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

77

Explanation

In `DeltaAQR, sin 30^@ = (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(u cos theta) + 3M xx 0 = (M + 3M)v implies v = (M u cos theta)/(4M) = (u cos theta)/(4)`
At R, T = 0 but `v != 0`
`:. 4 Mg cos 60^@ = (4 MV^2)/(l) implies V^2 = (gl)/2 = 50/3`
Applying energy conservation principle for P and R
`Delta K = Delta U = 0`
`implies 1/2 (4 M) v_R^2 - 1/2 (4M )v_P^2 + (4M)gh = 0`
`=1/2 (4M) (u^2 cos^2 theta)/(16) = (4M) g xx (10/3 + 5/3) + 1/2 (4 M) (50/3)`
`implies cos^2 theta = 0.75 " "implies " "cos theta = 0.86 implies theta = 30^@`
`R/2 = (u^2 sin 2theta)/(2g) = (50 xx 50 xx sqrt(3))/(2 xx 10 xx 2) = 108.25 m " " H = (u^2 sin^2 theta)/(2g) = (50 xx 50 xx 1)/(2 xx 10 xx 4) = 31.25 m`.