Download Question #1. 1. A tennis ball of mass m = 0.080 kg and speed v = 45

Question #1.
1. A tennis ball of mass m = 0.080 kg and speed v = 45 m/s strikes a floor at a 58° angle
and rebounds with the same speed at 35° (Figure 1). What is the impulse (magnitude
and direction) given to the ball?
J
58°
35°
Figure 1

 
p y  mv f inal  mv initial  m v sin 35 o  m  v sin 58 o



 mv sin 35 o  sin 58  0.08 45 0.574  0.848   0.986 ˆj kgm / s ,
upward.


 mv sin 35 o  sin 58  0.08 45 0.574  0.848   5.12 ˆj kgm / s , upward.

 
p x  mv f inal  mv initial  m  v cos 35 o  m  v cos 58 o



 mv  cos 35 o  cos 58 o  0.08 45  0.819  0.530   1.04 ˆi kgm / s
Therefore, the net impulse =
 1.0412  5.122
 5.22 kgm / s
 5.22 
o
And the direction is tan 1 
  78 or 78o North of West (102o)
  1.04 
R  T  20  50 g  R  70g  T  686  466  221 N
2. A 100 cm diameter disk accelerates uniformly about its center from 100 rpm to 300 rpm in 6.0
seconds. Determine:
a)
Its angular acceleration, and
rev  2π  1 min 


  1.47 rad / s
ω0  100

min  1rev  60 s 

rev  2π  1 min 


  31.42 rad / s
ω   300

min  1rev  60 s 

The angular acceleration.
b)
The radial and tangential components of the linear acceleration of a point on the edge of the
wheel 4.0 seconds after it has started accelerating.
To find the components of the acceleration,
the instantaneous angular velocity is needed.
The instantaneous radial acceleration is given by aR   2 r.
The tangential acceleration is given by
atan   r .
3. A 20.0-kg uniform plank is supported by the floor at one end and by a vertical rope at the other as
shown in Figure 2. A 50.0-kg mass person stands on the plank at a distance three-fourths of the
length plank from the end on the floor.
T
120O
R
30O
Figure 2
a)
What is the tension, T in the rope?
Taking the torque of the plank at the floor:
T sin 60 L   20g cos 30 L 2  50g cos 30 3L 4
T
b)
10g cos 30  150 g cos 30 4
 466 N
sin 60
What is the magnitude of the force of floor on the plank, R?
R  T  20  50 g  R  70g  T  686  466  221 N
4. Two objects attract each other gravitationally with a force of 2.5 x 10-10 N when they are
0.25 m apart. Their total mass is 4.00kg. Find their individual masses.
(G = 6.67 x 10-11 N.m2/kg2)
a)
Two objects attract each other gravitationally with a force of 2.5 x 10-10 N when they are 0.25
m apart. Their total mass is 4.00kg. Find their individual masses.
(G = 6.67 x 10-11 N.m2/kg2)
m1  m
m2  4.00 kg  m.
F=G
=G
2.5 x 10-10 N = ( 6.67 x 10-11 N.m2/kg2 )
0 = m2 – 4m +0.234
m1= 3.94 kg
m2 =0.059 kg
b)
Planet Z-34 has a mass equal to one-third that of Earth and a radius equal to one-third that of
Earth. With g representing, as usual, the acceleration due to gravity on the surface of Earth,
what is the acceleration due to gravity on the surface of Z-34?
gE 
GM E
rE 2
gZ 
GM Z
rZ 
2

G31 ME 


2
1
r
3 E

1
3
GM E
1 2
r
9 E
3
GM E
rE2
 3g E