MHT CET 2019 Physics Solved Paper

When two masses strike the rod, then angular momentum imparted to rod,
L1+L2‌‌=2mv(L)+m(2v)(2L)
‌‌=6mvL
Now, after striking of masses to rod the angular momentum of complete rod about the centre C ,
L‌rod ‌=10
where, ω= angular velocity of the rod and ‌‌I= moment of inertia of the rod.
The moment of inertia of rod I=‌

Ml2

12

=‌

8m(6L)2

12

=24mL2
∴M=8m and I=6L (given)
Now, moment of inertia of two masses after the striking to \/ rod
l1=2m(L)2=2mL2
and ‌‌I2=m(2⟂)2=4mL2
∴ The net moment of the inertia about the centre of rod,
I=24mL2+2mL2+4mL2
⇒=30mL2
By putting this value in Eq. (i), we get
L‌rod ‌=30ML2ω
From the law of conservation of angular momentum,
L1+L2‌‌=L‌od ‌
⇒6mvL‌‌=30mL2ω
⇒ω‌‌=‌

v

5L

Hence, the anguler velocity of the rod is ‌

v

5L

.