A uniform rod of mass m and length L is at rest on a smooth horizontal surface. A ball of mass m, moving with velocity v0, hits the rod perpendicularly at its one end and sticks to it. The angular velocity of rod after collision is

Question

A uniform rod of mass m and length L is at rest on a smooth horizontal surface. A ball of mass m, moving with velocity v0, hits the rod perpendicularly at its one end and sticks to it. The angular velocity of rod after collision is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-6hbtfddrwsm4auxe.jpg" /> (A) (6 v0/5 L) (B) (12 v0/5 L) (C) (2 v0/5 L) (D) (3 v0/7 L)

Tardigrade Admin · Accepted Answer

mv0× ((L/4))=((7/48) m L2 + m × (L2/16))× ω ⇒ ω =(6 v0/5 L)

Q. A uniform rod of mass $m$ and length $L$ is at rest on a smooth horizontal surface. A ball of mass $m,$ moving with velocity $v_{0},$ hits the rod perpendicularly at its one end and sticks to it. The angular velocity of rod after collision is

$\frac{6 v _{0}}{5 L}$

$\frac{12 v _{0}}{5 L}$

$\frac{2 v _{0}}{5 L}$

$\frac{3 v _{0}}{7 L}$

$m v_{0} \times (\frac{L}{4}) = (\frac{7}{48} m L^{2} + m \times \frac{L ^{2}}{16}) \times ω$
$\Rightarrow ω = \frac{6 v _{0}}{5 L}$

Q. A uniform rod of mass m and length L is at rest on a smooth horizontal surface. A ball of mass m, moving with velocity v0​, hits the rod perpendicularly at its one end and sticks to it. The angular velocity of rod after collision is

5L6v0​​

5L12v0​​

5L2v0​​

7L3v0​​