BCLN PHYSICS 11 - Rev Sept/17
Physics 11 Final Review
Name
1. What is the momentum of a 0.046 kg golf ball, travelling at 22 m/s?
p = mv
p = 0.046 kg x 22 m/s = 1.0 kgŸm/s
2. A railroad car of mass 12 000 kg is travelling at a speed of 6.0 m/s when it collides with an
identical car at rest. The two cars lock together. What is their common speed after the
collision?
𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚!&# 𝑣!&#
12000 kg x 6.0 m/s + 0 = (12,000 kg + 12,000) V2
72,000 = 24,000 V2
V2 = 72,000/24,000 = 3.0 m/s
3. A 4.2 kg rifle shoots a 0.050 kg bullet at a speed of 3.00 x 102 m/s. At what speed does the
rifle recoil?
𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚! 𝑣! + 𝑚! 𝑣!
0.050 kg x - 3.00 x 102 m/s + 0 = (0.050kg + 4.2 kg) V2
- 15 = 4.25 V2
V2 = - 3.5 m/s
4. A 0.250 kg ball of Plasticine™ moving 5.0 m/s overtakes and collides with a 0.300 kg
ball of Plasticine™, travelling in the same direction at 2.0 m/s. The two balls of
Plasticine™ stick together on collision. What is their speed after the collision?
𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚!&# 𝑣!&#
0.250 kg x 5.0 m/s + 0.300 x 2.0 m/s = (0.250 + 0.300) V2
1.85 = 0.55 V2
V2 = 1.85 / 0.55 = 3.4 m/s
5. A 53 kg skateboarder on a 2.0 kg skateboard is coasting along at 1.6 m/s. (See figure
below) If he collides with a stationary skateboarder of mass 43 kg (also on a 2.0 kg
skateboard), and the two skateboarders coast off in the same direction that the first
skateboarder was travelling, what velocity will the combined skateboarders now have?
𝑚! 𝑣! + 𝑚! 𝑣! = 𝑚!&# 𝑣!&#
55kg x 1.6m/s + 0 = (55kg + 45kg) V2
88 = 100 V2
V2 = 88/100 = 0.88 m/s
6. What impulse is needed to change the speed of a 10.0 kg object from 12.6 m/s to 25.5 m/s in a
time of 5.00 s? How much force is needed?
𝑰 = ∆(𝒌𝒈 ×𝒎/𝒔)
I = 10.0 kg x (25.5 m/s – 12.6 m/s) = 10 x 12.9 = 129 Ns
F = I/t
F = 129/5.00s = 25.8 N
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BCLN PHYSICS 11 - Rev Sept/17
7. a) What impulse must be imparted by a baseball bat to a 145 g ball to change its velocity from
40.0 m/s to -50.0 m/s? (Why is there a minus sign here?) ball has reversed direction
b) If the collision between the baseball and the bat lasts 1.00 ms, what force was exerted on the
ball?
𝑰 = ∆(𝒌𝒈 ×𝒎/𝒔)
I = 0.145 kg x (40.0 m/s – – 50.0 m/s) = 0.145 x 90 = – 13.1 Ns
F = I/t
F = – 13.1 / .001s = =1.31 x 104 N
8. If an impulse of 2.0 Ns is applied to a 0.046 kg golf ball, what speed will it have immediately
after the impulse is applied?
𝑰 = ∆(𝒌𝒈 ×𝒎/𝒔)
2.0 Ns = 0.046 x v
v = 2.0/0.046 = 43 m/s
9. The force of gravity on a box of apples is 98.0 N. How much work will you do
(a) if you lift the box from the floor to a height of 1.2 m?
W = Fd
W = 98.0 N x 1.2 m = 1.2 x 102 J
(b) if you carry the box horizontally a distance of 2.0 m?
No work, 0 J
10. The force of gravity on a box is 100.0 N. The coefficient of friction between the floor and the box
is 0.250. How much work is done when the box is pushed along the floor, at a steady speed, for a
distance of 15.0 m?
W = Fd
100N x 0.25 = 25N x 15.0m = 375 J
11. How much energy is consumed by a 100.0 W light bulb, if it is ‘left on’ for 12.0 h?
12.0 h x 60 x 60 = 43,200 s
P = ΔE / t
ΔE = P x t = 100 x 43200 = 4.32 x 106 J
12. A motor does 25 MJ (megajoules) of work in one hour. One hour equals 3600 seconds
(a) What is the power rating of the motor?
P = ΔE / t
P = 25,000,000 / 3600 = 6.9 x 103 W
(b) How many horsepower is this motor, if 1 HP = 750 W?
6900 W / 750 W = 9.2 HP
13. Discuss the scientific accuracy of this statement:
“I used a ramp to get my motorbike up on my truck, and the ramp saved me a lot of work!”
The ramp does not save work. For work to have scientific meaning there must be force acting
on an object and the object must move a distance in the direction of the force. Using the ramp
increases the distance and therefore lowers the energy required to load the bike.
14. A pendulum bob is moving 1.8 m/s at the bottom of its swing. To what height above the bottom
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BCLN PHYSICS 11 - Rev Sept/17
of the swing will the bob travel?
mgh = ½ mv2
h = v2 / 2 x 9.8
gh = ½ v2
h = 3.24 / 19.6 = 0.17m
masses cancel so
h = (1.8)2 / 19.6
15. The initial speed of a golf ball, struck with a driver by professional golfer Tiger Woods, has been
measured to be 285 km/h. From what vertical height would you have to drop a golf ball, if it is to
reach this same speed as it hits the ground? Assume that air friction can be ignored.
285 km/h divided by 3.6 = 79.2 m/s
a = 9.8 m/s2
d = v2 / 2a
d = (79.2)2 / 19.6 = 320 m
16. A skier has 60 kJ of gravitational potential energy when at the top of the hill. Assuming no
friction, how much kinetic energy does she have when she is one-third of the way down the hill?
Ep = Ek
60kJ / 3 = 20 kJ
17. The head of a golf club imparts a certain amount of kinetic energy to the ball upon impact. Let
this be Ek. If the golfer lightens the mass of the club head by 1/3, and increases the club head
speed so that it is 3 times it previous speed, how much kinetic energy will be imparted to the ball
now?
There will be 3 times the kinetic energy 3Ek. If his speed (v) before was 1.00 m/s then 3 times
that is 3.0 m/s.
18. A steel rod is at a temperature of 25oC. To what Celsius temperature must you raise it, in order to
double its Kelvin temperature?
K = OC + 273
O
C = K – 273
25 OC + 273 k = 298 x 2 = 596 – 273 = 323 OC
19. A certain metal has a specific heat capacity of 420 J/kg/Co, while water has a specific heat
capacity of 4200 J/kg/Co. A kilogram of the metal and a kilogram of water are both at a
temperature of 98oC. If both are allowed to cool to 18oC, which will give off more heat to the
atmosphere, and how much more will it release?
4200 J/kg/ OC divided by 420 J/kg/ OC = 10
than the metal.
The water will give off 10 times more heat
20. What is the efficiency of a 1500 W kettle if it supplies heat at the rate of 1400 W to the water in
it?
% Efficiency = (work out / work in) x 100% 1400/1500 x 100% = 93 %
21. A 1200 W kettle warms 800.0 g of water from 20oC to 99oC in 4.0 min. How efficient is the
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kettle?
E=Pxt
t = 4 x 60 = 240 s
800.0 g = 0.80 kg
E = 1200 W x 240 s = 288,000 J
Eh of water = Eh = m c Δt
BCLN PHYSICS 11 - Rev Sept/17
= 0.80 kg x 4200(water heat capacity) x (99-20) = 265,440 J
% Efficiency = (work out / work in) x 100% 265,440/288,000 x 100% = 92 %
22. If 24 kJ of energy will warm 0.600 kg of a metal from 20° up to 220°C, what is the specific heat
capacity of the metal?
E = 24 kJ = 2.4 x 104 J
m = 0.600 kg Δt = tf – ti = 220 - 20 = 200 OC
Eh = m c Δt
c = E / m Δ t c = 2.4 x 104 J / 0.600 kg x 200 OC
C = 24000 / 120 = 200 J/kg/ OC
23. A golfer wishes to hit his drives further by increasing the kinetic energy of the golf club when it
strikes the ball. Which would have the greater effect on the energy transferred to the ball by the
driver ⎯ doubling the mass of the club head or doubling the speed of the club head? Explain.
Because of the equation Ek = ½ m v2 doubling the speed of the club head would
increase the kinetic energy. If the mass is double it would double the kinetic energy but
if the velocity is doubled then it would quadruple the kinetic energy because the
velocity is squared. 22 = 4
24. How much work must be done to increase the speed of a 12 kg bicycle ridden by a 68 kg rider
from 8.2 m/s to 12.7 m/s? total mass = 12kg + 68 kg = 80 kg
Intitial Ek = ½ mv2 = ½ (80kg) (8.2 m/s)2 = 2690 J
Final Ek = ½ mv2
= ½ (80kg) (12.7 m/s)2 = 6452 J
6452 J – 2690 J = 3762 J = 3.8 x 103 J
25. A vehicle moving with a speed of 90 km/h (25 m/s) loses its brakes but sees a 'runaway' hill near
the highway. If the driver steers his vehicle into the runaway hill, how far up the hill (vertically)
will the vehicle travel before it comes to a stop? (Ignore friction.) If friction is taken into account,
will the vertical distance the vehicle moves be less or greater than the 'ideal' distance you just
solved for, neglecting friction? Explain.
mgh = ½ mv2
masses cancel so
gh = ½ v2
h = v2 / 2 x 9.8
h = (25m/s)2 / 19.6 h = 625 / 19.6 = 32 m
If friction is taken into account then the height is less since energy is lost due to friction.
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BCLN PHYSICS 11 - Rev Sept/17
26. What is the current in the ammeter A shown below?
(ans: 2.0A)
1/Rp = 1/6 + 1/6 = 2/6
Rp = 6/2 = 3 Ω
3Ω+3Ω=6Ω
V = IR
12.0V = 6 Ω x I
I = 12 / 6 = 2.0 A
27. What is the voltage V of the power supply shown
below?
1/Rp = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
Rp = 12/3 = 4 Ω
4 Ω + 8 Ω = 12 Ω
V = IR
amps = 4.0 + 2.0 = 6.0A
V= 12 Ω x 6 = 72V
(ans: 72V)
28. Consider the circuit shown below;
a. What is the voltage across the 8.0 Ω resistor (between 1 and 2)? (ans: 16V)
Resistors in series: 4 Ω + 8 Ω = 12 Ω
1/Rp = 1/12 + 1/6 = 1/12 + 2/12 = 3/12
Rp = 12/3 = 4 Ω
4 Ω + 5 Ω +11 Ω = 20 Ω
V = IR
120.V = I x 20 Ω
I = 120 / 20 = 6.0 A
Therefore 4.0 A across 6.0 Ω resistor and 2.0 A
across both 4.0 Ω & 8.0 Ω resistors in series
2.0 A x 8.0 Ω = 16V
b. How much power is dissipated in the 5.0 Ω resistor? (ans: 180W)
P = I2 R
P = (6.0A)2 x 5 Ω
= 180 W
29. What is the internal resistance of the
battery shown below?
V = IR
I = 4.0 A
V = 12.0 V
Total R = V/I = 12.0 V/ 4.0A = 3 Ω
3 Ω – 2.50 Ω = 0.50 Ω
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BCLN PHYSICS 11 - Rev Sept/17
30. A flashlight contains a battery of two cells in series, with a bulb of resistance 12.0 Ω. The internal
resistance of each cell is 0.260 Ω. If the potential difference across the bulb is 2.88 V, what is the
EMF of each cell? (ans: 1.5V)
V=IR
V = 12.0 Ω + 0.26 Ω + 0.26 Ω = 12.52 Ω
I = 2.88 V / 12. Ω (bulb) = 0.24 A
VT = ( I r)
= 0.24 A x 12.52 Ω = 3.00 V
divided by 2 because 2 batteries in series
so volts add up = 1.5V
31. A 1500 W kettle is connected to a 110 V source. What is the resistance of the kettle element?
(ans: 8.1 Ω)
P = VI
1500W = 110V x I
I = 1500 / 110 = 13.6 A
R = P / I2
R = 1500 W / (13.6A)2 = 1500 W / 185 A = 8.1 Ω
32. A power station delivers 500 kW of power to a village through lines with a total resistance of
5.00 Ω. The input voltage is 400,000 V. If the input voltage was reduced to 10,000 V, by what
factor would the power wasted as heat in the lines be multiplied? (ans: 1600 times as much wasted
power)
P = VI
500,000 = 400,000V x I
I = 500,000 / 400,000V = 1.25 A
PLOST = I2 R PLOST = (1.25A)2 x 5.0 Ω = 7.8 W
P = VI
500,000 = 10,000V x I
I = 500,000 / 10,000V = 50 A
PLOST = I2 R PLOST = (50 A)2 x 5.0 Ω = 12,500 W
12,500 W / 7.8 W = 1600 times
33. What is the efficiency of a transmission line if 20.0 A, input at 120 V, delivers 2.0 x 103 W of
power? (ans: 83%)
V=IR
P = I2 R
R=V/I
R = 120 V / 20 A = 6 Ω
2
P = (20.0 A) x 6 Ω = 2,400 W
Efficiency = (Work output ÷ Work input) x 100%
(2,000W/2,400W) x 100 = 83%
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BCLN PHYSICS 11 - Rev Sept/17
Some Units 1-3 Review
34. A caterpillar travels across the length of a 2.00 m porch in 6.5 minutes. What is the average
speed of the caterpillar? Time = 6.5 x 60 = 390 s
V = d/t
v = 2.00 m / 390 s = 5.13 x 10-3 m/s
35. A cross-country race car driver sets out on a 1 hour, 100 km race. At the halfway marker (50 km), the pit
crew radios that the car had averaged a speed of only 80 km/h.
a. How long did it take the driver to travel the first 50 km?
v = d/t t = d/v
t = 50 km / 80km/h = 0.625 h
b. How fast must the driver drive over the remaining distance in order to average 100 km/h for the entire
race ?
1 h - .625 h = .375 h v = d/t
v = 50km / .375h
= 133 km/h
36. Match the graphs below. Match the appropriate displacement-time graph with the corresponding
velocity-time graph drawn below. Note: Some v-t graphs may be used more than once, or not at
all
a.
1.
graph # 5.
2.
b.
graph # 3.
3.
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BCLN PHYSICS 11 - Rev Sept/17
c.
graph # 1.
4.
d.
V
t
graph # 2.
5.
e.
graph # 3.
37. When a jet airliner lands on a runway it accelerates at -1.9 m/s2 for 55 s until it stops. How fast was the
plane moving when it touched down on the runway?
a = v/t
v=axt
v = 1.9 m/s2 x 55s = 104.5 m/s
38. A team of students competing in the ‘Egg Drop’ competition at the Physics Olympics at UBC, drop their
egg from a ledge, three floors above the ground.
a. If it takes the egg 2.4 seconds to reach the ground, calculate the height of the ledge in meters.
Vi = 0
t = 2.4s
a = - 9.8 m/s2 d = ?
d = vot + ½ at2
d = ½ at2
d = 4.9 x 5.76 = 28.2 m
d = .5 x 9.8 m/s2 x (2.4s)2
b. How fast was the egg traveling the instant before it hit the ground?
Vf = ? t = 2.4s a = -9.8 m/s2
Vf = Vi + at
v=axt
v = -9.8 m/s2 x 2.4s = - 23.5 m/s
c. The students had planned to protect the egg by dropping it into a garbage can filled with
confetti, 0.85 m deep. Assuming that the egg stopped just as it reached the bottom of the can,
calculate the rate of deceleration of the egg.
Vf = 23.5 m/s a = 9.8 m/s2 d= 0.85
Vf2 = Vi2 + 2ad
V2 = 2ad
a = v2/2d a = (23.5)2 /2x0.85m = 325 m/s2
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BCLN PHYSICS 11 - Rev Sept/17
39. The velocity of a jogger is plotted over a 12 second duration and the v-t graph is plotted below:
a. Determine the total displacement of the jogger for these 12 seconds.
1st 4 seconds: (4s x 4 m/s )/2 = 8m
next 3 seconds: (3s x 4 m/s) = 12m
next 2 seconds: (2s x 4 m/s)/2 = 4m
last 3 seconds: (3s x – 6 m/s)/2 = -9m
8 + 12 + 4 – 9 = 15m
b. Construct a d-t graph that will represent the jogger’s motion.
d(m)
time (s)
2
4
6
8
10
12
40. A tragic accident occurred on the Sea-to-Sky highway: a car passed over the edge of the road,
over a cliff and into the ocean. The police measured that the horizontal distance from the ocean
to the highway to be 110m, and that the road was 60 m above the ocean. Use these measurements
to determine if the car was travelling over the posted 80 km/hr speed limit at the time of the
accident.
1) use vertical to calculate time
a = -9.8 m/s2 t = ? d = -60m
d = ½ a t2
-60m = ½ (9.8m/s2) t2
-60m = .5(9.8) t2
t2 = -60/-4.9
t = √12.24 = 3.5 s
2) use time to calculate velocity during horizontal
v = d/t
v = 110m / 3.5 s = 31.43 m/s x 3.6 = 113 km/h yes, car was travelling over
speed limit.
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BCLN PHYSICS 11 - Rev Sept/17
41. Isaac is attempting to cross a river in a kayak. The river flows due east at 1.9 m/s. Isaac points
the kayak due north and paddles at 2.4 m/s (relative to the water). The river is 38 m wide at this
location.
a. Determine the resultant velocity of the boat - both magnitude and direction.
1.9 m/s E
a2 + b2 = c2 (2.4)2 + (1.9)2 = c2
c2 = 5.76 + 3.6
c = √9.37
c = 3.1 m/s
tan-1 θ = 1.9/2.4 = 38O East of North
2.4 m/s N
θ
V=?
b. Determine the time for Isaac to cross the river.
v = d/t
t = d/v t = 38m / 2.4 m/s (Isaac’s velocity) = 16 s
c. How far downstream will the boat be when Isaac reaches the opposite shore?
v = d/t
d=vxt
d = 1.9 m/s (river’s velocity) x 16s = 30 m
42. An Astronaut has a mass of 86 kg. Determine the mass and the weight of this astronaut on the
moon where the gravitational field is one-sixth that of the Earth.
9.8 m/s2 x 86kg = 140 N / 6 = 23N (answer key shows 140N)
43. A physics teacher jumps out of an airplane without a parachute. Seconds later his 82.5-kg body
experiences 118 N of air resistance. Determine his acceleration at this instant in time.
9.8 m/s2 x 82.5kg = 808.5 N - 118N = 690.5 N / 82.5 = 8.37 m/s2 down
44. Maty is attempting to drag her 32.6-kg Shepard across the wooden floor by applying a horizontal
force. What force must she apply to move the dog with a constant speed of 0.95 m/s? The
coefficient of friction between the dog and the floor is 0.72.
Constant speed means all the forces are balanced. Friction force will equal applied force.
F applied = µFn µ = 0.72 Fn = 9.8m/s2 x 32.6kg = 319.48 N
F = 0.72 x 319.48 N= 230 N
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BCLN PHYSICS 11 - Rev Sept/17
45. A 20 kg block rests on a frictionless table. A cord attached to the block extends horizontally to a
pulley at the edge of the table. A 10 kg mass hangs at the end of the cord.
a. Clearly draw and label the force vectors acting on each object.
Fn
Ft
A
Ft
B
mg
mg
b. Calculate the acceleration of the block and mass.
F = ma
a = F/m
a = Box B 10 kg x 9.8 m/s2 = 98/ (both masses 10 + 20) = 98/30 = 3.27 m/s2
c. Calculate the tension in the cord.
FTENSION
= ma = Box A 20kg x 3.3 = N/kg = 65.3 N
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BCLN PHYSICS 11 - Rev Sept/17
ANSWERS
1.
2.
3.
4.
5.
6.
7.
8.
9.
1.0 kg m/s
3.0 m/s
- 3.6 m/s
3.4 m/s
0.88 m/s
129kg m/s or 129 Ns, 25.8 N
a) -13.1 kgm/s, or -13.1 Ns (Ball reverses direction.), b) - 1.31 x 104 N
43 m/s
2
a) 1.2 x 10 J, b) 0J
10. 375 J
11. 4.32 x 106 J
12. (a) 6.9 x 103 W, (b) 9.3 HP
13. The ramp reduces effort force at the expense of effort distance, which increases. The ramp does not save
you work!
14. 0.17 m (or 17 cm)
15. 320 m
16. 20 kJ
17. 3Ek
18. 323oC
19. Water gives off 10 times as much heat.
20. Efficiency = 93%
21. Efficiency = 92%
o
22. 200 J/kg/C
23. Ek = ½ mv2 , Doubling mass will double kinetic energy, but doubling speed will quadruple kinetic energy.
24. 3.8 x 103 J
25. Without considering friction, ½mv2 = mgh.
2
(25 m /s)
v2
= 32 m
h=
= 2(9.8 m /s2)
2g
With friction, the vertical height will be less.
26. 2.0 A
27. 72 V
28. 16V, 180W
29. 0.50 Ω
30. 1.5 V
31. 8.1 Ω
32. 1600x
33. 83%
34. diagram
35. 0.645 m
36. 5.0 x 106 m, or 5000 km!
37. (a) 0.20 Hz, (b) 5.0 s
38. (a) refraction, (b) frequency, (c) Wave speed is reduced by about one half.
39. change in frequency of a wave caused by a source moving in relation to the receiver
40. 335 m/s
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BCLN PHYSICS 11 - Rev Sept/17
41. (a) D, (b) C
42. ∠r = 21°, The beam will leave the glass at 35°.
43. 9. ∠ic = 41.1°
44. 10. n = 1/sin45° = 1.41
45. 1.4
8
46. 2.06 x 10 m/s
47.
48.
49.
50.
51.
52.
53.
54.
Place the filament between f and 2f.
1.7 m (real image)
2.0 m
(a)
Do is greater than 2f.
(b)
Do is between f and 2f.
(c)
Do = 2f
(d)
Do is much less than f.
Inverted, diminished, real image forms between C and F.
22.5 cm
Inverted, enlarged, real image forms beyond C
Di = - 3.75 cm (virtual image)
Hi
Di = 3.75cm = 0.375
Ho = D
10.00cm
o
(Image is diminished and right side up.) Hi = 1.9 cm.
Image is on the same side of the lens as object.
55. vavg. = 5.13 x 10-3 m/s
56. a. t = 0.625 h, b. v = 133 km/h
57. a-5, b-3, c-1, d-2, e-3
58. v = 104.5 m/s
59. a. d = 28.2 m, b. v = -23.5 m/s, c. a = +325 m/s2
60. a. Total Area = 15 m
b.
61. 113 km/h
62. a. 3.1 m/s, 38° east of north, b. 16s, c. 30m
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BCLN PHYSICS 11 - Rev Sept/17
63.
64.
65.
66.
mass = 86kg, weight = 140N
8.37m/s2 down
230N
a. refer to tutorials, b. 3.27 m/s2, c. 65.3N
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