A stick of length L and mass M lies on a fric | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the stick have mass $M$ and length $L$. The ball of mass $m$ hits the end of the stick with velocity $v$ and comes to rest.$1$. Conservation o

Vedclass · Accepted Answer

$m = M/4$

Vedclass · Answer

(D) Let the stick have mass $M$ and length $L$. The ball of mass $m$ hits the end of the stick with velocity $v$ and comes to rest.$1$. Conservation of linear momentum: $mv = MV$,where $V$ is the velocity of the center of mass of the stick. So,$V = \frac{mv}{M}$.$2$. Conservation of angular momentum about the center of mass of the stick: $mv(L/2) = I\omega$,where $I = \frac{ML^2}{12}$ is the moment of inertia of the stick about its center. Thus,$mv(L/2) = \frac{ML^2}{12} \omega$,which gives $\omega = \frac{6mv}{ML}$.$3$. Conservation of kinetic energy (elastic collision): $\frac{1}{2}mv^2 = \frac{1}{2}MV^2 + \frac{1}{2}I\omega^2$.Substituting $V$ and $\omega$: $\frac{1}{2}mv^2 = \frac{1}{2}M(\frac{mv}{M})^2 + \frac{1}{2}(\frac{ML^2}{12})(\frac{6mv}{ML})^2$.$\frac{1}{2}mv^2 = \frac{1}{2}\frac{m^2v^2}{M} + \frac{1}{2}(\frac{ML^2}{12})(\frac{36m^2v^2}{M^2L^2})$.$mv^2 = \frac{m^2v^2}{M} + \frac{3m^2v^2}{M}$.$1 = \frac{m}{M} + \frac{3m}{M} = \frac{4m}{M}$.Therefore,$m = M/4$.

Similar Questions

State the law of conservation of angular momentum and explain it with the example of a girl sitting on a swivel chair.

$A$ person standing on a rotating table holds weights in their outstretched hands and is rotating. They suddenly pull their hands in close to their body. Which statement is true?

$A$ thin uniform rod of mass $M$ and length $L$ is rotating about a perpendicular axis passing through its centre with a constant angular velocity $\omega$. Two objects,each of mass $\frac{M}{3}$,are attached gently to the two ends of the rod. The rod will now rotate with an angular velocity of

Vedclass Test Series

Exam Paper Generator

Online Exam Module