PROBLEM (1) (25 points) A A driver wants to jump across a 90-m wide canyon with his
B
y
car. The car starts from rest at point A, travels with a constant acceleration, moves horizontally off the cliff on the higher
g
side through point B into the air and lands on the edge of the
x
C
120 m
cliff at point C, on the lower side. Express your answers to
75 m
two significant figures. (g = 10 m/s2)
90 m
(a) (6 pts) Find the time it takes the car to go from B to C.
Let's take the origin at point B. 1 y f " y i = v yi t + a y t 2 where v yi = 0 2
2("45 m) "10 m/s 2 t = 3.0 s
2(y f " y i )
t= !
t=
!
ay !
a y = "g = "10 m/s 2 !y f " y i = "(120 m " 75 m) = "45 m
! ! (b) (6 pts) Find the speed at which the car moves off the cliff at B.
x f " x i = v xi t
v xi = !
x f " xi t
=
90 m 3.0 s
v xi = 30 m/s
! (c) (7 pts) Find!the car's velocity just before it lands at C. Express your answer in unit vectors.
" " ! v f = v xf i + v yf j " " " " ! v f = (30 i " 30 j) m/s = 30( i " j) m/s
v xf = v xi = 30 m/s
v yf = v yi + a y t = 0 + ("10 m/s 2 )(3.0 s) = "30 m/s !
!
!
! (d) (6 pts) Assuming the acceleration of the car from A to B was 5.0 m/s2, find the distance between A and B.
v xf2 = v xi2 + 2a x (x f " x i ) x f " xi =
!
v xf2
" v xi2
2a x
=
v xf2
"0
2a x
Notice: Alternative solution is possible. 2
=
(30 m/s) 2(5.0 m/s 2 )
x f " x i = 90 m
! !
Phys 105 First Midterm Examination
Page 1
Friday, 15-Jul, 2011
PROBLEM (2) (25 points) An ant (a small insect) makes a trip on a graph paper. It starts from the origin and walks along the x axis a distance of 8.0 cm in 2.0 s. It then turns left 37° and walks in a straight line another 10 cm in 1.8 s. Finally, it turns another 53° to the left and walks another 6.0 cm in 1.2 s and finishes the trip. Express your answers to two significant figures. (cos 37° = sin 53° = 0.80; sin 37° = cos 53° = 0.60)
! ! ! (a) (6 pts) Using the x and y axes given below, draw the ant's three successive displacements ( "r1 , "r2 , and "r3 ) and the ! resultant displacement ( "r ). In the figure, label the displacements and show the angles mentioned.
y
!
!
! !r 3 53°
! !r
! !r1
! !r 2 37°
x
O (b) (6 pts) Find the final position vector of the ant. Express your answer in unit vectors.
" " " " ! r f = 8.0 i cm + (8.0 i + 6.0 j) cm + 6.0 j cm " " ! r f = (16 i +12 j) cm
! ! ! ! r f = "r1 + "r2 + "r3 " ! "r1 = 8.0 i cm
" " ! "r2 = [10(cos 37°) + 10(sin 37°)] cm = (8.0 i + 6.0 j) cm ! " ! "r3 = 6.0 j cm !
! !
! (c) (6 pts) Find ! the ant's average velocity during the trip. Express your answer in unit vectors.
! v avg
! ! ! ! "r r f # ri v avg = = "t "t " " " " (16 i +12 j) cm " 0 16 i cm 12 j cm = = + (2.0 + 1.8 + 1.2) s 5.0 s 5.0 s
" " ! v avg = (3.2 i + 2.4 j) cm/s
!
!
! (d) (7 pts) If the ant walks to its starting point in a straight line in 5.0 s, find its average speed during the round trip. v avg =
d "t
d = total distance traveled ! ! ! ! !d = "r1 + "r2 + "r3 + "r4 ! ! # & "r4 = r f = % (16) 2 + (12) 2 ( cm = 20 cm $ '
!
"t = (2.0 + 1.8 + 1.2) s + 5.0 s = 10 s 44 cm v avg = 10 s !
v avg = 4.4 cm/s
! !
d = 8.0 cm +10 cm + 6.0 cm + 20 cm = 44 cm
!
!
Phys 105 First Midterm Examination
Page 2
Friday, 15-Jul, 2011
PROBLEM (3) (25 points) An airplane with a speed of 300 km/h relative to the air heads due north (N) in a wind blowing at 80.0 km/h toward 37.0° south of west (S of W). Express your answers to three significant figures. (sin 37.0° = 0.600 and cos 37.0° = 0.800) (a) (6 pts) Let P represents the plane, A represents the air, and G represents the ground. Using the axes given below, draw a ! ! ! vector diagram representing the vector equation v PG = v PA + v AG .
N ! v PA
! v PG
!
W
E
37.0°
! v AG S (b) (7 pts) Find the velocity of the plane relative to the ground. Choose the positive x axis to be to the east (E) and the positive y axis north (N), and express your answer in unit vectors.
N y ! v PA
! v PG ! v AG, x
W ! v AG,y 37.0° ! v AG
E
x
!
S
! ! ! v PG = v PA + v AG " " ! ! v PG,x = v AG,x = "v AG cos 37.0° i = "(80.0 km/h)(0.800) i " ! v PG,x = "64.0 i km/h ! " " ! ! ! v PG,y = v PA " v AG,y = 300 j " v AG sin 37.0° j " " " ! v PG,y!= [300 j " (80.0)(0.600) j ] km/h = 252 j km/h " " ! v PG = ("64.0 i + 252 j) km/h !
! in a south of west (S of W) direction at an angle !, the plane's (c) (12 pts) If the wind blows at the same speed 80.0 km/h velocity relative to the ground deviates by the maximum angle " from ! due north (N), as shown in the figure. Find sin !.
N
The angle " is largest for an angle ! such that: d $ 80 cos " ' d (tan # ) = 0 ---> & )= 0 d" % 300 # 80sin " ( d"
! v PG
"
Quotient rule:
" W
E
!
!
! d (tan # ) = d" ($80 sin " )(300 $ 80 sin " ) $ (80 cos " )($80 cos " ) ! =0 (300 $ 80 sin " ) 2
! v AG S
tan " =
v PG,x v AG cos # 80 cos # = = v PG,y v PA $ v AG sin # 300 $ 80sin #
"24000sin # + 6400sin 2 # + 6400 cos 2 # =0 (300 " 80sin # ) 2 !
"24000 sin # + 6400(sin 2 # + cos 2 # ) = 0 6400 4 ---> sin " = sin " = = 0.267 24000 15
! !
d " u % 1 " du dv % $ '= 2 $ v(u ' dx # v & v # dx dx &
!
! Phys 105 First Midterm Examination
Page 3
! Friday, 15-Jul, 2011
PROBLEM (4) (25 points)
! F
Consider the system of the two blocks with masses m and ! 2m, shown in the figure. In the absence of friction a force F
2m
g
pulls the blocks at a constant velocity up the planes. Express
!2
your answers in terms of some or all of the given quantities ! and related constants as needed.
m
(a) (6 pts) Draw a free-body diagram of each block.
! n
!1
! n
! T m
! T
!1 ! mg
2m
! F
!2 ! 2mg
! (b) (6 pts) Find the magnitude of force F . Consider the system of two blocks. Take the positive ! direction to be in the direction of motion. ! ! ! a = 0 ---> " F = ma = 0
F " (mg sin #1 + 2mg sin # 2 ) = 0
!
F! = mg sin "1 + 2mg sin " 2 F = mg(sin "1 + 2 sin " 2 )
! !
! (c) (6 pts) ! The force F is now removed. Find the magnitude and direction of the acceleration of the blocks. Take the positive direction to be in the direction of motion. ! ! ! F mg sin "1 + 2mg sin " 2 " ---> a = a= m + 2m 3m g(sin "1 + 2sin " 2 ) ; down the planes a= 3 ! ! !
(d) ! (7 pts) In the absence of force F , find the tension in the string connecting the blocks.
*$ sin "1 ' 2sin " 2 T = mg,&sin "1 # )# / 3 ( 3 . +%
Apply Newton's second ! law to the block with mass m: ! ! " F = ma
$ g(sin "1 + 2sin " 2 ) ' mg sin "1 # T = ma = m& )( % 3 ! $ mg(sin "1 + 2sin " 2 ) ' T = mg sin "1 # & )( % 3
$ 2sin "1 2sin " 2 ' T = mg& # ) % 3 3 ( 2mg T= (sin "1 # sin " 2 ) 3
!
!
! !
! Phys 105 First Midterm Examination
Page 4
Friday, 15-Jul, 2011