Answer:
Momentum of the ball just after the release is 1.94 kg m/s
Explanation:
As we know that initially the spring and compressed against the ball
So here we will have
[tex]\frac{1}{2}mv^2 = \frac{1}{2}kx^2[/tex]
now we have
[tex]v = \sqrt{\frac{k}{m}} x[/tex]
Here we know that
mass = 0.5 kg
k = 120 N/m
x = 0.25 m
[tex]v = \sqrt{\frac{120}{0.5}}(0.25)[/tex]
[tex]v = 3.87 m/s[/tex]
now we can use formula of momentum here
[tex]P = mv[/tex]
[tex]P = (0.5)(3.87)[/tex]
[tex]P = 1.94 kg-m/s[/tex]
