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# A rod AB of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits the end ‘A’ of the rod with a velocity V0 in a direction perpendicular to AB. The collision is elastic. After the collision the particle comes to rest. (a) Find the ratio m/M. (b) A point P on the rod is at rest immediately after collision. Find the distance AP. (c) Find the linear speed of the point P after a time pL/3V0 after the collision.

## 2 Answers

9 years ago

Dear Jay,

after collision let linear and angular velocity of centre of rod be V and w. Let moment of inertia of rod perpendicular to rod through centre of rod be I,

mv0=MV,(v0*L/2)m=Iw

(m/2)(v0)2=(I/2)w2+(M/2)V2, solve the three equations to obtain V,w and M.

b)(lAPl-(L/2))w=V

c)at t=(pi)L/3V0 AP would have rotated by w*t

(linear speed)2=(V+((lAPl-(L/2))sin(2700-wt)))2+((lAPl-(L/2))cos(2700-wt))2

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CHETAN MANDAYAM NAYAKAR

8 months ago
Dear Student,
Please find below the solution to your problem.

after collision let linear and angular velocity of centre of rod be V and w. Let moment of inertia of rod perpendicular to rod through centre of rod be I,
mv0=MV,(v0*L/2)m=Iw
(m/2)(v0)2=(I/2)w2+(M/2)V2, solve the three equations to obtain V,w and M.

b)(lAPl-(L/2))w=V
c)at t=(pi)L/3V0 AP would have rotated by w*t
(linear speed)2=(V+((lAPl-(L/2))sin(2700-wt)))2+((lAPl-(L/2))cos(2700-wt))2

Thanks and Regards

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