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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 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 ?  Badiuddin askIITians.ismu Expert
147 Points
11 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

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.

All the best.

Regards, Kushagra Madhukar
10 months 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