Question

A spring gun shoots out a plastic ball at speed v. The spring is then compressed twice the distance it was on the first shot.a) By what factor is the spring's potential energy increased? Explain.b) By what factor is the ball's speed increased? Explain.

Answer 1

a) By a factor 4

The spring potential energy is given by

$U=\frac{1}{2}kx^2$

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:

$x'=2x$

So the new elastic potential energy will be

$U'=\frac{1}{2}k(2x)^2=4(\frac{1}{2}kx^2)=4U$

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:

$U=K=\frac{1}{2}mv^2$

where

m is the mass of the ball

v is its initial speed

Re-arranging the equation we have

$v=\sqrt{\frac{2U}{m}}$

We said that the potential energy has increased by a factor 4, so

U' = 4U

So the new speed of the ball will be

$v'=\sqrt{\frac{2(4U)}{m}}=2(\sqrt{\frac{2U}{m}})=2v$

So, the speed will double.