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

The momentum is 1.94 kg m/s.

Step-by-step 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 = (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 = (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,

`(1)/(2)mv^2 = (1)/(2)kx^2,`

solving for
`v` we get:

`v = x \sqrt{(k)/(m) }.`

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

`p =mx \sqrt{(k)/(m) }.`

Putting in numbers we get:

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

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