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Physics 101 Midterm-I Solutions Outline - Winter 2024, Study notes of Physics

Solutions for the written problems and multiple-choice questions in the physics 101 midterm-i exam for the winter 2024 semester. The solutions cover topics such as constants, equations, conversions, vectors, trigonometric functions, translational kinematics, projectile motion, and circular motion. The document also includes extra credit problems related to these topics.

Typology: Study notes

2022/2023

Uploaded on 03/14/2024

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quan-b-nguyen ๐Ÿ‡บ๐Ÿ‡ธ

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SOLUTIONS(OUTLINE(
PHYS(101(-(Winter(2024((((MIDTERM-I(
(
Name:(___________________________________________________( Recitation(section:(________________(
1. Your solutions to written problems must have adequate detail, and the answers must contain
appropriate units to get full credit. Read the problems carefully.
2. There are four multiple-choice questions and two numerical problems + one extra credit
multiple-choice question and one extra credit numerical problem.
Prob.
Score
1
2
3
4
Total
pf3
pf4
pf5

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Download Physics 101 Midterm-I Solutions Outline - Winter 2024 and more Study notes Physics in PDF only on Docsity!

SOLUTIONS OUTLINE

PHYS 101 - Winter 2024 MIDTERM-I

Name: ___________________________________________________ Recitation section: ________________

1. Your solutions to written problems must have adequate detail, and the answers must contain

appropriate units to get full credit. Read the problems carefully.

2. There are four multiple-choice questions and two numerical problems + one extra credit

multiple-choice question and one extra credit numerical problem.

Prob. Score

Total

Constants, Equations, Conversions

g = 9.80 m/s

2

Quadratic Equation

Roots of ๐‘Ž๐‘ฅ

!

  • ๐‘๐‘ฅ + ๐‘ = 0

ยฑ

#$ยฑโˆš$

! #&'(

!'

Vectors

)

Simple trigonometric functions

!

  • ๐‘

!

sin ๐œƒ =

'

,

cos ๐œƒ =

$

,

tan ๐œƒ =

'

$

Translational Kinematics - 1D

'-,)

) !

#) "

/ !

#/ "

)

0/

'-,)

  • !#

#- "#

/ !

#/ "

)

0 -

0/

1

)

/

2

)

(t) = ๐‘ฃ

)

/

2

Special case: Constant acceleration โ€“ 1 - D

)

)

4

!

)

!

  • ๐‘ฃ

3

)

! = ๐‘ฃ

!

  • 2 ๐‘Ž )

3

3

4

!

)

Projectile Motion

3

3

3

3

4

!

!

)

3

3

3

3

Circular motion

5'

!

5

& 6

! 7

8

!

[III] ( 8 pts) A jogger runs along a straight and level road for a distance of 5.0 km and then runs

back to her starting point. It took the jogger 2.0hrs to complete the round trip. For the round

trip, which one of the following statements is true?

[a] Her average speed is 5.0km/hr and her average velocity is 5.0km/hr.

[b] Her average speed is 5.0km/hr and her average velocity is 0.0km/hr.

[c] Her average speed is 0.0km/hr and her average velocity is 0.0km/hr.

[IV] ( 8 pts) A careless driver drives his car S through red traffic light and

continues to travel at a constant speed. A police car P starts from rest and

moves with constant acceleration to chase S as shown in the speed( v ) vs.

time( t ) graph. The police car P will catch up with S at:

[a] tc [b] before tc [c] after tc

[IV] (Extra Credit)(5pts) A ball B1 is thrown up from the ground at a velocity, vo , such that it will reach a

maximum height H. Simultaneously, another ball B2 is dropped from rest at height H.

The speed of ball B2 when it hits the ground is:

[a] < vo [b] vo. [c] > vo

The ball B2 takes time T to hit the ground. The time is taken as t=0 when B2 is released. The balls cross paths

at time:

[a] < T/2 [b] T /2 [c] > T /

The balls cross paths at the position above the ground at:

[a] < H /2 [b] H /2 [c] > H /

S

P

v

t t c

o

PROB. 2 (28pts) A jogger runs from P to O in three straight line segments ๐ด

, and ๐ถ

in 80.0s as

shown in the diagram, where Pa = 40.0m, ab = 80.0m, and bO = 50.0m.

[a] (21pts) Express the three displacement vectors ๐ด

, and ๐ถ

in their component form in the

unit vector (๐šคฬ‚ , ๐šฅฬ‚ )^ notation. (Example:๐ด

!

"

[b] (7pts) What is the average speed of the jogger for the trip PabO?

A = 40. 0 (sin 53

o ห† i โˆ’ cos 53

o ห† j ) m = ( 32. 0

i โˆ’ 24. 0

j ) m

B = 80. 0 (cos 60

o (^) ห†

i + sin 60

o (^) ห†

j ) m = ( 40. 0

i + 69. 3

j ) m

C = โˆ’ 50. 0 (cos 37

o ห† i + sin 37

o ห† j ) m = (โˆ’ 40. 0

i โˆ’ 30. 0

j ) m

ave. speed =

total dist.

time

( 40. 0 + 80. 0 + 50. 0 ) m

  1. 0 s

= 2. 1 m / s

P

o 120

o

o

x

y

o

A

B

C

a

b

53

Prob. 4 (Extra Credit ) (5pts) While filming a chase scene in a

movie, a stuntwoman is to run horizontally off the edge of one

rooftop and land on a second roof that is 1.50m away

(horizontally) and 1.25m lower. Find the minimum speed the

stuntwoman should have by modeling her as a point particle.

Let t be the time it takes to free-fall a distance of 1.25m.

ร— ( 9. 8 ๐‘š/๐‘ 

$

$

๐‘ก = H

Minimum speed required for the stuntwoman to travel a horizontal distance of 1.5m in 0.5s is:

.2+

1.25m

1.5m