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A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius 0.600 m and mass 1.60 kg and is initially at rest. A bullet with mass 0.0200 kg is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of 0.600 m from the axle. After being struck by the bullet, the disk rotates at 4.00 rad/s. What is the horizontal velocity of the bullet just before it strikes the disk?

Answer :

Answer:

v = 98.4 m/s

Explanation:

As we know that the total angular momentum of the system is conserved

so here we will have

[tex]L_{bullet} + L_{disk} = constant[/tex]

[tex]MvR = (\frac{1}{2}mR^2 + MR^2)\omega[/tex]

now plug in all data

[tex]0.02 v(0.600) = (\frac{1}{2}(1.60)(0.600)^2 + 0.02(0.600)^2)4[/tex]

[tex]0.012 v = 1.18[/tex]

[tex]v = \frac{1.18}{0.012}[/tex]

[tex]v = 98.4 m/s[/tex]

snehashish65

The horizontal velocity of the bullet just before it strikes the disk is 648 m/s.

Given data:

The radius of disk is, [tex]r = 0.600 \;\rm m[/tex].

The mass of disk is, m = 1.60 kg.

The mass of bullet is, m' = 0.0200 kg.

The distance from the axle is, L = 0.600 m.

The angular speed of disk is, [tex]\omega = 4.00 \;\rm rad/s[/tex].

Applying conservation of angular momentum as,

Angular momentum of bullet = Angular momentum of disk

[tex]L = L'\\m'vL = \dfrac{1}{2}(m+m')r \omega^2[/tex]

Here, v is horizontal velocity of the bullet just before it strikes the disk.

Solving as,

[tex]0.0200 \times v \times 0.600 = \dfrac{1}{2}(1.60+0.0200) \times 0.600 \times 4.00^2\\v = 648 \;\rm m/s[/tex]

Thus, the horizontal velocity of the bullet just before it strikes the disk is 648 m/s.

Learn more about angular momentum here:

https://brainly.com/question/15104254?referrer=searchResults

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