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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 ?

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 ?

Grade:Upto college level

3 Answers

Badiuddin askIITians.ismu Expert
147 Points
11 years ago

Dear akash

 

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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

 

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Sakshi
20 Points
3 years ago
I have a doubt in this ques isnt the linear momentum should be mv=MVInstead of what you`ve written.....🙇
Kushagra Madhukar
askIITians Faculty 629 Points
10 months ago
Dear student,
Please find the solution to your problem.
 
The center of mass of the combined system after the small mass sticks to the rod is
xcm = [2m(L/2) + m(0)]/3m = L/6 from the bottom of rod where the mass strikes.
 
Now, no external force or torque acts on the system. Both angular and linear momentum are conserved.
 
Hence, Pi = Pf
mv = 3mvcm
or, vcm = v/3
 
Li = Lf (about center of mass)
mvL/6 = Iw
Where I = 2mL2/12 + 2m(L/2 – L/6)2 + m(L/6)2 = (6 + 8 + 1)/36 mL2 = 15/36 mL2 = 5/12 mL2
Hence, mvL/6 = 5/12 mL2 w
Or, w = 2v/5L
 
Now, Ki = ½ mv2
Kf = ½ (3m) vcm2 + ½ Iw2
    = ½ (3m) (v2/9) + ½ (5mL2/12) (2v/5L)2
    = 1/5 mv2
 
Hence, Loss of KE = Ki – Kf = ½ mv2 – 1/5 mv2 = 3/10 mv2
 
Thanks and regards,
Kushagra

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