A ballof mass 100 g was thrownvertically upwards with a velocity of 49 m*s^(-1). At the same time another identical ball was dropped, to fall along the same path from a height of 98 m. After some time, the two balls collided and got stuck together. This combined mass then fell to the ground. How long was the combined mass in motion?

A ballof mass 100 g was thrownvertically upwards with a velocity of 49

1.	A ballof mass 100 g was thrownvertically upwards with a velocity of 49 m*s^(-1). At the same time another identical ball was dropped, to fall along the same path from a height of 98 m. After some time, the two balls collided and got stuck together. This combined mass then fell to the ground. How long was the combined mass in motion?
Answer» Solution :Suppose the balls COLLIDE at a height h above theground in time `t_1` .Analysing theupward motion of the1st ball , we get, `h=49 t_1-9.8 t_1^2//2` …….(1) Analysing the downward motion of the 2ND ball , we get, `98-h=9.8 t_1^2//2` .......(2) Fromequations (1) and (2) , `98=49 t_1 or, t_1=2s` At the time of collision let the velocityof the 1st ball be`v_1` and that of the 2nd ball be `v_2`. `therefore v_1=49-9.8xx2=29.4ms^(-1)` (upward) `and v_2 =9.8xx2=19.6 ms^(-1)` (downward) After collision, let THEVELOCITY of thecombined mass be V. Then ACCORDING tothe law ofconservation of momentum, `0.1xx29.4-0.1xx19.6=2xx0.1xxV` [indicating downward motion with a negative sign] or, `V=4.9m*s^(-1)` (upward) From equation (1) ,we get, `h=49xx2-9.8xx(2)^2//2=78.4m` If the COMBINED mass was inmotion for a time t, then from the relation `h=ut +1/2"gt"^2` we get, `-78.4=4.9 t-1/2xx9.8t^2, or, t^2-t-16=0` `therefore t=1/2(1 pm sqrt(1+4xx1xx16))` As t cannot be negative , we have ,t =4.53 s