:


T = ma

T = 810 x 2.5

T = 2025 N



9.8 m/s2

14.9 m/s2

3.1 m/s2

6.4 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2




23 kg

31 kg

6.0 kg

15 kg

A.

B. C.

D.

Solution:


and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg


2

4

1/4

1/2

A.

B. C.

D.

Solution:



it for 4.7 s?

14 kg•m/s

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s









A.

B. C.

D.

Solution:





speed of the cart?

5.0 m/s

3.0 m/s

0 m/s

2.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s


6.13 x 105 J

4.47 x 105 J

1.66 x 105 J

7.79 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J






point?

3.57 m/s

3.60 m/s

2.17 m/s

2.21 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s


1.76 J

1.84 J

0.074 J

0.081 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J

16. A 55.0 kg athlete steps off a 10.0 m high platform and drops onto a trampoline. As the

trampoline stretches, it brings him to a stop 1.00 m above the ground.

How much energy must have been momentarily stored in the trampoline when he came to rest?

4 850 J

5 390 J

0 J

539 J

A.

B. C.

D.


16. Solution:

The potential energy at top will be converted to kinetic energy and potential energy.

The kinetic energy will be the energy stored and will equal the change in potential from the top

to the lowest position. Kinetic energy Ek = mg x 9

Estored

= 4851 J

20. A car moving at 80 km/h skids 50 m with locked brakes. How far will the car skid with locked

brakes if it is travelling at 160 km/h?

100 m

200 m

50 m

71 m

A.

B. C.

D.

Solution:

∆Ek = W

∆Ek = Fd

d = ∆Ek÷F

d = 0.5 mv2 ÷ F, so d α v2

If the velocity is increased by a factor of 2, then distance will be increased by a factor of 4.

The new distance will be 50 x 4 = 200 m.

22. A machine can lift a 50 kg object at a constant speed a distance of 10 m in a time of 2.0 s. What

is the power of the machine?

1.0 x 103 W

2.5 x 103 W

9.8 x 101 W

2.5 x 102 W

A.

B.

C.

D.

Solution:

P=W÷t, where W = Fxd

P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W




86.36



9.8 m/s2

14.9 m/s2

3.1 m/s2

6.4 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2



8.0 N

16 N

2.0 N

6.0 N

A.

B. C.

D.


5. Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N



1.0 x 103 W

2.5 x 103 W

9.8 x 101 W

2.5 x 102 W

A.

B.

C.

D.

Solution:


P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W




86.36



8.0 N

16 N

2.0 N

6.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N

13. What is the kinetic energy of a 5.0 kg rock after it falls 20 m from rest?

100 J

49 J

5.0 J

980 J

A.

B. C.

D.


13. Solution:

Ek = E

p

Ek = mgh

Ek = 980 J



1.0 x 103 W

2.5 x 103 W

9.8 x 101 W

2.5 x 102 W

A.

B.

C.

D.

Solution:


P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W




90.91



16 N

2.0 N

6.0 N

8.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N









three intervals?





A.

B. C.

D.

Solution:







59.09



5.9 x 103 N

7.9 x 103 N

0 N

2.0 x 103 N

A.

B.

C.

D.

Solution:


T = ma

T = 810 x 2.5

T = 2025 N








A.

B. C.

D.

Solution:





speed of the cart?

5.0 m/s

3.0 m/s

0 m/s

2.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s



6.13 x 105 J

4.47 x 105 J

1.66 x 105 J

7.79 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J





point?

3.57 m/s

3.60 m/s

2.17 m/s

2.21 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s





4 850 J

5 390 J

0 J

539 J

A.

B. C.

D.

Solution:




Estored

= 4851 J



100 m

200 m

50 m

71 m

A.

B. C.

D.

Solution:

∆Ek = W

∆Ek = Fd

d = ∆Ek÷F

d = 0.5 mv2 ÷ F, so d α v2





11 m/s

14 m/s

7.3 m/s

8.7 m/s

A.

B. C.

D.


21. Solution:




0.5mv2 = 270 v = 7.3 m/s








three intervals?





A.

B. C.

D.

Solution:







36.36



7.9 x 103 N

5.9 x 103 N

2.0 x 103 N

0 N

A.

B.

C.

D.

Solution:


T = ma

T = 810 x 2.5

T = 2025 N



14.9 m/s2

3.1 m/s2

6.4 m/s2

9.8 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2




31 kg

23 kg

15 kg

6.0 kg

A.

B. C.

D.

Solution:


and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg



16 N

2.0 N

6.0 N

8.0 N

A.

B. C.

D.


5. Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N


1/2

1/4

4

2

A.

B. C.

D.

Solution:



it for 4.7 s?

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

14 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s








A.

B. C.

D.


8. Solution:





speed of the cart?

3.0 m/s

0 m/s

2.0 m/s

5.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s




6.0 m/s 36 J

6.0 m/s 18 J

4.2 m/s 36 J

4.2 m/s 18 J

A.

B. C.

D.

Solution:



pi = p


pf = m

4v

4 - m

2v

2 , where p

f = 0

v2 = m

4v

4/m

2

v2 = 4 x 3 / 2

v2 = 6 m/s


Ek = 0.5 x 2 x 62

Ek = 36 J




7.5 m/s south

4.0 m/s north

4.0 m/s south

7.5 m/s north

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2



4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J

13. What is the kinetic energy of a 5.0 kg rock after it falls 20 m from rest?

49 J

100 J

980 J

5.0 J

A.

B. C.

D.

Solution:

Ek = E

p

Ek = mgh

Ek = 980 J






point?

3.60 m/s

2.17 m/s

2.21 m/s

3.57 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s


0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J




5 390 J

0 J

539 J

4 850 J

A.

B. C.

D.


16. Solution:




Estored

= 4851 J

17. A 200 N force acts on a 10.0 kg block and moves it 5.0 m in 2.5 seconds. How much work is

being done?

1490 J

40 J

510 J

1000 J

A.

B. C.

D.

Solution:

W = Fd cosø

F is in the direction of the displacement

W = 200 x 5.0

W = 1000 J

18. What is the minimum work done when a 65 kg student climbs an 8.0 m-high stairway in 12 s?

6 200 J

420 J

520 JN

5 100 J

A.

B. C.

D.

Solution:

W = Fd, where F = mg

W = mgd

W = 65 x 9.8 x 8

W = 5096 J



2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J



100 m

50 m

71 m

200 m

A.

B. C.

D.


20. Solution:

∆Ek = W

∆Ek = Fd

d = ∆Ek÷F

d = 0.5 mv2 ÷ F, so d α v2





8.7 m/s

7.3 m/s

14 m/s

11 m/s

A.

B. C.

D.

Solution:




0.5mv2 = 270 v = 7.3 m/s



2.5 x 103 W

1.0 x 103 W

2.5 x 102 W

9.8 x 101 W

A.

B.

C.

D.

Solution:


P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W








three intervals?





A.

B. C.

D.


23. Solution:







77.27



9.8 m/s2

14.9 m/s2

3.1 m/s2

6.4 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2



8.0 N

16 N

2.0 N

6.0 N

A.

B. C.

D.


5. Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N



7.5 m/s north

7.5 m/s south

4.0 m/s north

4.0 m/s south

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2






4 850 J

5 390 J

0 J

539 J

A.

B. C.

D.

Solution:




Estored

= 4851 J








three intervals?





A.

B. C.

D.

Solution:







90.91



31 kg

23 kg

15 kg

6.0 kg

A.

B. C.

D.

Solution:


and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg





0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J




77.27



16 N

2.0 N

6.0 N

8.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N








A.

B. C.

D.


8. Solution:





7.5 m/s south

4.0 m/s north

4.0 m/s south

7.5 m/s north

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2






0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J




2.5 x 103 W

1.0 x 103 W

2.5 x 102 W

9.8 x 101 W

A.

B.

C.

D.

Solution:


P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W




63.64



7.9 x 103 N

5.9 x 103 N

2.0 x 103 N

0 N

A.

B.

C.

D.

Solution:


T = ma

T = 810 x 2.5

T = 2025 N








A.

B. C.

D.

Solution:





speed of the cart?

3.0 m/s

0 m/s

2.0 m/s

5.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s





6.0 m/s 36 J

6.0 m/s 18 J

4.2 m/s 36 J

4.2 m/s 18 J

A.

B. C.

D.

Solution:



pi = p


pf = m

4v

4 - m

2v

2 , where p

f = 0

v2 = m

4v

4/m

2

v2 = 4 x 3 / 2

v2 = 6 m/s


Ek = 0.5 x 2 x 62

Ek = 36 J


4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J

18. What is the minimum work done when a 65 kg student climbs an 8.0 m-high stairway in 12 s?

6 200 J

420 J

520 JN

5 100 J

A.

B. C.

D.

Solution:

W = Fd, where F = mg

W = mgd

W = 65 x 9.8 x 8

W = 5096 J




2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J



8.7 m/s

7.3 m/s

14 m/s

11 m/s

A.

B. C.

D.

Solution:




0.5mv2 = 270 v = 7.3 m/s




72.73



16 N

2.0 N

6.0 N

8.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N








A.

B. C.

D.


8. Solution:




4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J





0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J



100 m

50 m

71 m

200 m

A.

B. C.

D.


20. Solution:

∆Ek = W

∆Ek = Fd

d = ∆Ek÷F

d = 0.5 mv2 ÷ F, so d α v2










three intervals?





A.

B. C.

D.

Solution:







81.82


1/2

1/4

4

2

A.

B. C.

D.

Solution:






point?

3.60 m/s

2.17 m/s

2.21 m/s

3.57 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s



8.7 m/s

7.3 m/s

14 m/s

11 m/s

A.

B. C.

D.


21. Solution:




0.5mv2 = 270 v = 7.3 m/s








three intervals?





A.

B. C.

D.

Solution:







86.36



8.0 N

16 N

2.0 N

6.0 N

A.

B. C.

D.

Solution:


Fnet

= T2

T2 = ma

T2 = 3 x 2 = 6.0 N

For the 1 kg mass:

Fnet

= T -T2

T-T2 = ma

T = ma + T2

T = 1 x 2 + 6

T = 8 N


2

4

1/4

1/2

A.

B. C.

D.

Solution:




it for 4.7 s?

14 kg•m/s

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s




45.45



14.9 m/s2

3.1 m/s2

6.4 m/s2

9.8 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2



31 kg

23 kg

15 kg

6.0 kg

A.

B. C.

D.

Solution:


and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg


1/2

1/4

4

2

A.

B. C.

D.


6. Solution:



it for 4.7 s?

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

14 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s








A.

B. C.

D.

Solution:





speed of the cart?

3.0 m/s

0 m/s

2.0 m/s

5.0 m/s

A.

B. C.

D.

Solution:

∆p = 0

∆pcart

= ∆pboy

(m∆v)cart

= (m∆v)boy

∆vcart

= (m∆v)boy

÷mcart

∆vcart

= (75 x 2.0) ÷ 30

∆vcart

= 5.0 m/s


4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.


12. Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J





0.081 J

0.074 J

1.84 J

1.76 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J

17. A 200 N force acts on a 10.0 kg block and moves it 5.0 m in 2.5 seconds. How much work is

being done?

1490 J

40 J

510 J

1000 J

A.

B. C.

D.

Solution:

W = Fd cosø

F is in the direction of the displacement

W = 200 x 5.0

W = 1000 J



2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

A.

B.

C.

D.


19. Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J



2.5 x 103 W

1.0 x 103 W

2.5 x 102 W

9.8 x 101 W

A.

B.

C.

D.

Solution:


P = (mg) x d ÷ t

P = 50 x 9.81 x 10 ÷ 2

P = 2.45 x 103 W








three intervals?





A.

B. C.

D.

Solution:







90.91



7.5 m/s north

7.5 m/s south

4.0 m/s north

4.0 m/s south

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2






1.76 J

1.84 J

0.074 J

0.081 J

A.

B. C.

D.


15. Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J




63.64



14.9 m/s2

3.1 m/s2

6.4 m/s2

9.8 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2


it for 4.7 s?

16 kg•m/s

0.62 kg•m/s

3.9 kg•m/s

14 kg•m/s

A.

B. C.

D.

Solution:

impulse = ∆p

∆p = F∆t

∆p = 13.6 N•s








A.

B. C.

D.

Solution:






7.5 m/s south

4.0 m/s north

4.0 m/s south

7.5 m/s north

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2



4.47 x 105 J

6.13 x 105 J

7.79 x 105 J

1.66 x 105 J

A.

B.

C.

D.

Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J






point?

3.60 m/s

2.17 m/s

2.21 m/s

3.57 m/s

A.

B. C.

D.

Solution:

Ek = E

p

v = √(2gh)

v = 2.21 m/s



2.5 x 105 J

4.4 x 103 J

1.5 x 105 J

2.4 x 105 J

A.

B.

C.

D.

Solution:

W = ∆Ek

W = 0.5 m (v2final

- v2initial

)

W = 0.5 x 250 (452 - 102)

W = 2.4 x 105 J








three intervals?





A.

B. C.

D.

Solution:







90.91



4.0 m/s north

4.0 m/s south

7.5 m/s north

7.5 m/s south

A.

B. C.

D.

Solution:

∆p=0

pf = p


m2v

2 = m

5.2v'

5.2 - m

2v'

2

v'5.2

= (m2v

2 + m

2v'

2) ÷ m

5.2

v'5.2

= (2 x 15 + 2 x 4.5) ÷ 5.2

v'5.2





0 J

539 J

4 850 J

5 390 J

A.

B. C.

D.


16. Solution:




Estored

= 4851 J




72.73



9.8 m/s2

14.9 m/s2

3.1 m/s2

6.4 m/s2

A.

B.

C.

D.

Solution:

Fnet

= F23

- F12

ma = m23

g - m12

g

a = (23 x 9.8 - 12 x 9.8)÷35

a = 3.08 m/s2



23 kg

31 kg

6.0 kg

15 kg

A.

B. C.

D.

Solution:


and Fnet

= mtotal

a.

m8.2

g = mtotal

a

8.2 x 9.8 = (8.2 + m) 3.5

m = 14.8 kg


6.13 x 105 J

4.47 x 105 J

1.66 x 105 J

7.79 x 105 J

A.

B.

C.

D.


12. Solution:


Et = E

p + E

k + E

h where E

h = 0

Et = 850 x 9.81 x 20 + 0.5 x 850 x 382

Et = 7.8 x 105 J





1.76 J

1.84 J

0.074 J

0.081 J

A.

B. C.

D.

Solution:

∆Ep = E

h

∆Ep = mg∆h

∆Ep = 0.074 J



11 m/s

14 m/s

7.3 m/s

8.7 m/s

A.

B. C.

D.

Solution:




0.5mv2 = 270 v = 7.3 m/s









three intervals?





A.

B. C.

D.

Solution:





School District #23 Science 10 INDIVIDUAL FEEDBACK … · INDIVIDUAL FEEDBACK Report 200047085 ......

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Transcript of School District #23 Science 10 INDIVIDUAL FEEDBACK … · INDIVIDUAL FEEDBACK Report 200047085 ......