First slide

Rotational motion

Question

A homogenous rod of length $l = η x$ and mass M is lying on a smooth horizontal floor. A bullet of mass m hits the rod at a distance x from the middle of the rod at a velocity v₀ perpendicular to the rod and comes to rest after collision. If the velocity of the farther end of the rod just after the impact is in the opposite direction of v₀, then

Difficult

A

$η > 3$
B

$η < 3$
C

$η > 6$
D

$η < 6$

Solution

Let after collision velocity of rod be v and angular velocity be $ω$ , then
$\begin{array}{l} m v_{0} = m v + M V \Rightarrow m v_{0} x = m v x + \frac{M ℓ^{2}}{12} ω \\ m v_{0} = m v + \frac{M ω η^{2} x}{12} \end{array}$
From above equations,
$\frac{M ω η^{2} x}{12} = M V \Rightarrow ω η^{2} x = 12 V$
$Velocity of farther end of the rod = V - \frac{l ω}{2}$
$= \frac{ω η^{2} x}{12} - \frac{ω η x}{2} = \frac{ω η x}{2} (\frac{η}{6} - 1)$
If this velocity is opposite to v₀, then
$\frac{η}{6} - 1 < 0 \Rightarrow η < 6$

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