# 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 then1..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 ?

148 Points
14 years ago

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

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Sakshi
20 Points
6 years ago
I have a doubt in this ques isnt the linear momentum should be mv=MVInstead of what you`ve written.....🙇
4 years ago
Dear student,

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