PHYS 100 Midterm Review: Exam Format, Questions Covered, and Strategies, Exercises of Physics

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Physics 100 Midterm Review, Slide 1

PHYS 100 Midterm Exam

Review Session

(^

)^

(^

2 1 0

2 2 2

2 0

(^00)

0

x

x^

x x

x

x

x

x

v^

v^

a t x^

x^

v^

a t

v^

v

a^ x

x

=^

=^

+^

=^

+^

− x y

net on A

A^ A

F

m a=

^

Physics 100 Midterm Review, Slide 2

Midterm Exam

TODAY (Mar 9):

Review Lecture

Next Tuesday (Mar 13)

Midterm Exam7pm 144 LoomisIf you have a conflict with this time, youcan Sign Up in Gradebook before 10pm Mar12 for Conflict Exam (5:15pm 136 Loomis)

Practice Exams

Old PHYS 211 Exams available from our homepage

Office Hours (Mon Mar 12)

1pm – 4pm in 271 Loomis

Physics 100 Midterm Review, Slide 4

What does the exam cover?

• ~30% Conceptual

Questions

• ~70% Calculations

• 3/5 Numerical– 2/5 Symbolic

Question Types

• ~50% Kinematics

• Graphs– 1-D kinematics– 2-D kinematics– Relative motion

• ~50% Newton’s Laws

• Applying Newton’s Laws– Friction– UCM

Topics (there’s overlap)

Physics 100 Midterm Review, Slide 5

Major Ideas of PHYS 100 KINEMATICS

•^

Definitions^ –^

Graphs– Vector Pictures

-^

Constant Acceleration Eqns^ –^

1-D problems– 2-D problems•^

Projectile Motion ( )

(^

2 1

0

2 2 2

0

(^00)

0

x

x^

x x

x

x

x

x

v^

v^

a t

x^

x^

v^ t

a t

v^

v

a^

x^

x

=^

=^

+^

=^

+^

Newton’s Second Law

Fnet on A,x

=^ m

aAx

F^1 x^

+^ F^2

+… = x

m^ aA

x

Fnet on A,y

=^ m

aAy

F^1 y^

+^ F^2

+… = y

m A

ay

net on A

A^ A

F

m a=

^

x y

DYNAMICS

Physics 100 Midterm Review, Slide 7

Practice Problems

We’ll do some here.There are lots more in:

•ONLINE PRACTICE EXAMS

–available from “Practice Exams” link onHomepage

•CheckPoints•Online Quizzes•Discussion Problems•Worked examples

Physics 100 Midterm Review, Slide 8

Sp06 #13)

A box of mass

M^ is pulled across a horizontal floor with constant

velocity by a rope that makes an angle

θ^ with the horizontal. The tension in

the rope is

T^ and the coefficient of kinetic friction between the box and the

floor is

μ. The magnitude of the normal force between the floor and the box

is^ N

The net force on the box is(a)^

in the +x direction. (b)^

in the -x direction. (c)^

zero.

^ a

Fnet

BB

Physics 100 Midterm Review, Slide 10

Sp06 #14) Which one of the following relationships is true? (a)^

T^ cos

θ^ -^ μ

N^ = 0

(b)^

N^ -^

Mg^

(c)^

T^ -^

Mg^

(d)^

T^ sin

θ^ -^ N

cos

θ^ = 0

(e)^

T^ cos

θ^ -^ μ

Mg^

Mg^

N

f =^ μ

N

x:^

Tcos

θ^ –^

μN = 0

y:^

Tsin

θ^ + N - Mg = 0

Note: N

∫^ Mg

BB

Physics 100 Midterm Review, Slide 11

Sp06 #8)

A stuntwoman jumps from the rooftop of a building with an initial

velocity

v= 6.0 m/s^0

at an angle

θ^ = 30°

with respect to the ground. She lands on a

mattress as shown. The height of the building is

H^ = 5.0 m.

Ignore air resistance

and the thickness of the mattress.

For what length of time

t^ is she in the air?

(a)

t^ = 3.08 sec

(b)

t^ = 2.71 sec

(c)

t^ = 2.58 sec

(d)

t^ = 2.03 sec

(e)

t^ = 1.36 sec

Strategy Poll:

Which strategy works and is the most efficient?

A) Use the fact that the y-velocity at the top of the trajectory is zero to find the

time to the top of the flight (using v

= vy

+a0y t).^ y

Next find the time it takes to

go from this point to hitting the ground using y = y

+v 0 0y

+1/2a

(^2) t.^ y Add the two

times to together to get the total time in the air. B) Use the distance-time kinematic equation y = y

+v 0 0y

t+1/2a

(^2) ty in the y-direction

to find the time to go from the initial height to hitting the ground. C) Use the velocity-distance equation v

2 =vy

2 +2a0y

(y-yy

) to find the woman’s final y- 0

velocity when she hits the ground. Use this and the initial velocity in thevelocity-time equations v

= vy

+a0y t to get the flight time.y

BB

Physics 100 Midterm Review, Slide 13

Sp06 #8)

A stuntwoman jumps from the rooftop of a building with an initial

velocity

v= 6.0 m/s^0

at an angle

θ^ = 30°

with respect to the ground. She lands on a

mattress as shown. The height of the building is

H^ = 5.0 m

. Ignore air resistance

and the thickness of the mattress.

She is in air for

1.36 s.

WHEN DID SHE REACH HER HIGHEST POINT? (a)

t^ = 0.15 sec

(b)

t^ = 0.3 sec

(c)

t^ = 0.45 sec

(d)

t^ = 0.68 sec

(e)

t^ = 0.9 sec

at

v

v^

+= o (^) gt

v v^

o y^

−

=

θ sin

FOLLOW

• UP

2

1 att 2

v

x

x^

o o^

v^ y

g v t^

o top

θ sin =

ttop

BB

Physics 100 Midterm Review, Slide 14

Sp06 #8)

A stuntwoman jumps from the rooftop of a building with an initial

velocity

v= 6.0 m/s^0

at an angle

θ^ = 30°

with respect to the ground. She lands on a

mattress as shown. The height of the building is

H^ = 5.0 m

. Ignore air resistance

and the thickness of the mattress.

She is in air for

1.36 s.

HOW FAR DID SHE GO? (a)

D^ = 4.2 m

(b)

D^ = 5.2 m

(c)

D^ = 7.1 m

(d)

D^ = 8.4 m

(e)

D^ = 9.1 m

at

v

v^

+= o

FOLLOW

• UP

2

1 att 2

v

x

x^

o o^

D =?

(^0) = a^ x

(^

cos

)

6

/^ *cos(30 ) *1.

o D^

v^

t^

m^

s^

s^

m

θ

=^

=^

=



BB

Physics 100 Midterm Review, Slide 16

F07 #22) A student who can swim with a speed of

3 m/s

in still water wants to get

to the other side of a

400 m

wide river whose waters flows downstream at

2 m/s

If she swims in such a way that her path as viewed by someone on the shore isstraight across the river, how long does it take her to get to the other side?(a)^

179 s (b)^

112 s (c)^

208 s (d)^

92 s (e)^

156 s

Solve and vote… we will then discuss

2 m/s 3 m/s

m/s 5

s

m 400 m/s 5

= t

BB

Physics 100 Midterm Review, Slide 17

A man holds a

10 kg

mass suspended at rest above the ground by pushing it against

a wall. If he pushes with a force of

90 N

directed

60 degrees

above the

horizontal, what is the force of friction exerted by the wall on the mass? Thecoefficient of static friction between the block and the wall is

a)^

20 N

b) 30 Nc)^

36 N

d) 40 Ne) 78.5 N

Which strategy can be used to solve this problem?^ A) Use f

=^ μs

N^ to find the static friction force.s

Use F

= Ma in thenet

x-direction to find the normal force to plug in. The accelerationsare a

=0 and ax

=0.y

B) Use f

=^ μs

N^ to find the static friction force.s

Use F

= Ma in thenet

x-direction to find the normal force to plug in. The accelerationsare a

=0 and ax

=g downward.y

C) Use F

= Ma in the y-direction to find the static friction force.net The accelerations are a

=0 and ax

=0.y

D) Use F

= Ma in the y-direction to find the static friction force.net The accelerations are a

=0 and ax

=g downward.y

x y

BB