PHYS 218 Practice Problems for the Final Exam Solutions

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PHYS 218

Practice Problems for the Final Exam Solutions

Newton’s Laws

a. The racket exerts a force on the ball and the ball exerts a force on the racket.

b. The two skaters each exert a force on each other. (They will both start to move as a result.)

c. Force of tree on car and force of car on tree. (They will both suffer damage.)

d. Each car exerts a force on the other car. (They will both be damaged as a result¿)

e. Force of person on wall and force of wall on person. (The force of the wall on the person has a vertical component that is equal and opposite to the force of gravity, but it is not a reaction force to gravity.)

f. Force of hammer on nail and nail on hammer.

g. Force of string on mass and force of mass on string. (The force of the string on the mass is equal in magnitude but opposite in direction to the force of gravity on the mass, but those forces are not an action-reaction pair.)

h. Force of bird on pole and force of pole on bird.

Motion in One Dimension

Problem 1

The only force acting on you is the force of the rocket (the contact force of the seat on your back, to be specific). You start from rest at t = 0. So:

X F = ma

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F _rocket = m ∆v

∆t = 75 kg × 500 ^m _s − 0 ^m _s

2.4 min − 0 min = 1935 N Problem 2

There are two ways to solve this problem. The kinematic method, and the energy method. As usual, when it is available, the energy method is the easiest option.

Kinematic Method : x _f = x ₀ + v _i t + ¹ ₂ at ² . This is the kinematic equation we can apply if (and only if) a is constant.

F _f = −µ _k N = −µ _k M _car g

This is a constant force. It is also the only force acting on the car in the horizontal direction. So P F _x = F _f = M _car a = constant and we can apply that equation, with a = −µ k g after canceling M car .

x _f − x ₀ = v _i t − 1 2 µ _k gt ²

Unfortunately, there are two unknown quantities, µ k and t. So we will need one more equation to solve the problem. That equation is v _f = v i + at or

v f − v _i = −µ k gt t = − ∆v

µ _k g

We can plug that into our first equation to obtain

∆x = − v i ∆v µ _k g − 1

2 µ k g( ∆v µ s g ) ²

∆x = − v i ∆v µ _k g − 1

2 ∆v ² µ _k g

∆x = 1

µ _k g × (−v _i ∆v − 1 2 ∆v ² ) µ _k = 1

g∆x × (−v _i ∆v − 1 2 ∆v ² ) Now, ∆v = −v i , so:

µ _k = 1

g∆x × (v _i ² − 1

2 v ² _i ) = v _i ²

2g∆x = 0.51

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Energy Method : The only force acting on the car with a component along its direction of motion is the frictional force. So the only work done on the car is work due to the frictional force. The potential energy of the car does not change, so the change in total energy will be equal to the change in kinetic energy.

W = ∆E = E _f − E _i F f ∆x = −µ k M car g∆x = 0 − 1

2 M car v ² _i = − 1

2 M car v ² _i µ k M car g∆x = 1

2 M car v _i ² µ s = v ² _i

2∆x = 0.51

Motion in Two and Three Dimensions

X ~ F = m~a

P F _x = ma x P F _y = ma y

0 = ma _x −mg = ma _y =⇒ a _y = −g

x f = v x t 0 = y i − ¹ ₂ gt ²

↓ ←− t = q

2y

i

g

x _f = v _x q

2y

i

g t = 2.6 s

x _f = 45 m

v f

x

= v i

x

+ a x t v f

y

= v i

y

+ a y t

v _f

_x

= v i

x

v _f

_y

= −gt

v _f

_x

= 20 ^m _s v _f

_y

= −22 ^m _s

& .

|~v| = p20 ² + (−22) ^{2 m} _s

|~v| = 30 ^m _s

θ = tan ⁻¹ (− ²² ₂₀ ) = −0.83 rad θ = −48 °

Note that you could have figured out |~ v| using energy methods, but not the

angle or time to impact.

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Uniform Circular Motion and Rotational Motion

a. T = 60 s 60 rev = 1 s b. ω = 2πf = 2π

T = 2π s ⁻¹ c. a c = v ²

r = ω ² r = 4π ² s ⁻¹ × 0.30 m ≈ 11.8 ^m _s

₂

d. 0, the angular velocity is constant

Gravitation

To a good approximation, the gravitational force of the Sun is the only force on the Earth. To a good approximation, the Earth travels around the Sun along a circular path at a constant speed. Let r Earth be the distance between the Sun and the Earth.

F Sun = m Earth a Earth

F Sun = m Earth

v _Earth ² r Earth

Now, v _Earth = ^distance _time = ^2πr _{1 year}

^Earth

, so F _Sun = m _Earth 4π ² r _Earth

1 year ² = 3.5 × 10 ²² N On the other hand,

F _Sun = G M _Sun m _Earth r _Earth ² M Sun = F _Sun r ² _Earth

Gm _Earth = 2.0 × 10 ³⁰ kg

This is, roughly, how the mass of the Sun was first calculated.

Work and Energy

Problem 1

a. 0 J. The refrigerator moves horizontally at a constant speed, so

W _total = ∆E = ∆K + ∆U = 0 + 0 = 0.

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b. There are two forces (in the horizontal direction) on the refrigerator. The force you are exerting, and the force of friction. W _tot = 0 =⇒ W _{f riction} =

−W _you .

W _you = F _you ∆x = 300 N × 3.0 m = 900 J W _{f riction} = −900 J

The time given was completely irrelevant. Energy has nothing to do with time and everything to do with distance. Also notice that we were able do figure out the work done by the frictional force without knowing the mass of the fridge or the coefficient of friction, by exploiting the Work-Energy theorem.

Problem 2

We will treat gravity using potential energy. Then no forces do work on the skier, so ∆E = ∆K + ∆U = 0.

K _f − K _i + U _f − U _i = 0 1

2 mv _f ² − 0 + 0 − mgh = 0 v _f = p

2gh = 63 m s

Momentum and Collisions

Let v i be the initial velocity of the smaller object of mass m, and let v _f and V _f be the final velocities of the objects of mass m and M , respectively.

Momentum is conserved during this collision and, because the collision is elastic, so is kinetic energy.

p _f = p _i

mv f + M V f = mv i + M × 0 = mv i

V f = mv _i − mv _f M K _f = K _i 1

2 mv ² _f + 1

2 M V _f ² = 1

2 mv _i ² + 1

2 M 0 ²

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mv ² _f + M V _f ² = mv _i ² mv _f ² + M ( mv _i − mv _f

M ) ² = mv ² _i (m + m ²

M )v _f ² − (2 m ²

M v i )v f + ( m ²

M − m)v ² _i = 0 v _f = (2 ^m _M

²

v _i ) ±

q

(2 ^m _M

²

v _i ) ² − 4(m + ^m _M

²

)( ^m _M

²

− m)v ² _i 2(m + ^m _M

²

)

v f =

(2 ^m _M

²

v _i ) ± q

4m ² v _i ² 2(m + ^m _M

²

) v f =

m

²

M ± m

m

²

M + m v i

. &

v _f = v _i v _f =

m

²

M − m

m

²

M + m v _i v _f = v _i v _f = m − M

m + M v _i V f = ^mv

ⁱ

_M ^−mv

ⁱ

V f =

mv

i

−m m − M m + M ^v

ⁱ

M

V _f = 0 V _f = _M ^m (1 − m − M m + M )

The left option does not change the velocity of either object. This does conserver both energy and momentum, but it represents the smaller object passing through the larger object without interacting with it at all. That is impossible (and contrary to the statement of the problem) so we should reject that option.

The right option must then be the correct answer.

v f = m − M

m + M v i = − 2 kg 7 kg 12 m

s = − 24 7

m

s ≈ −3.4 m s

V _f = m

M (1 − m − M

m + M )v i = 2.5 kg 4.5 kg

9 kg 7 kg 12 m

s = 5

7 × 12 m s = 60

7 m

s ≈ 8.6 m

s

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Angular Momentum

Problem 1

See solution to Problem Set 6.

Problem 2

X τ = Iα

All three objects experience the same torques. So the object with the smallest moment of inertia will have the largest angular acceleration at all times and will win the race.

I _cyl = ¹ ₂ M R ² , I _hollow = ² ₃ M R ² , and I _solid = ³ ₅ M R ² . So the cylinder will win.

Rotational Energy

See solution to Problem Set 5.

Fluids

Let location 1 be the water tower, and location 2 be the house. P ₁ = P _atm in a normal water tower. h 2 − h ₁ is the given 30 m. And v 1 ≈ 0 since the water tower, which can serve a whole town, empties very very slowly.

P 1 + ρgh 1 + 1

2 ρv ₁ ² = P 2 + ρgh 2 + 1 2 ρv ² ₂ P 2 − P _atm + ρg(h 1 − h ₂ ) = 1

2 ρv ² ₂ P atm + ρg(h 1 − h ₂ ) − P atm + ρg(h 1 − h ₂ ) = 1

2 ρv ₂ ² 2gρ(h 1 − h ₂ ) = 1

2 ρv ₂ ² v ₂ = p

4g(h ₁ − h ₂ ) = 34 m

s

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Harmonic Motion

f _pend = _2π ¹ q

g

L , so T _pend = 2π q

L

g . Thus to double the period you should quadruple (×4) the length of the swinging arm.

The frequency of a pendulum (at small enough angles to undergo simple harmonic motion) is independent of the amplitude of the motion. So altering the amplitude will not change the period.

Elasticity and the Various Moduli

F

A = Y ∆L L 0

F = Y A∆L L 0

= 1000 N

∆L L 0

= 6 × 10 ⁻⁴

Waves and Sound

Problem 1

The general equation we’ll need is v _wave = f λ. Which is equivalent to f = ^v

PHYS 218 Practice Problems for the Final Exam Solutions

PHYS 218

Practice Problems for the Final Exam Solutions

Newton’s Laws

a. The racket exerts a force on the ball and the ball exerts a force on the racket.

b. The two skaters each exert a force on each other. (They will both start to move as a result.)

c. Force of tree on car and force of car on tree. (They will both suffer damage.)

d. Each car exerts a force on the other car. (They will both be damaged as a result¿)

e. Force of person on wall and force of wall on person. (The force of the wall on the person has a vertical component that is equal and opposite to the force of gravity, but it is not a reaction force to gravity.)

f. Force of hammer on nail and nail on hammer.

g. Force of string on mass and force of mass on string. (The force of the string on the mass is equal in magnitude but opposite in direction to the force of gravity on the mass, but those forces are not an action-reaction pair.)

h. Force of bird on pole and force of pole on bird.

Motion in One Dimension

Problem 1

The only force acting on you is the force of the rocket (the contact force of the seat on your back, to be specific). You start from rest at t = 0. So:

X F = ma

F rocket = m ∆v

∆t = 75 kg × 500 m s − 0 m s

2.4 min − 0 min = 1935 N Problem 2

There are two ways to solve this problem. The kinematic method, and the energy method. As usual, when it is available, the energy method is the easiest option.

Kinematic Method : x f = x 0 + v i t + 1 2 at 2 . This is the kinematic equation we can apply if (and only if) a is constant.

F f = −µ k N = −µ k M car g

This is a constant force. It is also the only force acting on the car in the horizontal direction. So P F x = F f = M car a = constant and we can apply that equation, with a = −µ k g after canceling M car .

x f − x 0 = v i t − 1 2 µ k gt 2

Unfortunately, there are two unknown quantities, µ k and t. So we will need one more equation to solve the problem. That equation is v f = v i + at or

v f − v i = −µ k gt t = − ∆v

µ k g

We can plug that into our first equation to obtain

∆x = − v i ∆v µ k g − 1

2 µ k g( ∆v µ s g ) 2

∆x = − v i ∆v µ k g − 1

2

∆v 2 µ k g

∆x = 1

µ k g × (−v i ∆v − 1 2 ∆v 2 ) µ k = 1

g∆x × (−v i ∆v − 1 2 ∆v 2 ) Now, ∆v = −v i , so:

µ k = 1

g∆x × (v i 2 − 1

2 v 2 i ) = v i 2

2g∆x = 0.51

W = ∆E = E f − E i F f ∆x = −µ k M car g∆x = 0 − 1

2 M car v 2 i = − 1

2 M car v 2 i µ k M car g∆x = 1

2 M car v i 2 µ s = v 2 i

2∆x = 0.51

Motion in Two and Three Dimensions

X ~ F = m~a

P F x = ma x P F y = ma y

0 = ma x −mg = ma y =⇒ a y = −g

x f = v x t 0 = y i − 1 2 gt 2

↓ ←− t = q

2y

g

x f = v x q

2y

g t = 2.6 s

x f = 45 m

v f

= v i

+ a x t v f

= v i

+ a y t

v f

= v i

v f

= −gt

v f

= 20 m s v f

= −22 m s

& .

|~v| = p20 2 + (−22) 2 m s

|~v| = 30 m s

θ = tan −1 (− 22 20 ) = −0.83 rad θ = −48 °

Note that you could have figured out |~ v| using energy methods, but not the

angle or time to impact.

Uniform Circular Motion and Rotational Motion

a. T = 60 s 60 rev = 1 s b. ω = 2πf = 2π

T = 2π s −1 c. a c = v 2

r = ω 2 r = 4π 2 s −1 × 0.30 m ≈ 11.8 m s

d. 0, the angular velocity is constant

F _rocket = m ∆v

∆t = 75 kg × 500 ^m _s − 0 ^m _s

Kinematic Method : x _f = x ₀ + v _i t + ¹ ₂ at ² . This is the kinematic equation we can apply if (and only if) a is constant.

F _f = −µ _k N = −µ _k M _car g

This is a constant force. It is also the only force acting on the car in the horizontal direction. So P F _x = F _f = M _car a = constant and we can apply that equation, with a = −µ k g after canceling M car .

x _f − x ₀ = v _i t − 1 2 µ _k gt ²

Unfortunately, there are two unknown quantities, µ k and t. So we will need one more equation to solve the problem. That equation is v _f = v i + at or

v f − v _i = −µ k gt t = − ∆v

µ _k g

∆x = − v i ∆v µ _k g − 1

2 µ k g( ∆v µ s g ) ²

∆x = − v i ∆v µ _k g − 1

∆v ² µ _k g

µ _k g × (−v _i ∆v − 1 2 ∆v ² ) µ _k = 1

g∆x × (−v _i ∆v − 1 2 ∆v ² ) Now, ∆v = −v i , so:

µ _k = 1

g∆x × (v _i ² − 1

2 v ² _i ) = v _i ²

W = ∆E = E _f − E _i F f ∆x = −µ k M car g∆x = 0 − 1

2 M car v ² _i = − 1

2 M car v ² _i µ k M car g∆x = 1

2 M car v _i ² µ s = v ² _i

P F _x = ma x P F _y = ma y

0 = ma _x −mg = ma _y =⇒ a _y = −g

x f = v x t 0 = y i − ¹ ₂ gt ²

x _f = v _x q

x _f = 45 m

v _f

v _f

v _f

= 20 ^m _s v _f

= −22 ^m _s

|~v| = p20 ² + (−22) ^{2 m} _s

|~v| = 30 ^m _s

θ = tan ⁻¹ (− ²² ₂₀ ) = −0.83 rad θ = −48 °

T = 2π s ⁻¹ c. a c = v ²

r = ω ² r = 4π ² s ⁻¹ × 0.30 m ≈ 11.8 ^m _s

v _Earth ² r Earth

Now, v _Earth = ^distance _time = ^2πr _{1 year}

, so F _Sun = m _Earth 4π ² r _Earth

1 year ² = 3.5 × 10 ²² N On the other hand,

F _Sun = G M _Sun m _Earth r _Earth ² M Sun = F _Sun r ² _Earth

Gm _Earth = 2.0 × 10 ³⁰ kg

W _total = ∆E = ∆K + ∆U = 0 + 0 = 0.

b. There are two forces (in the horizontal direction) on the refrigerator. The force you are exerting, and the force of friction. W _tot = 0 =⇒ W _{f riction} =

−W _you .

W _you = F _you ∆x = 300 N × 3.0 m = 900 J W _{f riction} = −900 J

K _f − K _i + U _f − U _i = 0 1

2 mv _f ² − 0 + 0 − mgh = 0 v _f = p

Let v i be the initial velocity of the smaller object of mass m, and let v _f and V _f be the final velocities of the objects of mass m and M , respectively.

p _f = p _i

V f = mv _i − mv _f M K _f = K _i 1

2 mv ² _f + 1

2 M V _f ² = 1

2 mv _i ² + 1

2 M 0 ²

mv ² _f + M V _f ² = mv _i ² mv _f ² + M ( mv _i − mv _f

M ) ² = mv ² _i (m + m ²

M )v _f ² − (2 m ²

M v i )v f + ( m ²

M − m)v ² _i = 0 v _f = (2 ^m _M

v _i ) ±

(2 ^m _M

v _i ) ² − 4(m + ^m _M

)( ^m _M

− m)v ² _i 2(m + ^m _M

(2 ^m _M

v _i ) ± q

4m ² v _i ² 2(m + ^m _M

v _f = v _i v _f =

M + m v _i v _f = v _i v _f = m − M

m + M v _i V f = ^mv

_M ^−mv

−m m − M m + M ^v

V _f = 0 V _f = _M ^m (1 − m − M m + M )