MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: R

There are no items in this cart.
Continue Shopping
Menu
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


 


7168-1431_7662_untitled.JPG


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.
 
Regards,
Askiitians Experts
Badiuddin


 


 

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: R 2,600
  • View Details
Get extra R 520 off
USE CODE: MOB20
  • Complete Physics Course - Class 11
  • OFFERED PRICE: R 2,800
  • View Details
Get extra R 560 off
USE CODE: MOB20

Ask Experts

Have any Question? Ask Experts

Post Question

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