A ball of mass 100 gm is projected vertically upwards from the ground with a velocity of `49m//sec`. At the same time another identical ball is dropped from a height of 98 m to fall freely along the same path as that followed by the first ball. After some time the two balls collide and stick together and finally fall to the ground. Find the time of flight of the masses.

Answer

C

Explanation

For the ball thrown up
`v_1^2-u_1^2=2a_1s_1`
`:. v_1^2-2401=19.6h…..(i)`
`s_1=u_1t_1+1/2a_1t_1^2`
`h=49t-4.9t^2…..(ii)`


For the ball dropped from height
`v_2^2-u_2^2=2a_2s_2`
`:. v_2^2=19.6(98-h)…..(iii)`
`s_2=u_2t_2+1/2a_2t_2^2`
`98-h=4.9t^2.....(iv)`
From (ii) and (iv)
`98-(49t-4.9t^2)=4.9t^2 :. 98-49t=0`
`:. t= 2sec`
`:. h=49xx2-4.9xx2^2=78.4m` (from (ii))
`Substituting this value of h in (i) and (ii), we get
`v_1^2-2401=-19.6xx78.4 v_2^2=19.6(98-78.4)`
`v_1^2=864.36 rArr v_2^2=384.16`
`v_1=29.4 m//s rArr v_2=19.6 m//s`
At point C where the two bodies collide, thereafter
both bodies stick and behave as a single body.
Thus, we apply conservation of linear momentum, which
gives
`m_1v_1-m_2v_2=2mv`
`:. v=(v_1-v_2)/2=(29.4-19.6)/2=4.9 m//s`
`u=4.9 m//s,s_1=-78.4, a_1=-9.8m//s^2 , t=?`
`s=ut+1/2at^2 rArr -78.4=4.9t-4.9t^2`
`:. t^2-t-16=0 rArr t=(1+-sqrt(1+64))/2=4.53`
`Total time =4.53+2=6.53sec.`