SPH4U Unit 2 Key Questions
Daniel Ko
17. The process in which electricity is generated
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A)
Given – D = 10.0 m
F = 100.0N , ∠ = 45o
Ff = 30.0N , ∠ =180o
Required – W done by the child
Analysis and solution W = Fdcos∠
W = (100.0 N)(cos45)(10.0m)
W = 707.11
W = 707 J
Paraphrase – The child did 707 J of work on the sled
Required – W done by friction
Analysis and solution W = Fdcos∠
W = (30.0 N)(cos180)(10.0m)
W = -300 J
Paraphrase – The friction did –300 J of work on the sled
Required – W total
Analysis and solution –
Wtotal = W boy + W friction
Wtotal = (707 J) + (-300 J)
Wtotal = 407 J
Paraphrase – Therefore, the net work done on the sled is 407 J.
18b)
Given – m = 47.0 kg , W = 407 J , v1 = 0
Required – v2
Analysis and solution W = ∠Ek
W = Ek2 - Ek2
W = ½mv22-0 J
407 J = ½(47.0kg)v2f
(v1 was not included because it is 0)
Vf = √2(407 J) / 47 kg
Vf = 4.16m/s = 4 m/s
Paraphrase – Therefore, the final speed of the sled would be 4 m/s
19a)
Given M = 47.0 kg
H = 10.0 m
Required – Eg
Analysis and solution Eg = mgh
Eg = (47.0kg)(9.8N/kg2)(10.0m)
Eg = 4606 J
Paraphrase – Therefore, the child on the sled gains 4606 J of total mechanical energy.
B) Etotal = 4606 J = Ek
Ek = 1/2mv2
4606 J = ½(47.0kg)v2
V= √(2(4606 J) / (47.0 kg)
V = 14 m/s
Paraphrase – Therefore, the child's speed at the bottom of the hill is 14 m/s
C) I believe that there is so much more than just a calculation like I explained in unit 1 with
similar questions in regards to unexpected conditions. Obstacles, the weather, the wind blowing
can all have an impact on the friction, external pushes and energy in a real-life situation.
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Lesson 6
20a)
Given – k = 225N/m , x = 12.0cm = 0.12 (hookes law)
Required – F
Analysis and solution –
F = kx
F = (225N/m)(0.12cm)
F = 27 N
Paraphrase – Therefore, 27 N force must be exerted on the spring to compress it to 12.0cm
B) Given – x = 12.0cm = .12cm , k = 225N/m
Required – Ee
Analysis and solution Ee = ½kx2
Ee = ½(225N/m)(.12cm)2
Ee = 1.62 J
Paraphrase – Therefore, 1.62 J of work is needed on the spring to compress it to 12.0cm
C) The amount of elastic energy stored in the spring while compressed would be equal to the
amount needed on it (1.62 J of work). The answer would be the same as question B.
D) Given – Etotal = 1.62 J , K = 225N/m , m = 1.5kg
Required – Maximum v
Analysis and solution – The maximum speed of the block will occur when kinetic energy is the
greatest, which is when the spring passes its equilbrum position (x=0, therefore there is no
EBE)
Etotal = Ek = ½mv2
1.62 J = ½(1.5)v2
V = √[2(1.62 J) / 1.5kg]
V = 1.469 = 1.47m/s
Paraphrase – Therefore, the maximum speed the box will attain at is 1.47m/s
21a)
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I) Where maximum speed will kick in will be when a object gets to a point where kinetic energy
is the greatest. When this happens there is no more EPE (elastic potential energery) because it
has all transferred to kinetic energy.
II) Minimum speed occurs at a point of maximum compression and maximum extension, the
mass momentarily stops to change directions. With the kinetic energy being at 0 J at these
points all energy is in form of EPE (given that velocity is at 0m/s [minimum speed]).
III)
B) Let's take the strings on a guitar for an example. They are held vertically following the shape
on the guitar. So how does damping work on this? When we flick the strings to make a noise,
the strings make a movement and create a sound, the sound is based off the movement and
vibration of the flick to the guitar. The movement is an example of damping and until the
vibration is done and goes back to it's original position.
Lesson 7
22.
Fnett=mv2-mv1
Fnet(0.40s) = (1200kg)(8.0m/s) - (1200kg)(20.0m/s)
Fnet(0.40s) = -14400
-14400/.40s = -36000
Fnet = -36000
Therefore, the average net force acting on the car during the collision is –36000
B) If the car in the test hits a bridge support directly instead of the barrels, the average net force
is going to higher due to the fact a bridge support is a very solid rough object. A barrel is going
to react differently during the collision because its going to move during the impact. A bridge
support will remain and its solid interior and exterior is going to have a difference in the time
during the collision. Therefore, collision timing will be quicker and more damage to the car.
C) Assuming the fact the question asks in regards of antifreeze in Canada is obviously in
regards of the issues that we will face during the winter seasons. It's important to use antifreeze
because just water alone in the barrels will freeze up during the cold Canadian weather giving it
a solid interior during the collision with the car. This is kind of similar to the bridge compared to
the barrel as the barrel is softer inside in regards to the bridge support which is all solid inside
and out. Now that the barrel is solid inside its going to make the collision timing quicker and
have more damage to the car.
4
23.
Pto = Ptf
0+
0=(2000)v+(25)(250)
24.
A) In earths frame of reference, the carts are initially shown below
Cart 1 [] -> +2.0 m/s
-5.0 m/s <- [] Cart 2
To switch to a reference frame where cart 2 is at rest, I must add +5.0m/s to each velocity.
Cart 1 [] -> +7.0 m/s
0m/s <- [] cart 2
Now, solve the problem in this new reference frame.
The moving cart is mass 1 and the stationary cart is mass 2
V1f = m1-m2/m1+m2(v1o)
= .80-.60/.80+.60(7m/s)
= 1 m/s
Now I must switch back to earth's rotation by adding the –5.0m/s, the opposite of what I added
before. Cart 1 will move at 1m/s + (-5.0m/s) = -4 m/s = 4 m/s [W] after the collision
For cart 2 V2f = 2m1 / m1 + m2 (v1o)
= 2(.80) / .80 + .60 (7)
= 8m/s
Now I must switch back to the earth's rotation by adding the –5.0m/s, the opposite of what I
added before. Cart 2 will move at 8m/s + (-5.0m/s) = 3.0m/s, therefore V2f = 3.0 m/s [E]after the
collision.
B)
First, I must find how fast the carts are moving at the point where the spring is at maximum
compression. At this point, the two carts must have the same velocity.
PTo = PTf
M1v1o + m2v2o = (m1+m2)vf
(.80kg)(2.0m/s) + (.60kg)(-5.0m/s) = (.80kg + .60kg)vf
1.4kgm/s = (1.4kg)vf
Vf = 1 m/s
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At a maximum compression, the two carts are moving at 1 m/s.
Now use conversation of energy to find the maximum compression.
½m1v1o2+1/2m2v2o2=½(m1+m2)v2f2+½kx2
½(.80kg)(2.0m/s)2+½(.60kg)(5.0m/s)2=½(.80kg+.60kg)(1)2+½(1200)x2
9.1 = 600.7x2
I don’t know how to solve for the rest of this question please help
25.
To find the velocity I must also find the total kinetic energy
Eto = Etf
Eto = (m1+m2)gh
Eto = (65kg+45kg)(9.8m/s)(30)
Eto = 32340 J
The total kinetic energy is 32340 J
Now I can find the final velocity.
Ektf = ½(m1 + m2)v22f
32340 J = ½(65kg+45kg)v22f
32340 J = 55v22f
=588v22f
V22f =√588
V22f = 24.24 m/s
The final velocity of the skiers after the collision is 24.24m/s
Lesson 8
26.
A)
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Fnetx = m (v2x - v1x)
Fnetx = (.152kg)(-48.864m/s - 32.0m/s)
Fnetx = -12.291 = 12.3N
Fnetx = m (v2x - v1x)
Fnetx = (.152kg)(17.785m/s - 0)
Fnetx = -2.703 = 2.7
F = √(12.32 + 2.72)
F = 12.592 = 12.6
Tan∠ = 2.7 / 12.6
∠ = 12.38 = 12.4
The impulse experienced by the ball is 12.6kgm/s [W ∠12.4 N]
B) Fnet t = ∠P
Fnet (.00200s) = (12.6kgm/s)
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Fnet = 6300 N
The average net force of the ball is 6300 N
27.
Drawing the given velocity vectors
For the x-component
Ptox = Ptfx
M1v1ox + m2v2ox = (m1 + m2)vfx
(20g = .020kg)
(.800kg)(-3m/s)+(.02kg)(0)=(.82kg)vfx
(-2.4)=(.82kg)vfx
Vfx = -2.926 = 2.9 m/s
For the y-component
Ptoy = Ptfy
M1v1oy + m2v2oy = (m1 + m2)vfy
(.800kg)(0)+(.02kg)(260m/s)=(.82kg)vfy
Vfy = 6.341 = 6.3m/s
Finding the final velocity using Pythagorean theorem
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V3f = √(2.9m/s2 + 6.3m/s2)
= 6.935 = 6.9m/s
To find the direction
Tan-1∠ = 6.3/2.9
∠ = 65.28 = 65.3
The final velocity after hitting the target is 6.9m/s [W ∠65.3 N]
28.
Ptox = Ptfx
9
M1v10 + m2v2ox = m1v1fx + m2v2fx
(.30kg)(-5m/s) + (.40kg)(0) = (.30kg)(-3.64m/s)+(.40kg)v2fx
v2fx = -1.0 m/s
Ptoy = Ptfy
M1v10 + m2v2oy = m1v1fy + m2v2fy
m(0) + m(0) = (.30kg)(2.1m/s)+(.40kg)v2fy
V2fy = -1.6m/s
Finding the final velocity using Pythagorean theorem
V2f = √(1.0m/s2+1.6m/s2)
V2f = 1.886 = 1.9 m/s
To find the direction
Tan-1∠ = 1.6m/s / 1.0 m/s
∠ = 57.994 = 58
The final velocity of puck B 1.9m/s [W ∠58 S]
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