A bullet of mass M is fired with a velocity `50m//s` at an angle with the horizontal. At the highest point of its trajectory, it collides head-on with a bob of mass 3M suspended by a massless string of length `10//3` metres and gets embeded in the bob. After the collision, the string moves through an angle of `120^@`. Find
(i) the angle `theta`,
(ii) the vertical and horizontal coordinates of the initial position of the bob with respect to the point of firing of the bullet. Take `g=10 m//s^2`

Answer

A, B, C, D

Explanation

(i) In `DeltaAQR`
`sin30^@=(QR)/(10//3), QR=5/3`
`u=50 m//s` (Given )
At the highest point P, the velocity of the bullet=`u cos theta`

Applying conservation of linear momentum at the highest
point
`M(ucos theta)+3Mxx0=(M+3M)v`
`v=(Mu cos theta)/(4M)=(ucos theta)/4`
Applying energy conservation principle for P and R
K.E. of the bullet-mass system at `P=P.E.` of the bullet-mass
system at R
`1/2(4M)v^2=(4M)gh`
`1/2(4M) (u^2cos^2 theta)/16=4Mgxx(10/3+5/3)`
`cos^2 theta=(9.8xx5xx2xx16)/(50xx50) rArr theta=37^@`
(ii) `R/2=(u^2sin 2 theta)/(2g)=(50xx50sin3xx37^@)/(2xx9.8)=122.6m`
`H=(u^2sin^2 theta)/(2g)=(50xx50xx(sin37^@)^2)/(2xx9.8)=46m`