Download Solutions to semester2 practice problems - Head

Physics 9 - Semester 2 Exam review
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Practice problems (Chapters 2 - 37)
1. A car is at rest, then it accelerates at a constant rate of 4 m/s/s for 5 seconds.
a. how fast is the car moving after exactly 5 seconds?
a = ∆v/∆t
==> v = 20m/s
2. A ball is dropped (v = 0) from a height of 45 meters.
a. how long does it take the ball to reach the ground?
d = (1/2)at2 ==> t = 3 sec
b. what is the speed of the ball the instant before it hits the ground?
a = ∆v/∆t
==> v = 30 m/s
3. A ball is thrown horizontally at 15 m/s from a window 45 meters high.
a. how long does it take the ball to reach the ground?
d = (1/2)at2 ==> t = 3 sec
b. how far from the building does the ball land?
dx = (vx )t
==> dx = 45 m
4. A 2 kg object is on earth.
a. what is the weight of the object?
mg = 20N
b. what is the mass of the object?
m = 2 kg
c. what is the force of gravity acting on the object?
mg = 20N
d. what is the acceleration when it falls (ignoring air resistance)?
a = 10 m/s/s
e. what is the value of “g” for this object?
g = 10 m/s/s
5. A 1 kg ball and a 10 kg ball of the same size are dropped from 50 meters above the
earth without air resistance.
a. After falling exactly 2 seconds,which ball feels the greater force of gravity?
the 10 kg ball
b. After falling exactly 2 seconds,which ball experiences the greater acceleration?
both the same
c. After falling exactly 2 seconds,which ball experiences the greater net force?
the 10 kg ball
d. Which ball reaches the ground first?
both the same
e. Which ball has a greater speed the instant before it reaches the ground?
both the same
6. A 1 kg ball and a 10 kg ball of the same size are dropped from the same height on earth
with air resistance present.
a. After falling exactly 2 seconds,which ball feels the greater force of gravity?
the 10 kg ball
b. After falling exactly 2 seconds,which ball experiences the greater acceleration?
the 10 kg ball
c. After falling exactly 2 seconds,which ball experiences the greater net force?
the 10 kg ball
d. Which ball reaches the ground first?
the 10 kg ball
e. Which ball has a greater speed the instant before it reaches the ground?
the 10 kg ball
7. How much net force is needed to keep a 1,000 kg car moving at a constant velocity of
15 m/s?
net Force = ma = (1000 kg)(0 m/s/s) = zero ... none!
[note the correction - Thanks Erica!]
8. A car has a maximum acceleration of 4 m/s/s.
a. what is the maximum acceleration if the engine is replaced by one that
produces twice the force?
2 x F ==> 2 x a
so: 8 m/s/s
b. what is the maximum acceleration if, instead of replacing the engine, the car’s mass is
doubled by loading it with cargo?
2 x m ==> (1/2) x a
so: 2 m/s/s
9. A 50 kg rocket is being pushed upward by a thrust of 3,000 N, while at the same time
being pulled downward by the force of gravity.
a. Determine the weight of the rocket
Wt. = mg = 500N
b. Determine the net force acting on the rocket
net F = the sum of the forces acting on the rocket
net F = Thrust - Weight = 3,000 N - 500 N = 2,500 N
c. Determine the acceleration of the rocket
net F = ma
2,500 N = (50 kg)(a)
so: a = 50 m/s/s
10. A girl exerts a force of 30 N to push a 20 kg box across the floor at a constant speed
of 2 m/s.
a. determine the acceleration of the box
a = ∆v/∆t = 0/∆t
==> a = 0
b. determine the net force on the box
net F = ma = 0
==> net F = 0
c. determine the amount of friction present as the box slides
push - friction = net F = 0
==> friction = 30 N
d. how hard must the girl push to accelerate the box at a constant 3 m/s/s?
net F = ma = (20 kg)(3 m/s/s) = 60 N
push - friction = 60 N
push - 30 N = 60 N
==> push = 90 N
11. If a second crate was stacked on top of the crate in #10, how much friction would be
present?
double the wt ==> double the friction so: friction = 60 N
12. A 20 kg box is suspended by two vertical strings. How much tension is in each string?
each tension = (1/2)(weight) = 100N
13. A 20 kg box is suspended by two strings angled at 45 degrees.
a. What is the vertical component of tension in each string? 100 N
b. What is the horizontal component of tension in each string?
100 N
c. What is the actual tension in each string?
141 N (pyth. thm)
14. A 2 kg box is pulled by a horizontal rope and accelerated at 3 m/s/s on a frictionless
surface. Determine the tension in the rope.
net F = T = ma = 6 N
15. Two 2 kg boxes connected by a rope are pulled by a second rope and accelerated at
3 m/s/s on a frictionless surface. Determine the tension in each rope.
pull
Left Tension = ma = (2 kg)(3 m/s/s) = 6 N
Right Tension = ma = (4 kg)(3 m/s/s) = 12 N
16. A 2 kg glider rests on a horizontal air track is attached by a string that runs over a
pulley and a hanging 0.5 kg mass. Determine the accelerations of the glider and
hanging mass.
accel. of the system = (net force on the system)/(mass of the system)
a = (5 N)/(2.5 kg) = 2 m/s/s [note the correction! - thanks Erica!]
17. Identify two “action/reaction pairs” involved when a hammer hits a tent stake into the ground.
1. hammer pushes stake down/stake pushes hammer up
2. stake pushes ground down/ground pushes stake up
(also: Earth pulls stake down/stake pulls Earth up)
18. A 2 kg box sits at rest on a table.
a. determine the force of gravity acting on the box
mg = 20 N
b. determine the support force acting on the box
20 N
c. are these two forces an action/ reaction pair?
No!! (A/R pairs act on differnet objects and therefore never cancel)
19. Imagine a Honda Odyssey loaded with bricks (5,000 kg) moving at 20 m/s collides
head-on, bumper to bumper with an empty Honda Odyssey (2,500 kg) at rest.
a. Which car experiences a greater force?
Neither: the cars experience equal and opposite forces!
b. Which car experiences a greater acceleration?
the less massive car
c. Which car experiences more damage
Neither: the cars experience equal and opposite forces!
20. If action/reaction pairs are equal and opposite, why does a football accelerate when
you kick it?
action/reaction pairs don’t cancel (see #18 c)
21. A 2 kg ball (initially at rest) is accelerated to a velocity of +30 m/s (due north).
a. What is the initial momentum? the final momentum?
initial p = mv = 0
final p = mv = +60 kgm/s [north]
b. What is the change in momentum?
∆p = m∆v = +60 kgm/s [north]
c. What is the impulse experienced by the ball?
I = ∆p = +60 kgm/s [north]
22. When a 2 kg ball of clay moving at + 20 m/s hits a wall and sticks,
a. what is the initial momentum of the ball of clay?
p = mv = +40 kgm/s
b. what is the change in momentum of the ball when it strikes the wall?
∆p = final p - initial p = - 40 kgm/s (note the neg. sign!)
c. what is the impulse experienced by the ball of clay when it hits the wall?
impulse = ∆p = - 40 kgm/s
d. what is the impulse experienced by the wall when the ball hits it?
impulse = +40 kgm/s
e which object experiences a greater force during the collision?
equal and opposite forces!
f. is this collision inelastic or elastic?
inelastic
23. A 100 kg skater moving to the right at 6 m/s colides and hangs onto a 50 kg skater moving
to the left at 6 m/s.
a. What is the total momentum before the collision? ... after the collision?
before: m1v1 + m2v2 = +600 kgm/s + (-300 kgm/s) = +300 kgm/s
after: conservation of momentum = +300 kgm/s
b. What is the final velocity of the pair of skaters?
p = mv
==> +300 kgm/s = (150 kg)(v) ==> v = +2m/s
c. How much impulse experienced by each skater?
100 kg skater:
I = ∆p = final p - inital p = (100kg)(+2m/s) - (100kg)(+6m/s)
I = - 400 kgm/s
50 kg skater:
I = + 400 kgm/s
d. If the actual collision lasts 0.2 seconds, how much force does each skater experience?
100 kg skater:
I = Ft
- 400 kgm/s = (F)(0.2 sec)
F = - 2000 N
50 kg skater:
F = + 2000 N
24. A box on the floor weighing 200N is lifted 2 meters into the air, and held there.
a. How much work is done on the box?
W = Fd = (200N)(2m) = 400 J
b. What is the change in potential energy?
PE = mgh = 400J
or W = ∆PE = 400J
c. If the box is dropped from this height, what will it’s kinetic energy be the instant
before it hits the ground?
initial PE = final KE
==> KE = 400J
25. A 2 kg ball is thrown horizontally at a speed of 20 m/s from a window 25m high.
a. What is the total mechanichal energy of the ball
the instant after it is thrown?
TME = PE + KE = 900 J
half way to the ground?
TME = constant
==> 900J
the instant before it hots the ground?
TME = 900 J
b. What is the speed of the ball the instant before it hits the ground?
TME = KE = 900J ==>v = 30m/s
c. Does the mass of the ball affect your answers to (a) or (b)?
mass does effect the answer to (a)
mass does not effect answer to (b)
26. A 1,000 kg car moving at 2 m/s skids to a stop in 4 meters. Find the force of friction present.
W = ∆E
(F)(d) = ∆KE = (1/2)mv2
(F)(4m) = (1/2)(1,000 kg)(2)2
==> F = 500N
27. An inclined plane has dimensions: vertical ht = 2 m & diagonal length = 8 meters.
a. Determine the theoretical mechanical advantage of this simple machine
TMA = (input dist.)/(output dist.) = 8/2
==> TMA = 4
b. How much force is needed to slide a block of ice weighing 200N up the 8 meter
incline? (no friction)
Force = (200N)/4 = 50 N
c. Later, you notice it takes a 75 N force to slide a 200N wooden crate up the 8 meter
incline. How much friction is present?
75 N - 50 N = 25 N
d. What is the actual mechanical advantage in (c)?
AMA = (output force)/(input force) = (200 N)/(75 N) = 8/3 = 2.67
e. What is the efficiency of the inclined plane in (b)? in (c)?
in (b): 100%
(anytime there is no friction!)
in (c): eff. = AMA/TMA = 2.67/4 = 2/3 = 0.67 or 67%
28. A 60 ohm light bulb is connected to a 120 V outlet.in your house.
a. how much current flows through the bulb?
I = V/R = 2A
b. is the current AC or DC?
AC
(in your house wiring)
c. what is the power of the bulb?
P = IV = (2A)(120V) = 240 W
29. How much energy (in kWh) do you use to keep a 100W bulb lit for 20 hours?
P = E/t
so: E = (P)(t) = (0.1kW)(10h) = 1kWh
or: E = (P)(t) = (100J/s)(36000 s) = 3,600,000 J
30. A flashlight uses a single 6V battery connected to a single 60 ohm bulb.
a. How much current flows thru the bulb?
I = V/R = 0.1 A
b. How many coulombs pass thru the bulb every 10 seconds?
1 coulomb
c. How many joules of energy does the battery give each coulomb that passes thru it?
6 Joules
31. A 4 ohm and 8 ohm resistor are all connected in SERIES to a 6 volt battery.
a. what is the total resistance of the circuit?
12 ohms
b. How much current flows thru the battery and each resistor?
I = V/R = 6V/12 ohms = 0.5 A
c. How much does the voltage drop across each resistor?
8 ohm:
V = IR = (0.5A)(8 ohms) = 4V
4 ohm:
V = IR = (0.5A)(4 ohms) = 2V
32. A 3 ohm and 6 ohm resistor are all connected in PARALLEL to a 6 volt battery.
a. How much does the voltage drop across each resistor?
6 V each
b. How much current flows thru the battery and each resistor?
3 ohm:
I = V/R = 6V/3 ohms = 2A
6 ohm:
I = V/R = 6V/3]6 ohms = 1A
c.. What is the total resistance of this circuit?
total I = (total V)/R
3A = 6V/R ==> total R = 2 ohms
33. Draw magnetic the field lines around a bar magnet
Look in your notes!
(remember that the field lines point from N to S outside a magnet, S to N inside)
34. Draw the magnetic field lines around a wire carrying current out of the page.
Look in your notes!
(LH rule: clockwise concentric circles)
35. An electron moves between the poles of a horseshoe magnet. Determine the direction of
force on the electron.
N
S
LH rule#2: Force is out of the page (toward you!)
36. The primary coil of a transformer is plugged into the wall (120V AC), while the secondary coil
is connected to a cell phone. There are 1000 turns in the primary coil and 100 turns in the
secondary coil.
a. What is the voltage in the secondary coil?
1/10 the turns ==> 1/10 the V
so: V = 12V
b. If the curent in the primary coil is 1 amp, what is the current in the secondary coil?
1/10 the turns ==> 10 times the I
so: I = 10 A
37. Describe how a DC motor works, how a speaker works, how a generator works,
how a microphone works, how a transformer works.
Look in your notes!
(remember that motors and speakers take electical energy to produce KE)
(remember that generators and microphones take KE and turn it into electrical energy)