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A rod of lenght 2m and mass 2 kg is lying on smooth horizental x-y plane along Y- axis with its center at origin O as shown in the figure . An impulse J of magnitude 10 Ns is applied perpendicular to AB at A.
Q.1The distance of point P from point A which is at rest just after the impact is :
(A) 2/3M
(B)4/3M
(C)1/2M
(d) 3/4M
Q.2 Coordinate of point A of the rod after time t=pi/30
(A)(root3/2+pi/9,1/2)
(b) (c)(1/2+pi/6,0) (D)(1+pi/6,0)
(c)(1/2+pi/6,0)
(D)(1+pi/6,0)
Let I be moment of inertia about axis perpendicular to x-y plane and passing through O.
(Linear) velocity of O=10Ns/2kg=5m/s
angular velocity=10*1/I
let answer be 'l'
5=(l-1)*angular velocity
consider A={x(t),y(t)},y(t)=t*1m*angular velocity*cos(ang.vel.*time))
x(t)=t*1m*angular velocity*sin(ang.vel.*time)+t*linear velocity
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