Physics I
Homework VI
Chapter 6: 8, 11, 20, 27, 48, 60, 70
CJ
6.8. 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  2.0jˆ ) m/s at t  1 s. The ball is at its highest point at t  2 s, so vy  0 m/s.
The horizontal velocity is constant in projectile motion, so vx  2.0 m/s at all times. Thus v  2.0iˆ m/s at t  2 s. We can
2
see that the y-component of velocity changed by vy  –2.0 m/s between t  1 s and t  2 s. Because ay is constant, vy
changes by –2.0 m/s in any 1-s interval. At t  3 s, vy is 2.0 m/s less than its value of 0 at t  2 s. At t  0 s, vy 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  4.0jˆ ) m/s
At t  1 s,
v0  ( 2.0iˆ  2.0jˆ ) m/s
At t  2 s,
v0  ( 2.0iˆ  0.0jˆ ) m/s
At t  3 s,
v0  ( 2.0iˆ – 2.0jˆ ) m/s
(b) Because vy is changing at the rate –2.0 m/s per s, the y-component of acceleration is ay  –2.0 m/s2. But ay  –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 
  63.4 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.
6.11. 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   21 ay  t1  t0  , we obtain
2
 2.0 10
2



m  0 m  0 m  21 9.8 m/s2  t1  0 s  t1  0.0639 s
2
(b) Using x1  x0  v0 x  t1  t0   21 ax  t1  t0  ,
2
 50 m  0 m  v0 x  0.0639 s  0 s  0 m
 v0 x  782 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.
6.20. Model: Assume the particle model for the rocket-powered hockey puck and use the constant-acceleration
kinematic equations.
Solve: (a) The force and acceleration of the hockey puck are
F  F cos iˆ  F sin ˆj  ma
a
F cos ˆ F sin ˆ
i
j
m
m
From kinematics:
x  x0  v0 x  t  t0   21 ax  t  t0   0 m  0 m  21 ax t 2  21 ax t 2
2
y  y0  v0 y  t  t0   21 ay  t  t0   0 m  v0t  21 ayt 2  v0t  21 ayt 2
2
From the x-equation t  2x ax . Substituting this into the y-equation, we get
y  v0
2x 1 2x
2 mx
 ay
 v0
 x tan
ax 2 ax
F cos
(b) Substituting the known values into the expression from part (a),
y   2.0 m/s
2 1.0 kg  x
 2.0 N  cos45
 x tan45  2.38 m/s
when   45 and when x and y are in meters.
The equation for   –45 is
y  2.38 m/s x  x
x x
6.27. Model: The golf ball is a particle following projectile motion.
Visualize:
(a) The distance traveled is x1  v0xt1  v0 cos  t1. The flight time is found from the y-equation, using the fact that the ball
starts and ends at y  0:
y1  y0  0  v0 sin t1  21 gt12  (v0 sin  21 gt1 ) t1  t1 
2v0 sin
g
Thus the distance traveled is
x1  v0 cos 
2v0 sin 2v02sin cos

g
g
For   30°, the distances are
( x1 )earth 
2v02sin cos 2(25 m/s) 2sin30cos30

 55.2 m
gearth
9.80 m/s2
( x1 )moon 
2v02sin cos 2v02sin cos
2v02sin cos


6

 6( x1 )earth  331.2 m
1
gmoon
gearth
6 gearth
The golf ball travels 331.2 m – 55.2 m  276 m farther on the moon than on earth.
(b) The flight times are
(t1 )earth 
2v0 sin
 2.55 s
gearth
(t1 )moon 
2v0 sin 2v0 sin
 1
 6(t1 )earth  15.30 s
gmoon
g
6 earth
The ball spends 15.30 s – 2.55 s  12.75 s longer in flight on the moon.
6.48. Model: The model rocket is treated as a particle and its motion is determined by constant-acceleration kinematic
equations.
Visualize:
Solve: As the rocket is accidentally bumped v0 x  0.5 m/s and v0 y  0 m/s . On the other hand, when the engine is fired
Fx  max  ax 
Fx
20 N

 40 m/s2
m 0.5 kg
(a) Using y1  y0  v0 y  t1  t0   21 ay  t1  t0  ,
2
0 m  40 m  0 m 
1
2
 9.8 m/s  t
2
2
1
 t1  2.857 s
The distance from the base of the wall is


x1  x0  v0 x  t1  t0   21 ax  t1  t0   0 m   0.5 m/s 2.857s  21 40 m/s2  2.857 s  165 m
2
2
(b) The x- and y-equations are
y  y0  v0 y  t  t0   21 ay  t  t0   40  4.9t 2
2
x  x0  v0 x  t  t0   21 ax  t  t0   0.5t  20t 2
2
Except for a brief interval at t  0, 20t2 >> 0.5t. Thus x  20t 2 , or t2  x/20. Substituting this into the y-equation gives
y  40 – 0.245x
This is the equation of a straight line, so the rocket follows a linear trajectory to the ground.
6.60. Model: Let the ground frame be S and the car frame be S. S moves relative to S with a velocity V along the xdirection.
Solve: The Galilean transformation of velocity is v  v  V where v and v are the velocities of the raindrops in frames S
and S. While driving north, V   25 m/s iˆ and v  v cos ˆj  v sin iˆ . Thus,
R
R
v  v  V   vR sin  25 m/s iˆ  vR cos ˆj
Since the observer in the car finds the raindrops making an angle of 38 with the vertical, we have
vR sin  25 m/s
 tan38
vR cos
While driving south, V    25 m/s iˆ , and v  vR cos ˆj  vR sin iˆ . Thus,
v   vR sin  25 m/s iˆ  vR cos ˆj
Since the observer in the car finds the raindrops falling vertically straight, we have
vR sin  25 m/s
 tan0  0  vR sin  25 m/s
vR cos
Substituting this value of vR sin into the expression obtained for driving north yields:
25 m/s  25 m/s
50 m/s
 tan38  vR cos 
 64.0 m/s
vR cos
tan38
Therefore, we have for the velocity of the raindrops:
 vR sin    vR cos    25 m/s   64.0 m/s
2
2
2
tan 
2
 vR2  4721 m/s   vR  68.7 m/s
2
vR sin 25 m/s

   21.3
v R cos 64 m/s
The raindrops fall at 68.7 m/s while making an angle of 21.3 with the vertical.
6.70. Model: The airplanes are modeled as particles, and are undergoing relative motion according to the Galilean
transformations of position and velocity. We designate Uri’s plane as frame S and the earth as frame S. Frame S moves
relative to frame S with velocity V .
Visualize:
Solve: According to the Galilean transformation of velocity v  v  V , where v is the velocity of Val’s plane relative to
the Earth, v is the velocity of Val’s plane relative to Uri’s plane, and V is the velocity of Uri’s plane relative to the earth.
We have
v    500 mph  cos30iˆ   500 mph  sin30 ˆj
V    500 mph  cos20iˆ   500 mph  sin20 ˆj
v  v  V   36.8 mph  iˆ   421 mph  ˆj
 421 
  85
 36.8 
  tan1 
The fuselage of Val’s plane points 30 north of west. Val sees her plane moving in a direction 85 north of east. Thus the
angle between the fuselage and the direction of motion is
  180  30  85  65