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Tuesday, September 16, 2008

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

A) 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.

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2 comments:

G said...

Could you esxplain the 98-S part better, please? Thanks!

Anonymous said...

Thanks for the answer man, I have been searching for the answer to this question (JEE-85) forever