spring physics?

Question

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 1.55 m from the edge of the table; see Figure 8-46. Bobby compresses the spring 1.10 cm, but the center of the marble falls 28.0 cm short of the center of the box. How far should Rhoda compress the spring to score a direct hit?

Answer 1

Since there is no figure I will make some assumptions:

First, I will assume that the box is on the floor and the height of the top of the table to the center of the box is h.

Also, I will assume that there is no resistance in the system.

So, when the marble rolls to the edge of the table, it has only horizontal speed. After the lip of the edge it begins to fall.

The speed of the marble is related to the compression distance of the spring through conservation of energy and Hooke's Law:

.5*m*v^2=.5*k*x^2

at the edge of the table the marble will fly the distance
d=v*t

and fall the height, h
h=.5*g*t^2

What we know about Bobby's attempt:

1.55 - .280=v1*t1
t1=(1.55-.280)/v1

h=.5*9.81*((1.55-.280)/v1)^2
and
.5*m*v1^2=.5*k*x1^2
m*v1^2=k*.011^2

the second attempt to score the direct hit:
1.55=v2*t2
t2=1.55/v2

m*v2^2=k*x2^2

h=.5*9.81*(1.55/v2)^2

since the h is constant

((1.55-.280)/v1)^2=(1.55/v2)^2
doing some algebra:

v1^2/v2^2=(1.55-.280)/1.55
flipping
v2^2/v1^2=1.55/(1.55-.280)


dividing the Hooke's Law equations:
v1^2/v2^2=0.011^2/x2^2
so
x2^2/0.011^2=v2^2/v1^2
x2=sqrt((0.011^2)*
1.55/(1.55-.280))

x2=0.01215m
or 1.215 cm

j