Question

Please help!!!

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 it is fired?

Answer 1

Momentum of the ball just after the release is 1.94 kg m/s

Step-by-step explanation:

As we know that initially the spring and compressed against the ball

So here we will have

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

now we have

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

Here we know that

mass = 0.5 kg

k = 120 N/m

x = 0.25 m

`v = \sqrt{(120)/(0.5)}(0.25)`

`v = 3.87 m/s`

now we can use formula of momentum here

`P = mv`

`P = (0.5)(3.87)`

`P = 1.94 kg-m/s`