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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 ?

akash kashyp , 16 Years ago

Grade Upto college level

3 Answers

Badiuddin askIITians.ismu Expert

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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Badiuddin

Last Activity: 16 Years ago

Sakshi

I have a doubt in this ques isnt the linear momentum should be mv=MVInstead of what you`ve written.....🙇

Last Activity: 8 Years ago

Kushagra Madhukar

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

x_cm = [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, P_i = P_f

mv = 3mv_cm

or, v_cm = v/3

L_i = L_f (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) v_cm² + ½ Iw²

= ½ (3m) (v²/9) + ½ (5mL²/12) (2v/5L)²

= 1/5 mv²

Hence, Loss of KE = K_i – K_f = ½ mv² – 1/5 mv² = 3/10 mv²

Thanks and regards,

Kushagra

Last Activity: 5 Years ago

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