Question

Please help! I'm not sure what equation or the process to do this question.

A 120 N/m spring is compressed 0.25m and is used to launch a 0.5kg ball. What is the momentum of the ball immediately after is fired?

Answer 1

Answer:

The momentum is 1.94 kg m/s.

Explanation:

To solve this problem we equate the potential energy of the spring with the kinetic energy of the ball.

The potential energy $U$ of the compressed spring is given by

$U = \dfrac{1}{2} kx^2$,

where $x$ is the length of compression and $k$ is the spring constant.

And the kinetic energy of the ball is

$K.E = \dfrac{1}{2}mv^2.$

When the spring is released all of the potential energy of the spring goes into the kinetic energy of the ball; therefore,

$\dfrac{1}{2}mv^2 = \dfrac{1}{2}kx^2,$

solving for $v$ we get:

$v = x \sqrt{\dfrac{k}{m} }.$

And since momentum of the ball is $p=mv$,

$p =mx \sqrt{\dfrac{k}{m} }.$

Putting in numbers we get:

$p =(0.5kg)(0.25m) \sqrt{\dfrac{(120N/m)}{0.5kg} }.$

$\boxed{p=1.94kg\: m/s}$