Question

A uniform rod of length l and mass 2m rests on a smooth horizontal table. a point mass m moving horizontally at right angles to the rod with velocity ν collides with one end of the rod and sticks it. then

Answer 1

The linear and angular momentum will remain conserved due to there being no external force applied to the rod

so:

MV = M(WL/2) + 2MV1

V = WL/2  + 2V1 .

V1 = V/2 - WL/4            (a)

And with the 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 (a)

V1 = V/5

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

 

Now we can calculate the loss in K.E if we put in the given values.

Answer 2

Answer:

Explanation: Watch this