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PHYS 117 Midterm Test Package Solutions Fall 2019

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©CourseCram
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
2019 PRACTICE 1
(note that this official β€œpractice” exam is mainly assembled from real past midterm and final exam
questions, so it’s great practice!)
Q1
C)
DISPLACEMENT
We can see the components of the final displacement
𝑑π‘₯ = (βˆ’3) + 6 = 3 π‘π‘™π‘œπ‘π‘˜π‘ 
𝑑𝑦 = 4 π‘π‘™π‘œπ‘π‘˜π‘ 
so the final displacement is
|𝑑| = βˆšπ‘‘π‘₯2 + 𝑑𝑦2 = √32 + 42 = 5 π‘π‘™π‘œπ‘π‘˜π‘ 
πœƒ = tanβˆ’1 (
𝑑𝑦
4
) = tanβˆ’1 ( ) = 53°
𝑑π‘₯
3
From the diagram we see this is north of east. The answer is C)
2
A)
AVERAGE SPEED, PROJECTILE
Note that we are asked for the average speed. The average velocity of the motion would be zero, as
it ends up back where it started. For speed
it starts at 29.4 m/s
At the highest point it has a vertical speed of 0 m/s, and this is the only direction its moving so the
speed is 0 m/s
it comes back down to the ground at the same speed as it launched, by symmetry, so the speed is
29.4 m/s
So the average speed when it’s going up is
|π‘£π‘Žπ‘£π‘” π‘”π‘œπ‘–π‘›π‘” 𝑒𝑝 | =
𝑠𝑝𝑒𝑒𝑑 π‘Žπ‘‘ π‘ π‘‘π‘Žπ‘Ÿπ‘‘ + 𝑠𝑝𝑒𝑒𝑑 π‘Žπ‘‘ π‘‘π‘œπ‘ |𝑣0 | + 0 29.4
=
=
= 14.7 π‘š/𝑠
2
2
2
Likewise when coming down
|π‘£π‘Žπ‘£π‘” π‘π‘œπ‘šπ‘–π‘›π‘” π‘‘π‘œπ‘€π‘› | =
(𝑠𝑝𝑒𝑒𝑑 π‘Žπ‘‘ π‘‘π‘œπ‘) + (𝑠𝑝𝑒𝑒𝑑 π‘Žπ‘‘ 𝑒𝑛𝑑) 0 + |𝑣0 | 29.4
=
=
= 14.7 π‘š/𝑠
2
2
2
It spends the same amount of time going up and coming down, so
|π‘£π‘Žπ‘£π‘” π‘‘π‘œπ‘‘π‘Žπ‘™ | =
14.7 + 14.7
= 14.7 π‘š/𝑠
2
1
©CourseCram
Q3
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
FORCES, FRICTION, NORMAL FORCE
This is an β€œall for a” problem: when we have a set of blocks all moving together, we can find the
acceleration π‘Ž of all of them by looking at all of them, then find the value we want.
Looking at the whole assembly, the total frictional force will be 3 × 3 = 9 𝑁
The total mass is 3 × 3 = 9 π‘˜π‘”
So in the x-direction
Σ𝐹π‘₯ = π‘šπ‘‘ π‘Ž
+18 βˆ’ 9 = 9π‘Ž
π‘Ž = 1 π‘š/𝑠 2
We want to find the normal force exerted on the middle cube by the left cube. So we must draw a
diagram not including the left cube, so that its force is external on our diagram.
It’s simplest to look at the middle and right cube together. That way the force between the middle
and right cubes is internal and doesn’t have to be included. This leaves us with a mass of two cubes
π‘š2 = 2 × 3 = 6 π‘˜π‘”
The total frictional force due to 2 cubes is 2 × 3 = 6 𝑁 to the left
We now have
Σ𝐹π‘₯ = π‘š2 π‘Ž
+𝑁𝑙𝑒𝑓𝑑 βˆ’ 6 = 6(1)
𝑁𝑙𝑒𝑓𝑑 = 12 𝑁
2
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Q4
A)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
NORMAL FORCE, ELEVATOR
The reading on the scale is the normal force.
The person is riding up, and the elevator is slowing down, so the acceleration is opposite to the
direction of motion – it is downwards, and
π‘Žπ‘¦ = βˆ’2.5 π‘š/𝑠 2
Σ𝐹𝑦 = π‘šπ‘Ž
+𝑁 βˆ’ π‘šπ‘” = π‘šπ‘Ž
𝑁 βˆ’ 70(9.8) = 70(βˆ’2.5)
𝑁 = 511 β‰ˆ 510 𝑁
Q5
D)
RELATIVE MOTION
The boat tries to move 8.0 m/s north, but the current washes it east at 2.0 m/s. So the resultant
velocity is
|π‘£π‘Ÿπ‘’π‘ π‘’π‘™π‘‘ | = √82 + 22 = 8.2 π‘š/𝑠
2
πœƒ = tanβˆ’1 ( ) = 14° πΈπ‘Žπ‘ π‘‘ π‘œπ‘“ π‘π‘œπ‘Ÿπ‘‘β„Ž
8
So this angle is 90 βˆ’ 14 = 76° π‘›π‘œπ‘Ÿπ‘‘β„Ž π‘œπ‘“ π‘’π‘Žπ‘ π‘‘
3
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Q6
A)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
PROJECTILE MOTION
The speed is the total speed due to both x and y components
|π‘£π‘“π‘–π‘›π‘Žπ‘™ | = βˆšπ‘£π‘₯2 + 𝑣𝑦2
The wheel starts with the same velocity as the helicopter
𝑣0π‘₯ = 54 π‘š/𝑠
𝑣0𝑦 = 0 π‘š/𝑠
There is no horizontal acceleration so the horizontal component does not change
𝑣π‘₯ = 𝑣0π‘₯ = 54 π‘š/𝑠
The vertical component when hitting the ground can be found by the equations of motion
2
𝑣𝑦2 = 𝑣0𝑦
+ 2π‘Ž(𝑦 βˆ’ 𝑦0 )
𝑣𝑦2 = 0 + 2(βˆ’9.8)(0 βˆ’ 100)
𝑣𝑦2 = 1960
Taking the square root can give a positive or negative answer. We know that this component will be
negative.
𝑣𝑦 = βˆ’44.3 π‘š/𝑠
and so
|𝑣| = √542 + (βˆ’44.3)2
= 69.8 π‘š/𝑠 β‰ˆ 70 π‘š/𝑠
4
©CourseCram
Q7
A)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
FRICTION (on a slope)
The block does NOT slide, so the force definitely doesn’t involve kinetic friction, so the answer is
NOT B).
The force holding the block in place is the force of static friction. But this is NOT necessarily the
maximum force of static friction! We’re not told that the block is β€œabout to slide” or that we’re β€œat
the steepest angle”, so we can NOT say anything about πœ‡π‘  . Remember that
0 < 𝑓𝑠 < 𝑓𝑠 π‘šπ‘Žπ‘₯ = πœ‡π‘  𝑁
Looking at the free body diagram
The block is not sliding along the tilted x-axis, so
Σ𝐹π‘₯ = 0
βˆ’π‘“π‘  + π‘šπ‘” sin πœƒ = 0
𝑓𝑠 = π‘šπ‘” sin πœƒ
So the answer is A)
BONUS: If we’d been told β€œthe slope is increased until the block is just about to start sliding”, THEN
we could say the static friction was maximum, and
𝑓𝑠 = 𝑓𝑠 π‘šπ‘Žπ‘₯ = πœ‡π‘  𝑁 = πœ‡π‘  π‘šπ‘” cos πœƒ
5
©CourseCram
Q8
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
FRICTION
The sled is being pulled at a constant speed in a constant direction, so it has constant velocity, and
so the acceleration is zero. So the total force is zero.
We’re doing friction so we look in the N direction first
Σ𝐹𝑦 = 0
+𝑁 βˆ’ π‘Š cos πœƒ = 0
𝑁 βˆ’ 150 cos 28 = 0
𝑁 = 132.4 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
then
Σ𝐹π‘₯ = 0
βˆ’120 + π‘Š sin πœƒ + π‘“π‘˜ = 0
βˆ’120 + 150 sin 28 + πœ‡π‘˜ 𝑁 = 0
βˆ’120 + 150 sin 28 + πœ‡π‘˜ (132.4) = 0
πœ‡π‘˜ = 0.37
6
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Q9
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
SPRING FORCE, SLOPE
The reading in the spring scale is the force in the spring. The system is in equilibrium so Σ𝐹 = 0
Looking at the block
Σ𝐹π‘₯ = 0
βˆ’π‘šπ‘” sin πœƒ + 𝐹𝑠 = 0
βˆ’10(9.8) sin 60 + 𝐹𝑠 = 0
𝐹𝑠 = 84.9 𝑁
Q10
C)
TENSION, FRICTION.
We have two blocks and we want to find forces involved in them. It’s easiest to look at each block.
The 5 kg block is moving to the right with an acceleration
π‘Ž5 = +π‘Ž
the hanging 9 kg weight is falling with the same size of acceleration downwards,
π‘Ž9 = βˆ’π‘Ž
The hanging weight is simpler so we look at that first.
Σ𝐹𝑦 = π‘š9 π‘Ž9
+𝑇 βˆ’ π‘š9 𝑔 = π‘š9 (βˆ’π‘Ž)
𝑇 βˆ’ 9(9.8) = 9(βˆ’π‘Ž)
One equation with two unknowns, so we rewrite as one unknown equals everything else
𝑇 = 88.2 βˆ’ 9π‘Ž
Looking at the 5 kg block
7
©CourseCram
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
We have friction so we look at the normal force direction first
Σ𝐹𝑦 = 0
+𝑁 βˆ’ π‘š5 𝑔 = 0
𝑁 βˆ’ 5(9.8) = 0
𝑁 = 49 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
then
Σ𝐹π‘₯ = π‘š5 π‘Ž5
+𝑇 βˆ’ π‘“π‘˜ = π‘š5 (+π‘Ž)
But we know the tension
(88.2 βˆ’ 9π‘Ž) βˆ’ πœ‡π‘˜ 𝑁 = π‘š5 π‘Ž
88.2 βˆ’ 9π‘Ž βˆ’ 0.1(49) = 5π‘Ž
83.3 = 14π‘Ž
π‘Ž = 5.95 π‘š/𝑠 2
and so
𝑇 = 88.2 βˆ’ 9(5.95)
= 34.65 β‰ˆ 35 𝑁
8
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Q11
C)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 1
NEWTON’S THIRD LAW
By Newton’s third law if the daughter exerts a force on the father, the father exerts exactly the same
force magnitude on the father. So the answer cannot be A) or B)
The effect of the force is given by newton’s second law
𝐹 = π‘šπ‘Ž
If the father weights twice as much as the daughter, he has twice the mass. And from
π‘Ž=
𝐹
π‘š
if the daughter has half the mass, she will experience twice the acceleration. The answer is C
Q12
A)
FORCES, WEIGHT, FRICTION
The force of kinetic friction is constant. The slope is constant. Everything is constant, and so the
acceleration is constant, and sot he object will continue to speed up at a constant rate.
9
PREP101
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
2019 PRACTICE 2 MCQ
(note that this β€œpractice exam” is made entirely of REAL past PHYS117 final exam questions, from
2011 and 2007, so it is certainly excellent practice!)
Q1
C)
CIRCULAR MOTION
For orbital motion, the net force towards the center of the circle is the centripetal force. Here
Σ𝐹𝑔 = 𝐹𝐢
πΊπ‘€π‘š π‘šπ‘£ 2
=
π‘Ÿ2
π‘Ÿ
𝐺𝑀
= 𝑣2
π‘Ÿ
𝐺𝑀
𝑣=√
π‘Ÿ
= √(
6.67 × 10βˆ’11 × 5.68 × 1026
)
100,000 × 103
= 1.95 x 104 m/s
= 19.5 km/s
Q2
D)
WORK, KINETIC ENERGY
By conservation of energy
𝐸𝑖 + π‘Šπ‘›π‘ = 𝐸𝑓
𝐾𝐸𝑖 + 𝑃𝐸𝑔𝑖 + π‘Šπ‘›π‘ = 𝐾𝐸𝑓 + 𝑃𝐸𝑔𝑓
So if we want the change in kinetic energy we rearrange to get
𝐾𝐸𝑓 βˆ’ 𝐾𝐸𝑖 = π‘Šπ‘›π‘ + 𝑃𝐸𝑔𝑖 βˆ’ 𝑃𝐸𝑔𝑓
We set the original height to zero, so
𝐾𝐸𝑓 βˆ’ 𝐾𝐸𝑖 = π‘Šπ‘›π‘ βˆ’ 𝑃𝐸𝑔𝑓
The work done by non-conservative forces is the work done by the 150 N force and the work done
by friction
π‘Šπ‘›π‘ = π‘Š150 + π‘Šπ‘“
π‘Š150 = 𝐹𝑑
= 150(12)
= 1800 𝐽
π‘Šπ‘“ = βˆ’π‘“π‘‘
10
PREP101
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
= βˆ’πœ‡π‘π‘‘
= βˆ’πœ‡π‘šπ‘” cos πœƒ 𝑑
= βˆ’0.1(20)(9.8) cos 30 (12)
= βˆ’203.7 𝐽
So we have
π‘Šπ‘›π‘ = 1800 + (βˆ’203.7)
= 1596.3
and so
𝐾𝐸𝑓 βˆ’ 𝐾𝐸𝑖 = 1596.3 βˆ’ π‘šπ‘”β„Ž
= 1596.3 βˆ’ 20(9.8)(12 sin 30)
= 420 𝐽
ALTERNATE METHOD
Work energy theorem
Work done = change in energy
Wnc = Ef - Ei
W150 - Wfriction = Ξ”KE + Ξ”PE
Ξ”KE = 150(12) – Ffr(12) - Ξ”PE
We have
Wfriction = Ffr d
= ΞΌN d
= ΞΌmgcosΞΈ d
= 0.1(20)(9.80)cos30(12)
= 203.7 J
and
Ξ”PE = mgh
= 20 (9.80) 12sin30
= 1176 J
Ξ”KE = 150(12) – 203.7 - 1176
= 420 J
11
PREP101
Q3
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
CONSERVATION OF ENERGY, CIRCULAR MOTION
The ball is going in a circle so the force towards the center must always equal Fc=mv2/r
If the string is just short of going slack at the top, the tension is zero.
and we have
Ξ£Fy = -mg = -mv2/L
vtop = √(Lg)
Moving from the top to the bottom – we’re talking about changing velocities and heights, so it’s
conservation of energy
Etop = Ebottom
KE + PE = KE
½ mvtop2 + mg2L = ½ mvbottom2
½ (Lg) + 2Lg = ½ vb2
vb = √(5Lg)
12
PREP101
Q4
C)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Scaling (with Gravitational Force)
A scaling question, so the fastest method is ratios!
π‘Ÿπ‘Žπ‘‘π‘–π‘œ =
𝑛𝑒𝑀
π‘œπ‘™π‘‘
𝐺𝑀 π‘š
𝑝
𝐹𝑔𝑝 ( π‘Ÿπ‘2 )
= 𝐺𝑀𝐸 π‘š
𝐹𝑔
2
π‘ŸπΈ
𝐺(
𝐹𝑔𝑝
=
600
(
1
)𝑀𝐸 π‘š
100
2
1
( π‘ŸπΈ )
4
)
𝐺𝑀𝐸 π‘š
π‘ŸπΈ2
𝐹𝑔𝑝
=
600
1
100
1 2
(4)
𝐹𝑔𝑝 = 96 𝑁
ALTERNATE METHOD
Writing the old value then the new one
On earth, his weight is
π‘Š = 𝐹𝑔 =
𝐺𝑀𝐸 π‘š
= 600
π‘ŸπΈ2
On the other planet, he weighs
π‘Šπ‘ = 𝐹𝑔𝑝 =
=
𝐺𝑀𝑃 π‘š
π‘Ÿπ‘2
𝐺(0.01𝑀𝐸 )π‘š
(0.25π‘ŸπΈ )2
= 0.16
𝐺𝑀𝐸 π‘š
π‘ŸπΈ2
= 0.16(600)
= 96 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
Q5
A)
Impulse
β€œAverage force” means it’s an impulse question
I = FavgΞ”t = Ξ”p
5000Ξ”t = pnew - pold
5000Ξ”t = 0.14(37) – 0.14(-42.3)
Ξ”t = 2.22 x 10-3 s
13
PREP101
Q6
A)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Momentum (2D)
Explosion means conservation of momentum. Choose the 1kg piece to be shooting straight up.
Then we have
px before= px after
0 = -4v4cos30 + 5v5cos 30
v5 = 0.8 v4
py before= py after
0 = + 1(10) – 4v4sin30 – 5v5sin30
0 = 10 – 2 v4 – 2.5(0.8)v4
10 = 4v4
v4= 2.5 m/s
v5 = 0.8(2.5) = 2 m/s
14
PREP101
Q7
E)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Power (Ski lift)
Power = W/t
W=E
The energy given to each passenger is
E = KE + PE
= ½ mv2 + mgh
= ½ 190 (6)2 + 190 (9.80) 150
= 2.83 x 105 J
Then the power is
P = 2.83 x 105/30
= 9.43 x 103 W
A difference of only 0.01 is usually a rounding error. The answer is E)
Q8
A)
Equations of motion
The car is moving at v0 = 25 m/s, and decelerates at a = - 8.0 m/s2 to final velocity of v = 0 m/s.
We can use the equation of motion without time
𝑣 2 = 𝑣02 + 2π‘Žπ›₯π‘₯
0 = 252 + 2(βˆ’8)π›₯π‘₯
π›₯π‘₯ = 39.1 π‘š
15
PREP101
Q9
B)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Spring Potential Energy
Comparing an old and a new value, so a quick way of doing this is with ratios
π‘Ÿπ‘Žπ‘‘π‘–π‘œ =
𝑛𝑒𝑀
π‘œπ‘™π‘‘
1
2
𝑃𝐸𝑠𝑛𝑒𝑀 2 π‘˜π‘₯𝑛
=1
π‘ƒπΈπ‘ π‘œπ‘™π‘‘
π‘˜π‘₯π‘œ2
2
100
π‘₯𝑛2
=
(0.12)2
72
π‘₯𝑛 = 0.141 π‘š
= 14.1 π‘π‘š
ALTERNATE METHOD
We can also do this by writing down the old value, then the new one
The energy stored in a spring is given by
1
𝑃𝐸𝑠 = π‘˜π›₯π‘₯ 2
2
If 72 J is stored by 0.12 m of compression
1
72 = π‘˜(0.12)2
2
72 = 0.0072 π‘˜
π‘˜ = 10,000 𝑁/π‘š
So to store 100 J, we require
1
𝑃𝐸𝑠 = π‘˜π›₯π‘₯ 2
2
1
100 = (10,000)π›₯π‘₯ 2
2
100 = 5,000 π›₯π‘₯ 2
π›₯π‘₯ = 0.141 π‘š
16
PREP101
Q10
C)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Power
The average power is given by
𝑃=
π‘Š π›₯𝐸
=
𝑑
𝑑
Here the change in energy is the change in kinetic energy
π›₯𝐸 = 𝐾𝐸𝑓 βˆ’ 𝐾𝐸𝑖
1
= π‘šπ‘£ 2 βˆ’ 0
2
1
= 1000(25)2
2
= 312,500
= 313 π‘˜π½
And the power is
𝑃=
=
π›₯𝐸
𝑑
312,500
7.5
= 41,666 π‘Š
= 41.7 π‘˜π‘Š
Q11
E)
Impulse
We have the length of time of the collision, and are asked for the "average force" - these are
keywords which tell us it is an impulse question
𝐼 = π›₯𝑝 = πΉπ‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π›₯𝑑
𝑝𝑓 βˆ’ 𝑝𝑖 = πΉπ‘Žπ‘£π‘” π›₯𝑑
π‘šπ‘£π‘“ βˆ’ π‘šπ‘£π‘– = πΉπ‘Žπ‘£π‘’ π›₯𝑑
Remember to include the signs of the directions for velocity - up is positive, down is negative
0.3(4.2) βˆ’ 0.3(βˆ’4.5) = πΉπ‘Žπ‘£π‘’ 0.03
πΉπ‘Žπ‘£π‘” = 87 𝑁
17
PREP101
Q12
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Conservation of momentum, Forces, Equations of motion
A block grinding to a halt over friction is a classic conservation of energy problem
𝐸𝑖 + π‘Šπ‘›βˆ’π‘ = 𝐸𝑓
𝐾𝐸𝑖 + 𝑃𝐸𝑖 + π‘Šπ‘›βˆ’π‘ = 𝐾𝑓 + 𝑃𝐸𝑓
There are no heights or springs so the potential energies are both zero.
It grinds to a halt so the final kinetic energy is zero.
𝐾𝐸𝑖 + π‘Šπ‘›βˆ’π‘ = 0
The non-conservative force is zero, so we have
π‘Šπ‘›βˆ’π‘ = βˆ’π‘“π‘˜ 𝑑
= βˆ’πœ‡π‘˜ 𝑁𝑑
= βˆ’πœ‡π‘˜ π‘šπ‘”π‘‘
and we have
1
π‘šπ‘£π‘–2 βˆ’ πœ‡π‘˜ π‘šπ‘”π‘‘ = 0
2
1 2
𝑣 βˆ’ πœ‡π‘˜ 𝑔𝑑 = 0
2 𝑖
1 2
𝑣 βˆ’ 0.5(9.8)(2.1) = 0
2 𝑖
𝑣𝑖 = 4.54 π‘š/𝑠
This is the initial speed for the energy section of the problem, so it’s the final speed for the next part
of the problem, the collision.
Then we can look at the collision. The large lump of clay starts from rest, and so has zero velocity
and momentum
π‘π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ = π‘π‘“π‘–π‘›π‘Žπ‘™
0.1𝑣1𝑖 + 0 = (0.1 + 0.5)4.54
𝑣1𝑖 = 27.2 π‘š/𝑠
ALTERNATE METHOD
We can also work out the speed from forces and equations of motion, but as we see it takes longer
When the two lumps of clay stick together, that is a perfectly inelastic collision, so that is
conservation of momentum.
18
PREP101
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
The resulting lump is moving and then comes to a halt, so that is equations of motion.
The force that halts it is friction, so that is forces.
𝛴𝐹 = π‘šπ‘Ž
βˆ’π‘“π‘˜ = (0.1 + 0.5)π‘Ž
βˆ’πœ‡π‘  𝑁 = 0.6π‘Ž
The clay is on a horizontal surface with no angled forces, so N = mg
βˆ’πœ‡π‘  π‘šπ‘” = 0.6π‘Ž
βˆ’0.5(0.6)9.8 = 0.6π‘Ž
π‘Ž = βˆ’4.9 π‘š/𝑠 2
Now we can look at the lump grinding to a halt. We are given distances and no time, so we use the
equation of motion without time.
𝑣 2 βˆ’ 𝑣02 = 2π‘Žπ›₯π‘₯
02 βˆ’ 𝑣02 = 2(βˆ’4.9)(2.1)
𝑣0 = 4.54 π‘š/𝑠
19
PREP101
Q13
D)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Centripetal force
Because the object is moving in a circle, the total force towards the center of the circle must equal
the centripetal force. At the bottom of the circle, the center is above it, in the +y direction, so we
have
𝛴𝐹𝑦 = +𝐹𝑐
π‘šπ‘£ 2
𝑇 βˆ’ π‘šπ‘” =
π‘Ÿ
𝑇 = π‘šπ‘” +
= 0.4(9.8) +
π‘šπ‘£ 2
π‘Ÿ
0.4(4)2
0.5
𝑇 = 16.7 𝑁
20
PREP101
Q14
A)
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 MCQ
Scaling (with Gravitational Force)
We are asked to find the new acceleration due to gravity, given the new planet's values compared to
Earth. This is a classic ratio question.
The gravitational acceleration is given by
𝑔=
𝐺𝑀
= 9.8
π‘Ÿ2
where M and r are the mass and radius of Earth.
First we write down the values we're given:
The new planet has three times the radius of Earth, so we have
π‘Ÿπ‘›π‘’π‘€ = 3π‘Ÿ
The new planet has the same average density as Earth, so we have
πœŒπ‘›π‘’π‘€ = πœŒπΈπ‘Žπ‘Ÿπ‘‘β„Ž
𝑀𝑛𝑒𝑀
4
3
πœ‹π‘Ÿπ‘›π‘’π‘€
3
=4
3
𝑀
πœ‹π‘Ÿ 3
𝑀𝑛𝑒𝑀 𝑀
=
(3π‘Ÿ)3 π‘Ÿ 3
𝑀𝑛𝑒𝑀
=𝑀
27
𝑀𝑛𝑒𝑀 = 27𝑀
Now we can work out the new acceleration due to gravity using the ratio
𝑔𝑛𝑒𝑀
=
𝑔
𝐺𝑀𝑛𝑒𝑀
2
π‘Ÿπ‘›π‘’π‘€
𝐺𝑀
π‘Ÿ2
𝑔𝑛𝑒𝑀
=
9.8
𝑀𝑛𝑒𝑀
2
π‘Ÿπ‘›π‘’π‘€
𝑀
π‘Ÿ2
𝑔𝑛𝑒𝑀
=
9.8
27𝑀
(3π‘Ÿ)2
𝑀
π‘Ÿ
𝑔𝑛𝑒𝑀 27
=
9.8
9
𝑔𝑛𝑒𝑀 = 29.4 π‘š/𝑠 2
21
PREP101
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 long answer
2019 PRACTICE 2 long answer
1
Spring Forces, Conservation of Energy, Momentum
a)
A force of 2 N is required to compress the spring by 0.25 cm = 0.0025 m
So we have
|𝐹𝑠 | = |π‘˜π›₯π‘₯|
2 = π‘˜(0.0025)
π‘˜ = 800 𝑁/π‘š
b)
The bullet and block are moving at one point, and later have compressed the spring and
stopped moving. This is two different locations with no time mentioned - this is a conservation of
energy question. Kinetic energy has been converted into spring potential energy. There are no
frictional or other external forces, so there is no external work done.
𝐸𝑖 + π‘Šπ‘›βˆ’π‘ = 𝐸𝑓
𝐸𝑖 = 𝐸𝑓
𝐾𝐸𝑖 = 𝑃𝐸𝑠𝑓
1
1
π‘šπ‘£ 2 = π‘˜(π›₯π‘₯)2
2
2
π‘šπ‘£ 2 = π‘˜(π›₯π‘₯)2
(0.01 + 0.99)𝑣 2 = 800(0.15)2
𝑣 2 = 18
𝑣 = 4.24 π‘š/𝑠
c)
We have objects colliding and sticking together, so we are dealing with perfectly inelastic collisions.
π‘π‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ = π‘π‘“π‘–π‘›π‘Žπ‘™
The block is initially not moving, so its initial momentum is zero.
π‘šπ‘π‘’π‘™π‘™π‘’π‘‘ 𝑣0 + 0 = (π‘šπ‘π‘’π‘™π‘™π‘’π‘‘ + π‘šπ‘π‘™π‘œπ‘π‘˜ )𝑣
0.01𝑣0 = (0.01 + 0.99)4.24
𝑣0 = 424 π‘š/𝑠
22
PREP101
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 long answer
2
CONSERVATION OF ENERGY, CONSERVATION OF MOMENTUM
a)
C of E
The block changes speed over a certain distance – perfect for conservation of energy or equation of
motion. Conservation of energy is always faster
πΈπ‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ + π‘Šπ‘›βˆ’π‘ = πΈπ‘“π‘–π‘›π‘Žπ‘™
The initial energy is all kinetic, the work done by friction is negative, and the final energy is zero
1
π‘šπ‘£ 2 + (βˆ’π‘“π‘˜ 𝑑) = 0
2
The force of friction is
π‘“π‘˜ = πœ‡π‘˜ 𝑁
= πœ‡π‘˜ π‘šπ‘”
And we have
1
π‘šπ‘£ 2 βˆ’ πœ‡π‘˜ π‘šπ‘”π‘‘ = 0
2
1 2
𝑣 βˆ’ πœ‡π‘˜ 𝑔𝑑 = 0
2
1 2
𝑣 βˆ’ 0.25(9.8)(9.5) = 0
2
v = 6.82 m/s
b)
C of MOMENTUM
Things colliding – this is a momentum question
pbefore= pafter
mbulletvbullet +mblock(0) = (mbullet + mblock)6.82
0.015vbullet = 1.115(6.82)
vbullet = 507 m/s
23
PREP101
Q3
PHYS117 MIDTERM SOLUTIONS
2019 PRACTICE 2 long answer
CIRCULAR MOTION
Circular motion – so the net force towards the center of the circle is mv2/r. Here β€œtowards the
center of the circle” is in the negative x direction.
We are told the length of the lower rope is 1 m, so we can find the radius of the circle from
r = 1 cos30
Draw an FBD.
In a rope question like this it’s fastest to do the y direction first
Ξ£Fy= 0
Tsin60 + Tsin30 – mg = 0
1.366𝑇 βˆ’ 2.5(9.8) = 0
𝑇 = 17.94 𝑁
Moving on to x
Σ𝐹π‘₯ =
π‘šπ‘£ 2
π‘Ÿ
2.5𝑣 2
𝑇 cos 60 + 𝑇 cos 30 =
1 cos 30
1.366(17.94) =
2.5𝑣 2
cos 30
𝑣 = 2.91 π‘š/𝑠
24
PHY117 Midterm Solutions
Q1
B)
2011 Q1
Equations of motion
The distance is how much further the cheetah will travel than the gazelle in 30 seconds.
We find distance from
π·π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ = 𝑠𝑝𝑒𝑒𝑑 × π‘‘π‘–π‘šπ‘’
but we need the speed in the same units as the time. We convert from km/hr to m/s by
π‘˜π‘š/β„Žπ‘Ÿ =
1000 π‘š
1000
1
=
π‘š/𝑠 =
π‘š/𝑠
60(60) 𝑠 3600
3.6
The cheetah travels
π‘‘π‘β„Žπ‘’π‘’π‘‘π‘Žβ„Ž =
100
× 30 = 833.3 π‘š
3.6
π‘‘π‘”π‘Žπ‘§π‘’π‘™π‘™π‘’ =
80
× 30 = 666.7 π‘š
3.6
The gazelle travels
The difference is
833.3 βˆ’ 666.7 = 166.6 π‘š β‰ˆ 167 π‘š
ALTERNATE METHOD
The cheetah is moving 100 - 80 = 20 km/hr faster than the gazelle. So the difference in distance over
30 seconds is
𝑑=
20
× 30 = 166.7 π‘š β‰ˆ 167 π‘š
3.6
25
PHY117 Midterm Solutions
Q2
E)
2011 Q2
Projectile
There are two ways to do this, a fast way and a long way.
The fast method is to remember that projectile motion is symmetric: the rock will come down with
the same speed as at goes up. So when we throw a rock up at +15 m/s, it will fall back past us at -15
m/s. We only need to work out when that happens, and if we throw the second rock at the same
time with the same velocity they will hit the ground at the same time.
When will the upwards-thrown rock pass the top of the building? We find that from
1
Δ𝑦 = 𝑣0𝑦 𝑑 βˆ’ 𝑔𝑑 2
2
1
𝑦 βˆ’ 𝑦0 = 𝑣0𝑦 𝑑 βˆ’ 𝑔𝑑 2
2
1
100 βˆ’ 100 = 15𝑑 βˆ’ (9.8)𝑑 2
2
0 = 15𝑑 βˆ’ 4.9𝑑 2
We don't need to do quadratic - we can divide by t to find
0 = 15 βˆ’ 4.9𝑑
4.9𝑑 = 15
𝑑=
15
= 3.06 𝑠
4.9
ALTERNATE METHOD
The simpler, but longer method is to find the time taken for both rocks to hit the ground then find
the difference.
For the upward rock, the rock starts at y0 = 100 and an initial vertical velocity of v0y = 15 m/s. It hits
the ground when y = 0.
1
Δ𝑦 = 𝑣0𝑦 𝑑 βˆ’ 𝑔𝑑 2
2
Remember
Δ𝑦 = 𝑦 βˆ’ 𝑦0
which in this case is
Δ𝑦 = 0 βˆ’ 100
1
2
0 βˆ’ 100 = (+15)𝑑𝑒𝑝 βˆ’ (9.8)𝑑𝑒𝑝
2
2
4.9𝑑𝑒𝑝
βˆ’ 15𝑑𝑒𝑝 βˆ’ 100 = 0
Solving for tup and taking the positive value, we find
𝑑𝑒𝑝 = 6.30 𝑠
Then we do the same thing for the second rock, which starts with (-15 m/s)
26
PHY117 Midterm Solutions
2011 Q2
1
Δ𝑦 = 𝑣0𝑦 𝑑 βˆ’ 𝑔𝑑 2
2
1
2
0 βˆ’ 100 = (βˆ’15)π‘‘π‘‘π‘œπ‘€π‘› βˆ’ (9.8)π‘‘π‘‘π‘œπ‘€π‘›
2
2
4.9π‘‘π‘‘π‘œπ‘€π‘›
+ 15π‘‘π‘‘π‘œπ‘€π‘› βˆ’ 100 = 0
π‘‘π‘‘π‘œπ‘€π‘› = 3.24 𝑠
So the difference, how much later you should throw the second rock, is
𝑑𝑒𝑝 βˆ’ π‘‘π‘‘π‘œπ‘€π‘› = 6.30 βˆ’ 3.24 = 3.06 𝑠
As you can see, this method is much longer, but it gives the same answer.
27
PHY117 Midterm Solutions
Q3
C)
2011 Q3
Equations of motion
This motion has two stages: in the first stage it is accelerating upwards, and in the second it
continues in free-fall under the influence of gravity until it reaches its maximum height.
Stage 1: Accelerating
The rocket is accelerating at 30 m/ss for 1.0 s. We are told it is launched from the ground, so assume
its initial velocity was zero. The height reached is
1
Δ𝑦 = 𝑣0𝑦 𝑑 + π‘Žπ‘‘ 2
2
1
Δ𝑦1 = 0 + 30(1)2
2
Δ𝑦1 = 15 π‘š
At this point it has a new velocity, given by
𝑣1 = 𝑣0 + π‘Žπ‘‘
= 0 + 30(1)
= 30 π‘š/𝑠
Stage 2: Free-fall
Even though the rocket is in "free-fall", it has a large upward velocity of 30 m/s, and will continue
moving up until acceleration due to gravity causes it to start falling down. The maximum height of
any projectile is given by the point where the vertical velocity is zero, so we use the formula
𝑣 2 = 𝑣02 + 2π‘ŽΞ”π‘₯
The starting velocity of stage 2 is v1, the final velocity is zero, the change in position is the change in
height over the second stage, Δ𝑦2 , and the acceleration is (-g), so we have
𝑣 2 = 𝑣12 + 2(βˆ’π‘”)Δ𝑦2
0 = 302 + 2(βˆ’9.8)Δ𝑦2
Δ𝑦2 = 45.9 π‘š
So the total height covered is
𝑦 = Δ𝑦1 + Δ𝑦2
= 15 + 45.9
= 60.9 β‰ˆ 61 π‘š
28
PHY117 Midterm Solutions
Q4
A)
2011 Q4
Pythagoras
Moving east-west and north-south, like moving horizontally and vertically, are two components of
the total displacement. So we can use Pythagoras to find the total distance.
The taxi moves a total of 5 blocks east, and a total of 7 blocks north, so we have
π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ = √52 + 72
= 8.6 π‘π‘™π‘œπ‘π‘˜π‘ 
29
PHY117 Midterm Solutions
Q5
D)
2011 Q5
Projectile Motion
Projectile motion is just the equations of motion in both x and y.
To find out about the impact of a projectile, we could use the y-component equations of motion to
find the time of impact, then the time to find out whatever else we want.
But there's a faster way! We only want to know the velocity at impact, and we already have an
equation to find the velocity at another location.
2
𝑣𝑦2 = 𝑣0𝑦
+ 2π‘ŽΞ”π‘¦
The launch vertical velocity component is
𝑣0𝑦 = 𝑣0 sin πœƒ
= 15 sin 53
The acceleration is downwards due to gravity
π‘Ž = βˆ’9.8
And the distance is
Δ𝑦 = 𝑦 βˆ’ 𝑦0
= 0 βˆ’ 35
= βˆ’35
So we have
𝑣𝑦2 = (15 sin 53)2 + 2(βˆ’9.8)(βˆ’35)
= 829.5
𝑣𝑦 = 28.8 π‘š/𝑠 β‰ˆ 29 π‘š/𝑠
30
PHY117 Midterm Solutions
Q6
C)
2011 Q6
Relative motion
The speed of B as observed from A is given by
𝑣𝐡/𝐴 = 𝑣𝐡 βˆ’ 𝑣𝐴
The easiest way to do this is by breaking it up into x and y components.
All the units are in mph, but since we are only dealing with speeds we can leave the units like that
throughout. (If we wanted to find distances or momenta, we would need to convert to m/s)
x components
𝑣𝐴π‘₯ = 𝑣𝐴 cos πœƒπ΄
= 400 cos 45
= 282.8 π‘šπ‘β„Ž
𝑣𝐡π‘₯ = 500 cos 90
=0
or you can just say that vB is entirely vertical, so the horizontal component is zero.
y components
𝑣𝐴𝑦 = 𝑣𝐴 sin πœƒπ΄
= 400 sin 45
= 282.8 π‘šπ‘β„Ž
𝑣𝐡𝑦 = 500 sin 900
= 500
or you can just say that vB is entirely vertical and so equals 500.
So we have
𝑣𝐡/𝐴π‘₯ = 0 βˆ’ 282.8
= βˆ’282.8 π‘šπ‘β„Ž
𝑣𝐡/𝐴𝑦 = 500 βˆ’ 282.8
= 217.2 π‘šπ‘β„Ž
The total speed is then found by pythagoras
|𝑣𝐡 | = √(βˆ’282.8)2 + (217.2)2
= 356.6 β‰ˆ 357 π‘šπ‘β„Ž
31
PHY117 Midterm Solutions
Q7
D)
2011 Q7
Scaling (with Energy)
The rabbit is being scaled by a factor of 2. So length values scale by 2, area values scale by 22 = 4, and
volume values scale by 22 = 8. We check each case individually. But first, remember that this is a
great place to practice the TRUE/FALSE/MAYBE method for multiple choice we discussed in class!
A) The force exerted by a muscle depends on its cross-section area, so this scales by 4. TRUE
B) The distances all scale by 2. TRUE
C) The gravitational potential energy is given by
𝑃𝐸𝑔 = π‘šπ‘”β„Ž
The mass depends on the volume, so it scales by 8.
The gravitational acceleration is a constant which doesn't change.
But what about the height? We can tell this from conservation of energy. What's more, this is a
special case we can remember: this saves us a lot of work if we see the problem again!
The gravitational potential energy comes from the initial kinetic energy when the rabbit jumps. The
initial kinetic energy comes from the work done by the muscles. The work done by muscles is given
by
π‘Š = 𝐹. 𝑑
We know the force increases by a factor of 4. The distance increases by a factor of 2. So the total
work done increases by a factor of 8, and the total energy of the rabbit increases by a factor of 8.
We've already seen the mass in the gravitational potential energy term increases by a factor of 8, so
the height must remain unchanged.
Likewise, if we look at the kinetic energy
1
𝐾𝐸 = π‘šπ‘£ 2
2
we see that the mass increases by a factor of 8, so if the total energy also only increases by a factor
of 8, the velocity must remain unchanged.
C) is TRUE
D) We know the mass depends on volume, and increases by a factor of 8, not 4. THIS IS FALSE.
E) This is TRUE (see the discussion in section C)
32
PHY117 Midterm Solutions
Q8
E)
2011 Q8
Forces
There is no friction on the table, so the 5 kg block will fall down and the 15 kg block will be pulled to
the right.
There are two ways of dealing with this, either looking at the blocks separately or together.
Separately
Both blocks are moving at the same rate, with the same speed, and size of acceleration a.
Looking at the forces on the top block. It will slide to the right with acceleration a.
We have
Ξ£πΉπ‘‘π‘œπ‘ π‘₯ = π‘šπ‘Ž
𝑇 = π‘šπ‘‘π‘œπ‘ π‘Ž
𝑇 = 15π‘Ž
Looking at the hanging block, it will fall down with acceleration (-a) (negative because it is going
down).
33
PHY117 Midterm Solutions
2011 Q8
Ξ£πΉβ„Žπ‘Žπ‘›π‘” 𝑦 = π‘šβ„Žπ‘Žπ‘›π‘” (βˆ’π‘Ž)
+𝑇 βˆ’ π‘šβ„Žπ‘Žπ‘›π‘” 𝑔 = βˆ’π‘šβ„Žπ‘Žπ‘›π‘” π‘Ž
15π‘Ž βˆ’ 5𝑔 = βˆ’5π‘Ž
20π‘Ž = 5𝑔
π‘Ž=
5(9.8)
20
π‘Ž = 2.45 π‘š/𝑠 2
and
𝑇 = 15(2.45) = 36.75 𝑁 β‰ˆ 37 𝑁
ALTERNATE METHOD: SHORTCUT
We can also view the whole system as a single mass moving along a single axis, along the rope. We
take the positive direction moving to the right and down along the rope.
This gives us
Σ𝐹 = π‘šπ‘‘π‘œπ‘‘π‘Žπ‘™ π‘Ž
+π‘šβ„Žπ‘Žπ‘›π‘” 𝑔 = (π‘šπ‘‘π‘œπ‘ + π‘šβ„Žπ‘Žπ‘›π‘” )π‘Ž
5(9.8) = (5 + 15)π‘Ž
5(9.8) = 20π‘Ž
π‘Ž=
5(9.8)
20
π‘Ž = 2.45 π‘š/𝑠 2
Then we look at one block, say the top block, and find
Ξ£πΉπ‘‘π‘œπ‘ π‘₯ = π‘šπ‘‘π‘œπ‘ π‘Ž
+𝑇 = 15(2.45)
= 36.75 𝑁 β‰ˆ 37 𝑁
34
PHY117 Midterm Solutions
Q9
B)
2011 Q9
Forces and Friction
Note that we're given the weight in Newtons, so that's already a force. We don't need to multiply by
g.
We want to move the top five books, so object we're moving has the weight of five books, 5(5) = 25
N
This experiences friction along the bottom, between the fifth and bottom books, in the opposite
direction to the intended motion. (It would also be possible to start the fifth book only sliding out
from between the other books, but that would take more force, and we're asked for the minimum
force so we don't have to worry about that.)
The minimum force required to start motion in this case is the maximum static friction
𝐹 = 𝑓𝑠
= πœ‡π‘  𝑁
We find the Normal force between the fifth and sixth books by looking in the y direction for the five
stacked books. They're not accelerating up or down so their acceleration in y is zero.
Σ𝐹𝑦 = 0
π‘βˆ’π‘Š =0
𝑁=π‘Š
𝑁 = 25
And we have
𝐹 = πœ‡π‘  𝑁
= 0.2(25)
= 5.0 𝑁
35
PHY117 Midterm Solutions
Q10
C)
2011 Q10
Forces and friction along a slope
When the mass is sliding we have the kinetic frictional force, so we use πœ‡π‘˜ = 0.520
The frictional force is given by
π‘“π‘˜ = πœ‡π‘˜ 𝑁
We need the normal force between the slope and the block. We can remember this formula (faster
for use in multiple choice questions!) or we can look at the forces. We take tilted axes along the
slope. Since the block is not accelerating off or into the slope, in the tilted y direction we have
Σ𝐹𝑦 = 0
𝑁 βˆ’ π‘šπ‘” cos πœƒ = 0
𝑁 = π‘šπ‘” cos πœƒ
As you can see, just remembering that formula for a block on a slope with no angled forces (a
common situation in 117) is a faster technique.
Now we have
π‘“π‘˜ = πœ‡π‘˜ π‘šπ‘” cos πœƒ
= 0.520 (10)(9.8) cos 25
= 46.2 𝑁
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PHY117 Midterm Solutions
Q11
A)
2011 Q11
Forces, friction, and Equations of Motion
Work out what you need to do by looking at the keywords in the question.
The "minimum stopping distance" means he must come to a stop over a distance. He's changing his
velocity from 15.0 m/s to 0 m/s, and we only care about distance, so we use
𝑣 2 = 𝑣02 + 2π‘ŽΞ”π‘₯
0 = 152 + 2π‘ŽΞ”π‘₯
We have two unknowns, so we need another equation.
But before we go, we rearrange to get what we want:
2π‘ŽΞ”π‘₯ = βˆ’152
The minimum stopping distance is then
Ξ”π‘₯ =
βˆ’152
2π‘Ž
Now, we need some other equation. Friction is mentioned, so that is what we use. The maximum
force he can exert on the crate is the maximum force of static friction (if he applies a larger force,
the crate will slide and the crystal will be in trouble). So the maximum force is
𝐹 = 𝑓𝑠
π‘šπ‘Ž = πœ‡π‘  𝑁
Since the crate is on a level surface, the normal force is the weight, N = mg, and we have
π‘šπ‘Ž = πœ‡π‘  π‘šπ‘”
π‘Ž = πœ‡π‘  𝑔
π‘Ž = 0.4(9.8)
= 3.92 π‘š/𝑠 2
This acceleration must be in the negative directino, because the crate is slowing down.
π‘Ž = βˆ’3.92 π‘š/𝑠 2
We plug that back into the first equation and we have
Ξ”π‘₯ =
βˆ’152
2(βˆ’3.92)
= 28.7 π‘š
37
PHY117 Midterm Solutions
Q12
B)
2011 Q12
Work and friction
The work done is given by
π‘Š = 𝐹. 𝑑
The force is friction. Normally we find the frictional force from
π‘“π‘˜ = πœ‡π‘˜ 𝑁
but here they haven't given us the coefficient of friction. There must be another way to calculate the
force of friction. Well, if they don't give us the friction values, we must look at the force equation.
Σ𝐹 = π‘šπ‘Ž
πΉπ‘Žπ‘π‘π‘™π‘–π‘’π‘‘ βˆ’ π‘“π‘˜ = π‘šπ‘Ž
200 βˆ’ π‘“π‘˜ = 55(2)
π‘“π‘˜ = 200 βˆ’ 55(2)
π‘“π‘˜ = 90 𝑁
This is in the opposite direction to the 10 m distance travelled, so the dot product is negative, from
𝑓. 𝑑 = 𝑓𝑑 cos πœƒ
So we have
π‘Š = βˆ’90(10)
= βˆ’900 𝐽
38
PHY117 Midterm Solutions
Q13
A)
2011 Q13-14
Work
The keyword is "constant velocity". Constant velocity means zero change in acceleration, zero
change in kinetic energy. And since the cart is moving horizontally, there is no change in potential
energy. So there's no change in total energy.
And if there's no change in total energy, there is no work done.
Q14
C)
Conservation of energy
Since there is no friction, the total energy of this system is conserved. The total energy at the point
where the bolck is released is equal to the energy at x = 0.
When the block is released, its speed is zero.
When the block is at x =0, the spring potential energy is zero.
So we have
πΈπ‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ + π‘Šπ‘›π‘ = πΈπ‘“π‘–π‘›π‘Žπ‘™
There are no nonconservative forces acting, so
πΈπ‘–π‘›π‘–π‘‘π‘–π‘Žπ‘™ + 0 = πΈπ‘“π‘–π‘›π‘Žπ‘™
πΈπ‘Ÿπ‘’π‘™π‘’π‘Žπ‘ π‘’ = 𝐸π‘₯=0
𝑃𝐸𝑠 π‘Ÿπ‘’π‘™π‘’π‘Žπ‘ π‘’ + πΎπΈπ‘Ÿπ‘’π‘™π‘’π‘Žπ‘ π‘’ = 𝑃𝐸𝑠 π‘₯=0 + 𝐾𝐸π‘₯=0
𝑃𝐸𝑠 π‘Ÿπ‘’π‘™π‘’π‘Žπ‘  + 0 = 0 + 𝐾𝐸π‘₯=0
1 2 1
π‘˜π‘₯ = π‘šπ‘£ 2
2
2
π‘˜π‘₯ 2 = π‘šπ‘£ 2
(4 × 102 )(0.05)2 = 5𝑣 2
𝑣 2 = 0.2
𝑣 = 0.447 π‘š/𝑠
39
PHY117 Midterm Solutions
Q15
E)
2003 Q1
Forces
The reading on the scale is the normal force acting on the person from the scale.
Looking at the forces acting on the person
Σ𝐹𝑦 = π‘šπ‘Ž
A key point is finding the acceleration. The person is riding down, so the velocity is negative. But
they are slowing down, so their acceleration is in the opposite direction, so the acceleration is
positive.
𝑁 βˆ’ π‘šπ‘” = π‘šπ‘Ž
𝑁 βˆ’ 80(9.8) = 80(2.4)
𝑁 = 80(9.8) + 80(2.4)
= 976 𝑁 β‰ˆ 980 𝑁
40
PHY117 Midterm Solutions
2003 Q1
2003 midterm
Q1
B)
Scaling
When we double the linear dimensions, the lengths are multiplied by 2, the areas and multiplied by
4, and the volumes (and so masses) are multiplied by 8. We check each answer. Note that this is a
great question for practicing the TRUE/FALSE/MAYBE system we described in class.
A) The weight will increase by 8. FALSE
B) Why would it need to eat more? Because it's losing more energy in the form of heat through its
skin, the surface area, which has increased by a factor of 4. In our simple scaling this looks TRUE.
C) The strength of the muscles depends on their cross sectional area, which has increased by 4. The
weight depends on the mass, and so the volume, which has increased by 8. So the new
strength:weight ratio is 4:8. Which is the same as 1:2. It now has half as much strength (1) as weight
(2). The strength:weight ratio hasn't doubled, it has halfed! FALSE
D) The surface area has increased by 4. FALSE.
E) How likely is it to break bones? The bones' strength depends on their area, which has increased by
4. But the weight they must support has increased by 8, which is twice as much. The animal is now
MORE likely to break bones. FALSe.
41
PHY117 Midterm Solutions
Q2
A)
2003 Q2
Equations of motion
We have two stones, and we want to know when they cross, so we want to know when y1 = y2.
The first stone starts from the bottom of the cliff and is thrown up, so y10 = 0 and v10 = +9 m/s
The second stone starts at the top of the cliff and is thrown down, so y20 = 6 and v20 = - 9 m/s
We use
1
Δ𝑦 = 𝑣0 𝑑 + π‘Žπ‘‘ 2
2
1
𝑦 βˆ’ 𝑦0 = 𝑣0 𝑑 βˆ’ 𝑔𝑑 2
2
𝑦 = 𝑦0 + 𝑣0 𝑑 βˆ’ 4.9 𝑑 2
The stones cross when
𝑦1 = 𝑦2
𝑦10 + 𝑣10 𝑑 βˆ’ 4.9𝑑 2 = 𝑦20 + 𝑣20 𝑑 βˆ’ 4.9𝑑 2
𝑦10 + 𝑣10 𝑑 = 𝑦20 + 𝑣20 𝑑
0 + 9𝑑 = 6 βˆ’ 9𝑑
18𝑑 = 6
𝑑 = 0.333 𝑠
Now we know when theye cross, we can find where that happens by looking at either stone. Looking
at the first
𝑦1 = 𝑦0 + 𝑣10 𝑑 βˆ’ 4.9𝑑 2
= 0 + 9(0.333) βˆ’ 4.9(0.333)2
= 2.46 π‘š
42
PHY117 Midterm Solutions
Q3
B)
2003 Q3
Equations of Motion
The tile is accelerating linearly due to gravity, with a = - g.
The average speed of the tile as it passes the window is found from
π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ 𝑠𝑝𝑒𝑒𝑑 =
π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ 1.6
=
= 8 π‘š/𝑠
π‘‘π‘–π‘šπ‘’
0.2
This is moving downwards, so the velocity is
π‘Žπ‘£π‘’π‘Ÿπ‘Žπ‘”π‘’ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ = βˆ’8 π‘š/𝑠
Since it's accelerating linearly, this is the velocity of the block at the midpoint of the window, 0.8 m
below the top of the window. (If it was firing boosters or something we couldn't say this).
We know the tile started at a velocity of 0, at rest at the top of the building. So to find the distance
between the points (we don't care about time) we can use
𝑣 2 = 𝑣02 + 2π‘ŽΞ”π‘¦
(βˆ’8)2 = 02 + 2(βˆ’9.8)Δ𝑦
βˆ’64 = βˆ’19.6Δ𝑦
Δ𝑦 = 3.27 π‘š
This is the distance from the roof to the midpoint of the window. So the distance from the roof to
the top of the window is
π‘‘π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ = 3.27 βˆ’ 0.8
= 2.47 β‰ˆ 2.5 π‘š
43
PHY117 Midterm Solutions
Q4
A)
2003 Q4-5
Projectile motion
To find the angle of impact with the ground, we need the x and y components of the velocity. Then
πœƒπ‘–π‘šπ‘π‘Žπ‘π‘‘ = tanβˆ’1 (
𝑣𝑦
)
𝑣π‘₯
For projectile motion the x component of the velocity does not change, so the impact x component
equals the release x component. Which is
𝑣0π‘₯ = 𝑣0 cos πœƒ
= 97.5 cos 50
= 62.7 π‘š/𝑠
and so
𝑣π‘₯ = 62.7 π‘š/𝑠
The y component of the release velocity is
𝑣0𝑦 = 𝑣0 sin πœƒ
= 97.5 sin 50
= 62.7 π‘š/𝑠
We know how far it falls before hitting the ground (and don't care about time), so we can find y
component of the impact velocity with
2
𝑣𝑦2 = 𝑣0𝑦
+ 2π‘ŽΞ”π‘¦
BE CAREFUL! Remember that
Δ𝑦 = 𝑦 βˆ’ 𝑦0
That's the NEW height minus the OLD height, so we have
Δ𝑦 = 0 βˆ’ 732
= βˆ’732
It's an easy mistake to forget the sign!
𝑣𝑦2 = 74.72 + 2(βˆ’9.8)(βˆ’732)
= 19927 π‘š/𝑠
𝑣𝑦 = βˆ’141 π‘š/𝑠
NOTE that the square root of a number can be positive or negative. We know this velocity will be
negative because the crate is falling down.
This makes sense: the package started by going up, because it kept going with the velocity of the
plane when it was released, but gravity causes it to fall downward.
Now we have
44
PHY117 Midterm Solutions
πœƒ = tanβˆ’1 (
= tanβˆ’1 (βˆ’
2003 Q4-5
𝑣𝑦
)
𝑣π‘₯
141
)
62.7
= βˆ’66.0°
The package strikes with a velocity pointing 66.0° below the horizontal.
Q5
E)
Relative motion (plane)
In a plane problem we draw the resultant vector in the direction we want to go. This is due west.
Then we draw the wind pointing towards the pointed end of that arrow. This is the wind. This wind is
blowing FROM the south, so it is blowing north.
We must then fly to the bottom of the wind arrow, as shown
We can instantly see we must fly south of west, which cancels out some of the answers.
To find the angle, we use
πœƒ = sinβˆ’1 (
π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’
)
β„Žπ‘¦π‘π‘œπ‘‘π‘’π‘›π‘’π‘’π‘ π‘’
= sinβˆ’1 (
𝑣𝑀𝑖𝑛𝑑
)
π‘£π‘π‘™π‘Žπ‘›π‘’
= sinβˆ’1 (
38
)
245
= 8.92°
8.92° south of west, answer E)
45
PHY117 Midterm Solutions
Q6
E)
2003 Q6
Scaling (with the Force of gravity)
A classic scaling question
The weight of an object is the force of gravity acting on it. This is given by
𝐹𝑔 =
πΊπ‘€π‘š
π‘Ÿ2
If we have
𝑀𝐴 = 0.6𝑀𝐡
and
𝐹𝐴𝑔 = 𝐹𝐡𝑔
then we have
𝐺𝑀𝐴 π‘š 𝐺𝑀𝐡 π‘š
=
π‘Ÿπ΄2
π‘Ÿπ΅2
𝑀𝐴 𝑀𝐡
= 2
π‘Ÿπ΄2
π‘Ÿπ΅
0.6𝑀𝐡 𝑀𝐡
= 2
π‘Ÿπ΄2
π‘Ÿπ΅
0.6
1
2 = 2
π‘Ÿπ΄
π‘Ÿπ΅
Rearrange to get our ratio
0.6 =
π‘Ÿπ΄2
π‘Ÿπ΅2
π‘Ÿπ΄
= √0.6
π‘Ÿπ΅
= 0.77
46
PHY117 Midterm Solutions
Q7
C)
2003 Q7
Forces (elevator)
The reading on the scale is the normal force exerted between the scale and the box.
This force is given by looking at the forces on the box
Ξ£πΉπ‘π‘œπ‘₯ 𝑦 = π‘šπ‘Ž
𝑁 βˆ’ π‘šπ‘” = π‘šπ‘Ž
𝑁 = 60(9.8) + 60π‘Ž
To find the normal force N we need to know the acceleration of the box. This is given by looking at
the forces on the entire system, the elevator and scale with box
Ξ£πΉπ‘‘π‘œπ‘‘π‘Žπ‘™ 𝑦 = π‘šπ‘‘π‘œπ‘‘π‘Žπ‘™ π‘Ž
𝑇 βˆ’ π‘šπ‘‘π‘œπ‘‘π‘Žπ‘™ 𝑔 = (815 + 60)π‘Ž
9410 βˆ’ (815 + 60)9.8 = 875π‘Ž
835 = 875π‘Ž
π‘Ž = 0.954 π‘š/𝑠 2
and so
𝑁 = 60(9.8) + 60(0.954)
= 645 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
47
PHY117 Midterm Solutions
Q8
E)
2003 Q8
Frictional forces
This is a problem with two parts. That doesn't make it complicated, we just do one part, then the
other.
First part: top block only slides
To make the top block just begin to move, the total horizontal force is zero (think of it as the force
just exactly balancing the maximum friction, so that things are about to move)
𝛴𝐹π‘₯ = 0
+πΉπ‘‘π‘œπ‘ βˆ’ 𝑓𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 = 0
πΉπ‘‘π‘œπ‘ = 𝑓𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛
= πœ‡π‘  𝑁𝑏𝑒𝑑𝑀𝑒𝑒𝑛
The system is horizontal with no angled forces, so the normal force is just the weight of the top
block.
𝑁𝑏𝑒𝑑𝑀𝑒𝑒𝑛 = π‘šπ‘”
πΉπ‘‘π‘œπ‘ = πœ‡π‘  π‘šπ‘”
We are given the applied force, so we know
47 = πœ‡π‘  π‘šπ‘”
Now we look at the second situation
48
PHY117 Midterm Solutions
2003 Q8
Second part: bottom block only slides
Now we are applying a force to the bottom block, which must equal the maximum static frictional
force. This is the total frictional force with the ground AND with the upper block. The coefficient of
static friction is the same in each case.
𝛴𝐹π‘₯ = 0
+πΉπ‘π‘œπ‘‘π‘‘π‘œπ‘š βˆ’ 𝑓𝑠 π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ βˆ’ 𝑓𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 = 0
πΉπ‘π‘œπ‘‘π‘‘π‘œπ‘š = 𝑓𝑠 π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ + 𝑓𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛
= πœ‡π‘  π‘π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ + πœ‡π‘  𝑁𝑏𝑒𝑑𝑀𝑒𝑒𝑛
We already know the maximum static friction between the blocks.
𝑓𝑠 𝑏𝑒𝑑𝑀𝑒𝑒𝑛 = πœ‡π‘  π‘šπ‘” = 47
To find the Normal force with the ground, again, that equals the total weight pressing down from
above. This is the weight of two blocks.
π‘π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ = 2π‘šπ‘”
And we have
𝑓𝑠 π‘”π‘Ÿπ‘œπ‘’π‘›π‘‘ = πœ‡π‘  (2π‘šπ‘”)
= 2(πœ‡π‘  π‘šπ‘”)
= 2(47)
= 94 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
And we have a total of
πΉπ‘π‘œπ‘‘π‘‘π‘œπ‘š = 94 + 47 = 141 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
49
PHY117 Midterm Solutions
Q9
D)
2003 Q9
Forces
It's important to include every force acting on the motorcycle. They've given us propulsion and air
resistance, but the motorcycle is going up a ramp so there will also be the component of weight
acting along the slope, given by
𝐹𝑔π‘₯ = π‘šπ‘” sin πœƒ
= 292(9.8) sin 30
= 1431 π‘π‘’π‘€π‘‘π‘œπ‘›π‘ 
This will be acting down the slope, against the direction of the propulsive force. We have
Σ𝐹π‘₯ = π‘šπ‘Ž
π‘ƒπ‘Ÿπ‘œπ‘π‘’π‘™π‘ π‘–π‘œπ‘› βˆ’ π‘Žπ‘–π‘Ÿ π‘Ÿπ‘’π‘ π‘–π‘ π‘‘π‘Žπ‘›π‘π‘’ βˆ’ π‘Šπ‘₯ = π‘šπ‘Ž
3150 βˆ’ 250 βˆ’ 1431 = 292π‘Ž
1470 = 292π‘Ž
π‘Ž = 5.03 π‘š/𝑠 2
50
PHY117 Midterm Solutions
Q10
D)
2003 Q10
Forces (ramp)
The blocks are connected by a string, so they all move the same amount, they all have the same
magnitude of velocity and acceleration.
We can find this acceleration by looking at each block, or we can use a shortcut by considering all
the blocks along the rope as one system. We take the positive direction as up the slope, so that A is
moving to the left and B is falling down.
Now the force equation for the blocks is
Ξ£πΉπ‘Žπ‘™π‘œπ‘›π‘” π‘Ÿπ‘œπ‘π‘’ = π‘šπ‘Žπ‘™π‘™ π‘π‘™π‘œπ‘π‘˜π‘  π‘Ž
+π‘šπ΅ 𝑔 βˆ’ π‘šπ΄ 𝑔 sin πœƒ βˆ’ π‘“π‘˜π΄ = (π‘šπ΄ + π‘šπ΅ )π‘Ž
We also have mB = 2mA, so
2π‘šπ΄ 𝑔 βˆ’ π‘šπ΄ 𝑔 sin πœƒ βˆ’ π‘“π‘˜π΄ = (π‘šπ΄ + 2π‘šπ΄ )π‘Ž
2π‘šπ΄ 𝑔 βˆ’ π‘šπ΄ 𝑔 sin πœƒ βˆ’ π‘“π‘˜π΄ = 3π‘šπ΄ π‘Ž
We can find the friction acting on block A from
π‘“π‘˜π΄ = πœ‡π‘˜ 𝑁𝐴
There are no angled forces, so the Normal force with the slope equals the component of the weight
into the slope, from
Σ𝐹𝐴𝑦 = 0
𝑁𝐴𝑦 βˆ’ π‘Šπ΄π‘¦ = 0
𝑁𝐴𝑦 = π‘šπ΄ 𝑔 cos πœƒ
so we have
π‘“π‘˜π΄ = πœ‡π‘˜ π‘šπ΄ 𝑔 cos πœƒ
Plugging that back in, we get
2π‘šπ΄ 𝑔 βˆ’ π‘šπ΄ 𝑔 sin πœƒ βˆ’ πœ‡π‘˜ π‘šπ΄ 𝑔 cos πœƒ = 3π‘šπ΄ π‘Ž
2𝑔 βˆ’ 𝑔 sin πœƒ βˆ’ πœ‡π‘˜ 𝑔 cos πœƒ = 3π‘Ž
π‘Ž=
𝑔
(2 βˆ’ sin πœƒ βˆ’ πœ‡π‘˜ cos πœƒ)
3
51
PHY117 Midterm Solutions
Q11
A)
2003 Q11
Forces and Pythagoras
What's happening here? The block has a weight acting down, but is swinging out to the side. The
train must be accelerating, providing a force which swings the block out to the side.
The block is swinging back to the left, so the train must be accelerating to the right.
We can find the values we want from Pythagoras. The total force on the block is the reading on the
spring scale, 12 N.
The horizontal force would give us the horizontal acceleration, but we can't find that without the
angle. We can find the angle from other things we do know.
The vertical component is the weight of the block, mg = 1(9.8) = 9.8 N.
So the angle can be found from
π‘šπ‘”
πœƒ = cos βˆ’1 ( )
𝑇
= cos βˆ’1 (
9.8
)
12
= 35.2°
Now that we have the angle, we can find the horizontal component of the force acting on the block
𝐹π‘₯ = 12 sin 35.2
= 6.92 π‘π‘’π‘€π‘‘π‘œπ‘›
This is the force swinging the block to the side, so
𝐹π‘₯ = π‘šπ‘Žπ‘₯
6.92 = (1)π‘Žπ‘₯
π‘Žπ‘₯ = 6.92 π‘š/𝑠 2 β‰ˆ 7 π‘š/𝑠 2
The block is swinging back to the left, so the train must be accelerating to the right.
52
PREP101
Q12
C)
PHYS117 MIDTERM SOLUTIONS
2003 Q12-13
Forces (3rd Law)
This is the classic question of Newton's third law. If A exerts a force on B, B exerts the SAME force on
A. It doesn't matter what difference in masses exists. So the answer is C or D.
The acceleration is given by
𝐹 = π‘šπ‘Ž
π‘Ž=
𝐹
π‘š
Since the father has twice the mass, he will have half the resulting acceleration. This gives answer C).
Q13
A)
Friction
If we have increased the angle until the object starts to move, the component of weight along the
slope exceeded the maximum static frictional force. The object now experiences kinetic friction.
Since the kinetic friction and weight components are constant, the total force is constant, the total
acceleration is constant, and the object will continue to speed up at a constant rate.
53
COURSE CRAM
Q1
E)
Midterm Supplement Solutions
Extra Torque Questions
Statics, torque and forces
First we draw a free body diagram of the ladder.
There will be normal forces where it contacts the wall and the floor.
There will be weight forces at the center of the ladder, and the location of the man.
The frictional force is static, because the point of contact is not slipping, and to the
right, to prevent the ladder slipping out to the left. (The only other horizontal force on
the ladder is to the left, the normal force with the wall).
We always start with torques, and we always take the torques at the point with the
most unknowns.
Taking the torques around the base of the ladder
π›΄πœ = 0
βˆ’π‘šπ‘š 𝑔(2.5 sin 30) βˆ’ π‘šπΏ 𝑔(5 sin 30) + 𝑁𝑀 (10 sin 60) = 0
βˆ’80(9.8)(2.5 sin 30) βˆ’ 20(9.8)(5 sin 30) + 𝑁𝑀 (10 sin 60) = 0
and we have
𝑁𝑀 = 168
After the torques, we always look at the forces
𝛴𝐹π‘₯ = 0
𝑓𝑠 βˆ’ 𝑁𝑀 = 0
𝑓𝑠 βˆ’ 168 = 0
𝑓𝑠 = 168 𝑁
To two significant figures in scientific notation, 168 β‰ˆ 1.7 × 102 = 170 𝑁
COURSE CRAM
Q2
C)
Midterm Supplement Solutions
Extra Torque Questions
Torque
First we draw a free body diagram. Don't forget that the stick has a 1 N weight, which
will be at the midpoint!
We take the torque around the point with the most unknowns that we don't want, so
we take it around the left end. The torque equation is then
π›΄πœ = 0
βˆ’2(0.1) βˆ’ 1(0.5) βˆ’ 3(0.5) βˆ’ 3(0.6) + 𝑇100 1 = 0
𝑇100 = 4 𝑁
BONUS PRACTICE
Then if we wanted to find the tension in the other rope
Σ𝐹𝑦 = 0
𝑇0 βˆ’ 2 βˆ’ 1 βˆ’ 3 βˆ’ 3 + 4 = 0
𝑇0 = 5 𝑁
COURSE CRAM
Midterm Supplement Solutions
Extra Torque Questions
Q3
Torque
D)
When the bucket is "just" raising, the total torque on the crank is zero.
Draw the free body diagram of the cran
k.
π›΄πœ = 0
+π‘‡π‘Ÿ βˆ’ 𝐹𝑅 = 0
The tension in the cable is provided by the weight of the bucket.
Looking at the bucket, which has zero acceleration because the crank is only "just"
moving,
𝛴𝐹𝑦 = 0
𝑇 βˆ’ π‘šπ‘” = 0
𝑇 = π‘šπ‘”
and in the torque equation we have
π‘šπ‘”π‘Ÿ βˆ’ 𝐹𝑅 = 0
23(9.8)0.08 βˆ’ 𝐹(0.25) = 0
𝐹 = 72.1 𝑁
COURSE CRAM
Midterm Supplement Solutions
Extra Torque Questions
Q4
E)
Equilibrium means Ξ£F = 0 and Στ = 0. Look at each, taking Ο„ around the center (if it’s
in equilibrium, Στ will be zero around any point)
1)
Ξ£F = -F – 2F + 3F = 0
Στ = +F(r) – 2F(r)
2)
Ξ£F = +3F – F – 2F = 0
Στ = -3F(r) -2F(r) = -5Fr
3)
Ξ£F = F – 2F + F = 0
Στ = -F(r) + F(r) = 0
Only 3) is in equilibrium
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