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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 |
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Answer» `30^(@)`  `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 velocitiesof 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` |
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