Post on 19-May-2018
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 Page 1 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 Page 2 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 Page 3 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 Page 4 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 Page 5 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200058974
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
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?
8.0 N
16 N
2.0 N
6.0 N
A.
B. C.
D.
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
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.
MTERM2 Page 2 Version 2
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.
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?
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
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 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:
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
MTERM2 Page 3 Version 2
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
12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?
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:
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
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.
MTERM2 Page 4 Version 2
14. If there were no friction, what would the speed of the bob be as it first passed through its lowest
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
15. How much heat energy is produced because of friction?
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
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?
2.4 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
MTERM2 Page 5 Version 2
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?
11 m/s
14 m/s
7.3 m/s
8.7 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 Page 6 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200059671
90.91
6. When the speed of a truck is doubled its momentum changes by a factor of:
1/2
1/4
4
2
A.
B. C.
D.
Solution:
momentum is directly proportional to velocity, so doubling v doubles p
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 Page 1 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200059841
72.73
3. A cable joins the two masses shown below and passes over the frictionless peg 'P'. What is the
acceleration of the system?
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
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 and its kinetic energy is different
Its momentum is the same and its kinetic energy is the same
Its momentum is different but its kinetic energy is the same
Its kinetic energy is different but its momentum is the same
A.
B. C.
D.
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.
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.
4.0 m/s north
4.0 m/s south
7.5 m/s north
7.5 m/s south
A.
B. C.
D.
MTERM2 Page 1 Version 0
11. 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
12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?
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:
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
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?
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
MTERM2 Page 2 Version 0
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.
MTERM2 Page 3 Version 0
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200068237
40.91
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?
5.9 x 103 N
7.9 x 103 N
0 N
2.0 x 103 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
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
MTERM2 Page 1 Version 2
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
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
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?
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
MTERM2 Page 2 Version 2
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.
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.
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?
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
12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?
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:
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 Page 3 Version 2
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?
3.57 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
15. How much heat energy is produced because of friction?
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.
MTERM2 Page 4 Version 2
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
MTERM2 Page 5 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200068333
86.36
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
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?
8.0 N
16 N
2.0 N
6.0 N
A.
B. C.
D.
MTERM2 Page 1 Version 2
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
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
MTERM2 Page 2 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200070267
86.36
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?
8.0 N
16 N
2.0 N
6.0 N
A.
B. C.
D.
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
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.
MTERM2 Page 1 Version 2
13. Solution:
Ek = E
p
Ek = mgh
Ek = 980 J
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
MTERM2 Page 2 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200078546
90.91
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.
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
MTERM2 Page 1 Version 1
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
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.
MTERM2 Page 2 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200089917
59.09
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?
5.9 x 103 N
7.9 x 103 N
0 N
2.0 x 103 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
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.
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.
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?
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
MTERM2 Page 1 Version 2
12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?
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:
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
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?
3.57 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
MTERM2 Page 2 Version 2
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.
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.
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?
11 m/s
14 m/s
7.3 m/s
8.7 m/s
A.
B. C.
D.
MTERM2 Page 3 Version 2
21. 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
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval I.
The power is highest in Interval III.
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.
MTERM2 Page 4 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 200092542
36.36
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
3. A cable joins the two masses shown below and passes over the frictionless peg 'P'. What is the
acceleration of the system?
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
MTERM2 Page 1 Version 1
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?
31 kg
23 kg
15 kg
6.0 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
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 Page 2 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
6. When the speed of a truck is doubled its momentum changes by a factor of:
1/2
1/4
4
2
A.
B. C.
D.
Solution:
momentum is directly proportional to velocity, so doubling v doubles p
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
MTERM2 Page 3 Version 1
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.
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.
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
MTERM2 Page 4 Version 1
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 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
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
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
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
MTERM2 Page 5 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
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?
5 390 J
0 J
539 J
4 850 J
A.
B. C.
D.
MTERM2 Page 6 Version 1
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
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
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
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
50 m
71 m
200 m
A.
B. C.
D.
MTERM2 Page 7 Version 1
20. 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.
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
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?
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=W÷t, where W = Fxd
P = (mg) x d ÷ t
P = 50 x 9.81 x 10 ÷ 2
P = 2.45 x 103 W
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
A.
B. C.
D.
MTERM2 Page 8 Version 1
23. 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.
MTERM2 Page 9 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300050963
77.27
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
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?
8.0 N
16 N
2.0 N
6.0 N
A.
B. C.
D.
MTERM2 Page 1 Version 2
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
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
MTERM2 Page 2 Version 2
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.
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
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval I.
The power is highest in Interval III.
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.
MTERM2 Page 3 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300064630
90.91
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?
31 kg
23 kg
15 kg
6.0 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
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.
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
MTERM2 Page 1 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300071844
77.27
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.
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
MTERM2 Page 1 Version 1
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.
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 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
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
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.
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
MTERM2 Page 2 Version 1
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?
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=W÷t, where W = Fxd
P = (mg) x d ÷ t
P = 50 x 9.81 x 10 ÷ 2
P = 2.45 x 103 W
MTERM2 Page 3 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300074382
63.64
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
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.
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.
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
MTERM2 Page 1 Version 1
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
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
MTERM2 Page 2 Version 1
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
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 Page 3 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300075780
72.73
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.
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
MTERM2 Page 1 Version 1
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.
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
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.
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
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
50 m
71 m
200 m
A.
B. C.
D.
MTERM2 Page 2 Version 1
20. 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.
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
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.
MTERM2 Page 3 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300077994
81.82
6. When the speed of a truck is doubled its momentum changes by a factor of:
1/2
1/4
4
2
A.
B. C.
D.
Solution:
momentum is directly proportional to velocity, so doubling v doubles p
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
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.
MTERM2 Page 1 Version 1
21. 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
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
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.
MTERM2 Page 2 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300080619
86.36
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?
8.0 N
16 N
2.0 N
6.0 N
A.
B. C.
D.
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
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
MTERM2 Page 1 Version 2
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?
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
MTERM2 Page 2 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300082738
45.45
3. A cable joins the two masses shown below and passes over the frictionless peg 'P'. What is the
acceleration of the system?
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
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?
31 kg
23 kg
15 kg
6.0 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
6. When the speed of a truck is doubled its momentum changes by a factor of:
1/2
1/4
4
2
A.
B. C.
D.
MTERM2 Page 1 Version 1
6. Solution:
momentum is directly proportional to velocity, so doubling v doubles p
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
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.
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.
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
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.
MTERM2 Page 2 Version 1
12. 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
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.
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
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
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.
MTERM2 Page 3 Version 1
19. Solution:
W = ∆Ek
W = 0.5 m (v2final
- v2initial
)
W = 0.5 x 250 (452 - 102)
W = 2.4 x 105 J
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?
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=W÷t, where W = Fxd
P = (mg) x d ÷ t
P = 50 x 9.81 x 10 ÷ 2
P = 2.45 x 103 W
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
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.
MTERM2 Page 4 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300083653
90.91
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
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.
15. How much heat energy is produced because of friction?
1.76 J
1.84 J
0.074 J
0.081 J
A.
B. C.
D.
MTERM2 Page 1 Version 2
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300085153
63.64
3. A cable joins the two masses shown below and passes over the frictionless peg 'P'. What is the
acceleration of the system?
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
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
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 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
Its momentum is different but its kinetic energy is the same
A.
B. C.
D.
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.
MTERM2 Page 1 Version 1
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 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
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
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
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.
MTERM2 Page 2 Version 1
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
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
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval III.
The power is highest in Interval I.
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.
MTERM2 Page 3 Version 1
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300089216
90.91
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.
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
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
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?
0 J
539 J
4 850 J
5 390 J
A.
B. C.
D.
MTERM2 Page 1 Version 0
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
MTERM2 Page 2 Version 0
School District #23 Science 10Science Dept.
INDIVIDUAL FEEDBACK Report 300092679
72.73
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
12. What is the total energy of a 850 kg sea plane flying at 38 m/s and 20 m above the ocean?
6.13 x 105 J
4.47 x 105 J
1.66 x 105 J
7.79 x 105 J
A.
B.
C.
D.
MTERM2 Page 1 Version 2
12. 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
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.
15. How much heat energy is produced because of friction?
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
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?
11 m/s
14 m/s
7.3 m/s
8.7 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 Page 2 Version 2
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 II.
The power is the same in each of the three intervals.
The power is highest in Interval I.
The power is highest in Interval III.
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.
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