a) By a factor 4
The spring potential energy is given by
[tex]U=\frac{1}{2}kx^2[/tex]
where
U is the potential energy
k is the spring constant
x is the compression of the spring
In this problem, the spring is compressed twice the initial distance, so:
[tex]x'=2x[/tex]
So the new elastic potential energy will be
[tex]U'=\frac{1}{2}k(2x)^2=4(\frac{1}{2}kx^2)=4U[/tex]
So, the spring potential energy will quadruple.
b) By a factor 2
When the gun shoots out the ball, the elastic potential energy is then converted into kinetic energy of the ball:
[tex]U=K=\frac{1}{2}mv^2[/tex]
where
m is the mass of the ball
v is its initial speed
Re-arranging the equation we have
[tex]v=\sqrt{\frac{2U}{m}}[/tex]
We said that the potential energy has increased by a factor 4, so
U' = 4U
So the new speed of the ball will be
[tex]v'=\sqrt{\frac{2(4U)}{m}}=2(\sqrt{\frac{2U}{m}})=2v[/tex]
So, the speed will double.
