1.

A particle of mass m is connected with a string of length 2 meter. Other end of the string is fixed with a point O at a height 1 m above the ground. The particle is thrown from some point in such a way that it strikes the ground (perfectly inelastic) with velocity v_(0) at an angle 37^(@) with vertical just below O.

Answer»

<P>P-2, Q-3, R-1, S-4
P-3, Q-2, R-4, S-1
P-4, Q-1, R-2, S-3
P-2, Q-3, R-4, S-1

Solution :(A) as the COLLISION is perfectly inelastic, velocity after collision is `v_(0)sin 37^(@)=(3)/(5)v_(0)` towards right till string becomes taut.
`R=(v^(2))/(a_(_|_))=(v_(0)^(2))/(g sin 37^(@))=(5v_(0)^(2))/(3g)`

(B)from figure, `cos theta=(1)/(2)rArr theta = 60^(@)`
after string becomes taut particle motion will seize along the string DUE to impulse. Just after impulse velocity of particle is `(3)/(5)v_(0)cos theta = (3)/(10)v_(0)` perpendicular to the string from conservation of mechanical energy , `(1)/(2)m((3)/(10)v_(0))^(2)=(1)/(2)m(sqrt(2G))^(2)+2mg(1+cos 60)` simplifying, `v_(0)=(20sqrt(2g))/(3)m//s`
(C ) `W_(G)=-2mg(1+cos 60^(@))=-3 mg`
`therefore - W_(G)=3 mg`
(D) At highest point velocity will be perpendicular to gravitational force
`therefore P = vec(F)vec(V)=0`


Discussion

No Comment Found

Related InterviewSolutions