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Grade 11Mechanics

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

Profile image of anita yogi
8 Years agoGrade 11
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1 Answer

Profile image of Arun
8 Years ago
Dear Anita
 
Let the 2 balls meet at time T.
Their displacements will be S and 98 - S meters.
 
S = gT^2/2
98 - S = 49T - gT^2/2
Adding the 2 equations:
98 = 49T
T = 2 seconds.
(and S = 20 m from top height) 
 
At t = 2 seconds, Velocity of balls have to be found.
v1 = g*2 seconds = 20 m/s downwards
v2 = 49 - g*2 = 29 m/s upwards
 
Now an inelastic collision occurs and we have to conserve the momentum.
0.1 kg (20 - 29) = 0.2 kg * V
V is 4.5 m/s upwards.
Now from here to final stop, time taken is to be found:
98 - 20 = 78 m = -4.5 t + gt^2/2
78 = -4.5t + 5t^2
t = 4.5 +- root (20.25 + 1560) / 10
t = 4.5 + 39.75 /10 =  4.425 seconds
 
So the balls collide after 2 seconds, and the lump of the masses takes a further 4.425 seconds to reach the ground.
 
Hence total time of flight = 2 + 4.425 = 6.425 sec
 
Regards
Arun (askIITians forum expert)