Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
akash kashyp Grade: Upto college level

a uniform rod of length  L and mass  2M on a smooth horizontal table .a point mass M moving horizontally at right angles to the rod with  velocity V collides with one end of rod and sticks it then

1..angular velocity of the system after collision is ?

2..the loss in k.E of the system as a whole as a result of  the collision is ?

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points

Dear akash



since there is no external force so linear momentum and angular momentum will remain conserved

MV = M(WL/2) + 2MV1

V = WL/2  + 2V1 .

V1 = V/2 - WL/4 ...........1

conservation of angular momentum

MVL/2 = MW(L/2)(L/2)  + I W

MVL/2 = MW(L/2)(L/2)  + 2ML2/12 W

V  =  WL/2  + WL/3

V   =5WL/6

W =6V/5L

from equation 1

 V1 = V/5

loss in K.E = 1/2MV2   - 1/2 2M V12 - 1/2 M (WL/2)2 - 1/2 I w2


put the value and calculate


Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed
solution very  quickly.
 We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

 All the best.
Askiitians Experts



7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details