Question

16/ In the figure given below, a bullet of mass m moving
with speed u strikes a rod AB of mass M and
length L at the end B and gets embedded into it.
If initially the rod is vertical and hinged at end A,
then angular speed of the system just after
collision will be
[NCERT Pg. 173]
A
B​

Answer 1

Given:

In the figure given below, a bullet of mass m moving with speed u strikes a rod AB of mass M and length L at the end B and gets embedded into it.

To find:

Angular speed of the system.

Calculation:

The whole Kinetic Energy of the bullet will be converted to the Rotational Kinetic Energy of the rod+bullet system.

Both the inertia of rod and that of rod will be considered while taking rotational kinetic energy.

Now, applying Conservation of Energy:

$\therefore \: \dfrac{1}{2} m {v}^{2} = \dfrac{1}{2} I { \omega}^{2}$

$= > \: m {v}^{2} = I { \omega}^{2}$

$= > \: m {v}^{2} = \bigg \{ \dfrac{(M) {L}^{2} }{3} +m{L}^{2} \bigg\}{ \omega}^{2}$

$= > \: m {v}^{2} = \bigg \{ \dfrac{(M + 3m) {L}^{2} }{3} \bigg\}{ \omega}^{2}$

$= > \: { \omega}^{2} = \dfrac{3m {v}^{2}}{(M + 3m) {L}^{2} }$

$= > \: \omega = \sqrt{ \dfrac{3m {v}^{2}}{(M + 3m) {L}^{2} }}$

So, final answer is:

$\boxed{ \bf{ \red{ \: \omega = \sqrt{ \dfrac{3m {v}^{2}}{(M + 3m) {L}^{2} }}}}}$

Answer 2

Answer:

answer is 3mv/(ML+3mL)

Explanation:

Net force of the system will be zero.... due to action reaction pair in case of striking. so net torque will be zero... i.e we can use conservation of angular momentum.