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.

Options

A. P-2, Q-3, R-1, S-4
B. P-3, Q-2, R-4, S-1
C. P-4, Q-1, R-2, S-3
D. P-2, Q-3, R-4, S-1

Answer

C

Explanation

(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`