A ball A of mass `m` is suspended by a thread of length `r = 1.2 m`. Another ball `B` of mass `2m` is projected from the ground with velocity it `u= 9 m//s` such that at the highest point of its trajectory it collides head-on elastically with ball `A`. It is observed that during subsequent motion, tension in the thread at the highest point is equal to `mg`.

The angle of projection (`theta`) of ball `B` is

Options

A. a. `30^(@)`
B. `60^(@)`
C. `45^(@)`
D. `75^(@)`

Answer

B

Explanation

Let velocity of ball `A` at highest point be `v` then first considering its `FBD` at this point

`mg+ T=(mv^(2))/r`
where` T=mg`
`implies v^(2)=2rg`
`implies v=sqrt(2rg)=2sqrt(6)m//s`
If the velocity of ball `A` just after the collisin is `v_(1)` then according to law of conservation of energy its `KE` at `A=(KE+PE)` at highest point
`1/2mv_(1)^(2)=1/2mv^(2)+mg(2r)`
`v_(1)=sqrt(6rg)` or `v_(1)=6sqrt(2)m//s`
If angle of project of ball `B` is `theta`, then horizontal component of its velocity will be `9 costheta` and it remains constant.

Now considering collision of balls `B` and `A`, let velocities of balls `A` and `B`, just after the collision, be `v_(1)` and `v_(2)` respectively as shown in figure.
Then according to the law of conservation of momentum
`mv_(1)+2mv_(2)=2m(9costheta)` ...........i
Since collision is elastic, therefore `e=1`
`(v_(1)-v_(2))/(9costheta)=1` or `v_(1)-v_(2)=9costheta` ........ii
Solving eqn i and ii we get
`costheta=1/(sqrt(2))` or `theta=45^(@)`
`v_(2)=3/(sqrt(2))ms^(-1)`
height of initial position of ball `A` (lowest position) is equal to maximum height ascened by the ball `B`. It is equal to
`H=((9sintheta)^(2))/(2g)=81/40m`
Height of point of suspension from the ground is
`H=r=129/40m`