Download Solution of Midterm 1

MIDTERM #1, PHYS 1211 (INTRODUCTORY PHYSICS), October 6, 2014
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This exam book has 6 pages including an equation sheet on the last page
PART I: MULTIPLE CHOICE QUESTIONS (question 1 to 5)
For each question circle the one!correct answer.
! ! !
!
!
1. (3 points) The vectors a , b and c are related by c = a ! b . Which diagram below
illustrates this relationship?
a) I
b) II c) III d) IV e) None of these
!
b
!
!b
!
!b
!
a
!
c ANSWER C III
!
a
2. (3 points) A cat runs in a straight line (the x-axis) from point A to point B to point C, as
shown below. The distance between point A and C is 5.00m, between point B and C is
10.0m, and the positive direction of the x-axis points to the right. The time to run from A
to B is 20.0s and from B to C is 8.00s. The average speed for the whole trip is closest to
a) -0.179 m/s b) 0.536 m/s c) 0.179 m/s d) 0.893 m/s e) -0.893 m/s
10 m
AB = 15m ,
Savg =
5m
!t AB = 20s ; BC = 10m ,
!t BC = 8s .
Average
speed
AB + BC
m
= 0.893 . ANSWER D. Note that the speed must always be positive.
!t AB + !t BC
s
3. (3 points) A golf ball is hit so that it leaves the ground at 60! above the horizontal and
feels no air resistance as it travels. Which of the following statement about the
subsequent motion of the ball while it is in the air is true? Only one correct answer!
a) Its speed is zero at its highest point.
b) Its velocity is zero at its highest point.
c) Its acceleration is always 9.8 m/s2 downward. ANSWER C
d) Its forward acceleration is always 9.8 m/s2.
e) Its acceleration is zero at its highest point.
1
4. (3 points) Shown here are the velocity and acceleration vectors for an object in several
different types of motion. In which case is the object slowing down and turning to the
right?
!
A) v
!
a
!
a
B) v
!
C) v
!
a
D)
v
a
!
E) v
a
ANSWER: A
5. (3 points) Identical guns fire identical bullets horizontally at the same speed from the
same height above level planes, one on the Earth ( gearth = 9.8mis !1 ) and one on the Moon
( gmoon = 1.62mis !1 ). Which of the following three statements is NOT TRUE? Note: only
one correct answer.
I. The horizontal distance traveled by the bullet is greater for the Moon.
II. The flight time is less for the bullet on the Earth.
III. The velocities of the bullets at impact with the ground are the same.
Due to the lower gravitational acceleration on the moon, it will remain in the air for a
longer period. Since in both cases the horizontal velocities, vox , are the same the
horizontal distance will be greater on the moon. So I and II are true, and III must be not
true. To verify note that the y-component of the velocity is vy2 = voy2 + 2g!y , where we
take down as positive, and !y > 0 is the height above the level plane. Since the initial
velocity is horizontal voy = 0 , and vy = 2g!y , which will be smaller on he moon due to
the smaller g on the moon. Hence the impact velocities are different. ANSWER III
PART II: FULL ANSWER QUESTIONS (question 6 to 9)
Do all four questions on the provided space. Show all works.
(
)
!
6. (12 points) Let r = ( !a + bt ) iˆ + ct 2 + dt 3 ˆj be the position of a particle moving in the
xy plane, where a = 2m , b = 2mis !1 , c = !3mis !2 and d = 1.1mis !3 .
a) Find the position of the particle at t = 1s and t = 3s , and hence find the average velocity
for the time interval t = 1s and t = 3s . Write your answers in unit vector notation.
!
2
3
At t = 1s, r (1s ) = ( !a + b (1s )) iˆ + c (1s ) + d (1s ) ˆj = 0iˆ ! (1.9m ) ˆj
(
!
At t = 3s, r ( 3s ) = ( !a + b ( 3s )) iˆ + ( c ( 3s )
)
2
+ d ( 3s )
3
) ˆj = ( 4m )iˆ + (2.7m ) ˆj
(1.5 point)
x ( 3s ) ! x (1s ) 4m
y ( 3s ) ! y (1s ) 4.6m
m
m
=
= 2.0 and vavg! y =
=
= 2.3
3s ! 1s
2s
s
3s ! 1s
2s
s
m
m
!
vavg = 2.0 iˆ + 2.3 jˆ (1.5 point)
s
s
!
b) At time t = 2s , find the magnitude and angle w.r.t. the positive x axis of the velocity v .
dx
dy
vx =
= b = 2mis !1 ; vy =
= 2ct + 3dt 3 = !6mis !2 t + 3.3mis !3 t 2 (1 point)
dt
dt
2
!1
At t = 2s, vx ( 2s ) = 2mis ; vy ( 2s ) = !6mis !2 ( 2s ) + 3.3mis !3 ( 2s ) = 1.2mis !1
!
v ( 2s ) = 2mis !1 iˆ + 1.2mis !1 ˆj (1 point)
vavg! x =
(
(
) (
)
(
) (
)
2
(
)
)
v=
( 2mis ) + (1.2mis )
!1 2
!1 2
= 2.3mis !1 ; ! = tan "1 (1.2 / 2 ) = 31o wrt +x (1 point)
!
c) At time t = 2.6s , find the acceleration a in unit vector notation.
vx = 2mis !1 ; vy = !6mis !2 t + 3.3mis !3 t 2
(
) (
)
(
)
dvy
dvx
= !6mis !2 + 6.6mis !3 t . (1.5 point)
= 0 ; ay =
dt
dt
At t = 2.6 s, ax ( 2.6s ) = 0 ;
ax =
(
)
ay ( 2.6s ) = !6.0mis !2 + 6.6mis !4 ( 2.6s ) = 11.16mis !2 .
!
a ( 2.6s ) = 11.6mis !2 ˆj . (1.5 point)
(
)
d) Find the time when the velocity is constant. Determine the velocity in unit vector
!
notation, at this time (when the velocity is constant). HINT: What should a be at this time?
!
The velocity is constant when the acceleration is zero, a = 0 , ax = 0 and ay = 0 . From
(
)
part c, ax = 0 always, and ay = !6mis !2 + 6.6mis !3 t = 0 . Solving for t,
( 6.6mis ) t = 6mis
!3
!2
" t = 0.91s . (1.5 point)
From part B the velocity is vx ( 0.91s ) = 2mis !1 ;
(
)
) (
(
)
)
vy ( 0.91s ) = ! 6.0mis !2 ( 0.91s ) + 3.3mis !3 ( 0.91s ) = !2.73mis !1 .
!
v ( 0.91s ) = 2mis !1 iˆ ! 2.73mis !1 ˆj . (1.5 point)
(
2
7. (10 points) A car moves along an x axis through a distance of 920 m, starting at rest (at x
= 0) and ending at rest (at x = 920 m). Through the first 1/4 of that distance, its
acceleration is +6.00 m/s2. Through the next 3/4 of that distance, its acceleration is -2.00
m/s2.
a) Draw a motion diagram of the path of the car that includes the direction of acceleration
and velocity.
!
!
(2 points)
a1
a2
+x
x1 = 0
v1 = 0
t1 = 0
230 m
x2 = 230m
v2 = ?
t2 = ?
690 m
b) What is its travel time through the 920 m?
1
1
From point 1 to point 2, x2 = x1 + v1t 2 + a1t 22 = a1t 22
2
2
2 ( 230m )
2x2
t2 =
=
= 8.75595s (2 points)
a1
6mis !2
3
x3 = 920m
v3 = 0
t3 = ?
The velocity at point 2 is v2 = v1 + a1t 2 = 0 + 6mis !2 " 8.756s = 52.535702mis !1 (2 points)
Since the question states that the car comes to rest at x3 = 920m , we know that v3 = 0 .
v2
52.535702mis !1
Hence v3 = 0 = v2 + a2 ( t 3 ! t 2 ) " t 3 = t 2 +
= 8.756s +
= 35s (2 points)
a2
2mis !2
Alternatively we can use the equation x = x0 + vot + 0.5at 2 , with x0 = 0 , x = 690m ,
a = !2mis !2 to
and
obtain
the
quadratic
equation,
vo = 52.5350702mis !1 ,
2
2
690 = 52.5350702t ! 0.5 " 2t or t ! 52.535702t + 690 = 0 , with solution
52.5350702 ± 52.535702 2 ! 4 " 690
= 26.268s , since 52.535702 2 ! 4 " 690 = 0 . This
2
gives a total time of 8.756s + 26.268s = 35s .
FINAL NOTE: From point 2 to 3, we can calculate the distance traveled
2
x = vot + 0.5a2t 2 = 52.53702mis !1 " 26.268s + 0.5 !2mis !2 ( 26.268s ) = 690m , so it does
t=
(
)
(
)
travel 690m. As for the final speed v3 = v2 ! a1t = 52.535702mis !1 ! 2mis !2 ( 26.268s ) = 0 , so
it comes to rest at x3 = 920m
c) What is its maximum speed?
Since it begins to decelerate once it reaches x2 = 230m , the top speed occurs at this point. It is
52.535702mis !1 (2 points)
8. (12 points) In the figure below, a ball is thrown horizontally from the top of a 20-m high
hill. It strikes the ground at an angle of 45°. With what speed was it thrown?
a) Calculate the time from when it was thrown to when it hits the ground.
Since the initial velocity is horizontal, the time it takes to hit the ground depends only on
the height above the ground and the acceleration due to gravity. Taking down to be
positive, and the launch point as yo = 0 , we have
1
1
y = gt 2 ! 20m = 9.8mis "2 t 2 ! t = 2.02s (4 points)
2
2
(
)
b) Calculate the y-component of the velocity, vy , when it strikes the ground.
Use vy = voy ! gt , with voy = 0 , vy = !gt = !9.8mis !2 ( 2.02s ) = !19.8mis !1 (4 points)
4
vy < 0 means that it is down.
c) What is the initial speed (when it was thrown horizontally) of the ball? Hint: when it
strikes the ground its velocity make an angle of 45! with the ground (horizontal). What
should the value of vx be in relation to vy ?
!
The final velocity is v = v iˆ + v ˆj . The angle shown in the diagram can be found by
x
! = tan "1
vy
vx
y
. Since ! = 45! , vx = vy = 19.8mis !1 . (2 points)
Basically vx = 19.8mis !1 if we take right as positive. Since vx is constant, and that the
m
initial velocity is horizontal, it is clear that the initial speed is simply 19.8 . (2 points)
s
9. (11 points) In the figure, you throw a ball toward a wall at speed 24.0 m/s and at angle
θ0 = 43.0˚ above the horizontal. The wall is distance d = 16.0 m from the release point of
the ball.
(a) Calculate the initial horizontal (x) and vertical (y) components of its velocity. Hence
calculate how far above the release point does the ball hit the wall?
m
m
vox = 24 cos 43! = 17.55
s
s
m
m
(1 point)
voy = 24 sin 43! = 16.37
s
s
The time to the wall can be calculated from the horizontal component by the equation
16m
d = vox t ! t =
= 0.91s (2 points). Taking up as positive, and launch point as
17.55mis "1
yo = 0 the y-component of the position is
1
1"
m%
2
y = voyt ! gt 2 = (16.37mis ) ( 0.91s ) ! $ 9.8 2 ' ( 0.91s ) = 10.84m (2 points)
2
2#
s &
So it hits the wall 10.84 above the release point.
(b) What are the horizontal and vertical components of its velocity as it hits the wall?
m
x-component is constant vox = 17.55 (1 point)
s
5
m "
m%
m
! $ 9.8 2 ' ( 0.91s ) = 7.45
(2 point)
s #
s &
s
(c) Based on the results of part b, plot a motion diagram of the trajectory that clearly
labels the x- and y- coordinate system, as well the direction of the velocity when the ball
hits the wall.
Since vy > 0 , the ball has not reached its maximum height, so the trajectory is as below:
y-component vy = voy ! gt = 16.37
+y
!
v
!
v
(3 points)
!
v
+x
6
Useful Equations
!
"
"
"
A = Ax i + Ay j + Az k ,
Scalar Product:
unit vector notation.
! !
Ai B = Ax Bx + Ay By + Az Bz = AB cos !
1 2
ax t , vx = v0 x + ax t , vx2 = v02x + 2ax ( x ! x0 )
2
x ( t 2 ) ! x ( t1 ) "x
v ( t ) ! vx ( t1 ) "vx
, aavg ! x = x 2
=
=
=
t 2 ! t1
"t
t 2 ! t1
"t
x = x0 + v0 x t +
vavg ! x
d2x
dx
dv
, ax = x , ax = 2
dt
dt
dt
total distance
average speed savg =
!t
2
g = 9.8m / s
vx =
Solution of quadratic equation, ax 2 + bx + c = 0 ! x =
2D kinematics:
"b ± b 2 " 4ac
2a
!
x ( t 2 ) ! x ( t1 ) "x
y ( t ) ! y ( t1 ) "y
!r
!
, vavg ! y = 2
vavg =
= vavg " x iˆ + vavg " y ˆj , vavg! x =
=
=
!t
t 2 ! t1
"t
t 2 ! t1
"t
!
v ( t ) ! vy ( t1 ) "vy
v ( t ) ! vx ( t1 ) "vx
!v
!
, aavg ! y = y 2
aavg =
= aavg " x iˆ + aavg " y ˆj , aavg ! x = x 2
=
=
!t
t 2 ! t1
"t
t 2 ! t1
"t
!
dx
dy
! dr
, vy =
v=
= vx iˆ + vy ˆj ; vx =
dt
dt
dt
!
dvy
d2x
d2y
dv
! dv
, ay = 2
a=
= ax iˆ + ay jˆ ; ax = x , ax = 2 ; ay =
dt
dt
dt
dt
dt
7

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