Physics 40 HW #4 Chapter 4 Key
NEATNESS COUNTS!
Solve but do not turn in the following problems from Chapter 4 Knight
Conceptual Questions: 8, 10, 11;
4.8. Anita is approaching ball 2 and moving away from where ball 1 was thrown, so ball 1 was thrown with the greater speed. This
can be determined numerically as well, treating Anita as a moving reference frame with respect to the ground, so
vAnita  vball  5 m/s. For ball 1, Anita measures 10 m/s  v1  5 m/s  v1  15 m/s. For ball 2, 10 m/s  v2  5 m/s  v2  5 m/s.
So ball 1 was thrown with greater speed.
4.10. Zach should throw his book outward and toward the back of
the car. The book has the same initial velocity as do Zach and the car,
so throw 1 or 2 will cause the book to land beyond the driveway in
the same direction as the car is traveling.
4.11. Since Zach and Yvette are traveling at the same speed they
share the same reference frame, so Zach should throw the book
straight to her (throw 2.)
Problems: 10, 11, 13, 37, 47, 53, 54. 79, 80
4.10. Model: Use the particle model for the puck.
Solve: Since the vx vs t and v y vs t graphs are straight lines, the
puck is undergoing constant acceleration along the x- and y- axes.
The components of the puck’s acceleration are
dvx vx 10 m/s  (  m/s) 


 2.0 m/s 2
dt
t
10 s  0 s
(10 m/s  0 m/s)
ay 
  m/s 2
(10 s  0 s)
ax 
The magnitude of the acceleration is a  ax2  a 2y  2.2 m/s2.
Assess: The acceleration is constant, so the computations above
apply to all times shown, not just at 5 s. The puck turns around at t  5 s in the x direction, and constantly accelerates in the y
direction. Traveling 50 m from the starting point in 10 s is reasonable.
4.11. Model: Assume the particle model for the ball, and apply the
constant-acceleration kinematic equations of motion in a plane.
Visualize:
Solve: (a) We know the velocity v1  (2.0iˆ  2.0 ˆj ) m/s at t  1 s The ball is at its highest point at t  2 s, so v y  0 m/s. The
horizontal velocity is constant in projectile motion, so vx  20 m/s at all times. Thus v2  20iˆ m/s at t  2 s We can see that the
y-component of velocity changed by v y  2.0 m/s between t  1 s and t  2 s Because a y is constant, v y changes by 2.0 m/s
in any 1-s interval. At t  3 s, v y is 2.0 m/s less than its value of 0 at t  2 s At t  0 s, v y must have been 2.0 m/s more than its
value of 2.0 m/s at t  1 s Consequently, at t  0 s,
v0  (2.0iˆ  4.0 ˆj) m/s ,
At t  3 s,
At t  1 s,
v (3)  (2.0iˆ  2.0 ˆj ) m/s
v (1)  (2.0iˆ  2.0 ˆj ) m/s
At t  2 s,
v (2)  (2.0iˆ  0.0 ˆj) m/s
(b) Because v y is changing at the rate 2.0 m/s per s, the y-component of acceleration is a y  2.0 m/s2. But a y   g for
projectile motion, so the value of g on Exidor is g  20 m/s2 
(c) From part (a) the components of v0 are v0 x  2.0 m/s and v0 y  4.0 m/s. This means
 v0 y 
1  4.0 m/s 
  tan 
  63 above x
 2.0 m/s 
 v0 x 
  tan 1 
Assess: The y-component of the velocity vector decreases from 2.0 m/s at t  1 s to 0 m/s at t  2 s This gives an acceleration of
2 m/s2. All the other values obtained above are also reasonable.
4.13. Model: The bullet is treated as a particle and the effect of air resistance on the motion of the bullet is neglected.
Visualize:
Solve: (a) Using y1  y0  v0 y (t1  t0 )  12 a y (t1  t0 )2 , we obtain
(2.0 102 m)  0 m  0 m  12 (9.8 m/s2 )(t1  0 s)2  t1  0.0639 s  0.064 s
(b) Using x1  x0  v0 x (t1  t0 )  12 ax (t1  t0 )2 ,
(50 m)  0 m  v0 x (0.0639 s  0 s)  0 m  v0 x  782 m/s  780 m/s
Assess: The bullet falls 2 cm during a horizontal displacement of 50 m. This implies a large initial velocity, and a value of 782 m/s
is understandable.
4.37. Model: Assume the spaceship is a particle. The acceleration is constant, so we can use the kinematic equations.
1
Visualize: We apply the kinematic equation sf  si  v0t  a(t )2 in each direction. t  35 min  2100 s.
2
Solve:
1
xf  6.0  105 km  (9.5 km/s)(2100 s)  (0.040 km/s2 )(2100 s)2 s  708000 km
2
1
yf  4.0  105 km  (0 km/s)(2100 s)  (0 km/s2 )(2100 s)2 s  400000 km
2
1
zf  2.0  105 km  (0 km/s)(2100 s)  (.020 km/s2 )(2100 s)2 s  156000 km
2
3
ˆ
ˆ
ˆ
Rounding to two sig figs gives r  (710i  400 j  160k ) 10 km.
f
Assess: The y-component didn’t change because there was no velocity or acceleration in the y-direction.
4.47. Model: We will use the particle model and the constant-acceleration kinematic equations in a plane.
Visualize:
Solve: The x-and y-equations of the ball are
x1B  x0B  (v0B ) x (t1B  t0B )  12 (aB ) x (t1B  t0B ) 2  65 m  0 m  (v0B cos30)t1B  0 m
2
y1B  y0B  (v0B ) y (t1B  t0B )  12 (aB ) y (t1B  t0B ) 2  0 m  0 m  (v0B sin30)t1B  12 ( g )t1B
From the y-equation,
v0B 
gt1B
(2sin 30)
Substituting this into the x-equation yields
2
gcos30t1B
2sin 30
 t1B  2.77 s
65 m 
For the runner:
t1R 
20 m
 2.50 s
8.0 m/s
Thus, the throw is too late by 0.27 s.
Assess: The times involved in running the bases are small, and a time of 2.5 s is reasonable.
4.53. Model: We define the x-axis along the direction of east and the y-axis along the direction of north.
Solve: (a) The kayaker’s speed of 3.0 m/s is
relative to the water. Since he’s being swept
toward the east, he needs to point at angle 
west of north. His velocity with respect to the
water is
vKW  (3.0 m/s,  west of north)  (3.0sin m/s)iˆ  (3.0cos m/s) ˆj
We can find his velocity with respect to the earth vKE  vKW  vWE , with vWE  (2.0 m/s)iˆ. Thus
vKE  ((30sin   20) m/s)iˆ  (3.0cos m/s) ˆj
In order to go straight north in the earth frame, the kayaker needs (vx )KE  0. This will be true if
sin  
2.0
 2.0 
   sin 1 
  41.8
3.0
 3.0 
Thus he must paddle in a direction 42° west of north.
(b) His northward speed is v y  3.0 cos(41.8) m/s  2.236 m/s. The time to cross is
t
100 m
 44.7 s
2.236 m/s
The kayaker takes 45 s to cross.
4.54. Model: Mike and Nancy coincide
at
t  0 s
Use subscripts B, M, N for the
ball, Mike, and Nancy respectively.
Solve: (a)
According to the Galilean
transformation of velocity vBN  vBM  vMN .
Mike throws the ball with velocity
vBM  (22 m/s)cos63iˆ  (22 m/s)sin63 ˆj, and vNM  (30 m/s)iˆ. Thus with respect to Nancy
vBN  vBM  vNM  (22cos63  30)iˆ m/s  (22sin63) ˆj m/s  (20.0iˆ  19.6 ˆj ) m/s
  tan 1
vy
| vx |
 tan 1
19.6 m/s
 44.4
20.0 m/s
The direction of the angle is 44.4 above the  x axis (in the second quadrant).
(b) With respect to Nancy
xBN  (20.0 m/s)t
and
yBN  0 m  (19.6 m/s)t  12 (9.8 m/s2 )t 2  (19.6 m/s)t  (4.9 m/s 2 )t 2
4.79. Model: Use the particle model for the arrow and the constant-acceleration kinematic equations.
Visualize:
Solve: Using v1y  v0 y  a y (t1  t0 ), we get
v1y  0 m/s  gt1  v1y   gt1
Also using x1  x0  v0 x (t1  t0 ) 
1 a (t  t )2 ,
0
2 x 1
60 m  0 m  v0 xt1  0 m  v0 x 
60 m
 v1x
t1
Since v1y /v1x   tan3.0  0.0524, using the components of v0 gives
 gt1
(0.0524)(60 m)
 0.0524  t1 
 0.566 s
(60 m/t1)
(9.8 m/s2 )
Having found t1, we can go back to the x-equation to obtain v0 x  60 m/0.566 s  106 m/s  110 m/s
Assess: In view of the fact that the arrow took only 0.566 s to cover a horizontal distance of 60 m, a speed of 106 m/s or 237 mph
for the arrow is understandable.
4.80. Model: Use the particle model for the
arrow and the constant-acceleration kinematic equations. We will assume that the archer shoots from 1.75 m above the slope (about
5 9).
Visualize:
Solve: For the y-motion:
y1  y0  v0 y (t1  t0 )  12 a y (t1  t0 )2  y1  1.75 m  (v0 sin 20)t1  12 gt12
 y1  1.75 m  (50 m/s)sin 20t1  12 gt12
For the x-motion:
x1  x0  v0 x (t1  t0 )  12 ax (t1  t0 )2  0 m  (v0 cos20)t1  0 m  (50 m/s)(cos20)t1
Because y1/x1   tan15  0.268,
1.75 m  (50 m/s)(sin 20)t1  12 gt12
(50 m/s)(cos 20)t1
 0.268  t1  6.12 s and 0.058 s (unphysical)
Using t1  612 s in the x- and y-equations above, we get y1  77.0 m and x1  287 m. This means the distance down the slope is
x12  y12  (287 m)2  (77.0 m)2  297 m.
Assess: With an initial speed of 112 mph (50 m/s) for the arrow, which is shot from a 15 slope at an angle of 20 above the
horizontal, a horizontal distance of 287 m and a vertical distance of 77.0 m are reasonable numbers.
Turn in the following. Show all your work clearly and neatly. Always draw a diagram, sketching the situation and
labeling variables define directions, etc Box final answers.
These are BRIEF INCOMPLET solutions. See solutions in glass case for complete solutions.
1. A particle moves in the xy plane with a constant acceleration given by a  4.0jˆ m/s2 . At t = 0, its
position and velocity are 10i m and (–2.0i + 8.0 j) m/s, respectively. What is the distance from the origin
to the particle at t = 2.0 s? What is the velocity of the particle? Sketch the displacement and velocity
vector equations as shown in the diagram 4.5 in the text.
Answer:
r  (6iˆ  8 ˆj )m,
v  2 ˆj m / s
r  10m
see keys in glass case for sketches.
2. A ball is thrown from an upper‐story window of a building. The ball is given an initial
velocity of 5.00 m/s at an angle of 20.0°above the horizontal. It strikes the ground 3.00 s later.
a)
b)
c)
d)
Sketch the event, labeling EVERYTHING, including the initial and final states.
How far horizontally from the base of the building does the ball strike the ground?
Find the height from which the ball was thrown.
Find the velocity that the balloon hits the ground with (both magnitude and direction.)
Answer:
R  14.1m,
H  39.0m, v f  (28.1m / s, 80.0)
3. A quarterback throws a football straight toward a receiver with an initial speed of 20.0m/s, at
an angle of 35.0° above the horizontal. At that instant, the receiver is 20.0 m in front (down
field) of the quarterback. With what constant speed should the receiver run in order to catch the
football at the level at which it was thrown?
Answer:
v  7.86m / s
5. A water balloon gun is held 1.40 m above the ground and pointed 25 degrees above the
horizontal. It shoots a water balloon that lands on the ground 4.4 m away in the horizontal
direction.
a) Sketch the event, labeling everything, givens, directions, including the initial and final states.
b) Find the total time the balloon travels in the air.
c) The initial speed of the water balloon.
d) Find the velocity that the balloon hits the ground with.
Answer:
t  0.84s, vi  5.78m / s, v f  (7.80m / s, 47.8)
4. A dive bomber has a velocity of 280 m/s at an angle below
the horizontal. When the altitude of the aircraft is 2.15 km, it
releases a bomb, which subsequently hits a target on the
ground. The magnitude of the displacement from the point of
release of the bomb to the target is 3.25 km.
a) Find the displacement in the x direction.
b) Find the launch angle as shown.
c) Find the time it takes the bomb to strike the target.
d) Find the velocity the bomb strikes the ground with, both
magnitude and direction and express it as an ijk vector too!
Answer:
When the bomb has fallen a vertical distance 2.15 km, it has traveled a horizontal distance x f given by
xf 
 3.25 km    2.15 km 
y f  x f tan  
2
gx
2
 2.437 km
2
f
2v cos2  i
2
i
2 150 m   2 437 m  tan  i
 9.8 m s   2 437 m 

2
2
2  280 m s  cos2  i
2

 2 150 m   2 437 m  tan  i   371.19 m  1  tan 2  i

 tan 2   6.565 tan  i  4.792  0
1
 tan  i   6.565 
2
 6.565
2
 4 1 4.792    3.283  3.945

Select the negative solution, since  i is below the horizontal.
 tan i  0.662 , i  33.5
Answers:
x f  2.437 km, i  33.5, t  10.43s, v f  (347m / s, 47.8)
6. A train travels due south at 30m/s (relative to the ground) in a rain that is blown toward the south by the
wind blowing due south. The path of each raindrop makes an angle of 70 degrees with the vertical, as
measured by an observer stationary on the ground. An observer on the train, however, sees the drops fall
perfectly vertically. Determine the speed of the raindrops.
Answer: vRG  31.9 m/ s
7. Tim in his Corvette accelerates at the rate of (3.50iˆ  2.70 ˆj )m / s 2 , while Jill in her Jaguar accelerates at
(2.50iˆ  1.50 ˆj )m / s 2 . They both start from rest at the origin of an xy coordinate system. After 15.0 s,
Find Tim’s and Jill’s position vectors, relative to the origin, expressed as ijk vectors. What is the distance
between them? What is Jill’s position, relative to Tim? Sketch all the vectors, labeling, etc.
Answer:
d  76.6m, rJT  (75.0iˆ  15.5 ˆj )m,
8. A ship is launched from shore is crossing a lake, heading 30.0 degrees west of north at 30.0 m/s
relative to the water. The velocity of the ship relative to the shore is 20.0 m/s due north. Find the velocity
(ijk vector and magnitude and direction) of the water relative to the shore both by graphical methods and
the method of components. Write the vector equation for the relative motion Express the direction of the
water relative to due east (the + x-axis.) Calculate the percent difference for the velocity from each
method, which should be within a few percent of each other.
Answer:
vWS  (16.2m / s,202)
9. A coast guard ship is traveling at a constant velocity of 4.20 m/s, due east, relative to the water. On
his radar screen the navigator detects an object that is moving at a constant velocity. The object is
located at a distance of 2310 m with respect to the ship, in a direction 32.0 degrees south of east. Six
minutes later, he notes that the object’s position relative to the ship has changed to 1120 m, 57.0 degrees
south of west. What are the magnitude and direction of the velocity of the object relative to the water?
Express the direction as an angle with respect to due west. Try solving both by methods of component
and by law of sines and cosines. Which is easier?
Answer: vOW  (3.04m / s,15.1)