6.2. WORKED EXAMPLES
147
2.20 m
Figure 6.12:
Spring propels marble off table and hits (or misses) box on the floor.
1.10 cm
v
(a)
(b)
Figure 6.13:
Marble propelled by the springgun: (a) Spring is compressed, and system has potential
energy. (b) Spring is released and system has kinetic energy of the marble.
Bobby compresses the spring 1.10 cm, but the center of the marble falls 27.0 cm
short of the center of the box. How far should Rhoda compress the spring to
score a direct hit?
[HRW5 8-36]
Let's put the origin of our coordinate system (for the motion of the marble) at the edge
of the table. With this choice of coordinates, the object of the game is to insure that the x
coordinate of the marble is 2.20 m when it reaches the level of the floor.
There are many things we are not told in this problem! We don't know the spring
constant for the gun, or the mass of the marble. We don't know the height of the table
above the floor, either!
When the gun propels the marble, the spring is initially compressed and the marble is
motionless (see Fig. 6.13(a).) The energy of the system here is the energy stored in the spring,
E
i
=
1
2
kx
2
, where k is the force constant of the spring and x is the amount of compression
of the spring.) When the spring has returned to its natural length and has given the marble
a speed v, then the energy of the system is E
f
=
1
2
mv
2
. If we can neglect friction then
mechanical energy is conserved during the firing, so that E
f
= E
i
, which gives us:
1
2
mv
2
=
1
2
kx
2
=
v =
k
m
x
2
= x
k
m