HW02 - 2D and 3D kinematics kinematics - Jan 25: 4P6, 4P17, 4P23, 4P28, 4P32, 4P35, 4P39, 4P67, 4P80; 5P8, 5P12. Chapter 4, problem 6: An electron’s position position is given by r (3.00 sm )ti
ˆ
t in seconds and (4.00 sm )t 2 j (2.00m)k , with ˆ
ˆ
2
in
meters. (a) In unit-vector notation, what is the electron’s electron’s velocity? At velocity? At t = 2.00 s, what is v (b) in unit-vector notation? we notation? we simply take a time-derivative, time-derivative, v
d r
t 2.00 s
(3.00 sm )i (8.00 sm )tj v(t ) (3.00 ms )i (16.00 ms ) j v(2.00 s) ; ˆ
dt
2
ˆ
ˆ
(1.1)
ˆ
and as (c) a magnitude and (d) an angle relative to the positive direction of the x-axis? The x-axis? The speed (magnitude of velocity) at t = = 2.00 s is,
v(2.00s) v(2. (2.00s) (3.00 m/s)2 ( 16.0 m/s)2 16.3 m/s ;
(1.2)
The angle the tangent-line makes with the +x axis is, (2.00 s)
tan 1
16.0 m / s tan1 79.4 ; v x (2.00s) 3 . 0 0 m / s
v y (2.00s)
(1.3)
th
WARNING: Since the y-component is negative, and the x-component is positive, the velocity points towards the 4 quadrant. That is the interpretation of the negative sign on the velocity. velocity. 2
2
xy-plane with acceleration acceleration components components a x = 4.0 m/s , and a y = -2.0 m/s . •• chapter 4, problem 17: 17 : A cart is propelled over an xy-plane Its initial velocity velocity has components v0x = 8.0 m/s and v0y = 12 m/s. In unit-vector notation, what is the velocity of the cart when it reaches its greatest y-coordinate? Solution: When the cart reaches its greatest y-coordinate, position is extremized. The derivative of an extremized extremiz ed function (evaluated at the critical, “extreming” “extreming” value of its argument) is 0; the derivative of position is velocity. So, we find the unknown time as, v0, y 12 ms a y 2.0 m2
at extreme
dy d * v ( t ) 0 y0 v0, yt 12 a y t 2 v0, y a y t * t * 6.0s; * t t dt t t dt y-value at time t s
*
(1.4)
*
Meanwhile, the velocity at any time is, time is, v(t )
v x (t )i v y (t ) j v0, x axt i v0, y a yt i 8.0 ms 4.0 sm t i 12 ms 2.0 sm t i ; ˆ
ˆ
ˆ
ˆ
ˆ
2
2
ˆ
(1.5)
Subsequently, the velocity at this critical time comes from plugging (1.4) into (1.5) (although, one can be lazy [encouraged] and realize “hey, the y-velocity y-velocity is 0, so I just have to evaluate the x-one”), x- one”),
v x (t * ) vx (6.0s)
8.0 m / s 4.0 m / s 2 6.0 s
32 m / s v(6.0 s) 32 ms i ˆ
(1.6)
• chapter 4, problem 23: 23 : A projectile is is fired horizontally from a gun that is 45.0 m above flat ground, emerging from from the gun with a speed of 250 m/s. (a) How long does the projectile remain in the air? (b) At what horizontal distance from the firing point does it strike the the ground? (c) What is the magnitude magnitude of the vertical vertical component of its velocity velocity as it strikes the the ground?
(a) The time of flight is, solve for t y h y0 v0, y t 12 ay t 2 0 0 12 ( g )t 2 t
2(h)
g
2h g
2(45.0 m) 9.80 m/s2
3.03 s.
(b) The horizontal distance traveled is given by motion at constant velocity at a time t, (1.7) ; let‟s see if we get anything interesting by plugging in the symbolic expression for t,
(1.7)
x x0 v0, xt 12 axt 2 0 v0, xt 0 v0, xt v0, x
(c) The projectile constantly accelerates at a y
2h g
2(45.0m)
(250 ms )
9.80 sm2
758m ;
(1.8)
g for a time (1.7), yielding,
v y v0, y ayt 0 (g )t gt (9.80 m/s2 )(3.03 s) 29.7 m/s ;
(1.9)
•• chapter 4, problem 28: In Fig. 4-34, a stone is pro jected at a cliff of height h with an initial speed of 42.0 m/s
directed at angle 0 60.0 above the horizontal. The stone strikes at A, 5.50 s after launching. Find (a) the height h of the cliff, (b) the speed of the stone just before impact at A, and (c) the maximum height H reached above the ground.
Solution: (a) compute y = h by noting the y-coordinate is given at all times by y y0
v0, yt 12 ayt 2 , in which we use
v0, y v0 sin 0 (as indicated in the schematic, taken from the book but augmented),
y y (t ) h y0 v0 sin 0t
1 2
gt 2 0 42.0 m s sin60.0 5.50 s 12 (9.81 sm2 )(5.50s ) 51.8m 2
(1.10)
(b) The horizontal motion is steady, so v x = v0 x = v0 cos 0, but the vertical component of velocity varies according to Eq. 4-23. Thus, the speed at impact, using the two components of v v 0
v v x 2 v y 2
2
at , is computed as, 2
2
2
a xt v0, y a yt 0 v0 cos 0 v0 sin 0 ( g )t 0, x
v
(1.11)
g 67.5 m.
(1.12)
2
2
42.0 sm cos 60.0 42.0 ms s in 60.0 9.81 sm (5.50 s) 27.4 m/s ; 2
(c) We use Eq. 4-24 with v y = 0 and y = H , which is a consequence of V
2
V0 2 2a d ,
v y 2 v0, y 2 2ay d y 02 (v0 sin 0 )2 2(g ) ( H 0) H
•• chapter 4, problem 32: You throw a ball toward a wall at speed 25.0 m/s and at angle 40.0° above the horizontal. The wall is distance d = 22.0 m from the release point of the ball. (a) How far above the release point does the ball hit the wall? What are the (b) horizontal and (c) vertical components of its velocity as it hits the wall? (d) When it
v0 sin 0 2 g
2
hits, has it passed the highest point on its trajectory?
The coordinate origin is at the release point (the initial position for the ball as it begins projectile motion in the sense of §4-5), and we let 0 be the angle of throw (shown in the figure). Since the horizontal component of the velocity of the ball is v x = v0 cos 40.0°, the time it takes for the ball to hit the wall is,
x x x0 v0, xt 12 axt 2 v0, xt 0 t
x v x ,0
22.0 m (25.0 m/s) cos 40.0
1.15 s;
(1.13)
(a) The vertical distance comes from an invocation of (1.13), but in the y-direction,
1
1
2
2
y (v0 sin 0 )t gt 2 (25.0 m/s)sin 40.0(1.15 s) (9.80 m/s 2 )(1.15 s) 2 12.0 m.
(1.14)
(b) The horizontal component of the velocity when it strikes the wall does not change from its initial value: v x = v0 cos 40.0° = 19.2 m/s. (c) The vertical component becomes,
v y v0, y ayt v0 sin 0 ( g )t (25.0 m/s)sin 40.0 (9.80 m/s2 )(1.15 s) 4.80 m/s ;
(1.15)
(d) Since v y > 0 when the ball hits the wall, it has not reached the highest point yet. •• chapter 4, problem 35: A rifle that shoots bullets at 460 m/s is to be aimed at a target 45.7 m away. If the center of the target is level with the rifle, how high above the target must the rifle barrel be pointed (e.g., the angle) so that the bullet hits dead center? We can draw a schematic: exaggerate the parabolic shape of the trajectory,
If the target is a distance d away, then its coordinates are x = d, y = 0. The projectile motion equations lead to,
x x0 d v0, xt 12 axt 2 (v0 cos0 )t 0; y y0 0 v0, yt 12 a yt 2 v0t sin 0 12 gt 2 ; By the x-equation of (1.16), the time at which the bullet strikes the target is t
(1.16)
d /(v0 cos 0 ) . Using this to eliminate t of the
y-equation leads to, 2
d 2 0 v0t sin 0 12 gt 2 v0 sin 0 12 g 2v0 sin 0 cos 0 gd 0 v0 cos 0 v0 cos 0 d
Using sin 0 cos 0
12 sin 2 0
1
(which follows from trigonometry ), we obtain,
(9.80 m/s2 )(45.7 m) sin v sin (20 ) gd 0 sin 0.0606; 2 2 v02 2 (460 m/s) If the gun is aimed at a point a distance above the target, then tan 0 d so that , 2 0
(1.17)
1
1 gd
1
1
d tan 0 (45.7 m) tan(0.0606) 0.0484 m 4.84 cm ;
(1.18)
(1.19)
Note that due to the downward gravitational acceleration, in order for the bullet to strike the target, the gun must be aimed at a point slightly above the target.
1
The same trigonometry from the footnote in Step 8 of my “Problem -Solving Algorithm” sheet! In fact, this problem runs completely parallel to this.
•• chapter 4, problem 39: In Fig. 4-37, a ball is thrown leftward from the left edge of the roof, at height h above the ground.
The ball hits the ground 1.50 s later, at distance d = 25.0 m from the building and at angle 60.0 with the horizontal. Draw a schematic before attempting anything,
(1.20) (a) Find h. (Hint: One way is to reverse the motion, as if on video.) Following the hint, we have the time-reversed problem with the ball thrown from the ground, toward the right, at 60° measured counterclockwise from a rightward axis. We see in this timereversed situation that it is convenient to use the familiar coordinate system with + x as rightward and with positive angles measured counterclockwise. (a) The x-equation (with x0 = 0 and x = 25.0 m) leads to 25.0 m = (v0 cos 60.0°)(1.50 s), so that v0 = 33.3 m/s. And with y0 = 0, and y = h > 0 at t = 1.50 s, we have y y0
v0 yt 12 gt 2 where v0 y = v0 sin 60.0°.
This leads to h = 32.3 m. What are the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown? (b) We find the x-component of the velocity, v x = v0 x = (33.3 m/s)cos 60.0° = 16.7 m/s (this is trivial). Subsequently, we use the y-velocity equation, 2 v y = v0 y – gt = (33.3 m/s)sin 60.0° – (9.80 m/s )(1.50 s) = 14.2 m/s. The magnitude of v is given by
| v | v x2 v y2 (16.7 m/s) 2 (14.2 m/s) 2 21.9 m/s. (c) The angle is
v y 1 14.2 m/s tan 40.4 . v 16.7 m/s x
tan 1
(d) Is the angle above or below the horizontal? (d) We interpret this result (“undoing” the time reversal) as an initial velocity (from the edge of the building) of magnitude 21.9 m/s with angle (down from leftward) of 40.4°. ••• Chapter 4, problem 67: A boy whirls a stone in a horizontal circle of radius 1.5 m and at height 2.0 m above ground level. The string breaks, and the stone flies horizontally and strikes the ground after traveling a horizontal distance of 10 m. What is the magnitude of the centripetal acceleration of the stone while in circular motion? This is a projectile problem “grafted onto” a centripetal acceleration problem. Do the projectile part first. Compute the amount of time the stone falls the height of 1.5 m, for 0 initial y-velocity,
y 1.5 m y0 v0, yt 12 a yt 2 0 0 12 ( g )t 2 t
2 y
g
2(2.0 m)
9.81 sm
We find from this (1.21) that we can compute the centripetal acceleration from this, given Also, take a moment to introduce the exact formula for a r in symbols/problem-givens,
2
4 3
0.553 s;
x 10 m, and ar
(1.21)
vtangential 2 r
v0, x 2 r
.
ar
vtan 2 r
v0, x 2
r
(x / t )2 r
x 2 2
t r
x 2
2 y 2 g
(
) r
g x 2
2 yr
(9.81 sm )(10 m)2 2
2(2.0 m)(1.5 m)
1.6 102
m s2
;
(1.22)
••Chapter 4, problem 80 (|| 4.79): A 200-m-wide river flows due east at a uniform speed of 2.0 m/s. A boat with a speed of 8.0 m/s relative to the water leaves the south bank pointed in a direction 30° west of north. Align our ground-coordinates so that east corresponds to + x and north corresponds to + y.
(1.23) The velocity of the boat is specified relative to the water ; we want to compute it relative to the ground, just to keep ourselves from getting confused. This is done by writing, vw
vw, xi vw, y j (vg , x 2 m s )i vg , y j ( v g cos120 2 ms )i v g sin120 j; ˆ
ˆ
In (1.24), we have the constraint that v w vw
ˆ
ˆ
ˆ
8.0 ms , which subsequently determines
8.0 m s
2
( v g cos120 2 ms )
v g
2
ˆ
(1.24)
v g ,
sin 2 120
(1.25)
What are the (a) magnitude and (b) direction of the boat’s velocity relative to the ground? (c) How long does the boat take t o cross the river? Solution: This is a classic problem involving two-dimensional relative motion. We align our coordinates so that
east corresponds to + x and north corresponds to + y. We write the vector addition equation as v BG
vBW vWG . We have
vWG (2.00 ) in the magnitude-angle notation (with the unit m/s understood), or vWG 2.0i in unit-vector notation. We
also have v BW
. 120 ) (80
where we have been careful to phrase the angle in the „standard‟ way (measured
counterclockwise from the + x axis), or v BW
(4.0i+6.9j) m/s. ˆ
ˆ
(a) We can solve the vector addition equation for v BG :
v BG vBW vWG (2.0 m/s)i (4.0i+6.9j) m/s ( 2.0 m/s)i (6.9 m/s) j. ˆ
ˆ
ˆ
ˆ
ˆ
Thus, we find | v BG | 7.2 m/s. 1
(b) The direction of v BG is tan [(6.9 m/s)/( 2.0 m/s)] 106 (measured counterclockwise from the + x axis), or 16° west of north. (c) The velocity is constant, and we apply y – y0 = v yt in a reference frame. Thus, in the ground reference frame, we have
(200 m) (7.2 m/s)sin(106)t t 29 s. Note: If a student obtains “28 s,” then the student has probably neglected to take the y component properly (a common mistake).
•• chapter 5, problem 8: A 2.00 kg object is subjected to three forces that give it an acceleration a (8.00
m 2 s
)i (6.00 sm2 ) j . If two of the three forces are ˆ
ˆ
F1
30.0 N i 16.0N j and ˆ
ˆ
F2
12.0 N i 8.00N j ˆ
ˆ
find the third force. ^
^
We note that m a = ( – 16 N) i + (12 N) j . With the other forces as specified in the problem, then Newton‟s second law gives the third force as,
F F F F ma ma F F 16i 12 j N (30.0 12.0)i (16.0 8.00) j N 34.0i 12.0 j N solve for F3
i
F3
ˆ
1
1
2
3
ˆ
ˆ
ˆ
ˆ
2
(1.26)
function of time t during the sliding. What is the angle
••• chapter 5, problem 12: Two horizontal forces F1 and F2
ˆ
between the constant directions of forces F1 and F2 ?
act on a 4.0 kg disk that s lides over frictionless ice, on
which an xy-coordinate system is laid out. Force F1 is in the positive direction of the x-axis, and has a magnitude of 7.0 N. Force F2 has a magnitude of 9.0 N. Figure 5-32 gives the x-component v x of the velocity of the disk as a
Draw a schematic of what we are to calculate,
(1.27) From the slope of the graph we find,
a x
v x (5 4) ms m 3.0 2 ; t (3 0) s s
(1.28)
nd
Using newton‟s 2 law in the x-direction (making it a scalar equation, in contrast to (1.26),
(4.0kg )(3.0 sm ) 7.0 N ma x F 1 1 56; (1.29) Fi F1, x F2,x F1 F2 cos max cos F cos 9.0N 2 solve for
1
2
